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FUNDAMENTAL PHYSICAL CONSTANTS Speed of light in vacuum C 2.997 924 58 108 [m/s] Avagadro’s number NA 6.022 141

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FUNDAMENTAL PHYSICAL CONSTANTS Speed of light in vacuum

C

2.997 924 58 108

[m/s]

Avagadro’s number

NA

6.022 141 99 1023

[molecule/mol]

Gas constant

R

8.314 472

[J/(mol K)]

Boltzmann’s constant (R/NA)

k

1.380 650 3 1023

(J/(molecule K)]

Faraday’s constant

F

9.648 534 15 104

[C/(mole)]

Elementary charge

Q

1.602 176 46 1019

[C]

4.803 204 19 1010

[esu]

Mass of a proton

m

1.672 621 58 1027

[kg]

Atomic mass unit

AMU

1.660 538 73 1027

[kg]

Atmospheric pressure (sea level)

P

1.013 25 l05

[Pa]

Gravitational acceleration (sea level)

g

9.806 55

[m/s2]

Pi

p

3.141 592 65

CONVERSION FACTORS ° ] 39.370 [in] 3.2808 [ft] 1 [m] 102 [cm] 1010 [A 1 [kg] 103 [g] 2.2046 [1bm] 0.068522 [slug] [K] [°C] 273.15 (5/9) [°R]; [°R] [°F] 459.67 1 [m3] 103 [L] 106 [cm3] 35.315 [ft3] 264.17 [gal] (U.S.) 1 [N] 105 [dyne] 0.22481 [lbf] 1 [atm] 1.01325 [bar] 1.01325 105 [Pa] 14.696[psi] 760 [torr] 1 [J] 107 [crg] 0.2.3885[cal] 9.4781 10-4[BTU] 6.242 1018 [eV] For electric and magnetic properties see Appendix D: Table D.2.

COMMON VALUES FOR THE GAS CONSTANT, R

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8.314

[J/(mol K)]

0.08314

[(L bar )/(mol K)]

1.987

[cal/(mol K)]

1.987

[BTU/(lbmol °R)]

0.08206

[(L atm)/(mol K)]

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SPECIAL NOTATION Properties Uppercase

Extensive

K : V, G, U, H, S, c

Lowercase

Intensive (molar)

k5

Circumﬂex, lowercase

Intensive (mass)

k k^ 5 5 ν^ , g^ , u^ , h^ , s^, c m

Pure species property

Ki : Vi, Gi, Ui, Hi, Si, c

k 5 ν, g, u, h, s, c n

Mixtures Subscript i

ki : νi, gi, ui, hi, si, c Bar, subscript i

Partial molar property

Ki : Vi, Gi, Ui, Hi, Si, c

As is

Total solution property

K : V, G, U, H, S, c k : ν, g, u, h, s, c

Delta, subscript mix

Property change of mixing:

DKmix : DVmix, DHmix, DSmix, c Dkmix : Dνmix, Dhmix, Dsmix, c

Other Dot Overbar

Rate of change

# # # # Q, W, n, V, c

Average

V2 , cp, c S

A complete set of notation used in this text can be found on page (vii)

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Engineering and Chemical Thermodynamics 2nd Edition Milo D. Koretsky School of Chemical, Biological, and Environmental Engineering Oregon State University

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VP & Publisher Associate Publisher Marketing Manager Associate Production Manager Designer Production Management Services

Don Fowley Dan Sayre Christopher Ruel Joyce Poh Kenji Ngieng Laserwords

The drawing on the cover illustrates a central theme of the book: using molecular concepts to reinforce the development of thermodynamic principles. The cover illustration depicts a turbine, a common process that can be analyzed using thermodynamics. A cutaway of the physical apparatus reveals a hypothetical thermodynamic pathway marked by dashed arrows. Using this text, students will learn how to construct such pathways to solve a variety of problems. The figure also contains a “molecular dipole,” which is drawn in the PT plane associated with the real fluid. By showing how principles of thermodynamics relate to concepts learned in prior courses, this text helps students construct new knowledge on a solid conceptual foundation. This book was set by Laserwords. Cover and text printed and bound by Courier Kendallville. This book is printed on acid free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2013, 2004 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate percopy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 070305774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return mailing label are available at HYPERLINK "http://www.wiley.com/go/returnlabel" www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative.

Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

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For Eileen Otis, mayn basherte

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► CHAPTER

Preface You see, I have made contributions to biochemistry. There were no courses in molecular biology. I had no courses in biology at all, but I am one of the founders of molecular biology. I had no courses in nutrition or vitaminology. Why? Why am I able to do these things? You see, I got such a good basic education in the fields where it is difficult for most people to learn by themselves. Linus Pauling On his ChE education

►AUDIENCE Engineering and Chemical Thermodynamics is intended for use in the undergraduate thermodynamics course(s) taught in the sophomore or junior year in most Chemical Engineering (ChE) and Biological Engineering (BioE) Departments. For the majority of ChE and BioE undergraduate students, chemical engineering thermodynamics, concentrating on the subjects of phase equilibria and chemical reaction equilibria, is one of the most abstract and difficult core courses in the curriculum. In fact, it has been noted by more than one thermodynamics guru (e.g., Denbigh, Sommerfeld) that this subject cannot be mastered in a single encounter. Understanding comes at greater and greater depths with every skirmish with this subject. Why another textbook in this area? This textbook is targeted specifically at the sophomore or junior undergraduate who must, for the first time, grapple with the treatment of equilibrium thermodynamics in sufficient detail to solve the wide variety of problems that chemical engineers must tackle. It is a conceptually based text, meant to provide students with a solid foundation in this subject in a single iteration. Its intent is to be both accessible and rigorous. Its accessibility allows students to retain as much as possible through their first pass while its rigor provides them the foundation to understand more advanced treatises and forms the basis of commercial computer simulations such as ASPEN®, HYSIS®, and CHEMCAD®.

►GOALS AND METHODOLOGY The text was developed from course notes that have been used in the undergraduate chemical engineering classes at Oregon State University since 1994. It uses a logically consistent development whereby each new concept is introduced in the context of a framework laid down previously. This textbook has been specifically designed to accommodate students with different learning styles. Its conceptual development, worked-out examples, and numerous end-of-chapter problems are intended to promote deep learning and provide students the ability to apply thermodynamics to real-world engineering problems. Two major threads weave throughout the text: (1) a common methodology for approaching topics, be it enthalpy or fugacity, and (2) the reinforcement of classical thermodynamics with molecular principles. Whenever possible, intuitive and qualitative arguments complement mathematical derivations. The basic premise on which the text is organized is that student learning is enhanced by connecting new information to prior knowledge and experiences. The approach is to introduce new concepts in the context of material that students already know. For example, the second law of thermodynamics is formulated analogously to the first law, as a generality to many observations of nature (as opposed to the more common approach of using specific statements about obtaining work from heat through thermodynamic cycles). Thus, the experience students have had in learning about the thermodynamic property energy, which they have already encountered in several classes, is applied to introduce a new thermodynamic property, entropy. Moreover, the underpinnings of the second law—reversibility, irreversibility, and the Carnot cycle—are introduced with the first law, a context with which students have more experience; thus they are not new when the second law is introduced.

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vi ► Preface

►LEARNING STYLES There has been recent attention in engineering education to crafting instruction that targets the many ways in which students learn. For example, in their landmark paper “Learnings and Teaching Styles in Engineering Education,”1 Richard Felder and Linda Silverman define specific dimensions of learning styles and corresponding teaching styles. In refining these ideas, the authors have focused on four specific dimensions of learning: sequential vs. global learners; active vs. reflective learners; visual vs. verbal learners; and sensing vs. intuitive learners. This textbook has been specifically designed to accommodate students with different learning styles by providing avenues for students with each style and, thereby, reducing the mismatches between its presentation of content and a student’s learning style. The objective is to create an effective text that enables students to access new concepts. For example, each chapter contains learning objectives at the beginning and a summary at the end. These sections do not parrot the order of coverage in the text, but rather are presented in a hierarchical order from the most significant concepts down. Such a presentation creates an effective environment for global learners (who should read the summary before embarking on the details in a chapter). On the other hand, to aid the sequential learner, the chapter is developed in a logical manner, with concepts constructed step by step based on previous material. Identified key concepts are presented schematically to aid visual learners. Questions about key points that have been discussed previously are inserted periodically in the text to aid both active and reflective learners. Examples are balanced between those that emphasize concrete, numerical problem solving for sensing learners and those that extend conceptual understanding for intuitive learners. In the cognitive dimension, we can form a taxonomy of the hierarchy of knowledge that a student may be asked to master. For example, a modified Bloom’s taxonomy includes: remember, understand, apply, analyze, evaluate, and create. The tasks are listed from lowest to highest level. To accomplish the lower-level tasks, surface learning is sufficient, but the ability to perform at the higher levels requires deep learning. In deep learning, students look for patterns and underlying principles, check evidence and relate it to conclusions, examine logic and argument cautiously and critically, and through this process become actively interested in course content. In contrast, students practicing surface learning tend to memorize facts, carry out procedures algorithmically, find it difficult to make sense of new ideas, and end up seeing little value in a thermodynamics course. While it is reinforced throughout the text, promotion of deep learning is most significantly influenced by what a student is expected to do. End-of-chapter problems have been constructed to cultivate a deep understanding of the material. Instead of merely finding the right equation to “plug and chug,” the student is asked to search for connections and patterns in the material, understand the physical meaning of the equations, and creatively apply the fundamental principles that have been covered to entirely new problems. The belief is that only through this deep learning is a student able to synthesize information from the university classroom and creatively apply it to new problems in the field.

►SOLUTION MANUAL The Solutions Manual is available for instructors who have adopted this book for their course. Please visit the Instructor Companion site located at www.wiley.com/college/koretsky to register for a password.

►MOLECULAR CONCEPTS While outside the realm of classical thermodynamics, the incorporation of molecular concepts is useful on many levels. In general, by the time undergraduate thermodynamics is taught, the chemical engineering student has had many chemistry courses, so why not take advantage of this experience! Thermodynamics is inherently abstract. Molecular concepts reinforce the text’s explanatory approach providing more access to the typical undergraduate student than could a mathematical derivation, by itself. 1

Felder, Richard M., and Linda K. Silverman, Engr. Education, 78, 674 (1988).

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Preface ◄ vii A molecular approach is also becoming important on a technological level, with the increased development of molecular based simulation and engineering at the molecular level with nanotechnology. Moreover, molecular understanding allows the undergraduate to form a link between the understanding of equilibrium thermodynamics and other fundamental engineering sciences such as transport phenomena. Finally, the research literature in cognitive science has shown that students can form persistent misconceptions in core engineering science topics, and a molecular approach is useful in mitigating these misconceptions. For example, in emergent processes, observed phenomena are not directly caused by macroscopic processes, but rather “emerge” indirectly from collective behavior of molecules. Concepts that are most difficult for students to learn often contain emergent processes which they mistake for direct causation. By including explanation at a molecular level, differences between emergent and direct phenomena can be explicitly addressed and the underlying causation is explained.

►THERMOSOLVER SOFTWARE The accompanying ThermoSolver software has been specifically designed to complement the text. This integrated, menu-driven program is easy to use and learning-based. ThermoSolver readily allows students to perform more complex calculations, giving them opportunity to explore a wide range of problem solving in thermodynamics. Equations used to perform the calculations can be viewed within the program and use nomenclature consistent with the text. Since the equations from the text are integrated into the software, students are better able to connect the concepts to the software output, reinforcing learning. The ThermoSolver software may be downloaded for free from the student companion site located at www.wiley.com/college/koretsky.

►ACKNOWLEDGMENTS First, I would like to acknowledge and offer thanks to those individuals who have provided thoughtful input: Stuart Adler, Connelly Barnes, Kenneth Benjamin, Bill Brooks, Hugo Caran, Chih-hung (Alex) Chang, Mladen Eic, John Falconer, Frank Foulkes, Jerome Garcia, Debbi Gilbuena, Enrique Gomez, Dennis Hess, Ken Jolls, P. K. Lim, Uzi Mann, Ron Miller, Erik Muehlenkamp, Jeff Reimer, Skip Rochefort, Wyatt Tenhaeff, Darrah Thomas, and David Wetzel. Second, I appreciate the effort and patience of the team at John Wiley & Sons, especially: Wayne Anderson, Dan Sayre, Alex Spicehandler, and Jenny Welter. Last, but not least, I am tremendously grateful to the students with whom, over the years, I have shared the thermodynamics classroom.

►NOTATION The study of thermodynamics inherently contains detailed notation. Below is a summary of the notation used in this text. The list includes: special notation, symbols, Greek symbols, subscripts, superscripts, operators and empirical parameters. Due to the large number of symbols as well as overlapping by convention, the same symbol sometimes represents different quantities. In these cases, you will need to deduce the proper designation based on the context in which a particular symbol is used.

Special Notation Properties

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Uppercase

Extensive

K : V, G, U, H, S, . . .

Lowercase

Intensive (molar)

Circumflex, lowercase

Intensive (specific)

K 5 v, g, u, h, s, c n K k^ 5 5 v^, g^, u^ , h^ , s^ , c m k5

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viii ► Preface Mixtures Subscript i

Pure species property

Ki : Vi, Gi, Ui, Hi, Si, c ki : vi, gi, ui, hi, si, c

Bar, subscript i

Partial molar property

Ki : Vi, Gi, Ui, Hi, Si, c

As is

Total solution property

K : V, G, U, H, S, . . . k : v, g, u, h, s, . . .

Delta, subscript mix

Property change of mixing:

DKmix : DVmix, DHmix, DSmix, c Dkmix : Dvmix, Dhmix, Dsmix, c

Other Dot

Rate of change

Overbar

Average

# # # Q, W, n# , V, c S

V2, cP, c

Symbols a, b . . ., i, . . . a, A A, B A

ai Ai b, B bf , Bf

bj cP cv ci Ci [i] COP

Di2j e, E

ek, EK ep, EP S E F F F

fi f^i f g, G

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Generic species in a mixture Helmholtz energy Labels for processes to be compared Area Activity of species i Species i in a chemical reaction Exergy Exalpy Element vector Heat capacity at constant pressure Heat capacity at constant volume Molal concentration of species i Mass concentration of species i Molar concentration of species i Coefficient of performance Bond i – j dissociation energy Energy Kinetic energy Potential energy Electric field Force Flow rate of feed Faraday’s constant Degrees of freedom Fugacity of pure species i Fugacity of species i in a mixture Total solution fugacity Gibbs energy

g h, H

| Dh s Hi i I I k, K

k k k K

kij Ki L m m MW n n

ni N

NA OF p P

pi

Gravitational acceleration Enthalpy Enthalpy of solution Henry’s law constant of solute i Interstitial Ionization energy Ionic strength Generic representation of any thermodynamic property except P or T Boltzmann’s constant Heat capacity ratio 1 cP/cv 2 Spring constant Equilibrium constant Binary interaction parameter between species i and j K-value Flow rate of liquid Number of chemical species Mass Molecular weight Number of moles Concentration of electrons in a semiconductor Intrinsic carrier concentration Number of molecules in the system or in a given state Avagadro’s number Objective function Concentration of holes in a semiconductor Pressure Partial pressure of species i in an ideal gas mixture

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Preface ◄ ix

Pisat

Saturation pressure of species i Heat Electric charge Distance between two molecules Gas constant Number of independent chemical reactions Stoichiometric constraints Entropy Time Temperature Temperature at the boiling point Temperature at the melting point Upper consulate temperature Internal energy Volume Flow rate of vapor

q, Q Q r R R s s, S t T

Tb Tm Tu u, U v, V V

V

Vacancy

V

Velocity Work Flow work Shaft work Non-Pv work Weight fraction of species i Quality (fraction vapor) Position along x-axis Mole fraction of liquid species i Mole fraction of solid species i Mole fraction of vapor species i Compressibility factor Position along z-axis Valence of an ion in solution Labels of specific states of a system Generic species in a mixture

S

w, W

wflow, Wflow ws , W S

w∗, W∗

wi x x

xi Xi yi z z z 1, 2 . . . 1, 2 . . .

Greek Symbols ai b bij E

wi w^ i w gi

giHenry’s

gm i g6

Polarizability of species i Thermal expansion coefficient Formula coefficient matrix Electrochemical potential Fugacity coefficient of pure species i Fugacity coefficient of species i in a mixture Total solution fugacity coefficient Activity coefficient of species i Activity coefficient using a Henry’s law reference state Molality based activity coefficient Mean activity coefficient of anions and cations in solution

h li G Gi Gij k mi mi mJT p P r ni

v j

Efficiency factor Lagrangian multiplier Molecular potential energy Activity coefficient of solid species i Molecular potential energy between species i and j Isothermal compressibility Dipole moment of species i Chemical potential of species i Joule-Thomson coefficient Phases Osmotic pressure Density Stiochiometric coefficient Pitzer acentric factor Extent of reaction

Subscripts a, b, . . ., i, . . . atm c C calc cycle exp f

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Generic species in a mixture Atmosphere Critical point Cold thermal reservoir Calculated Property change over a thermodynamic cycle Experimental Property value of formation (with D)

fus E H high ideal gas in inerts irrev

Fusion External Hot thermal reservoir High value (e.g. in interpolation) Ideal gas Flow stream into the system Inerts in a chemical reaction Irreversible process

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x ► Preface l low mix net out products pc r reactants

Liquid Low value (e.g. in interpolation) Equation of state parameter of a mixture Net heat or work transferred Flow stream out of the system Products of a chemical reaction Pseudocritical Reduced property Reactants in a chemical reaction

real gas rev rxn sub surr sys univ v vap z 0 1, 2 . . . 1, 2 . . .

Real gas Reversible process Reaction Sublimation Surroundings System Universe Vapor Vaporization In the z direction Environment Labels of specific states of a system Generic species in a mixture

Superscripts dep E ideal ideal gas molecular l o real

Departure function (with D) Excess property Ideal solution Ideal gas Molecular Liquid Value at the reference state Real fluid with intermolecular interactions

s sat v

a, b g ` (0) (1)

Solid At saturation Vapor Generic phases (in equilibrium) Volume exponential of a polytropic process At infinite dilution Simple fluid term Correction term

Operators d

d

e

Total differential Partial differential Difference between the final and initial value of a state property Gradient operator Integral

a, b a, b, a, k c A Aij A,B A, B A, B, C A, B, C, D, E B, C, D Br, Cr, Dr C6 Cn e Lij s

van der Waals or Redlich-Kwong attraction and size parameter, respectively Empirical parameters in various cubic equations of state Two-suffix Margules activity coefficient model parameter Three-suffix Margules activity coefficient model parameters (one form) Three-suffix Margules or van Laar activity coefficient model parameters Debye-Huckel parameters Empirical constants for the Antoine equation Empirical constants for the heat capacity equation Second, third and fourth virial coefficients Second, third and fourth virial coefficient in the pressure expansion Constant of van der Waals or Lennard-Jones attraction Constant of intermolecular repulsion potential of power r2n Lennard-Jones energy parameter Wilson activity coefficient model parameters Distance parameter in hard sphere, Lennard-Jones and other potential functions

' D =

ln log

P

a

Inexact (path dependent) differential Natural (base e) logarithm Base 10 logarithm Cumulative product operator Cumulative sum operator

Empirical parameters

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►

Contents

CHAPTER 1

Measured Thermodynamic Properties and Other Basic Concepts 1 Learning Objectives 1 1.1 Thermodynamics 2 1.2 Preliminary Concepts—The Language of Thermo 3 Thermodynamic Systems 3 Properties 4 Processes 5 Hypothetical Paths 6 Phases of Matter 6 Length Scales 6 Units 7 1.3 Measured Thermodynamic Properties 7 Volume (Extensive or Intensive) 7 Temperature (Intensive) 8 Pressure (Intensive) 11 The Ideal Gas 13 1.4 Equilibrium 15 Types of Equilibrium 15 Molecular View of Equilibrium 16 1.5 Independent and Dependent Thermodynamic Properties 17 The State Postulate 17 Gibbs Phase Rule 18 1.6 The PvT Surface and Its Projections for Pure Substances 20 Changes of State During a Process 22 Saturation Pressure vs. Vapor Pressure 23 The Critical Point 24 1.7 Thermodynamic Property Tables 26 1.8 Summary 30 1.9 Problems 31 Conceptual Problems 31 Numerical Problems 34 CHAPTER 2

The First Law of Thermodynamics 36 Learning Objectives 36 2.1 The First Law of Thermodynamics 37 Forms of Energy 37 Ways We Observe Changes in U 39

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2.2 2.3

2.4

2.5

2.6

2.7

2.8

2.9 2.10 2.11

Internal Energy of an Ideal Gas 40 Work and Heat: Transfer of Energy Between the System and the Surroundings 42 Construction of Hypothetical Paths 46 Reversible and Irreversible Processes 48 Reversible Processes 48 Irreversible Processes 48 Efficiency 55 The First Law of Thermodynamics for Closed Systems 55 Integral Balances 55 Differential Balances 57 The First Law of Thermodynamics for Open Systems 60 Material Balance 60 Flow Work 60 Enthalpy 62 Steady-State Energy Balances 62 Transient Energy Balance 63 Thermochemical Data For U and H 67 Heat Capacity: cv and cP 67 Latent Heats 76 Enthalpy of Reactions 80 Reversible Processes in Closed Systems 92 Reversible, Isothermal Expansion (Compression) 92 Adiabatic Expansion (Compression) with Constant Heat Capacity 93 Summary 95 Open-System Energy Balances on Process Equipment 95 Nozzles and Diffusers 96 Turbines and Pumps (or Compressors) 97 Heat Exchangers 98 Throttling Devices 101 Thermodynamic Cycles and the Carnot Cycle 102 Efficiency 104 Summary 108 Problems 110 Conceptual Problems 110 Numerical Problems 113

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xii ► Contents CHAPTER 3

Entropy and the Second Law Of Thermodynamics 127 Learning Objectives 127 3.1 Directionality of Processes/Spontaneity 128 3.2 Reversible and Irreversible Processes (Revisited) and their Relationship to Directionality 129 3.3 Entropy, the Thermodynamic Property 131 3.4 The Second Law of Thermodynamics 140 3.5 Other Common Statements of the Second Law of Thermodynamics 142 3.6 The Second Law of Thermodynamics for Closed and Open Systems 143 Calculation of Ds for Closed Systems 143 Calculation of Ds for Open Systems 147 3.7 Calculation of Ds for an Ideal Gas 151 3.8 The Mechanical Energy Balance and the Bernoulli Equation 160 3.9 Vapor-Compression Power and Refrigeration Cycles 164 The Rankine Cycle 164 The Vapor-Compression Refrigeration Cycle 169 3.10 Exergy (Availability) Analysis 172 Exergy 173 Exthalpy—Flow Exergy in Open Systems 178 3.11 Molecular View of Entropy 182 Maximizing Molecular Configurations over Space 185 Maximizing Molecular Configurations over Energy 186 3.12 Summary 190 3.13 Problems 191 Conceptual Problems 191 Numerical Problems 195 CHAPTER 4

Equations of State and Intermolecular Forces

209 Learning Objectives 209 4.1 Introduction 210 Motivation 210 The Ideal Gas 211 4.2 Intermolecular Forces 211 Internal (Molecular) Energy 211 The Electric Nature of Atoms and Molecules 212 Attractive Forces 213 Intermolecular Potential Functions and Repulsive Forces 223 Principle of Corresponding States 226 Chemical Forces 228 4.3 Equations of State 232 The van der Waals Equation of State 232

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4.4 4.5

4.6 4.7

Cubic Equations of State (General) 238 The Virial Equation of State 240 Equations of State for Liquids and Solids 245 Generalized Compressibility Charts 246 Determination of Parameters for Mixtures 249 Cubic Equations of State 250 Virial Equation of State 251 Corresponding States 252 Summary 254 Problems 255 Conceptual Problems 255 Numerical Problems 257

CHAPTER 5

The Thermodynamic Web

265 Learning Objectives 265 5.1 Types of Thermodynamic Properties 265 Measured Properties 265 Fundamental Properties 266 Derived Thermodynamic Properties 266 5.2 Thermodynamic Property Relationships 267 Dependent and Independent Properties 267 Hypothetical Paths (revisited) 268 Fundamental Property Relations 269 Maxwell Relations 271 Other Useful Mathematical Relations 272 Using the Thermodynamic Web to Access Reported Data 273 5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities 276 Relation of ds in Terms of Independent Properties T and v and Independent Properties T and P 276 Relation of du in Terms of Independent Properties T and v 277 Relation of dh in Terms of Independent Properties T and P 281 Alternative Formulation of the Web using T and P as Independent Properties 287 5.4 Departure Functions 290 Enthalpy Departure Function 290 Entropy Departure Function 293 5.5 Joule-Thomson Expansion and Liquefaction 298 Joule-Thomson Expansion 298 Liquefaction 301 5.6 Summary 304 5.7 Problems 305 Conceptual Problems 305 Numerical Problems 307

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Contents ◄ xiii CHAPTER 6

Phase Equilibria I: Problem Formulation

315

Learning Objectives 315 6.1 Introduction 315 The Phase Equilibria Problem 316 6.2 Pure Species Phase Equilibrium 318 Gibbs Energy as a Criterion for Chemical Equilibrium 318 Roles of Energy and Entropy in Phase Equilibria 321 The Relationship Between Saturation Pressure and Temperature: The Clapeyron Equation 327 Pure Component Vapor–Liquid Equilibrium: The Clausius–Clapeyron Equation 328 6.3 Thermodynamics of Mixtures 334 Introduction 334 Partial Molar Properties 335 The Gibbs–Duhem Equation 340 Summary of the Different Types of Thermodynamic Properties 342 Property Changes of Mixing 343 Determination of Partial Molar Properties 357 Relations Among Partial Molar Quantities 366 6.4 Multicomponent Phase Equilibria 367 The Chemical Potential—The Criteria for Chemical Equilibrium 367 Temperature and Pressure Dependence of μi 370 6.5 Summary 372 6.6 Problems 373 Conceptual Problems 373 Numerical Problems 377

CHAPTER 7

Phase Equilibria II: Fugacity

391 Learning Objectives 391 7.1 Introduction 391 7.2 The Fugacity 392 Definition of Fugacity 392 Criteria for Chemical Equilibria in Terms of Fugacity 395 7.3 Fugacity in the Vapor Phase 396 Fugacity and Fugacity Coefficient of Pure Gases 396 Fugacity and Fugacity Coefficient of Species i in a Gas Mixture 403 The Lewis Fugacity Rule 411 Property Changes of Mixing for Ideal Gases 412 7.4 Fugacity in the Liquid Phase 414 Reference States for the Liquid Phase 414 Thermodynamic Relations Between γi 422 Models for γi Using gE 428

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Equation of State Approach to the Liquid Phase 449 7.5 Fugacity in the Solid Phase 449 Pure Solids 449 Solid Solutions 449 Interstitials and Vacancies in Crystals 450 7.6 Summary 450 7.7 Problems 452 Conceptual Problems 452 Numerical Problems 454 CHAPTER 8

Phase Equilibria III: Applications 466 Learning Objectives 466 8.1 Vapor–Liquid Equilibrium (VLE) 467 Raoult’s Law (Ideal Gas and Ideal Solution) 467 Nonideal Liquids 475 Azeotropes 484 Fitting Activity Coefficient Models with VLE Data 490 Solubility of Gases in Liquids 495 Vapor–Liquid Equilibrium Using the Equations of State Method 501 8.2 Liquid 1 a 2 —Liquid 1 b 2 Equilibrium: LLE 511 8.3 Vapor–Liquid 1 a 2 — Liquid 1 b 2 Equilibrium: VLLE 519 8.4 Solid–Liquid and Solid–Solid Equilibrium: SLE and SSE 523 Pure Solids 523 Solid Solutions 529 8.5 Colligative Properties 531 Boiling Point Elevation and Freezing Point Depression 531 Osmotic Pressure 535 8.6 Summary 538 8.7 Problems 540 Conceptual Problems 540 Numerical Problems 544 CHAPTER 9

Chemical Reaction Equilibria 562 Learning Objectives 562 9.1 Thermodynamics and Kinetics 563 9.2 Chemical Reaction and Gibbs Energy 565 9.3 Equilibrium for a Single Reaction 568 9.4 Calculation of K from Thermochemical Data 572 Calculation of K from Gibbs Energy of Formation 572 The Temperature Dependence of K 574 9.5 Relationship Between the Equilibrium Constant and the Concentrations of Reacting Species 579

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xiv ► Contents

9.6

9.7

9.8

9.9 9.10

The Equilibrium Constant for a Gas-Phase Reaction 579 The Equilibrium Constant for a Liquid-Phase (or Solid-Phase) Reaction 586 The Equilibrium Constant for a Heterogeneous Reaction 587 Equilibrium in Electrochemical Systems 589 Electrochemical Cells 590 Shorthand Notation 591 Electrochemical Reaction Equilibrium 592 Thermochemical Data: Half-Cell Potentials 594 Activity Coefficients in Electrochemical Systems 597 Multiple Reactions 599 Extent of Reaction and Equilibrium Constant for R Reactions 599 Gibbs Phase Rule for Chemically Reacting Systems and Independent Reactions 601 Solution of Multiple Reaction Equilibria by Minimization of Gibbs Energy 610 Reaction Equilibria of Point Defects in Crystalline Solids 612 Atomic Defects 613 Electronic Defects 616 Effect of Gas Partial Pressure on Defect Concentrations 619 Summary 624 Problems 626 Conceptual Problems 626 Numerical Problems 628

B.4 Superheated Water Vapor 653 B.5 Subcooled Liquid Water 659 APPENDIX C

Lee–Kesler Generalized Correlation Tables 660 C.1 Values for z102 C.2 Values for z112

660 662 102

ideal gas

hTr,Pr 2 hTr,Pr

C.3 Values for B

R

RTc

112

ideal gas

C.4 Values for B C.5 Values for B

hTr,Pr 2 hTr,Pr

R

RTc

gas sTr,Pr 2 sideal Tr,Pr

R

C.6 Values for B

R

666

102

R

668 112

ideal gas

sTr,Pr 2 sTr,Pr

664

R

670

C.7 Values for log 3 w102 4 672 C.8 Values for log 3 w112 4 674 APPENDIX D

Unit Systems D.1 D.2

676

Common Variables Used in Thermodynamics and Their Associated Units 676 Conversion between CGS (Gaussian) units and SI units 679

APPENDIX E

ThermoSolver Software 680 APPENDIX A

Physical Property Data 639 A.1 Critical Constants, Acentric Factors, and Antoine Coefficients 639 A.2 Heat Capacity Data 641 A.3 Enthalpy and Gibbs Energy of Formation at 298 K and 1 Bar 643

E.1 Software Description 680 E.2 Corresponding States Using The Lee–Kesler Equation of State 683 APPENDIX F

References 685 F.1 Sources of Thermodynamic Data 685 F.2 Textbooks and Monographs 686

APPENDIX B

Steam Tables 647

Index

687

B.1 Saturated Water: Temperature Table 648 B.2 Saturated Water: Pressure Table 650 B.3 Saturated Water: Solid-Vapor 652

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► CHAPTER

1 Measured Thermodynamic Properties and Other Basic Concepts The Buddha, the Godhead, resides quite as comfortably in the circuits of a digital computer or the gears of a cycle transmission as he does on the top of a mountain or the petal of a flower. To think otherwise would be to demean the Buddha—which is to demean oneself. This is what I want to talk about in this Chautauqua. –Zen and the Art of Motorcycle Maintenance, by Robert M. Pirsig Learning Objectives To demonstrate mastery of the material in Chapter 1, you should be able to: ► Define the following terms in your own words: • Universe, system, surroundings, and boundary • Open system, closed system, and isolated system • Thermodynamic property, extensive and intensive properties • Thermodynamic state, state and path functions • Thermodynamic process; adiabatic, isothermal, isobaric, and isochoric processes • Phase and phase equilibrium • Macroscopic, microscopic, and molecular-length scales • Equilibrium and steady-state Ultimately, you need to be able to apply these concepts to formulate and solve engineering problems. ► Relate the measured thermodynamic properties of temperature and pressure to molecular behavior. Describe phase and chemical reaction equilibrium in terms of dynamic molecular processes. ► Apply the state postulate and the phase rule to determine the appropriate independent properties to constrain the state of a system that contains a pure species. ► Given two properties, identify the phases present on a PT or a Pv phase diagram, including solid, subcooled liquid, saturated liquid, saturated vapor, and superheated vapor and two-phase regions. Identify the critical point and

1

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2 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts triple point. Describe the difference between saturation pressure and vapor pressure. ► Use the steam tables to identify the phase of a substance and find the value of desired thermodynamic properties with two independent properties specified, using linear interpolation if necessary. ► Use the ideal gas model to solve for an unknown measured property given measured property values.

► 1.1 THERMODYNAMICS Science changes our perception of the world and contributes to an understanding of our place in it. Engineering can be thought of as a profession that creatively applies science to the development of processes and products to benefit humankind. Thermodynamics, perhaps more than any other subject, interweaves both these elements, and thus its pursuit is rich with practical as well as aesthetic rewards. It embodies engineering science in its purest form. As its name suggests, thermodynamics originally treated the conversion of heat to motion. It was first developed in the nineteenth century to increase the efficiency of engines—specifically, where the heat generated from the combustion of coal was converted to useful work. Toward this end, the two primary laws of thermodynamics were postulated. However, in extending these laws through logic and mathematics, thermodynamics has evolved into an engineering science that comprises much greater breadth. In addition to the calculation of heat effects and power requirements, thermodynamics can be used in many other ways. For example, we will learn that thermodynamics forms the framework whereby a relatively limited set of collected data can be efficiently used in a wide range of calculations. We will learn that you can determine certain useful properties of matter from measuring other properties and that you can predict the physical (phase) changes and chemical reactions that species undergo. A tribute to the wide applicability of this subject lies in the many fields that consider thermodynamics part of their core knowledge base. Such disciplines include biology, chemistry, physics, geology, oceanography, materials science, and, of course, engineering. Thermodynamics is a self-contained, logically consistent theory, resting on a few fundamental postulates that we call laws. A law, in essence, compresses an enormous amount of experience and knowledge into one general statement. We test our knowledge through experiment and use laws to extend our knowledge and make predictions. The laws of thermodynamics are based on observations of nature and taken to be true on the basis of our everyday experience. From these laws, we can derive the whole of thermodynamics using the rigor of mathematics. Thermodynamics is self-contained in the sense that we do not need to venture outside the subject itself to develop its fundamental structure. On one hand, by virtue of their generality, the principles of thermodynamics constitute a powerful framework for solving a myriad of real-life engineering problems. However, it is also important to realize the limitations of this subject. Equilibrium thermodynamics tells us nothing about the mechanisms or rates of physical or chemical processes. Thus, while the final design of a chemical or biological process requires the study of the kinetics of chemical reactions and rates of transport, thermodynamics defines the driving force for the process and provides us with a key tool in engineering analysis and design. We will pursue the study of thermodynamics from both conceptual and applied viewpoints. The conceptual perspective enables us to construct a broad intuitive foundation that provides us the ability to address the plethora of topics that thermodynamics

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1.2 Preliminary Concepts—The Language of Thermo ◄ 3

spans. The applied approach shows us how to actually use these concepts to solve problems of practical interest and, thereby, also enhances our conceptual understanding. Synergistically, these two tacks are intended to impart a deep understanding of thermodynamics.1 In demonstrating a deep understanding, you will need to do more than regurgitate isolated facts and find the right equation to “plug and chug.” Instead, you will need to search for connections and patterns in the material, understand the physical meaning of the equations you use, and creatively apply the fundamental principles that have been covered to entirely new problems. In fact, it is through this depth of learning that you will be able to transfer the synthesized information you are learning in the classroom and usefully and creatively apply it to new problems in the field or in the lab as a professional chemical engineer.

► 1.2 PRELIMINARY CONCEPTS—THE LANGUAGE OF THERMO In engineering and science, we try to be precise with the language that we use. This exactness allows us to translate the concepts we develop into quantitative, mathematical form.2 We are then able to use the rules of mathematics to further develop relationships and solve problems. This section introduces some fundamental concepts and definitions that we will use as a foundation for constructing the laws of thermodynamics and quantifying them with mathematics.

Thermodynamic Systems In thermodynamics, the universe represents all measured space. It is not very convenient, however, to consider the entire universe every time we need to do a calculation. Therefore, we break down the universe into the region in which we are interested, the system, and the rest of the universe, the surroundings. The system is usually chosen so that it contains the substance of interest, but not the physical apparatus itself. It may be of fixed volume, or its volume may change with time. Similarly, it may be of fixed composition, or the composition may change due to mass flow or chemical reaction. The system is separated from the surroundings by its boundary. The boundary may be real and physical, or it may be an imaginary construct. There are times when a judicious choice of the system and its boundary saves a great deal of computational effort. In an open system both mass and energy can flow across the boundary. In a closed system no mass flows across the boundary. We call the system isolated if neither mass nor energy crosses its boundaries. You will find that some refer to an open system as a control volume and its boundary as a control surface. For example, say we wish to study the piston–cylinder assembly in Figure 1.1. The usual choice of system, surroundings, and boundary are labeled. The boundary is depicted by the dashed line just inside the walls of the cylinder and below the piston. The system contains the gas within the piston–cylinder assembly but not the physical housing. The surroundings are on the other side of the boundary and comprise the rest of the universe. Likewise the system, surroundings, and boundary of an open system are labeled in Figure 1.2. In this case, the inlet and outlet flow streams, labeled “in” and “out,” respectively, allow mass to flow into and out of the system, across the system boundary. 1

For more discussion on deep learning vs. shallow learning in engineering education, see Philip C. Wancat, “Engineering Education: Not Enough Education and Not Enough Engineering,” 2nd International Conference on Teaching Science for Technology at the Tertiary Level, Stockholm, Sweden, June 14, 1997. 2 It can be argued that the ultimate language of science and engineering is mathematics.

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4 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts

Surroundings m Psurr

m

System

P1 T1 v1 =

V1 n

Boundary

State 1

Figure 1.1 Schematic of a piston–cylinder assembly. The system, surroundings, and boundary are delineated.

Properties The substance contained within a system can be characterized by its properties. These include measured properties of volume, pressure, and temperature. The properties of the gas in Figure 1.1 are labeled as T1, the temperature at which it exists; P1, its pressure; and v1, its molar volume. The properties of the open system depicted in Figure 1.2 are also labeled, Tsys and Psys. In this case, we can characterize the properties of the fluid in the inlet and outlet streams as well, as shown in the figure. Here n˙ represents the molar flow rate into and out of the system. As we develop and apply the laws of thermodynamics, we will learn about other properties; for example, internal energy, enthalpy, entropy, and Gibbs energy are all useful thermodynamic properties. Thermodynamic properties can be either extensive or intensive. Extensive properties depend on the size of the system while intensive properties do not. In other words, extensive properties are additive; intensive properties are not additive. An easy way to test whether a property is intensive or extensive is to ask yourself, “Would the value for this property change if I divided the system in half?” If the answer is “no,” the property is intensive. If the answer is “yes,” the property is extensive. For example, if we divide the system depicted in Figure 1.1 in half, the temperature on either side remains the same. Thus, the value of temperature does not change, and we conclude that temperature is intensive. Many properties can be expressed in both extensive and intensive forms. We must be careful with our nomenclature to distinguish between the different forms of these properties. We will use a capital letter for the extensive form of such a thermodynamic property. For example, extensive volume would be V of 3 m3 4 . The intensive form will be lowercase. We denote molar volume with a lowercase v 3 m3 /mol 4 and specific volume by v^ 3 m3 /kg 4 . On the other hand, pressure and temperature are always intensive and are written P and T, by convention. Surroundings

n in

in Tin Pin vin

Tsys Psys

out

System

n out

Tout Pout vout Boundary

Figure 1.2 Schematic of an open system into and out of which mass flows. The system, surroundings, and boundary are delineated.

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1.2 Preliminary Concepts—The Language of Thermo ◄ 5

Processes The thermodynamic state of a system is the condition in which we find the system at any given time. The state fixes the values of a substance’s intensive properties. Thus, two systems comprised of the same substance whose intensive properties have identical values are in the same state. The system in Figure 1.1 is in state 1. Hence, we label the properties with a subscript “1.” A system is said to undergo a process when it goes from one thermodynamic state to another. Figure 1.3 illustrates a process instigated by removing a block of mass m from the piston of Figure 1.1. The resulting force imbalance will cause the gas to expand and the piston to rise. As the gas expands, its pressure will drop. The expansion process will continue until the forces once again balance. Once the piston comes to rest, the system is in a new state, state 2. State 2 is defined by the properties T2, P2 and v2. The expansion process takes the system from state 1 to state 2. As the dashed line in Figure 1.3 illustrates, we have chosen our system boundary so that it expands with the piston during the process. Thus, no mass flows across the boundary and we have a closed system. Alternatively, we could have chosen a boundary that makes the volume of the system constant. In that case, mass would flow across the system boundary as the piston expands, making it an open system. In general, the former choice is more convenient for solving problems. Similarly, a process is depicted for the open system in Figure 1.2. However, we view this process slightly differently. In this case, the fluid enters the system in the inlet stream at a given state “in,” with properties Tin, Pin, and vin. It undergoes the process in the system and changes state. Thus, it exits in a different state, with properties Tout, Pout, and vout. During a process, at least some of the properties of the substances contained in the system change. In an adiabatic process, no heat transfer occurs across the system boundary. In an isothermal process, the temperature of the system remains constant. Similarly, isobaric and isochoric processes occur at constant pressure and volume, respectively.

m

m

Psurr

m

Process

P1 T1 V v1 = 1 n State 1

P2 T2 V v2 = 2 n

State 2

Figure 1.3 Schematic of a piston–cylinder assembly undergoing an expansion process from state 1 to state 2. This process is initiated by removal of a block of mass m.

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6 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts

Hypothetical Paths The values of thermodynamic properties do not depend on the process (i.e., path) through which the system arrived at its state; they depend only on the state itself. Thus, the change in a given property between states 1 and 2 will be the same for any process that starts at state 1 and ends at state 2. This aspect of thermodynamic properties is very useful in solving problems; we will exploit it often. We will devise hypothetical paths between thermodynamic states so that we can use data that are readily available to more easily perform computation. Thus, we may choose the following hypothetical path to calculate the change in any property for the process illustrated in Figure 1.3: We first consider an isothermal expansion from P1, T1 to P2, T1. We then execute an isobaric cooling from P2, T1 to P2, T2. The hypothetical path takes us to the same state as the real process—so all the properties must be identical. Since properties depend only on the state itself, they are often termed state functions. On the other hand, there are quantities that we will be interested in, such as heat and work, that depend on path. These are referred to as path functions. When calculating values for these quantities, we must use the real path the system takes during the process.

Phases of Matter A given phase of matter is characterized by both uniform physical structure and uniform chemical composition. It can be solid, liquid, or gas. The bonds between the atoms in a solid hold them in a specific position relative to other atoms in the solid. However, they are free to vibrate about this fixed position. A solid is called crystalline if it has a longrange, periodic order. The spatial arrangement in which the atoms are bonded is termed the lattice structure. A given substance can exist in several different crystalline lattice structures. Each different crystal structure represents a different phase, since the physical structure is different. For example, solid carbon can exist in the diamond phase or the graphite phase. A solid with no long-range order is called amorphous. Like a solid, molecules within the liquid phase are in close proximity to one another due to intermolecular attractive forces. However, the molecules in a liquid are not fixed in place by directional bonds; rather, they are in motion, free to move relative to one another. Multicomponent liquid mixtures can form different phases if the composition of the species differs in different regions. For example, while oil and water can coexist as liquids, they are considered separate liquid phases, since their compositions differ. Similarly, solids of different composition can coexist in different phases. Gas molecules show relatively weak intermolecular interactions. They move about to fill the entire volume of the container in which they are housed. This movement occurs in a random manner as the molecules continually change direction as they collide with one another and bounce off the container surfaces. More than one phase can coexist within the system at equilibrium. When this phenomenon occurs, a phase boundary separates the phases from each other. One of the major topics in chemical thermodynamics, phase equilibrium, is used to determine the chemical compositions of the different phases that coexist in a given mixture at a specified temperature and pressure.

Length Scales In this text, we will refer to three length scales: the macroscopic, microscopic, and molecular. The macroscopic scale is the largest; it represents the bulk systems we observe in everyday life. We will often consider the entire macroscopic system to be in a uniform thermodynamic state. In this case, its properties (e.g., T, P, v) are uniform throughout the

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1.3 Measured Thermodynamic Properties ◄ 7

system. By microscopic, we refer to differential volume elements that are too small to see with the naked eye; however, each volume element contains enough molecules to be considered as having a continuous distribution of matter, which we call a continuum. Thus, a microscopic volume element must be large enough for temperature, pressure, and molar volumes to have meaningful values. Microscopic balances are performed over differential elements, which can then be integrated to describe behavior in the macroscopic world. We often use microscopic balances when the properties change over the volume of the system or with time. The molecular 3 scale is that of individual atoms and molecules. At this level the continuum breaks down and matter can be viewed as discrete elements. We cannot describe individual molecules in terms of temperature, pressure, or molar volumes. Strictly speaking, the word molecule is outside the realm of classical thermodynamics. In fact, all of the concepts developed in this text can be developed based entirely on observations of macroscopic phenomena. This development does not require any knowledge of the molecular nature of the world in which we live. However, we are chemical engineers and can take advantage of our chemical intuition. Molecular concepts do account qualitatively for trends in data as well as magnitudes. Thus, they provide a means of understanding many of the phenomena encountered in classical thermodynamics. Consequently, we will often refer to molecular chemistry to explain thermodynamic phenomena.4 The objective is to provide an intuitive framework for the concepts about which we are learning.

Units By this time, you are probably experienced in working with units. Most science and engineering texts have a section in the first chapter on this topic. In this text, we will mainly use the Système International, or SI units. The SI unit system uses the primary dimensions m, s, kg, mol, and K. Details of different unit systems can be found in Appendix D. One of the easiest ways to tell that an equation is wrong is that the units on one side do not match the units on the other side. Probably the most common errors in solving problems result from dimensional inconsistencies. The upshot is: Pay close attention to units! Try not to write a number down without the associated units. You should be able to convert between unit systems. It is often easiest to put all variables into the same unit system before solving a problem. How many different units can you think of for length? For pressure? For energy?

►1.3 MEASURED THERMODYNAMIC PROPERTIES We have seen that if we specify the property values of the substance(s) in a system, we define its thermodynamic state. It is typically the measured thermodynamic properties that form our gateway into characterizing the particular state of a system. Measured thermodynamic properties are those that we obtain through direct measurement in the lab. These include volume, temperature, and pressure.

Volume (Extensive or Intensive) Volume is related to the size of the system. For a rectangular geometry, volume can be obtained by multiplying the measured length, width, and height. This procedure gives us the extensive form of volume, V, in units of 3 m3 4 or [gal]. We purchase milk and gasoline in volume with this form of units. 3

Some fields of science such as statistical mechanics use the term microscopic for what we call molecular. While this objective can often be achieved formally and quantitatively through statistical mechanics and quantum mechanics, we will opt for a more qualitative and descriptive approach reminiscent of the chemistry classes you have taken.

4

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8 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts Volume can also be described as an intensive property, either as molar volume, v 3 m3 /mol 4 , or specific volume, v^ 3 m3 /kg 4 . The specific volume is the reciprocal of density, r 3 kg/m3 4 . If a substance is distributed continuously and uniformly throughout the system, the intensive forms of volume can be determined by dividing the extensive volume by the total number of moles or the total mass, respectively. Thus, V n

(1.1)

V 1 5 r m

(1.2)

v5 and, v^ 5

If the amount of substance varies throughout the system, we can still refer to the molar or specific volume of a microscopic control volume. However, its value will change with position. In this case, the molar volume of any microscopic element can be defined: V v 5 lim ¢ ≤ V S Vr n

(1.3)

where Vr is the smallest volume over which the continuum approach is still valid and n is the number of moles.

Temperature (Intensive) Temperature, T, is loosely classified as the degree of hotness of a particular system. No doubt, you have a good intuitive feel for what temperature is. When the temperature is 90°F in the summer, it is hotter than when it is 40°F in the winter. Likewise, if you bake potatoes in an oven at 400°F, they will cook faster than at 300°F, apparently since the oven is hotter. In general, to say that object A is hotter than object B is to say TA . TB. In this case, A will spontaneously transfer energy via heat5 to B. Likewise if B is hotter than A, TA , TB, and energy will transfer spontaneously from B to A. When there is no tendency to transfer energy via heat in either direction, A and B must have equal hotness and TA 5 TB.6 A logical extension of this concept says that if two bodies are at equal hotness to a third body, they must be at the same temperature themselves. This principle forms the basis for thermometry, where a judicious choice of the third body allows us to measure temperature. Any substance with a measurable property that changes as its temperature changes can then serve as a thermometer. For example, in the commonly used mercury in glass thermometer, the change in the volume of mercury is correlated to temperature. For more accurate measurements, the pressure exerted by a gas or the electric potential of junction between two different metals can be used. Molecular View of Temperature On the molecular level, temperature is proportional to the average kinetic energy of the individual atoms (or molecules) in the system. All matter contains atoms that are in motion.7 Species in the gas phase, for example, move chaotically through space with finite 5

In Chapter 2, we will more carefully define heat.

6

This relation for temperature is often referred to as the “zeroth law of thermodynamics.” However, in the spirit of Rudolph Clausius, we will view thermodynamics in terms of two fundamental laws of nature that are represented by the first and second laws of thermodynamics. 7 Except in the ideal case of a perfect solid at a temperature of absolute zero.

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1.3 Measured Thermodynamic Properties ◄ 9

velocities. (What would happen to the air in a room if its molecules weren’t moving?) They can also vibrate and rotate. Figure 1.4 illustrates individual molecular velocities. The piston–cylinder assembly depicted to the left schematically displays the velocities of a set of individual molecules. Each arrow represents the velocity vector with the size of the arrow proportional to a given molecule’s speed. The velocities vary widely in magnitude and direction. Furthermore, the molecules constantly redistribute their velocities among themselves when they elastically collide with one another. In an elastic collision, the total kinetic energy of the colliding atoms is conserved. On the other hand, a particular molecule will change its velocity; as one molecule speeds up via collision, however, its collision partner slows down. Since the molecules in a gas move at great speeds, they collide with one another billions of times per second at room temperature and pressure. An individual molecule frequently speeds up and slows down as it undergoes these elastic collisions. However, within a short period of time the distribution of speeds of all the molecules in a given system becomes constant and well defined. It is termed the Maxwell–Boltzmann distribution and can be derived using the kinetic theory of gases. The right-hand side of Figure 1.4 shows the Maxwell–Boltzmann distributions for O2 at 300 K and 1000 K. The y-axis plots the fraction of O2 molecules at the speed given on the x-axis. At a given temperature, the fraction of molecules at any given speed does not change.8 In fact, the temperature of a gas is only strictly defined for an aggregate of gas molecules that have obtained this characteristic distribution. Similarly, for a microscopic volume element to be considered a continuum, it must have enough molecules for the gas to approximate this distribution. The distribution at the higher temperature has shifted to higher speeds and flattened out.

m T = 300 K

Pure gas

Fraction of gas

m

T = 1000 K

0

500

1000 1500 Speed (m/s)

2000

2500

Figure 1.4 A schematic representation of the different speeds molecules have in the gas phase. The left-hand side shows molecules flying around in the system. The right-hand side illustrates the Maxwell–Boltzmann distributions of O2 molecules at 300 K and 1000 K.

8

The macroscopic and the molecular scales present an interesting juxtaposition. At a well-defined temperature, there is one distinct distribution of molecular speeds. Thus, we say we have only one macrostate possible. However, if we keep track of all the individual molecules, we see there are many ways to arrange them within this one macrostate; that is, any given molecule can have many possible speeds. In Chapter 3, we will see that entropy is a measure of how many different molecular configurations a given macrostate can have.

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10 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts Kinetic theory shows the temperature is proportional to the average translational molecular kinetic energy, emolecular , which is related to the mean-square molecular velocity: K T

T1

T = T1

T =T2 > T1

T = T1 m

m Saturated vapor a

a a

a

a

a a

Saturated liquid

m

m

m

m

m

m

Saturated vapor a a a a a a a a a a a a Saturated liquid

b

Saturated vapor b a a b a b a a ba b b a b Liquid a

Saturated vapor a a b b a b a a a a a b a a b b a a Liquid a

1. The case depicted assumes ideal gas behavior and that species b does not condense into the liquid phase:

Figure 1.8 Graphical representation of the saturation pressure of pure a and the vapor pressure of a in a mixture of a and b. Two temperatures, T1 and T2 are shown.

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24 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts if the pressure is lower, it will be a single-phase vapor. At a higher temperature, the saturation pressure will be higher, as depicted for T2 in Figure 1.8. For example, at 303 K (30°C), water has a saturation pressure of 4.25 kPa. This incremental increase in temperature nearly doubles the saturation pressure. A schematic illustrating when we use vapor pressure is shown for the two systems depicted on the right in Figure 1.8, where the vapor phase contains a mixture of species a and b. The vapor pressure of species a represents its contribution to the total pressure of the mixture. The two temperatures shown, T1 and T2, are identical to those for pure species a on the left of the figure. For convenience, we assume species b is does not noticeably condense in the liquid and that the vapor behaves as an ideal gas.17 Then the vapor pressure of species a is identical to the corresponding saturation pressure at the same temperature. For example, now consider an open container of water sitting in a room at 293 K and 1 bar. Some of the water will evaporate and go into the air. The partial pressure of water at equilibrium with the air will be equal to the saturation pressure of pure a, 2.34 kPa. Since water is but one of many components in the mixture, we say water has a vapor pressure of 2.34 kPa. In contrast, the total pressure of the system is around 1 atm. The vapor pressure presented in Figure 1.8 depends only on the temperature of the water, not on the total pressure of the system. In other words, the vapor pressure of a is independent of how much b is present. While we can use saturation pressures to determine the vapor pressure in a given mixture, the term saturation pressure refers to the pure species. You should learn the difference between saturation pressure and vapor pressure because they are often confused.

The Critical Point A magnified view of the upper part of the Pv phase diagram is shown in Figure 1.9. Four isotherms are shown. Along all four isotherms, the volume increases as the pressure decreases. At the lowest two temperatures, the isotherms start in the liquid phase. In the liquid phase, the volume change is relatively small as the pressure drops. Along a given isotherm, the pressure decreases until it reaches the saturation pressure. This point is marked by the intersection with the left side of the liquid–vapor dome. At this point, any increase in volume leads to a two-phase liquid–vapor mixture, where the value of liquid volume is given by the intersection of the isotherm with the left side of the dome and the vapor volume is given by the intersection of the isotherm with the right side of the dome. The pressure remains constant in the two-phase region, since P and T are no longer independent. After complete vaporization, the pressure again decreases. The corresponding increase in volume of the vapor is noticeably larger than that of the liquid. As the temperatures of the isotherms increase, the saturated liquid volumes get larger and the saturated vapor volumes get smaller. Finally, at the critical point, located at the top of the liquid–vapor dome, the values of vl and vv become identical. The critical point represents a unique state and is identified with the subscript “c.” Thus, it is constrained by the critical temperature, Tc and the critical pressure, Pc. Values for these critical properties of many pure substances are reported in Appendix A. The critical point represents the point at which liquid and vapor regions are no longer distinguishable. The critical point is also labeled in the depictions in Figure 1.6.

17

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We will learn how to treat the more general case in Chapters 7 and 8.

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1.6 The PvT Surface and Its Projections for Pure Substances ◄ 25 P

T > Tc T = Tc Critical T < Tc point

Liquid

Liquid-vapor Vapor

v

Figure 1.9 Magnified view of the Pv diagram. Four isotherms are shown—two below the critical temperature (subcritical), one at the critical temperature, and one above the critical temperature (supercritical).

The critical isotherm goes through an inflection point at the critical point. Mathematically, this condition can be written as: 'P ≤ 50 'v Tc

(1.16)

'2P ≤ 50 'v2 Tc

(1.17)

¢

and,

¢

The partial derivatives in Equations (1.16) and (1.17) specify that we need to keep the temperature constant at its value at the critical point. The isotherm above the critical point is representative of a supercritical fluid. This isotherm continuously decreases in pressure as the volume increases. A supercritical fluid has partly liquidlike characteristics (e.g., high density) and partly vaporlike characteristics (compressibility, high-diffusivity). Not surprisingly, there are many interesting engineering applications for substances in this state. There can be confusion between the terms gas and vapor. We refer to a gas as any form of matter that fills the container; it can be either subcritical or supercritical. When we speak of vapor, it is gas that if isothermally compressed will condense into a liquid and is, therefore, always subcritical. EXAMPLE 1.2 Determination of Location of a TwoPhase System on a Phase Diagram

Consider a two-phase system at a specified T that contains 20% vapor, by mass, and 80% liquid. Identify the state on a Tv phase diagram. Explain why graphical determination of the state is termed the lever rule. SOLUTION The quality of the system, defined as the fraction of matter in the vapor, is 0.2. The molar volume can be written in terms of the quality according to Equation (1.15): v 5 vl 1 x 1 vv 2 vl 2 5 vl 1 0.2 1 vv 2 vl 2

(E1.2A) (Continued)

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26 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts

P T = T1 Liquid Liquid – vapor Vapor

P1

vI

Fraction vapor = 0.2

Fraction liquid = 0.8

vv

Fulcrum of the lever v

v1

Figure E1.2 The state of the system of Example 1.2 on a Pv phase diagram. The lever rule is indicated. The fraction of vapor is found by solving Equation (E1.2A): 0.2 5

v 2 vl vv 2 vl

(E1.2B)

Equation (E1.2B) can be interpreted as follows: The fraction of vapor is the ratio of the difference between the system volume and the volume of the other phase to the difference between vapor and liquid volumes. Similarly, the fraction of liquid is given by: 0.8 5

vv 2 v vv 2 vl

The fraction of liquid is the ratio of the difference between the volume of the other phase and the system volume to the difference between vapor and liquid volumes. This result is graphically presented in Figure E1.2. The overall composition of the system in the two-phase region is shown on top of the fulcrum. The intersection of the horizontal line with the liquid vapor dome gives the molar volumes of the liquid and vapor phases. The horizontal line is referred to as the tie line. The fraction of each phase present is obtained by taking the length of the line segment to the other phase and dividing by the total length of the line between one phase and the other. The line segment representing the liquid is four times greater than that representing the vapor.

►1.7 THERMODYNAMIC PROPERTY TABLES As we have seen, if we specify two independent intensive properties of a pure substance, the state of the system is constrained. Thus, any other thermodynamic property is restricted to only one possible value. Not surprisingly, for commonly used substances, such as water, tables of thermodynamic properties have been constructed that tabulate a set of useful properties. Appendix B reproduces a portion of the “steam tables,” where six intensive properties of water are tabulated.18 Recall that we use the term water to indicate the chemical species H2O in any phase: solid, liquid, or gas. 18

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J.H. Keenan, F. G. Keys, P. G. Hill, J. G. Moore, Steam Tables (New York: Wiley, 1969).

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1.7 Thermodynamic Property Tables ◄ 27

The tabulated thermodynamic properties include the measured properties T, P, and v^, as well as three other properties we will learn about in Chapters 2 and 3—the specific internal energy, u^ , the specific enthalpy, h^ , and the specific entropy, s^ . The values reported for the latter three properties are not reported as absolute values but rather as the change in that property relative to a well-defined reference state. The reference state used for the steam tables is as a liquid at the triple point of water. At this state, both internal energy and entropy are defined as zero.19 Similar property tables are available in the literature or the National Institute of Standards and Technology (NIST) website20 for many other common species as well, including Ar, N2 , O2, CH4, C2H4, C2H6, C3H8, C4H10, and several refrigerants (NH3, R-12, R-13, R-14, R-21, R-22, R-23, R-113, R-114, R-123, R-134a . . .). The reference state must be consistent when doing a thermodynamic calculation. Care should be used when taking values for a substance from different sources, since they sometimes use different reference states. The steam tables are organized according to the phase in which water exists in the state of interest. Figure 1.10 shows a PT diagram for water with the corresponding appendices where the thermodynamic property data are located. Appendices B.1 and B.2 report data for the saturated vapor–liquid region. Since the pressure and temperature are no longer independent in this two-phase region, if we specify either property, we fix the other. Appendix B.1 presents data for saturated vapor and liquid water at even intervals of temperature. We use this appendix when the value of temperature is known. Appendix B.2 also presents data for saturated water, but in terms of round numbers of pressure. For each of the other properties in the tables (v^, u^ , h^ , and s^ ), values are presented for the liquid, l; the vapor, v; and the difference between the vapor and the liquid; D 5 v 2 l. By analogy to Equation (1.15), any property value of the system is scaled by the quality. For example the specific internal energy can be found according to: u^ 5 1 1 2 x 2 u^ l 1 xu^ v 5 u^ l 1 xu^ D

(1.18)

P

Appendix B.5 Subcooled liquid

ate Appendix B.4 tur Appendices Superheated B.1 and B.2 vapor

Sa

Solid r

po

va id-

ol

ds

Triple point

te ura

t

Sa

vap

id-

qu d li

Critical point or

Appendix B.3 T

Figure 1.10 Illustration of the steam tables available for different phases of H2O. 19 In fact, the “third law” of thermodynamics specifies that the entropy of a perfect crystal is zero at a temperature of absolute zero. This principle allows absolute entropies to be defined and calculated. However, the entropies presented in the steam tables do not make use of the third law; thus, the values presented are relative. 20 P. J. Linstrom and W. G. Mallard, Eds., NIST Chemistry WebBook, NIST Standard Reference Database Number 69, March 2003, National Institute of Standards and Technology, Gaithersburg MD, 20899 (http://webbook.nist.gov/chemistry/fluid).

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28 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts Property values for superheated water vapor and subcooled liquid water are presented in Appendices B.4 and B.5, respectively. In these regions, T and P are independent, so if we specify both, we constrain the state of the system. The tables are organized first according to pressure, then by temperature at each specified pressure. With T and P specified, values of the other properties (v^, u^ , h^ , and s^ ) are reported. The data in the superheated steam tables begin with the saturated state, whereas the data in the subcooled water tables end at saturation. The usefulness of the steam tables is that in knowing any two independent properties of water, we may look up the values of any of the other properties to solve engineering prohlems. Often the states of a system do not fall exactly at a value reported in the steam tables. In this case, it is necessary to interpolate between two adjacent entries on a given table. If a linear relation is assumed, the unknown variable, y, is related to the known variable, x, by: y 2 ylow x 2 xlow 5 yhigh 2 ylow xhigh 2 xlow

(1.19)

where the values for “high” and “low” are read off the table. Solving for y gives: y 5 ylow 1 1 yhigh 2 ylow 2 ¢

x 2 xlow ≤ xhigh 2 xlow

(1.20)

If neither of the two independent properties used to specify the system falls on a value reported in the steam tables, we must first interpolate with respect to one property. Then we use those interpolated values to interpolate for the other property. This process is termed double interpolation. Determination of volume using double interpolation is illustrated in Example 1.3.

EXAMPLE 1.3 Double Interpolation of Steam Tables

Use linear interpolation to estimate the specific volume of water at P 5 1.4 MPa and T 5 333°C with data from the steam tables. SOLUTION Table E1.3 reports the appropriate entries for specific volume from the steam tables which bracket 1.4 MPa and 333°C. The specific volume at 333°C, v^ T5333, is found by applying Equation (1.20) as follows: v^ T5333 5 v^ T5320 1 1 v^ T5360 2 v^ T5320 2 a

333 2 320 b 360 2 320

(E1.3)

Applying Equation (E1.3), we get: v^ T5333 5 0.27414 3 m3 /kg 4 At 1.5 MPa, we get: v^ T5333 5 0.18086 3 m3 /kg 4 To find the specific volume at 1.4 MPa, we must now interpolate between these two values: v^ P51.4 5 v^ P51 1 1 v^ P51.5 2 v^ P51 2 a

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1.4 2 1 b 5 0.19951 3 m3 /kg 4 1.5 2 1

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1.7 Thermodynamic Property Tables ◄ 29

TABLE E1.3 Values of v^ from the Steam Tables P 5 1 MPa T 5 320°C T 5 360°C

EXAMPLE 1.4 Improvement in Interpolation of Example 1.3

m3 /kg

0.2678 0.2873 m3 /kg

P 5 1.5 MPa 0.1765 m3 /kg 0.1899 m3 /kg

Based on the ideal gas law, reestimate the specific volume of water at P 5 1. 4 MPa and T 5 333°C using data from the steam tables. SOLUTION If we apply the ideal gas model, we have: v5

RT P

The molar volume, v, is proportional to T. The first linear interpolation of Example 1.3 is consistent with this result, so again we have v^ T5333 5 0.18086 3 m3 /kg 4 . However, the ideal gas law shows molar volume is inversely proportional to pressure, so it is better to interpolate in (1/P). Thus, the second interpolation becomes:

v^ P51.4 5 v^ P51 1 1 v^ P51.5

1 1 2 P Plow 2 v^ P21 2 § ¥ 5 0.19418 3 m3 /kg 4 1 1 2 Phigh Plow

The difference in the value found in Example 1.2 and Example 1.4 is: %diff 5

0.19951 2 0.19418 3 100 5 2.7% 0.19418

This example illustrates the effectiveness of letting physical principles guide our mathematical procedures so that we can make better engineering estimates.

EXAMPLE 1.5 Determination of the State of a System using the Steam Tables

A rigid tank of volume 1.0 L contains 2.5 g of pure water at 70°C and is closed to the surroundings. The water is then heated and the temperature rises. (a) Determine the initial state of the system. (b) Determine the temperature at which the water in the tank is all vaporized. SOLUTION (a) We label the initial state, state 1, and the final state, state 2. We need two independent intensive properties to constrain the state of the system. In addition to temperature, we can find the specific volume of water in the tank: v^ 1 5

1000 g V1 1L 0.001 m3 m3 5 3 3 5 0.4 m1 2.5 g kg L kg

where we have put the specific volume in units consistent with data in the steam tables. From o the steam tables (Appendix B, Table B.1), we see that at a temperature of 70 C, the saturated (Continued)

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30 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts

m3 m3 R, and the saturate liquid volume is v^ l 5 0.001023 B R. kg kg Because the volume of the system is in between these two values, we have a mixture of saturated liquid and saturated vapor at 70°C. To constrain the system, we need to determine the quality (i.e., the proportion in the vapor): From Equation (1.15)

vapor volume is v^ v 5 5.042 B

x1 5

v^ 1 2 v^ l 5 0.079 v^ v 2 v^ l

Approximately 8% of the mass of water is in the vapor. (b) Because this process occurs in a closed system and the tank is rigid, the specific volume remains constant (i.e., v^ 2 5 v^ 1). We therefore wish to find the state at which the specific volume of saturated vapor is 0.4 m3 /kg. From Table B.1, the temperature table for saturated water from the steam tables, we see that this value of specific volume is between temperatures of 145°C and 150°C: T [°C]

v^ v B

m3 R kg

145

0.44632

150

0.39278

By linear interpolation, we get: T2 5 Tv^ 50.4 5 Tv^ 50.44632 1 1 Tv^ 50.39278 2 Tv^ 50.44632 2 ¢

0.4 2 0.44632 ≤ 5 145.8 3 °C 4 0.39278 2 0.44632

►1.8 SUMMARY The material in Chapter 1 forms the conceptual foundation on which we will construct our understanding of thermodynamics. We will formulate thermodynamics by identifying the state that a system is in and by looking at processes by which a system goes from one state to another. We are interested in both closed systems, which can attain thermodynamic equilibrium, and open systems. The state postulate and the phase rule allow us to identify which independent, intensive thermodynamic properties we can choose to constrain the state of the system. If we also know the amount of matter present, we can determine the extensive properties in the system. Thermodynamic properties are also called state functions. Since they do not depend on path, we may devise a convenient hypothetical path to calculate the change in their values between two states. Conversely, other quantities, such as heat or work, are path functions. The measured properties T, P, and v are especially useful in determining the thermodynamic state since we can measure them in the lab. Projections of the PvT surface, in the form of PT, Pv, and Tv phase diagrams, allow us to identify whether the system is in a single phase or is in phase equilibrium between two or three different phases. The pressures and temperatures of each of the phases in equilibrium are identical. Moreover, when a pure species contains two phases, T and P are not independent; therefore, the saturation pressure takes a unique value for any given temperature. The saturation pressure of a pure species can be related to its vapor pressure in a mixture. The ideal gas model allows us to relate P, v, and T for gases at low pressure or high temperature. On a molecular level, temperature is proportional to the average kinetic energy of the individual atoms (or molecules) in the system. Pressure can be viewed as the normal force per unit area exerted by the molecules as they elastically collide with a system boundary. Phase equilibrium

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1.9 Problems ◄ 31 can be described as a dynamic process on the molecular level, where the number of molecules leaving the surface of one phase is exactly balanced by the number arriving. Likewise, chemical reaction equilibrium can be described as the dynamic balance of forward and reverse reactions.

►1.9 PROBLEMS Conceptual Problems 1.1 You have a glass of water and a glass of wine, as shown in the figure. You perform the following processes. (1) transfer 1 teaspoon of water to the glass of wine and mix thoroughly; then (2) transfer 1 teaspoon of this contaminated wine to the water. Now both the water and the wine are contaminated. Which of the following is true? Explain. Hint: it may be useful to consider this problem in terms of an extensive property. (a) The volume of water contaminating the wine is greater than the volume of wine contaminating the water. (b) The volume of water contaminating the wine is equal to the volume of wine contaminating the water. (c) The volume of wine contaminating the water is greater than the volume of water contaminating the wine. 1 teaspoon

1 teaspoon 1

2

Water

Wine

1.2 You have a jar of 90 nickels and a jar of 90 pennies. You perform the following processes. (1) Transfer 10 nickels to the jar of pennies and mix thoroughly; then (2) transfer 10 coins from the contaminated pennies back to the jar with nickels. Which of the following is true? Explain. (a) The amount of pennies in the jar of mostly nickels is greater than the amount of nickels in the jar of mostly pennies. (b) The amount of pennies in the jar of mostly nickels is the equal to the amount of nickels in the jar of mostly pennies. (c) The amount of nickels in the jar of mostly pennies is greater than the amount of pennies in the jar of mostly nickels. 1.3 Shown in the following figure is a process from which Species A is isothermally compressed from 0.5 bar and 300 K to 1 bar. The insets of each state, which are of equal volume, contain a “molecular view” of species A.

Molecular view

A

A

Process

Molecular view

Ideal gas A

A

A

T1 = 300 K

T1 = 300 K

P1 = 0.5 bar

P2 = 1 bar

A

A A

A

A State 1

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State 2

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32 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts Next consider the open system shown in the following figure. Species A flows steadily through the system and expands through the valve from an inlet state at 5 bar and 300 K to an exit state at 1 bar. You may assume Species A acts as an ideal gas. In analogy to the closed system depicted earlier, the equal volume insets are shown in this figure. Fill in the corresponding “molecular views” of Species A. Explain your answer. T1 = 300 K P1 = 5 bar

P2 = 1 bar

Molecular view

Molecular view

valve

1.4 Go to the teaching or research labs at your university and determine three ways temperature is measured and three ways pressure is measured. 1.5 Consider a binary mixture of a light gas a with mass ma and a heavy gas b with mass mb at temperature T. How does the mean-square velocity of the two species compare? Which species, on average, moves faster? 1.6 Consider the system sketched below:

Cu Block Large reservoir of boiling water (100° C)

Large reservoir of ice water (0° C)

(a) After a short time, is this system in equilibrium? (b) After a long time? (c) After a very long time? 1.7 Consider a tightly capped water bottle containing two phases with a small amount of liquid water and saturated air. If the bottle is left in the sun on a hot day and the temperature increases, what happens to the amount of water in the liquid? Explain. 1.8 A rigid, sealed container initially contains pure water at 100°C. Some of the water is in the liquid phase, and some is in the vapor phase (i.e., as steam). Air is then injected into the system in an isothermal process at constant volume. What happens? 1.9 Using language a high school student could understand, explain the difference between saturation pressure and vapor pressure. 1.10 This question addresses the two piston-cylinder assemblies depicted on the left of Figure 1.8 that illustrate the concept of saturation pressure. The piston on the right is at twice the pressure of the system on the left; however, if you count molecules of species a, there are less than twice the number on the right (i.e., the number does not increase directly proportional to temperature). Is this a mistake? Explain.

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1.9 Problems ◄ 33 1.11 Sometimes a lid to a pot used for cooking fits “too well” and can be difficult to remove after the pot cools down. Why do you think this is happening? 1.12 I thoroughly inflated a bag of soccer balls last summer. However, when I brought them out to play this winter, they all were underinflated. Discuss the possible reasons. 1.13 Relative humidity is defined as the ratio of the mass of water in air divided by the mass of water at saturation. Compare the water content in the air on a day on which the temperature is 10°C with 90% relative humidity to a day at 30°C and 50% relative humidity. Which day has higher water content? 1.14 When a system contains regions that differ in physical structure or chemical composition, an overall value can be assigned to its properties. Consider the system, system 1, shown below. It contains na molecules in state a and nb molecules in state b. (a) Develop an expression for the extensive volume V1 in terms of na, nb, and the volumes of each homogeneous region Va and Vb. (b) Develop an expression for the intensive molar volume v1 in terms of na, nb and the molar volumes of each homogeneous region va and vb. (c) Generalize the result of part (a) to come up with an expression for any extensive property K1 in terms of na, nb, and the extensive properties Ka and Kb. (d) Generalize the result of part (b) to come up with an expression for the intensive form of the property in part (c), k1, in terms of na, nb, and the intensive properties ka and kb.

System 1 State b State a Va

Vb +

=

na

na nb

nb

1.15 Consider two systems of ideal gases. System I consists of pure gas A at a given pressure and temperature. System II contains a mixture of gases A and B at the same temperature and pressure. If the molecular weight of gas B is larger than gas A, how does the molar density 1 mol/cm3 2 of system I compare to system II? 1.16 Consider two systems of ideal gases. System I consists of pure gas A at a given pressure and temperature. System II contains a mixture of gases A and B at the same temperature and pressure. If the molecular weight of gas B is larger than gas A, how does the mass density 1 g/cm3 2 of system I compare to system II? 1.17 You can breathe in approximately 2 L of air into your lungs. What volume of helium do you think you can breathe in? Explain. 1.18 A “pressure cooker” is a device that allows food to be cooked at pressures that are higher than atmospheric pressure. Explain why this device changes how your food is cooked. 1.19 The ideal gas model is one example of an equation of state. Why do you think it is termed an equation of state? 1.20 Consider a system containing water in the following states. What phases are present? (a) P 5 10 3 bar 4 ; T 5 170 3 °C 4 (b) v^ 5 3 3 m3 /kg 4 ; T 5 70 3 °C 4 (c) P 5 60 3 bar 4 ; v^ 5 0.05 3 m3 /kg 4 (d) P 5 5 3 bar 4 ; s 5 7.0592 3 kJ/ 1 kg K 2 4

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34 ► Chapter 1. Measured Thermodynamic Properties and Other Basic Concepts

Numerical Problems 1.21 Estimate the speed at which the average oxygen molecule is moving in the room that you are in. 1.22 The Reamur temperature scale uses the normal freezing and boiling points of water to define 0°and 80°, respectively. What is the value of room temperature (22°C) on the Reamur scale? 1.23 At what temperature does water boil on the top of Mount Everest, elevation z 5 8848 m? Recall that the dependence of pressure with altitude is given by: P 5 Patm exp ¢2

MWgz RT

≤

where, Patm is atmospheric pressure, g is the gravitational acceleration, and MW is the molecular weight of the gas. 1.24 Water is cooled in a rigid closed container from the critical point to 10 bar. Determine the quality of the final state. 1.25 Using linear interpolation, estimate the specific volume of water under the following conditions using data from the steam tables: (a) P 5 1.9 3 MPa 4 ; T 5 250 3 °C 4 (b) P 5 1.9 3 MPa 4 ; T 5 300 3 °C 4 (c) P 5 1.9 3 MPa 4 ; T 5 270 3 °C 4 Look up the specific volumes of water that correspond to cases (a), (b), and (c) on the website http://webbook.nist.gov/chemistry/fluid/ and report their values. Comment on the agreement between the two sources. 1.26 Determine the mass of 1 L of saturated liquid water at 25°C. How do you think this value compares to the mass of 1L of subcooled liquid water at 25°C and atmospheric pressure? 1.27 Determine the temperature, quality, and internal energy of 5 kg of water in a rigid container of volume 1 m3 at a pressure of 2 bar. 1.28 A rigid container of volume 1 m3 contains saturated water at 1 MPa. If the quality is 0.10, what is the volume occupied by the vapor? 1.29 Use the data in the steam tables to plot the vapor–liquid dome on a Pv diagram. It is useful to plot v on a log scale. 1.30 Calculate the volume of water using the ideal gas model under the following conditions. Then report the percent error when compared to the values reported in the steam tables. (a) P 5 1.01 3 bar 4 ; T 5 100 3 °C 4 (b) P 5 1 3 bar 4 ; T 5 500 3 °C 4 (c) P 5 100 3 bar 4 ; T 5 500 3 °C 4 (d) P 5 100 3 bar 4 ; T 5 1000 3 °C 4 1.31 How many moles of air are in the room in which you are sitting? What is its mass? 1.32 Consider a gas at 20°C and 1 bar. The molecules may be considered to be hard spheres with a diameter of 3 Å. Estimate the percentage of the available volume they occupy. 1.33 You want to keep your house dry enough so that water does not condense on your walls at night. If the temperature gets down to 40°F at night, what is the maximum allowable density of water in the room during the day when the room is at 70°F? 1.34 Consider a rigid, thick-walled tube that is filled with H2O liquid and vapor at 0.1 MPa. After it is sealed, it is heated so that it passes through its critical point. What fraction of the mass in the tube is liquid? 1.35 As best as you can, estimate the specific volume of water at each of the following conditions. Justify your answer. (a) 2 bar and 200°C (b) 2 bar and 100°C

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1.9 Problems ◄ 35 1.36 A rigid container contains 1 kg of water at 90°C. If 200 g of the water are in the liquid phase and the rest is vapor, determine the pressure in the tank and the volume of the tank. 1.37 40 g of water are sealed in a 10 L container at 300°C. As accurately as you can, determine the pressure of the container. 1.38 A rigid container of 100 L contains saturated water at 100°C. The water is heated, and it reaches the critical point. Determine the initial mass of water in the tank and its quality. 1.39 A piston-cylinder assembly contains 0.5 kg of water at 50°C and 500 kPa. It is then isobarically heated until all the water is vaporized. What is the final temperature and volume?

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► CHAPTER

2 The First Law of Thermodynamics Learning Objectives To demonstrate mastery of the material in Chapter 2, you should be able to: ► State and illustrate by example the first law of thermodynamics—that is, the conservation of mass and energy—and its basic concepts, including conversion of energy from one form to another and the transfer of energy from the surroundings to the system by heat, work, and flow of mass. ► Write the integral and differential forms of the first law for (1) closed systems and (2) open systems under steady-state and transient (uniform-state) conditions. Convert these equations between intensive and extensive forms and between mass-based and molar forms. Given a physical problem, evaluate which terms in the equation are important and which terms are negligible or zero and determine whether the ideal gas model or property tables should be used to solve the problem. ► Apply the first law of thermodynamics to identify, formulate, and solve engineering problems for adiabatic and isothermal processes in the following types of systems: rigid tank, expansion/compression in a piston– cylinder assembly, nozzle, diffuser, turbine, pump, heat exchanger, throttling device, filling or emptying of a tank, and Carnot power and refrigeration cycles. ► Create an appropriate hypothetical path to solve these problems with available data. ► Describe the molecular basis for internal energy, heat transfer, work, and heat capacity. ► Describe the difference between a reversible process and an irreversible process, and, given a process, evaluate whether it is reversible or irreversible. ► Explain why it is convenient to use the thermodynamic property enthalpy for (1) streams flowing into and out of open systems and (2) closed systems at constant pressure. Describe the role of flow work and shaft work in opensystem energy balances. ► Describe the energy changes associated with sensible heat, latent heat, and chemical reaction on both a macroscopic and a molecular level. Calculate their enthalpy changes using available data such as heat capacity, enthalpies of vaporization, fusion and sublimation, and enthalpies of formation.

36

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2.1 The First Law of Thermodynamics ◄ 37

► 2.1 THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics states that while energy can be changed from one form to another, the total quantity of energy, E, in the universe is constant.1 This statement can be quantitatively expressed as follows: DEuniv 5 0

(2.1)

However, it is very inconvenient to consider the entire universe every time we need to do a calculation. As we have seen, we can break down the universe into the region in which we are interested (the system) and the rest of the universe (the surroundings). The system is separated from the surroundings by its boundary. We can now restate the first law by saying that the energy change of the system must be equal to the energy transferred across its boundaries from the surroundings. Energy can be transferred by heat, Q, by work, W, and, in the case of open systems, by the energy associated with the mass that flows into and out of the system. In essence, the first law then lets us be accountants of the energy in the system, by tracking the “deposits” to and “withdrawals” from the surroundings in much the same way as you would account for the balance of money in your bank account. We will consider explicit forms of the first law for closed and open systems shortly.

Forms of Energy The energy within a system can be transformed from one form to another.

►EXERCISE

Name the three common forms that energy is divided into. See if you can define each form: Energy is classified according to three specific forms: (1) The macroscopic kinetic energy, EK is the energy associated with the bulk (macroscopic) motion of the system S as a whole. For example, an object of mass m moving at velocity V has a kinetic energy given by: EK 5

1 S2 mV 2

(2.2)

(2) The macroscopic potential energy, EP, is the energy associated with the bulk (macroscopic) position of the system in a potential field. For example, an object in the Earth’s gravitational field has a potential energy given by: EP 5 mgz

(2.3)

where z is the height above the surface of the Earth and g is the gravitational constant.2 (3) The internal energy, U, is the energy associated with the motion, position, and chemical-bonding configuration of the individual molecules of the substances within the system. Energy is not an absolute quantity but rather is only defined relative to a reference state, so we must be careful to identify the particular reference state that we are using. As you read this text, what is your kinetic energy (assuming you are not riding the bus)? 1

Nuclear reaction presents an interesting case where energy and mass are coupled. However, we will not address this case in this text. 2 Potentials due to surface tension or electric or magnetic fields can also be included.

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38 ► Chapter 2. The First Law of Thermodynamics If you answer zero, you are correct in the context of a well-defined reference state, the Earth. However, if instead we had considered the sun to be the reference state, the answer would be quite different. The Earth is in motion with a velocity of 30,000 m/s around the sun, and your kinetic energy is on the order of 106 J! In the case of kinetic and S potential energy, we usually define EK (i.e., V ) as zero when there is no motion relative to the Earth and EP (i.e., z) as zero at the surface of the Earth. In fact, these reference states are so obvious they are sometimes implicitly assumed. In this text, we will be careful to identify references states explicitly. This effort will become useful in the case of U, where there is more than one convenient reference state. What is the reference state used for U in the steam tables? In your introductory physics course, you focused primarily on changes associated with the first two forms of energy. Since solving first-law problems involves proficiency in relating different forms of energy, it is instructive to review a typical example from mechanics that you may have seen in introductory physics. It is presented in the context in which we will approach problems in this text. You should realize that, as chemical engineers, the form of energy that we will primarily focus on is internal energy, which is not covered in the following example.

EXAMPLE 2.1 Typical Energy Problem in Mechanics

If a large stone is dropped from a cliff 10 m high, how fast will it be going when it hits the ground? SOLUTION

From Equation (2.1), we have: DE 5 DEK 1 DEP 5 0

(E2.1A)

We can define the process as throwing the stone off the cliff. Typically, we set up our problem by labeling the thermodynamic states between which our process is going. We can define state 1 as the initial state when the stone is at the top of the cliff and state 2 as when the stone hits the ground. Using Equations (2.2) and (2.3), Equation (E2.1A) becomes: 1 S 1 S a mV 22 2 mV12b 1 1 mgz2 2 mgz1 2 5 0 2 2 By convention, we define the change in a property, D, as “final − initial.” Now, using the reference states described above, we get: 0 0

1 S2 1 S2 a mV 2 2 mV1 b 1 1 mgz2 2 mgz1 2 5 0 2 2 or,

1 S2 mV2 2 mgz1 5 0 2 Finally, solving for the final velocity yields: S

V2 5 "2gz1 5 "2 1 9.8 3 m/s2 4 2 1 10 3 m 4 2 5 14 3 m/s 4 This value is equivalent to 31 miles/hr. Ouch! Note that our reference state of energy is arbitrary and if we had chosen different reference states, we would still get the same answer.

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2.1 The First Law of Thermodynamics ◄ 39

One philosophical comment: Energy is inherently a very abstract quantity; it is very hard to say exactly what energy is. However, you (hopefully!) are comfortable using it in the context of the problem above. The ability to apply the abstract property energy to solve engineering problems lies in your experience with it. This experience typically translates into relative comfort in including internal energy in the energy balance to solve first-law problems. Keep in mind that soon we will introduce thermodynamic properties with which you have less experience, such as entropy, S, and Gibbs energy, G. These properties are fundamentally no more challenging than energy to learn to work with; however, you may have an initial period of discomfort as you gain experience. The trick is that when you work with any thermodynamic property enough, you get used to it and become proficient at solving the type of problems for which it is useful.

Ways We Observe Changes in U As we mentioned, internal energy, U, is an important form of energy for chemical engineering applications. Originally internal energy was viewed simply as all forms of energy that are not associated with bulk motion or bulk position. However, it is instructive to take advantage of our knowledge of chemistry and consider a molecular perspective on internal energy. Internal energy encompasses all forms of molecular energy, including the kinetic and potential energies of the molecules themselves. A change in internal energy can present itself in several macroscopic manifestations; that is, molecular energy is noticed in the real world in different ways. Changes in internal energy can result in the following: 1. changes in temperature, for example, Tlow h Thigh 2. changes in phase, for example, solid h gas 3. changes in chemical structure, that is, chemical reaction 1 N2 1 3H2

h

2NH3 2

A change in internal energy that leads to a change in temperature is often termed sensible heat. Likewise, we often refer to a change in internal energy that results in phase transformations as latent heat. Let’s now examine how we can relate changes in molecular, chemical energy to the three macroscopic features described above. There are two general components of internal energy—molecular potential energy and molecular kinetic energy. Molecular potential energy can be either intermolecular (between different molecules) and intramolecular (within the same molecule) in character. Remember that we use the term molecular when we are describing what the species are doing on the atomic scale while we use the term macroscopic to describe behavior in the bulk (or molar) scale of the world in which we live. Like macroscopic kinetic energy, by molecular kinetic energy we mean motion; in this case, the motion of the individual molecules in a system. The type of motion depends on the phase the species are in. In the gas phase, for example, the molecules are flying around at significant speeds. This motion is referred to as translational motion because the individual molecules are going somewhere—that is, translating—even though the bulk of the gas may not be. Not all molecules have the same speed, but, at equilibrium, their speeds vary according to a Maxwell–Boltzmann distribution. Do you know how fast the average oxygen molecule that you are now breathing is moving? [ANSWER: Oxygen molecules at room temperature are moving, on average, as fast as a jet plane, that is approximately 450 m/s.] Additionally, diatomic and polyatomic molecules (as opposed to atoms) can vibrate and rotate, which provide additional modes of molecular kinetic energy—vibrational and rotational motion. As we saw in Chapter 1, the measured macroscopic property

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40 ► Chapter 2. The First Law of Thermodynamics temperature is representative of how fast the gas molecules are moving in the system (formally temperature is proportional to the mean-square velocity). However, the molecules’ speed is directly related to the molecular kinetic energy, which, in turn, is one part of the internal energy. Hence, as a gas increases in temperature, the average velocity of its molecules increases (the molecules move faster), so it has greater internal energy (see point 1 above). In contrast, solids do not have translational motion; their main mode of molecular kinetic energy is in the form of vibrations. The vibrations of the atoms in a solid are called phonons. Again, on the one hand, phonons represent part of the internal (or molecular) energy, and, on the other hand, they are directly related to the temperature of the solid. So the faster the solid is vibrating, the greater the temperature and the greater the internal energy. Consider next a phase change, such as the sublimation of a solid into a vapor. An example with which you may have experience is the case with CO2 (dry ice) at atmospheric temperature and pressure. The solid is held together by bonds between the molecules. Often the bonding in a solid results from electrostatic attraction of the molecules, that is, molecular potential energy. The attraction between molecules adds stability and decreases the molecular energy of the system. On the other hand, the molecules in the vapor are much farther away from one another and have little or no attraction. Thus, the vapor phase is representative of higher internal energy relative to the solid at the same temperature. To sublimate, the molecular energy of the attraction of the bonds must be overcome; that is, energy must be added to the system, resulting in higher internal energy. Similar arguments hold for the melting of solids and evaporation of liquids (see point 2 above). Finally, consider chemical reaction. In this case, the chemical bonds between the atoms in the molecules of the reactants are broken and replaced by the bonds of the products. For example, ammonia is produced by the reaction: N2 1 3H2

h

2NH3

A triple bond between N atoms and three single bonds between H atoms are replaced by six N i H bonds (three each for two molecules produced). The strength of a chemical bond is determined by the overlap of the valence electrons of the constituent atoms. Thus the energetics change as the atoms are rearranged, resulting in a change in the thermodynamic property, U. In this case, the product is lower in energy (a lot more stable), so U is reduced (see point 3 above).3

Internal Energy of an Ideal Gas We next explore the property dependence for the internal energy for an ideal gas. As we learned with the discussion of point 1 earlier, internal energy consists of two components, molecular kinetic energy and molecular potential energy, and temperature is directly related to one of them, the molecular kinetic energy. Because an ideal gas exhibits no intermolecular forces (see Section 1.3), its molecular potential energy is constant (assuming no chemical reactions). Therefore, barring chemical reaction, the internal energy, u, depends only on motion of the molecules, or the temperature. Hence, uideal gas 5 f 1 T only 2

(2.4)

Said another way, the internal energy for an ideal gas is independent of the position of the molecules. In Chapter 4, we will consider the thermodynamic properties of real 3

In Chapter 3, we will learn that thermodynamic entropy also plays a role in determining how far a reaction will proceed.

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2.1 The First Law of Thermodynamics ◄ 41

gases, where the molecules are close enough to experience the effect of intermolecular forces. In this case, two independent intensive properties are needed to specify the state of a system with constant composition, such as: ureal gas 5 u 1 T, v 2 or, ureal gas 5 u 1 T, P 2

EXAMPLE 2.2 Equivalent Energy Stored in u

In Example 2.1, we considered the potential energy of a stone at the top of a 10-m cliff. When it fell, it gained kinetic energy, resulting in a velocity around 31 miles/hr. Consider now an equivalent mass of water initially at 25ºC. How hot would the water end up if its internal energy increased by the same amount? SOLUTION If we write Equation (2.2) on a per-mass basis, we have: 1S De^ K 5 V22 5 98 3 J/kg 4 5 0.098 3 kJ/kg 4 2 We have used specific energy and converted the units to be consistent with the steam tables. Again we will use state 1 to denote the initial state and state 2 to denote the final state. Liquid water is subcooled at 25ºC and 1 atm; however, we do not expect the properties of a liquid to be significantly affected by pressure. Therefore, we can use the temperature tables for saturated liquid water at 25ºC (which is technically at a pressure of 0.03 atm).4 From Appendix B.1: u^ l,1 5 104.86 3 kJ/kg 4 The problem statement says the internal energy of the water increases by the same amount as the energy of the stone, that is, Du^ 5 u^ l,2 2 u^ l,1 5 De^ k 5 0.098 3 kJ/kg 4 So for the final state of water, we have: u^ l,2 5 Du^ 1 u^ l,1 5 104.96 3 J/kg 4 We can now go to the steam tables and determine at which temperature saturated water has this energy. Again we neglect the effect of the pressure difference between the subcooled state and the saturated state. Interpolating, we get: u^ l,2 1 at T2 2 2 u^ 1 1 at 25 3 °C 4 2 104.96 2 104.86 5 0.005 5 125.77 2 104.86 u^ l 1 at 30 3 °C 4 2 2 u^ l 1 at 25 3 °C 4 2 Finally, solving for the final temperature yields: T2 5 25 1 1 0.005 2 5 5 25.02°C The temperature of the water barely changes! Thus the energy stored in a stone 10 m up a cliff corresponds to a negligible amount of internal energy. This example illustrates that a large amount of energy is stored in u relative to the other forms of energy, and, consequently, why we are so interested in internal energy. As engineers, it provides us a large resource to be harvested. 4 We often use this trick to find the properties of subcooled water (or other substances) when the pressure is not appreciably different from the saturation pressure. It will serve you well to catalog your experiences of such tricks for reference for solving problems in the future.

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42 ► Chapter 2. The First Law of Thermodynamics

Work and Heat: Transfer of Energy Between the System and the Surroundings In the sciences we need to be very careful about how we use language and define terms such as work and heat. Both terms refer to the transfer of energy between the surroundings and system. In a closed system, the transfer of energy between the surroundings and the system can only be accomplished by heat or by work. Heat is the transfer of energy by a temperature gradient, whereas all other forms of energy transfer in a closed system occur via work. We generally associate work with something useful being done by (or to) the system. We will examine these terms in more detail below. Work There are many forms of work, for example, mechanical (expansion/compression, rotating shaft), electrical, and magnetic. The most common case of work in engineering thermodynamics is when a force causes a displacement in the boundary of a system. In the case of expansion, for example, the system needs to push the surroundings out of the way to increase the boundary; in this process, the system expends energy. Thus, the system exchanges energy with the surroundings in the form of work. The work, W, can be described mathematically by the line integral of the external force, FE, with respect to the direction of displacement, dx: W 5 3 FE # dx

(2.5)

In contrast to thermodynamic properties, the work on a system depends not only on the initial state, 1, and the final state, 2, of the system, but also on the specific path that it takes. Whenever we calculate the work, we must account for the real path that the system takes. Since work refers to the transfer of energy between the system and the surroundings, it has the same units as energy, such as joules, ergs, BTU, and so on. To complete the definition, we need to choose a sign convention for work. In this text, we will say work is positive when energy is transferred from the surroundings to the system and work is negative when energy is transferred from the system to the surroundings. The definition given by Equation (2.5) is consistent with this sign convention. You should be aware that this sign convention is arbitrary. We choose this convention to be consistent with today’s convention. However, when the first and second laws of thermodynamics were originally formulated, in the context of powering the steam engine, the opposite sign convention was used: Work from the system to the surroundings was defined as positive (since the engineering objective was to get work out of the system to power a train!). When you go to other sources, be careful to note which sign convention is chosen for work or you may get tripped up. A plot of FE vs. x for a general process is shown in Figure 2.1a. The work associated with the process in Figure 2.1a can be obtained from the area under the curve [which is equivalent to graphically integrating the expression in Equation (2.5)]. If the boundary of the system does not move, no work has been done, no matter how large the force is. If the external force is acting on a surface of cross-sectional area A, we can divide and multiply the terms on the right hand side of Equation (2.5) by A as follows: FE W 5 3 # d 1 Ax 2 5 3 PE # dV 5 3 PEdV cos u 5 23 PEdV A

(2.6)

where PE is the external pressure to the surface. The negative sign in Equation (2.6) results, since the external force and displacement vectors are in opposite directions.

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2.1 The First Law of Thermodynamics ◄ 43 PE

FE

1 1

2

2 Work

Work

(a)

x

(b)

V

Figure 2.1 Graphical determination of the value of work for a system that undergoes a process between states 1 and 2 by integrating: (a) FE vs. x; (b) PE vs. V.

Again, the work can be obtained from the area under the appropriate curve, as shown in Figure 2.1b. If Equation (2.6) is written on a molar basis (J/mol), we get: w 5 23 PEdν

(2.7)

Equation (2.7) is often encountered in thermodynamics; the work described by this equation will be referred to as “Pv work.” On a molecular scale, the energy transfer by Pv work can be understood in terms of momentum transfer of the molecules in the system when they bounce off the moving boundary, as discussed in Section 1.3. A piston– cylinder assembly is a common system that is used to obtain work (e.g., in your automobile). Example 2.3 illustrates how work is calculated for such a system.

EXAMPLE 2.3 Calculation of Pv Work in a Piston– Cylinder Assembly

Consider the constant pressure expansion that is illustrated in Figure E2.3. Initially the system contains 1 mole of gas A at 2 bar within a volume of 10 L. The expansion process is initiated by releasing the latch. The gas in the cylinder expands until the pressure of the gas matches the pressure of the surroundings. The final volume is 15.2 L. Calculate work done by the system during this process. SOLUTION The amount of work done can be calculated by applying Equation (2.6): V2

W 5 2 3 PEdV

(2.6)

V1

Since the external pressure is constant, it can be pulled out of the integral: V2

W 5 2PE 3 dV 5 2PE 1 V2 2 V1 2 5 21 bar B

105 Pa 1023 m3 R 1 15.2 2 10 2 LB R 5 2520 J bar L

V1

In this case, the value for work is negative since the system loses energy to the surroundings as a result of this process. The units of pressure and volume have been converted to their SI equivalents in this calculation 3 1 Pa m3 5 1 J 4 .

(Continued)

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44 ► Chapter 2. The First Law of Thermodynamics

PE = 1 bar

PE = 1 bar

Process

P1 = 2 bar V1 = 10 L 1 mol gas A Initial state (1)

P2 = 1 bar V2 = 15.2 L 1 mol gas A

Final state (2)

Figure E2.3 Example of a process in which energy is transferred from the system to the surroundings by Pv work: expansion of a gas in a piston–cylinder assembly. The surroundings are maintained at 1 bar.

Shaft work, Ws, is another important type of work encountered in engineering practice. Often a rotating shaft is used to deliver energy between the system and the surroundings. For example, consider the turbine shown in Figure 2.2. It is designed to convert the internal energy of the working fluid into useful work by means of a shaft. In this case, as the fluid passes through the turbine, it expands and cools, leading to the rotation of the shaft at the end of the turbine. A magnet placed on the end of the turbine also rotates. The changing magnetic field induces an electrical potential, which is used to generate a current that can charge a battery and store energy. Note that depending on how we draw the boundary to our system, the transfer of energy from the system to the surroundings (the shaft work) can be mechanical, magnetic, or electric. However, in all cases we are converting the internal energy of the fluid into useful work. In fact, when we generically use the term shaft work, it may indeed be any of these forms of work. Heat Heat, Q, refers to the transfer of energy between the surroundings and the system where the driving force is provided by a temperature gradient. Energy will transfer spontaneously from the high-temperature region to the low-temperature region. Sometimes this form of energy transfer is part of an engineering design (such as “heating” up your house on a cold day). Often, however, heat provides a path for the unwanted dissipation of energy (such as when your coffee gets cold or your soda gets warm). In the latter case, it is useful to try to insulate the system as well as possible to eliminate unwanted transfer of energy. In the ideal case, the transfer of energy by heat would be reduced to zero. We term such a process adiabatic 1 Q 5 0 2 . Unlike for work, the sign convention for heat has been historically invariant. A positive value indicates that energy is transferred from the surroundings to the system or, colloquially, the system is “heating up.” Alternatively, you may think of a positive value for Q to correspond with an increase in internal energy 1 DU 2 of the system (ignoring work). In Section 2.1 we referred to the effects of changes in U on the system. Those changes that manifested as changes in temperature were termed sensible heat while changes in phase were termed latent heat. This nomenclature is somewhat misleading, since we are referring to changes in the internal energy of the system and not

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2.1 The First Law of Thermodynamics ◄ 45 Fluid out Fluid in e in

Ws N S +

Fluid in Fluid out

i –

battery

e out

Figure 2.2 Schematic of a turbine converting the energy from a flowing fluid into shaft work. The rotating shaft is coupled to a battery by means of a magnet affixed to its end.

“heat,” which more precisely refers to transfer of energy across the system boundary in a specified way. However, this nomenclature is deeply rooted in the literature. There are three modes through which energy can be transferred by a temperature gradient: conduction, convection, and radiation. You will learn how to quantify the rates of these processes in your heat-transfer (or transport processes) class; however, the underlying mechanisms of each process will be described briefly here. It is easiest to think of conduction in terms of a solid body. If you expose the front side of a chunk of quartz glass, for example, to temperature Thigh and the back side to a temperature Tlow, energy will transfer through the glass. On a molecular scale, the phonons on the hot side (remember, phonons refer to the vibrations of the atoms in a solid) will be vibrating at a higher velocity, that is, with greater energy. However, these atoms are connected to those next to them on the crystal lattice and the region of high-energy lattice vibration will “spread out.” With time, the phonons on the front side will vibrate less vigorously (thus reducing their temperature) while the phonons on the back side increase in energy. The end result is a transfer of energy from Thigh to Tlow. The rate at which the energy transfers via conduction—that is, the rate at which the lattice vibration spreads out—is proportional to a property of the material called the thermal conductivity, k. Glass is not very conductive; it has a thermal conductivity of around 42 W m21 ºC21. On the other hand, metals tend to conduct well. In the case of metals, there is an additional mechanism for conduction of energy—drift of free electrons in the valence band. A thermally conducting material like copper may conduct energy an order of magnitude faster than glass, having a thermal conductivity of 385 W m21 ºC21. Wood, on the other hand, is an effective thermal insulator, having a thermal conductivity around 0.1 W m21 ºC21. Liquids and gases may also transfer energy through conduction. The thermal conductivity of liquids tends to be lower than that of solids, and gases have even lower thermal conductivities than liquids. Typical values range from 0.06–0.6 W m21 ºC21 for most liquids and 0.01–0.07 W m21 ºC21 for most gases. Can you provide a molecular explanation as to why gases have much lower thermal conductivity than solids? Convection is another mechanism by which energy can be transferred between the system and the surroundings via heat. Convection refers to the case of enhanced heat transfer through coupling with fluid flow. For example, consider the example of wanting to cool a bowl of hot soup. One way to enhance the transfer of energy (so that you can eat quickly and your tongue does not get burned) is by blowing on the soup in your spoon. This is an example of convection. When you blow on the soup, the flow of gas carries away hot molecules (moving with high velocity) and replaces them with colder

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46 ► Chapter 2. The First Law of Thermodynamics fluid. Thus, the temperature difference between the soup and the neighboring gas—that is, the driving force for energy transfer—is greater, and cooling occurs more quickly than by conduction alone. Clearly describing convection mathematically is more difficult than describing conduction. Convection depends not only on the conductive properties of the soup and the air but also on the type of flow patterns that are set up. Radiation refers to the transfer of energy by light. By light, we are referring to all wavelengths of electromagnetic radiation, not just the visible portion. All objects above absolute zero radiate light. On a molecular scale, radiation is associated with acceleration of charged particles (electrons and nuclei) near the surface of the object, due to vibration. Have you ever seen a piece of metal that is “red hot”? By being red hot, it is actually cooling by emitting photons in the red portion of the electromagnetic spectrum. Each photon that leaves carries some energy with it. The rate of heat transfer by radiation is a much stronger function of temperature than either conduction or convection. In # conduction and convection, the rate of heat transfer, Q, is proportional to temperature: # 1 conduction and convection 2 Q~T but for radiation: # Q ~ T4

1 radiation 2

Thus, at high temperature, radiation becomes the dominant mode of heat transfer. In this text, we will often lump all three modes of heat transfer together and just define the heat transfer for the system. We use both amounts of energy transferred, # Q [J], and rates of energy transferred Q 3 J/s or W 4 . You should be aware, however, what important modes may be present.

►EXERCISE

Consider a pot of boiling water. Sketch the system, surroundings, and boundary. Identify all the heat-transfer mechanisms associated with this system.

► 2.2 CONSTRUCTION OF HYPOTHETICAL PATHS Internal energy and volume are examples of thermodynamic properties. Because properties depend only on the initial and final states of the system, they are also called state functions. They are independent of the process; that is, they do not depend on path the system takes. (Conversely, heat and work are path dependent, and for these quantities we must follow the actual path the system takes.) To a large degree, thermodynamics is built on the path independence of properties. We can take advantage of this feature by constructing our own paths. We are free to choose any convenient path to calculate the change in a given property. If the path we use for calculation is different from the path the system actually undergoes, we call it a hypothetical path. The change in any property (e.g.Du) for a hypothetical path is the same as for the actual process as long as the system starts and ends in the same states as the actual process. The utility of this construct is enormous, and it is unique to thermodynamics. It allows us to use data like those available in the appendices of this textbook to solve many different problems. Hypothetical paths are constructed to make the calculation easier (or, in some cases, possible at all!). In fact, the ability to construct hypothetical paths between states allows for efficient collection and organization of experimental data. Often a path is constructed so that we can use available physical data. Much of the methodology of this textbook is based on using judicious choices of hypothetical paths to develop theory or obtain

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2.2 Construction of Hypothetical Paths ◄ 47

solutions to engineering problems of interest. We will not go through every possible path that may ever be useful. The goal is to get you to recognize when you can appropriately construct a hypothetical path and have you be able to develop the path and execute such a calculation on your own. Consequently, we will use hypothetical paths as we develop theory and illustrate their use with examples. However, undoubtedly the most useful way to learn about constructing hypothetical paths will be when you need to develop your own hypothetical paths in homework to solve the end-of-chapter problems. As an example of how we might construct such a path, consider a problem where we need to calculate the change in internal energy of a gas going from state 1 to state 2. Figure 2.3 shows the initial and final states on a temperature–volume plot. Drawing a diagram like that shown in Figure 2.3 can enable you to identify useful hypothetical paths. The process the system actually undergoes is depicted by the solid line. However, to solve this problem, we may use any path we want as long as the path starts at the initial state and ends at the final state. For example, we consider the calculation of Du for two hypothetical paths shown by the dashed lines in the figure. First, we consider a two-step hypothetical path, which we call path a. This path consists of first heating the gas at constant volume from T1 to T2 (labeled Step 1a) and then isothermally compressing the gas from v1 to v2 (labeled Step 2a). The calculations are simplified because only one property changes at a time. However, this path may not be good enough to allow us to calculate Du. Often the data that is available for the isothermal heating is for ideal gases (such as the ideal gas heat capacity that we will cover in Section 2.6). In that case, we may need to add another step to the hypothetical path, as shown in path b. In this case, we first expand the system to a very large molar volume, vlarge,where the gas behaves as an ideal gas (labeled Step 1b). The next two steps are similar to path a, where we heat at constant volume (Step 2b) and then compress isothermally (Step 3b). In such a case, we may need to have redrawn our hypothetical path from the choice that we first conceived. Coming up with clever hypothetical paths can turn into quite a creative endeavor and is a key to problem solving and to understanding thermodynamics.

Δu hypothetical path b Step 2b Step 1b

v1

Step 1a

State 1 (T1, v1)

Δuhypothetical path a Δuactual

v2

Step 2a

Molar volume (m3/mol)

Step 3b

vlarge

Actual path State 2 (T2, v2)

T1

T2 Temperature (K)

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Figure 2.3 Plot of a process that takes a system from state 1 to state 2 in Tv space. Three alternative paths are shown: the real path as well as two convenient hypothetical paths.

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48 ► Chapter 2. The First Law of Thermodynamics

►2.3 REVERSIBLE AND IRREVERSIBLE PROCESSES Reversible Processes The concept of a reversible process is crucial in understanding thermodynamics. A definition for reversible processes is as follows: A process is reversible if, after the process occurs, the system can be returned to its original state without any net effect on the surroundings. This result occurs only when the driving force is infinitesimally small.

We will see that to accomplish a reversible process we must be able to reverse the direction of the process at any point and go the other way by changing the driving force by an infinitesimal amount. For a process to be reversible, there can be no friction. We use driving force as a generic term that represents some type of influence on a system to change. If a gas undergoes an expansion process, we must be able to turn the process around at any point and compress it merely by changing the force on the piston by an infinitesimal amount. For example, if we examine the process depicted in Figure E2.3, we realize that this condition cannot be accomplished. If the piston is halfway up, it would take more than an infinitesimal force to turn it around and begin compressing the gas inside; hence, we label the process depicted in Figure E2.3 irreversible. In this case we consider force because it is the driving force for mechanical work, that is, momentum transfer. Likewise, if we are heating a gas, we must be able to turn the process around at any point and cool it merely by changing the temperature by an infinitesimal amount. Temperature difference is the driving force for energy transfer. Reversible processes are never completely realized in real life since they can be accomplished only by changing the driving force to a process by an infinitesimal amount. (Perhaps a pendulum comes close.) These processes represent an idealization. They represent a limiting case, that is, a process that is perfectly executed. However, in engineering these types of idealizations are often useful. For example, it is useful to know how much work we could get out of a system if a process could be executed reversibly. This value tells us the best that we could possibly do. We can then compare how well we really do, and see if it is worth focusing our efforts on improving the process. Good engineering is as much determining where to focus your resources as it is actual problem solving.

Irreversible Processes Real processes are not reversible. They have friction and are carried out with finite driving forces. Such processes are irreversible processes. In an irreversible process, if the system is returned to its original state, the surroundings must be altered. The work obtained in an irreversible process is always less than that obtained in the idealization of a reversible process. Why? To help solidify these abstract ideas, a concrete example is illustrative. We will compare the value of work for six processes. We will label these cases process A through process F. Three processes (A, C, and E) entail isothermal expansion of a piston–cylinder assembly between the same states: state 1 and state 2. The other three (B, D, and F) consist of the opposite process, isothermal compression between state 2 and state 1. An isothermal process results in the limit of fast heat transfer with the surroundings. We could perform a similar analysis on adiabatic processes where there is no energy transfer via heat between the system and the surroundings.

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2.3 Reversible and Irreversible Processes ◄ 49

Process A is illustrated in Figure 2.4. The system contains 1 mole of pure ideal gas. A 1020-kg mass sits on the piston. The surroundings are at atmospheric pressure. The molar volume in state 1 can be found from the area 1 0.1 m2 2 and the height (0,4 m): v1 5 ¢

Az m3 R ≤ 5 0.04 B n 1 mol

Since the piston is originally at rest, the pressure inside the piston can then be found by a force balance: P1 5

mg A

1 Patm 5 2 3 bar 4

where Patm is taken to be 1 bar. These two properties completely constrain the initial state. State 1 is labeled on the Pv diagram in Figure 2.4. Process A is initiated by removing the 1020-kg mass. The pressure of the piston is now greater than that exerted by the surroundings, and the gas within the piston expands. The expansion process continues until once again the pressures equilibrate. The piston then comes to rest in state 2 where the pressure is given by: P2 5 Patm 5 1 3 bar 4 The ideal gas law can be applied to this isothermal process to give: Pv 5 RT 5 const The volume of state 2 is then given by: v2 5

P1v1 m3 R 5 0.08 B P2 mol

State 2 is now constrained by two independent, intensive properties and is also labeled in Figure 2.4. To calculate the work, we must consider the “external” pressure upon which the gas must expand [see Equation (2.7) and discussion]. The piston only starts to move once the 1020-kg block is removed. Hence, at any volume larger than 0.04 m3, the external pressure is only that from the atmosphere. The path of external pressure vs. volume is illustrated in Figure 2.4. Note, we are not saying anything about the pressure in the system but rather we are graphically illustrating what external pressure the gas expands against. To find the work, we apply Equation (2.7): v2

v2

w 5 23 PEdv 5 2Patm3 dv 5 24000 3 J/mol 4 v1

(2.8)

v1

The same result can be found by graphically counting the area under the curve in the Pv diagram in Figure 2.4. The negative sign indicates we get 4 kJ of work out of the system from this expansion process. In our analysis, we have idealized the process to stop precisely where the forces balance. In reality, the piston may undergo damped oscillations in its path toward its

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50 ► Chapter 2. The First Law of Thermodynamics

Weightless, frictionless piston Patm

m = 1020 kg

A = 0.1 m2

Process A: Isothermal expansion

Patm

1 mol of pure, ideal gas

A = 0.1 m2 0.4 m

1 mol of pure, ideal gas

State 2

Fexternal / area or

"external" pressure (bar)

State 1 2

1

1

Process A

2 Work is given by the area under the curve

.04 Molar volume (m3/mol)

.08

Figure 2.4 Schematic of an isothermal expansion process (process A). The corresponding plot of the process on a PEv diagram is shown at the bottom.

final resting position in state 2. In that case, the kinetic energy of the piston causes it to overshoot its final equilibrium position, leading to a molar volume greater than 0.08 3 m3 /mol 4 . Once past this volume, the pressure in the system will be less than the pressure of the surroundings. At some point, the motion stops and the piston reverses direction, back toward the equilibrium position. It may again overshoot the equilibrium position, leading to a molar volume less than 0.08 3 m3 /mol 4 , where it is then again turned around and expands, and so on and so on. This process will inevitably contain a frictional dissipative mechanism that causes the piston to come to rest at state 2. Since the external pressure is the same during these oscillations, the contribution of the oscillating expansions and contractions to the work will exactly cancel, leading to the same value as calculated by Equation (2.8).5,6 In this text, we will ignore the oscillatory behavior of these types of processes and approximate them in the simpler context where the system does not overshoot its final equilibrium state. While it is only an approximation of the real behavior, this simplification proves useful in allowing us to compare irreversible processes with reversible processes. 5

We assume all the energy dissipation occurs within the system. On the other hand, if there were no dissipative mechanism, the piston would oscillate forever. Its kinetic energy could be accounted for by the difference in force between the system pressure and the external pressure. 6

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2.3 Reversible and Irreversible Processes ◄ 51

m = 1020 kg Patm

A = 0.1 m2 Process B: Isothermal compression

m = 1020 kg

Patm

0.8 m A = 0.1 m2 1 mol of pure, ideal gas

0.4 m

1 mol of pure, ideal gas State 1

Fexternal / area or

"external" pressure (bar)

State 2 1

2

Process B

2

1

Work is given by the area under the curve .08

.04 Molar volume (m3/mol)

Figure 2.5 Schematic of an isothermal compression process (process B). The corresponding plot of the process on a PEv diagram is shown at the bottom.

We next want to calculate the work needed to compress the gas from state 2 back to state 1. This process is illustrated in Figure 2.5 and is labeled process B. In process B, we drop the 1020-kg mass back on the piston, originally in state 2. The external pressure now exceeds the pressure of the gas initiating the compression process. The piston goes down until the pressures equilibrate, at state 1. The external pressure vs. molar volume is plotted in Figure 2.5. In this case, the external pressure consists of contributions from both the block and the atmosphere. Again, we are not representing the system pressure in this graph, but rather the force per area that acts on the piston. The work is found by Equation (2.7): v1

w 5 2 3 PEdv 5 2aPatm 1 v2

mg A

v1

b 3 dv 5 8000 c

J mol

d

v2

This value can also be found from the area under the curve. Comparing process A and process B, we see it costs us more work to compress the piston back to state 1 than we got from expanding it to state 2. The net difference in work 1 8000−4000 5 4000 3 J 4 2 in going from state 1 to state 2 and back to state 1 results in a “net effect on the surrounding.”

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52 ► Chapter 2. The First Law of Thermodynamics Examining our definition of a reversible process, we see that processes A and B are irreversible. Next we again consider expansion from state 1 to 2 (process C) and compression from state 2 to 1 (process D), but now we use two 510-kg masses instead of one larger 1020-kg mass. The expansion is carried out as follows: The system is originally in state 1 when the first 510-kg mass is removed. This causes the gas to expand to an intermediate state at a pressure of 1.5 bar and a molar volume of 0.053 m3 /mol. The second 510-kg mass is then removed, completing the expansion to state 2. These three states are shown in the Pv diagram in Figure 2.6. The expansion process is labeled process C and follows arrows from state 1 to the intermediate state to state 2. The work the system delivers to the surroundings is given by: v2

w 5 2 3 PEdv 5 2B aPatm 1

m 2g

A

v1

vint

v2

b 3 dv 1 Patm 3 dvR 5 24667 B v1

J mol

R

vint

Again, the work can be found graphically from the area under the curve. Process C is “better” than process A in that it allows us to extract more work from the system. Of course, we want to get the most work out of a system as possible. The compression process is the opposite of the expansion. With the system in state 2, a 510-kg block is placed on the piston until it compresses to the intermediate state, followed by placement of the second block. The work is found by: v1

w 5 2 3 PEdv 5 2B aPatm 1

m 2g

A

v2

vint

b 3 dv 1 aPatm 1 v2

mg A

v1

b 3 dvR 5 6667 B

J mol

R

vint

In analogy to the expansion process, process D is “better” than process B in that it costs us less work to compress the system back to state 1. When we have to put work into a system, we want it to be as small as possible. However, it still costs us more work to compress from state 2 to state 1 than we get out of the expansion, so these processes are still irreversible. We did “better” in both expansion and compression processes when we divided the 1020-kg mass into two parts. Presumably we would do better by dividing it into four parts, and even better by dividing it into eight parts, and so on. If we want to do the best possible, we can divide the 1020-kg mass into many “differential” units and take them off one at a time for expansion or place them on one at a time for compression. These processes are labeled process E and process F, respectively, and are illustrated in Figure 2.7. At each differential step, the system pressure is no more than 'mg/A different than the external pressure. Thus, to a close approximation: P 5 PE The process paths are illustrated in the Pv diagram in Figure 2.7. To find the work, we integrate over the external pressure. However, since the external pressure is equal to the system pressure, we get: v2

v2

v2

w 5 2 3 PEdv 5 2 3 Pdv 5 2 3 v1

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v1

J P1v1 v2 dv 5 2P1v1 ln ¢ ≤ 5 25545 c d v v1 mol

(2.9)

v1

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2.3 Reversible and Irreversible Processes ◄ 53

m = 510 kg

m = 510 kg

Patm

m = 510 kg A = 0.1 m2 m = 510 kg

Process C: Isothermal expansion

Patm

"external" pressure (bar)

1 mol of pure, ideal gas

0.4 m

1 mol of pure, ideal gas

Fexternal / area or

0.8 m

Process D: Isothermal compression

A = 0.1 m2

1

2

Process D

1

Process C

2 Work is given by the area under the curve

.04 Molar volume (m3/mol)

.08

Figure 2.6 Schematic of two-step isothermal expansion (process C) and compression (process D) processes. The corresponding plots of the process on a PEv diagram is shown at the bottom.

Similarly the work of compression is: v1

v1

v1

w 5 2 3 PEdv 5 2 3 Pdv 5 2 3 v2

v2

J P1v1 v1 dv 5 2 P1v1 ln ¢ ≤ 5 5545 c d v v2 mol

v2

In processes E and F, we can return to state 1 by supplying the same amount of work that we got from the system in the expansion process. Hence, we can go from state 1 to 2 and back to state 1 without a net effect on the surroundings. From our definition, we see that these processes are reversible. In a reversible process, we are never more than slightly out of equilibrium. At any point during the expansion, we could turn the process around the other way and compress the piston by adding differential masses instead of removing them. Moreover, we get more work out of the reversible expansion than the irreversible expansions. Similarly, the reversible compression costs us less work than the irreversible processes. The reversible case represents the limit of what is possible in the real world—it

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54 ► Chapter 2. The First Law of Thermodynamics

mT = 1020 kg mT = 1020 kg ∂m

Patm

∂m A = 0.1 m2 Process E: Isothermal expansion

Patm

0.4 m

"external" pressure (bar)

1 mol of pure, ideal gas

Fexternal / area or

0.8 m

Process F: Isothermal compression

A = 0.1 m2

2

1 mol of pure, ideal gas

1 Process F PE = P

1

Process E

.04 Molar volume (m3/mol)

2

.08

Figure 2.7 Schematic of infinitesimal-step, reversible isothermal expansion (process E) and compression (process F) processes. The corresponding plots of the processes on a PEv diagram is shown at the bottom.

gives us the most work we can get out or the least work we have to put in! Moreover, only in a reversible process can we substitute the system pressure for the external pressure. Why do we get less work out of the irreversible expansion (process A) than the reversible expansion (process E)? Work is the transfer of energy between the system and the surroundings. In this case, as the gas molecules bounce off the piston, their change in net z momentum between before and after a collision is determined by the movement of the piston. This change of momentum with time represents the net energy transferred between the system and surroundings; that is, it is the work. The irreversible expansion, process A, never has a mass on the piston; thus the molecules of the gas are running into something “smaller” and will not transfer as much energy. On the other hand, in the irreversible compression process, the mass on the piston is larger than the corresponding reversible process. It therefore imparts more energy to those molecules and costs more work. Those of you who are baseball fans may consider an analogy to the size of a hitter’s bat. The heavier the bat, the more energy can be transferred to the baseball. The greater the force the piston exerts on a given molecule as it rebounds off the piston, the more the molecule’s speed will increase and the higher its kinetic energy. If we sum up all the molecules, we see that the net energy transfer (work) is greater.

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2.4 The First Law of Thermodynamics for Closed Systems ◄ 55

Efficiency We can compare the amount of work required in an irreversible process to that of a reversible process by defining the efficiency factor, h. For an expansion process, we compare how much work we actually get out to the idealized, reversible process. Thus, the efficiency of expansion, hexp, is: hexp 5

wirrev wrev

(2.10)

For example, the efficiency of Process A would be: hexp 5

wA 24000 5 5 0.72 wE 25545

where wi represents the work of process i. Thus, we say process A is 72% efficient; in comparison, the two-block process C is 84% efficient. We have done better. On the other hand, to determine the efficiency of a compression process, hcomp we compare the reversible work to the work we must actually put in: hcomp 5

wrev wirrev

(2.11)

For example, the efficiency of process B would be: hcomp 5

wF 5545 5 5 0.69 wB 8000

or 69% efficient. In both cases, efficiencies are defined so that if we can operate a process reversibly, we would have 100% efficiency, while the real processes are less efficient. One strategy for actual, irreversible processes is to solve the problem for the idealized, reversible process and then correct for the irreversibilities using an assigned efficiency factor.

► 2.4 THE FIRST LAW OF THERMODYNAMICS FOR CLOSED SYSTEMS Integral Balances In this section, we consider energy balances for closed systems. In the next section, open systems will be treated. Figure 2.8 shows a schematic of a closed system that undergoes a process from initial state 1 to final state 2. In this figure, the system, surroundings, and boundary are delineated. In a closed system, mass cannot transfer across the system boundary. There are two ways in which to catalog the amount of material in the system— by mass or by moles. Each way can be convenient. For a pure species or a mixture of constant composition, the two forms are equivalent and can be interconverted using the molecular weight. When we address systems undergoing chemical reaction, care must be taken. While the total mass must be conserved, the number of moles or the mass of a particular component may change. In the absence of chemical reaction, the number of moles remains constant: n1 5 n2 Since mass cannot enter or leave a closed system, the changes in the energy within the system 1 D 5 final 2 initial 2 are equal to the energy transferred from the

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56 ► Chapter 2. The First Law of Thermodynamics +W

+Q System Surroundings property ΔEK = EK,2 − EK,1 ΔEP = EP,2 − EP,1 ΔU = U2 − U1

Boundary

Figure 2.8 Illustration of closed system and sign conventions for heat and work. All three forms of energy are considered.

surroundings by either heat or work. Figure 2.8 also illustrates the sign convention that we have defined for heat and work, namely, positive for energy transfer from the surroundings to the system. Writing down the first law in quantitative terms, we get:7 change in energy transferred c energy in s 5 c from surroundings s system to system DU 1 DEK 1 DEP 5 Q 1 W ('''')''''* (')'* property: depends on depends only path on state 1 and 2

(2.12a)

The properties on the left-hand side of Equation (2.12a) depend only on the initial and final states. They can be calculated using the real path or any hypothetical path we create. The terms on the right-hand side are process dependent and the real path of the system must be used. Since the composition of a closed system remains constant (barring chemical reaction), we can rewrite Equation (2.12a) using intensive properties by dividing through by the total number of moles:8 Du 1 DeK 1 DeP 5 q 1 w

(2.12b)

We often neglect macroscopic kinetic and potential energy. For this case, the extensive and intensive forms of the closed system energy balances become:

and,

DU 5 Q 1 W

(2.13a)

Du 5 q 1 w

(2.13b)

respectively. 7

Heat and work already refer to the amount of energy transferred; hence, it would be would be wrong to write them DQ or DW. We reserve the D for state function that depends just on the initial and final state of the system.

8

We will sometimes write balance equations on a molar basis, utilizing the appropriate intensive thermodynamic properties. For example, internal energy will be u [J/mol]. You should be able to convert any equation to a mass basis that uses the corresponding specific property, for example, u^ 3 J/kg 4 .

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2.4 The First Law of Thermodynamics for Closed Systems ◄ 57

Differential Balances The first law can also be written in differential form for each differential step in time during the process. Numerical solutions are obtained by integration of the resulting differential energy balance. Common forms of energy balance over a differential element can be written by analogy to the equations just presented. dU 1 dEK 1 dEP 5 dQ 1 dW

extensive 3 J 4

du 1 deK 1 deP 5 dq 1 dw

intensive 3 J/mol 4

or, neglecting kinetic and potential energy, extensive 3 J 4

dU 5 dQ 1 dW

(2.14)

intensive 3 J/mol 4

du 5 dq 1 dw

We use the exact differential d with the energy terms to indicate that they depend only on the final and initial states; in contrast, we use the inexact differential d with heat and work to remind us that we must keep track of the path when we integrate to get these quantities. The energy balances above are often differentiated with respect to time, yielding: # # dEK dEP dU 1 1 5Q1W dt dt dt

extensive 3 W 4

deK deP du 1 1 5 q# 1 w# dt dt dt

intensive 3 W/mol 4

where the rate of heat transfer and the rate of work [J/s or W] are denoted with a dot over the corresponding variable. Again, we often neglect kinetic and potential energy to give: # # dU 5Q1W dt du 5 q# 1 w# dt

►EXERCISE

Example 2.4 Closed System Energy Balance

extensive 3 W 4

(2.15)

intensive 3 W/mol 4

Consider the six processes depicted in Figures 2.4 through 2.7. What is the heat transferred in each case. Can you explain the difference in relation to the efficiency factor?

Consider a piston–cylinder assembly containing 10.0 kg of water. Initially, the gas has a pressure of 20.0 bar and occupies a volume of 1.0 m3. The system undergoes a reversible process in which it is compressed to 100 bar. The pressure volume relationship during this process is given by: Pv1.5 5 const (Continued)

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58 ► Chapter 2. The First Law of Thermodynamics

(a) (b) (c) (d)

What is the initial temperature? Calculate the work done during this process. Calculate the heat transferred during this process. What is the final temperature?

SOLUTION (a) We need two independent intensive properties to constrain the state of the system. Once values for these properties are determined, we can use the steam tables to find other properties. The specific volume of the initial state can be determined as follows: v^ 5

V m3 1.0 5 0.10 B R 5 m 10.0 kg

The values in the steam tables are in units of Pa, so we convert 20 bar to 2 MPa. If we look at Table B.4 (Appendix B), we see that at P 5 2 MPa to three significant figures the volume m3 at the saturation temperature is 0.100 B R. Therefore, looking at Table B.2, we see the kg temperature is 212.4°C (or slightly above). (b) To calculate the work it is useful to draw a schematic of the process. We define the initial state as state 1 and the final state as state 2, as shown in Figure E2.4.

H2O P1 = 20 [bar]

Process

v1 = 0.1 [m3/ kg] T1 = 212.4 [° C]

P2 = 100 [bar]

State 1

State 2

Figure E2.4 Initial and final states of the expansion process. Because this is a reversible process, we can write: w^ 5 23 PEdv^ 5 23 Pdv^ To find the upper limit on the integrand, we need to know the specific volume of the final state, v^ 2. We can calculate v^ 2 from the equation in the problem statement: Pv^ 1.5 5 const 5 P1v^ 1.5 ^ 1.5 1 5 P2v 2

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2.4 The First Law of Thermodynamics for Closed Systems ◄ 59

Solving for v^ 2 gives: v^ 2 5 ¢

P1v^ 1.5 1 ≤ P2

1/1.5

5 1 0.2 3 0.11.5 2 1/1.5 5 0.0342 B

m3 R kg

Now we can solve for work: 0.0342

w^ 5 23

0.1

0.0342

Pdv^ 5 23

0.1

0.0342

P1v^ 1.5 1 1 1 dv^ 5 2P1v^ 1.5 1 B 0.5 2 0.5 R v^ 1.5 v^ 2 v^ 1 0.1

5 284 B

kJ kg

R

The sign of work is positive because we are adding energy to the system during the compression. (c) To find heat, we can apply the first law: q 5 Du 2 w Because we solved for work in part (b), we need only to determine the internal energy in states 2 and 1 from the steam tables. In part (a), we found that state 1 is approximately saturated vapor at 20 bar. Looking at Table B.2, we find: u1 5 2600.3 B

kJ kg

R

For u2, we look at Table B.4 at a pressure of 10 MPa (100 bar). In this case, we need to interpolate: u2 2 u 1 T 5 500ºC 2 v2 2 v 1 T 5 500ºC 2 5 u 1 T 5 550ºC 2 2 u 1 T 5 500ºC 2 v 1 T 5 550ºC 2 2 v 1 T 5 500ºC 2 Solving for u2, u2 5 u 1 T 5 500ºC 2 1 3 u 1 T 5 550ºC 2 2 u 1 T 5 500ºC 2 4

B

v2 2 v 1 T 5 500ºC 2 R v 1 T 5 550ºC 2 2 v 1 T 5 500ºC 2

5 3045.8 1 3 3144.5 2 3045.8 4 B

5 3094.6 B

kJ kg

0.0342 2 0.03279 R 0.03564 2 0.03279

R

Thus, q 5 Du 2 w 5 u2 2 u1 2 w 5 210 B

kJ kg

R

Since the value of q is positive, heat transfers from the system to the surroundings. (d) To find T2, we must also interpolate. From Table B.4,we get: v2 2 v 1 T 5 500ºC 2 T2 5 500ºC 1 3 550ºC 2 500ºC 4 B R v 1 T 5 550ºC 2 2 v 1 T 5 500ºC 2 5 500ºC 1 3 50ºC 4 B 5 525ºC

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0.0342 2 0.03279 R 0.03564 2 0.03279

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60 ► Chapter 2. The First Law of Thermodynamics

►2.5 THE FIRST LAW OF THERMODYNAMICS FOR OPEN SYSTEMS Material Balance In open systems, mass can flow into and out of the system. A generic open system with two streams in and two streams out is shown in Figure 2.9. In setting up the balance equations, it is convenient to discuss rates: mass flow rates [kg/s] and rates of energy transfer [J/s or W]. We must keep track of the mass in the system since it can change with time. In the general case of many streams in and many streams out, we must sum over all the in and out streams. We first consider conservation of mass, for a nonreacting system. The accumulation of mass in the system is equal to the difference of the total rate of mass in minus the total rate of mass out. Thus the mass balance can be written: a

dm # # b 5 a min 2 a mout dt sys in out

(2.16)

where m# is the mass flow rate in [kg/s]. For a stream flowing through a cross-sectional S area A, at velocity V, the mass flow rate can be written as: S

AV m# 5 (2.17) v^ In many situations, it is convenient to express the amount of chemical species in terms of moles (molar basis) rather than mass. Mass can be converted to moles using the molecular weight.

Flow Work The energy balance for an open system contains all the terms associated with an energy balance for a closed system, but we must also account for the energy change in the system associated with the streams flowing into and out of the system. To accomplish this task, we consider the case of the generic open system illustrated in Figure 2.9. This open system happens to have two streams in and two streams out; however, the balances developed here will be true for any number of inlet or outlet streams. Let’s look at the contribution to the energy balance from the inlet stream labeled stream 1. Compared to the closed-system analysis we performed in Section 2.4, there are δ(Eflow)in = − PinAinδx in

PE = Pin

Imaginary piston

Stream 2 in Ws

min

In ûin Stream 1 (êK)in (êP)in ˆ flow = (PVˆ )in w

out

mout ûout (êK)out (êP)out ˆ flow = (PVˆ )out w

System

Qsys out

Figure 2.9 Schematic of an open system with two streams in and two streams out. The piston shown in the plot is hypothetical; it illustrates the point that flow work is always associated with fluid flowing into or out of the system.

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2.5 The First Law of Thermodynamics for Open Systems ◄ 61

two additional ways in which energy can be transferred from the surroundings. First, the molecules flowing into the system carry their own energy; typically the most important form of energy is the specific internal energy, u^ in, but the flowing streams can also have (macroscopic) kinetic energy, 1 e^ k 2 in, and potential energy, 1 e^ P 2 in, associated with them. S Inspection of Equations (2.2) and (2.3) shows these last two terms can be written as 12 V 2 and gz respectively. Second, the inlet stream adds energy into the system by supplying work, the so-called flow work. Flow work is the work the inlet fluid must do on the system to displace fluid within the system so that it can enter. It may be visualized by placing an imaginary piston in front of the material that is about to enter the system, as depicted in Figure 2.9. The imaginary piston acts like the real piston shown in Figure 2.5. The rate of work exerted by the fluid to enter the system is, therefore, given by: # dx S # v 1 Wflow 2 in 5 2PE Ain 5 Pin AinVin 5 Pin m in ^ in dt where Equation (2.17) was used. We have set PE 5 Pin and eliminated the negative sign, since the direction of velocity is the negative of the normal vector from the piston. If we divide by mass flow rate, we can write the flow work in intensive form: # 1 Wflow 2 in 1 w^ flow 2 in 5 5 Pinv^ in m# in We conclude that the flow work of any inlet stream is given by the term 1 Pv^ 2 in. By a similar argument, we can show the flow work of any outlet stream is given by: # 1 Wflow 2 out 5 2m# outPoutv^ out We can write the total energy transfer due to work in the system in terms of shaft work, Ws, and flow work, as follows: # # # # W 5 Wshaft 1 Wflow 5 BWs 1 a m# in 1 Pv^ 2 in 1 a m# out 1 2Pv^ 2 out R in

out

The shaft work is representative of the useful work that is obtained from the system. While shafts are commonly used to get work out of an open system, as discussed earlier, # Ws generically represents any possible way to achieve useful work; therefore, it does not include flow work! The flow work does not provide a source of power; it is merely the “cost” of pushing fluid into or out of the open system. We can now include the new ways in which energy can exchange between the system and surroundings in our energy balance for open systems. In the general case of many streams in and many streams out, we must sum over all the in and out streams. First, we consider a system at steady-state; that is, there is no accumulation of energy or mass in the system with time. The energy balance is written as [on a rate basis, by analogy to Equation (2.15)]: 1 1 0 5 a m# in(u^ 1 V2 1 gz)in 2 a m# out( u^ 1 V2 1 gz)out 2 2 in out S

steady state

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energy flowing into the system with the inlet steams

S

energy flowing out of the system with the outlet steams

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62 ► Chapter 2. The First Law of Thermodynamics # # 1Q 1 BWs 1 a m# in 1 Pv^ 2 in 1 a m# out 1 2Pv^ 2 out R in

out

flow work from the inlet streams

Rearranging, we get:

flow work from the outlet streams

# # 1S 1S 0 5 a m# in C 1 u^ 1 Pv^ 2 1 V2 1 gz D in 2 a m# out C 1 u^ 1 Pv^ 2 1 V2 1 gz D out 1 Q 1 Ws 2 2 in out

Enthalpy The inlet streams that flow into open systems always have terms associated with both internal energy and flow work; therefore, it is convenient to group these terms together (so that we don’t forget one). We give it the name enthalpy, h [J/mol]: h ; u 1 Pv

(2.18)

Since u, P, and v are all properties, this new group, enthalpy, is also a property. Thus, the additional energy associated with the flowing inlet stream is given by h^ in [as well as 1 S2 V and gz]. The term h^ in includes both the internal energy of the stream and the flow 2 work it adds to enter the system. Similarly, the combined internal energy and flow work leaving the system as a result of the exiting streams are given by h^ out. Enthalpy provides us a property that is a convenient way to account for these two contributions of flowing streams to the energy in open systems. As we will learn in the next section, enthalpy also is convenient to use with closed systems at constant pressure. We next develop the relationship between temperature and enthalpy for an ideal gas. Recall the internal energy of an ideal gas depends only on T. Application of the definition of enthalpy and the definition of an ideal gas gives: hideal gas 5 u 1 Pv 5 u 1 RT 5 f 1 T only 2 since Pv 5 RT for an ideal gas. Thus, like the internal energy, the enthalpy of an ideal gas depends only on T.

Steady-State Energy Balances In summary, the steady-state energy balances can be written: 1S 1S 0 5 a m# in (h^ 1 V2 1 gz)in 2 a m# out (h^ 1 V2 1 gz)out 2 2 in out # # 1 Q 1 Ws

(2.19)

Take a look at Equation (2.19). See if you can identify the physical meaning of each term. In cases where we can neglect (macroscopic) kinetic and potential energy, the steady-state, integral energy balance becomes: # # # 0 5 a minh^ in 2 a m# outh^ out 1 Q 1 Ws in

c02.indd 62

out

(2.20)

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2.5 The First Law of Thermodynamics for Open Systems ◄ 63

Transient Energy Balance Another useful form of the energy balance for open systems is for unsteady-state conditions. Unsteady-state is important, for example, in start-up as the equipment “warms up.” In the case when the inlet and outlet streams stay constant with time, the unsteadystate energy balance becomes: ¢

dEK dEP 1S 1S dU 1 1 ≤ 5 a m# in ah^ 1 V2 1 gzb 2 a m# out ah^ 1 V2 1 gzb dt dt dt sys 2 2 in in out out # # 1 Q 1 Ws (2.21)

In cases where we can neglect (macroscopic) kinetic and potential energy, the unsteady-state, integral energy balance becomes: a

# # dU # # b 5 a minh^ in 2 a mouth^ out 1 Q 1 Ws dt sys in out

(2.22)

In Equations (2.21) and (2.22), the left-hand side represents accumulation of energy within the system. There is no flow work associated with this term; hence, the appropriate property is U. On the other hand, the first two terms on the right-hand side account for energy flowing in and out of the system, respectively. These terms must account for both the internal energy and flow work of the flowing streams. In this case, h^ is appropriate. It is worthwhile for you to take a moment and reconcile the use of U and h^ above. It will save you many mistakes down the road!

►EXERCISE (a) Simplify Equations (2.19) and (2.21) for the case of one stream in and one stream out. (b) What are ways to find h of a system? (c) For the energy balance depicted in Equation (2.21), which terms correspond to accumulation? To energy in? To energy out?

EXAMPLE 2.5 Calculation of Work from the First Law

Steam enters a turbine with a mass flow rate of 10 kg/s. The inlet pressure is 100 bar and the inlet temperature is 500ºC. The outlet contains saturated steam at 1 bar. At steady-state, calculate the power (in kW) generated by the turbine. SOLUTION The steady-state energy balance is given by Equation (2.20): # # 0 5 a m# inh^ in 2 a m# outh^ out 1 Q 1 Ws in

(E2.5A)

out

For a turbine, there is one stream in and one stream out. Moreover, heat dissipation is negligible, # # since Q ,, Ws. If we label the inlet stream “1” and the outlet stream “2,” Equation (E2.5A) becomes: # 0 5 m# 1h^ 1 2 m# 2h^ 2 1 Ws

(E2.5B) (Continued)

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64 ► Chapter 2. The First Law of Thermodynamics

At steady-state, the mass balance can be written as: m# 1 5 m# 2 5 m#

(E2.5C)

Plugging in Equation (E2.5C) into (E2.5B) gives: # Ws 5 m# 1 h^ 2 2 h^ 1 2

(E2.5D)

We can look up values for specific enthalpy from the steam tables. For state 2, we use saturated steam at 1 bar 1 5 100 kPa 2 : h^ 2 5 2675.5 3 kJ/kg 4 while state 1 is superheated steam at 500ºC and 100 bar 1 5 10 Mpa 2 : h^ 1 5 3373.6 3 kJ/kg 4 Plugging these numerical values into Equation (E2.5D) gives: # Ws 5 10 3 kg/s 4 1 2675.5 2 3373.6 2 3 kJ/kg 4 5 26981 3 kW 4 Thus, this turbine generates approximately 7 MW of power. Note the negative sign, which indicates that we are getting useful work from the system. This is the equivalent power to that delivered by approximately 70 automobiles running simultaneously.

EXAMPLE 2.6 Calculation of Final Temperature for a Transient Process

Steam at 10 MPa, 450ºC is flowing in a pipe, as shown in Figure E2.6A. Connected to this pipe through a valve is an evacuated tank. The valve is opened and the tank fills with steam until the pressure is 10 MPa, and then the valve is closed. The process takes place adiabatically. (a) Determine the final temperature of the steam in the tank. (b) Explain why the final temperature in the tank is not the same as that of the steam flowing in the pipe. SOLUTION (a) This problem can be solved several ways. We first solve it with the fixed boundary illustrated in Figure (E2.6A). An alternative solution with a moving system boundary is then presented. Unsteady-State Analysis For the choice of system shown in Figure E2.6A, we must use an unsteady-state energy balance since the mass and energy inside the tank (system) increase with time. Let us define the initial state of the tank (empty), state 1, and the final state (filled to 10 MPa), state 2. The inlet flow stream will be designated by “in.” The conservation of mass is given by Equation (2.16). Since we have one stream in and no streams out, we get: a

dm # b 5 min dt sys

Separating variables and integrating gives: m2

t

# 3 dm 5 3 mindt m1 5 0

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0

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2.5 The First Law of Thermodynamics for Open Systems ◄ 65

Since state 1 identically has a value for m of 0, we get: t

m2 5 3 m# indt

(E2.6A)

0

The unsteady energy balance, in mass (specific) units, is given by Equation (2.22): 0 0 0 # # dU # ^ # ^ a b 5 minhin 2 mouthout 1 Q 1 Ws dt sys where the terms associated with flow out, heat, and work are zero. Separating variables and integrating both sides with respect to time from the initial empty state to the final state when the tank is at a pressure of 10 MPa gives: U2

t

t

# ^ # ^ 3 dU 5 3 minhindt 5 hin 3 mindt

U1 5 0

0

(E2.6B)

0

Steam 10 MPa. 450°C

Empty tank to be filled

System boundary

Figure E2.6A Flow of superheated steam from a supply line to fill an empty tank. The system boundary is indicated with dashed lines.

We have moved the specific enthalpy out of the integral, since the properties of the inlet stream remain constant throughout the process. Applying Equation (E2.6A) and integrating, Equation (E2.6B) becomes: m2u^ 2 5 m2h^ in

(E2.6C)

since state 1 identically has a value for U of 0. Equation (E2.6C) simplifies to: u^ 2 5 h^ in

(E2.6D)

From the steam tables, steam at P 5 10 MPa and T 5 450°C has a specific enthalpy: h^ in 5 3241 3 kJ/kg 4 According to Equation (E2.6D), the inlet specific enthalpy must be equal to the final specific internal energy of the system. Hence, the final state has two independent intensive properties (Continued)

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66 ► Chapter 2. The First Law of Thermodynamics specified: P 5 10 MPa and u^ 5 3241 3 kJ/kg 4 . From the steam tables, for steam at P 5 10 MPa and u^ 5 3241 3 kJ/kg 4 , the final temperature of the system is: T2 5 600°C Closed-System Analysis Alternatively, we can use a different choice of system to solve this problem. We consider the mass that starts in the pipe and eventually ends up in the tank as part of the system. The system boundary in this case is illustrated on the left side of Figure E2.6B. As mass flows into the tank, the boundary contracts. With this choice of system, no mass flow crosses the boundary; thus, we have a closed system. An equivalent closed system, in terms of our familiar piston–cylinder assembly, is shown on the right of Figure E2.6B. The adiabatic chamber is separated by a diaphragm that plays the same role as the valve in the original system. Above the diaphragm, the cylinder contains steam at 10 MPa and 450ºC, identical to the inlet steam. Below the diaphragm is a vacuum. The process is initiated by removing the diaphragm, and the system is taken from state “in” to state 2. An external pressure of 10 MPa, representative of the steam in the pipe outside the system boundary, acts against the piston until the pressures equilibrate. During the compression process, the surroundings transfer energy to the system via work. Can you see that the two processes depicted in Figure E2.6B are equivalent? A mass balance on either closed system depicted in Figure (E2.6B) gives: min 5 m2

(E2.6E) PE = 10 MPa

Steam 10 MPa 450°C

System boundary

Pin = 10 MPa Tin = 450°C Diaphragm

Equivalent Empty tank to be filled

Vacuum Q= 0

Figure E2.6B A closed system approach to solving the problem in Example 2.6. Likewise, the energy balance is: DU 5 m2u^ 2 2 minu^ in 5 Q 1 W

(E2.6F)

Since the process is adiabatic, Q 5 0. Work is given by: V

W52

3

PEdv 5 minPinv^ in

(E2.6G)

V1min v^ in

Substituting Equations (E2.6E) and (E2.6G) into (E2.6F) and rearranging gives: m2u^ 2 5 min 1 u^ in 1 Pinv^ in 2 5 minh^ in 5 m2h^ in which is identical to the result we obtained for the unsteady-state open system of Figure E2.6A., given by Equation (E2.6D).

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2.6 Thermochemical Data for U and H ◄ 67

(b) The fluid in the system receives flow work from the fluid behind it to get into the system. This work component adds energy from the incoming stream to the system, increasing T from 450ºC to 600ºC. The flow work of steam flowing into the system shown on the left of Figure E2.6B is equivalent to work done by an equivalent constant external pressure acting on the piston shown on the right. Thus, the closed-system analysis helps us see the importance of accounting for flow work in energy balances on open systems.

►2.6 THERMOCHEMICAL DATA FOR U AND H Heat Capacity: cv and cP In order to perform energy balances on both closed and open systems, it is necessary to be able to determine how the energy (or enthalpy) of the species in the system changes during a process. As we learned in Section 1.3, the internal energy, u, and the enthalpy, h, for a pure species are constrained by specifying two independent intensive properties. Moreover, since u is a thermodynamic property, we can specify a hypothetical path to calculate the change in internal energy, Du. It does not have to be the path of the actual process. Likewise for Dh. While any thermodynamic properties can be used, it is often convenient to choose measured properties (T, P, or v) as the independent properties. Temperature is almost always chosen as one of the independent properties, since it can be measured in the lab (or field), and there is a direct relationship between T and u; that is, temperature is a measure of the molecular kinetic energy, which is one component of u (see Section 2.1). In fact, for an ideal gas, it is only this component that contributes to u. The other independent property is typically also a measurable property, either P or v can be chosen according to convenience. Figure 2.3 illustrates a common hypothetical path used to calculate Du. In this case, T and v are chosen as independent properties. In step 1, we must know the temperature dependence of u to calculate Du, as we go from T1 to T2 at constant volume. This information is often obtained in the form of heat capacity (or specific heat). Similarly, the temperature dependence of h used to find Dh can be found through reported heat capacity values. Therefore, heat capacity data are crucial in this problem-solving methodology. In the next section, we will explore how heat capacities are experimentally determined and how they are reported. Heat Capacity at Constant Volume, cv To measure the heat capacity at constant volume, cv, an experimental setup as conceptually shown in Figure 2.10a can be used. This closed system consists of pure species A within a rigid container. The container is connected to a heat source (in this case, a resistance heater) and is otherwise well insulated. The experiment is conducted as follows: As a known amount of heat, q, is provided through the resistive heater, the temperature, T, of the system is measured. As heat is supplied, the molecules of A move faster, and the temperature increases. A typical data set for pure species A is also shown in Figure 2.10a. Since we can “sense” the result of the heat input with the thermocouple, this type of energy change is labeled sensible heat. This setup is known as a constantvolume calorimeter. Since there is no work done in this process, we can apply the first law [Equation (2.13b)] to this system and get: Du 5 q

c02.indd 67

closed system, const V

(2.23)

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68 ► Chapter 2. The First Law of Thermodynamics Note that in Figure 2.10a, the amount of heat supplied is plotted on the y-axis and the temperature measured is on the x-axis. However, Equation (2.23) shows heat input is identical to Du. We define the heat capacity at constant volume, cv, as: cv ; a

'u b 'T v

(2.24)

Hence, the slope of the curve gives us the heat capacity at any temperature. In these data, the heat capacity at T1 is less than that at T2. Typically, heat capacity changes with T. By taking the slope of this curve as a function of temperature, we can get: cv 5 cv 1 T 2 We can then fit the data to a polynomial expression of the form: cv 5 a 1 BT 1 CT2 1 DT22 1 ET3

Experiment:

(2.25)

Data: cv ≡ ∂u ∂T

T V = const

v

A A A

A A

q = Δu

A A A

Slope = cv at T2

A

A

Slope = cv at T1

q

T1

T2 T

(a) Constant-Volume Calorimeter: Determination of cv Experiment:

Data: cP ≡ ∂h ∂T

P = const

P

T A

A A

A A

A A

q = Δh

A Slope = cP at T2

A

A q

Slope = cP at T1

T1

T2 T

Pure substance, A (b) Constant-Pressure Calorimeter: Determination of cP

Figure 2.10 Schematics of the experimental determination of heat capacity. (a) constant-volume calorimeter to obtain cv; (b) constant-pressure calorimeter to determine cP.

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2.6 Thermochemical Data for U and H ◄ 69

Parameters a, B, C, D, and E are then tabulated and can be used any time we want to know how the internal energy of species A changes with temperature at constant volume. We can then find Du by integration: T2

T2

Du 5 3 cv dT 5 3 3 a 1 BT 1 CT2 1 DT22 4 dT T1

(2.26)

T1

Heat capacity should only be used for temperature changes between the same phase. When a phase change occurs, the latent heat must also be considered, as discussed shortly. Consider an ideal gas. Equation (2.24) shows us that if we increase T, we increase u. The amount the internal energy increases with temperature is quantified by the heat capacity according to Equation (2.26). The more energy it gains, the larger cv. On a molecular level, we may want to know how this increase in energy manifests itself. We can associate the increase in molecular kinetic energy and, therefore, in u with temperature to three possible modes in which the molecules can obtain kinetic energy. The first mode is related to the center-of-mass motion of the molecules through space. In Chapter 1, we saw that the Maxwell–Boltzmann distribution characterizes the velocities of the molecules at a given temperature. This translational energy contributes kT/2 per molecule (or RT/2 per mole) to the kinetic energy in each direction that the molecule moves. Since molecules translate through space in three directions, the translational motion contributes 3RT/2 per mole to the internal energy of the molecules. The contribution of translational motion to cv, given by the derivative of the internal energy with respect to temperature, is 3R/2. In fact, monatomic gases have heat capacities given by this value.9 Diatomic and polyatomic molecules can manifest kinetic energy in rotational and vibrational modes as well. Except at extremely low temperature, the additional kinetic energy due to rotational motion for linear and nonlinear molecules is RT and 3RT/2 per mole, respectively. The kinetic energy due to vibration is much more interesting. It is related to the specific quantized energy levels of the molecule. The distribution of these levels depends on temperature. To account for the vibrational modes of kinetic energy, we would need to resort to quantum mechanics. We will not formally address that problem here; however, it is useful to realize that the temperature dependence of the heat capacity indicated by Equation (2.25) manifests itself in the vibrational mode. At low temperature, the vibrational contribution goes to zero and the heat capacity is given by the translational and rotational modes only. At high temperature, where the vibrational motion is fully active, the contribution is R per mole. In summary, cv can be attributed to molecular structure and the ways in which each species exhibits translational, rotational, and vibrational kinetic energy. Heat capacities for gases, liquids, and solids can be intepreted in a similar way. Heat Capacity at Constand Pressure, cp The heat capacity at constant pressure, cP, is measured in a similar manner, only gas A is no longer held within a rigid container, but in a system that can expand as it is heated so as to keep the pressure constant. A conceptual representation of the experimental setup to measure cP is shown in Figure 2.10b. While the actual apparatus may look different, this depiction is in terms of the piston–cylinder assembly that we have previously examined. 9

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Except at very high temperature when electrons occupy excited states.

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70 ► Chapter 2. The First Law of Thermodynamics Since the system is now doing Pv work as it expands, the energy balance contains a term for work: Du 5 q 1 w 5 q 2 PDv

(2.27)

So Equation (2.27) can be rewritten as: Du 1 D 1 Pv 2 5 q since at constant pressure, DP 5 0, and: D 1 Pv 2 5 PDv 1 vDP 5 PDv Applying the definition for enthalpy [Equation (2.18)], we get: Dh 5 q

closed system, const P

(2.28)

Hence, in this case, an energy balance tells us the heat supplied at constant pressure is just equal to the change in the thermodynamic property, enthalpy. Therefore, we define the heat capacity at constant pressure as: cP ; a

'h b 'T P

(2.29)

Again, typical data for species A are presented in Figure 2.10b and can be fit to the polynomial form: cP 5 A 1 BT 1 CT2 1 DT22 1 ET3

(2.30)

The parameters A, B, C, D, and E are reported for some ideal gases in Appendix A.2. Heat capacity parameters at constant pressure of some liquids and solids are reported in this appendix. Enthalpy—A Second Common Use Recall that we “constructed” the property enthalpy to account for both the internal energy and flow work for streams flowing into and out of open systems. However, inspection of Equation (2.28) suggests a second common use of enthalpy. This equation holds, in general, for closed systems at constant P. In this case, it accounts for both the change in internal energy and the Pv work as the system boundary changes in order to keep pressure constant. In both cases, the property h couples internal energy and work. Therefore, experiments that are conveniently done in closed systems at constant P are reported using the thermodynamic property enthalpy. For example, the energetics of a chemical reaction, the so-called enthalpy of reaction, is reported in terms of the property Dhrxn. In this way, the experimentally measured heat can be related directly to a thermodynamic property. Relations between cP and cV By comparing Figure 2.10a to Figure 2.10b, we can estimate the difference in cv and cP for the different phases of matter. If species A is in the liquid or solid phase, its volume expansion upon heating should be relatively small; that is, the molar volumes of liquids and solids do not change much with temperature. Hence, the piston depicted

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2.6 Thermochemical Data for U and H ◄ 71

in Figure 2.10b will not move significantly and the value of work in Equation (2.27) will be small compared to q. Thus Equation (2.29) and Equation (2.24) are approximately equivalent and, consequently:10 cP < cv

liquids and solids

On the other hand, the volume expansion of a gas will be significant. For the case of an ideal gas, we can figure out the relationship between cP and cv by applying the ideal gas law to the definition for cP as follows: cP ; a

' 1 u 1 Pv 2 'h 'u 'RT 'u b 5 c d 5a b 1a b 5a b 1R 'T P 'T 'T P 'T P 'T P P

(2.31)

since Pv 5 RT for an ideal gas. However, for an ideal gas, the internal energy depends on temperature only; that is, the only change in molecular energy is in molecular kinetic energy. Therefore, a

'u du 'u 5a b b 5 'T P dT 'T v

(2.32)

Plugging Equation (2.32) into (2.31) gives: cP 5 cv 1 R

for ideal gases

Values of heat capacity for gases are almost always reported for the ideal gas state. Thus, when doing calculations using these data, you must choose a hypothetical path where the change in temperature occurs when the gas behaves ideally. Mean Heat Capacity For many gases, heat capacity data are often reported in terms of the mean heat capacity, cP. Use of cP eliminates the need for integration and can make the mechanics of problem solving easier. As its name suggests, the mean heat capacity is the average of cP between two temperatures. It is usually reported between 298 K and a given temperature, T. Hence, the enthalpy change becomes: Dh 5 cP 1 T 2 298 2 Solving for cP gives: T

e cPdT

cP 5

298

T 2 298

(2.33)

Note that Equation (2.33) is also, by definition, the mathematical average of the continuous function cP over temperature between 298 K and T.

10

c02.indd 71

We will revisit the relation between cP and cv of liquids in Chapter 5, Problems 5.20 and 5.21.

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72 ► Chapter 2. The First Law of Thermodynamics

EXAMPLE 2.7 Heat Input Calculations Using Different Data Sources

Consider heating 2 moles of steam from 200ºC and 1 MPa to 500ºC and 1 MPa. Calculate the heat input required using the following sources for data: (a) Heat capacity (b) Steam tables SOLUTION (a) Since this process occurs at constant pressure, the system will expand as T increases. In accordance with the discussion above, enthalpy is the appropriate property to calculate the heat input. The extensive version of Equation (2.28) can be written as: Q 5 nDh If we assume water is an ideal gas, we use the values of heat capacity given in Appendix A.2 to calculate Dh: ideal gas

cP

R

5 A 1 BT 1 DT22 5 3.470 1 1.450 3 1023T 1

0.121 3 105 T2

Using the definition of heat capacity, we get the following integral expression: 773

T2

Dh 5 3 cPdT 5 R 3 3 A 1 BT 1 DT22 4 dT 473

T1

Integrating, Dh 5 R c AT 1

B 2 D 773 T 2 d 2 T 473

Substituting in values: Dh 5 8.314 3 3.470 1 300 2 1 0.725 3 1023 1 7732 2 4732 2 20.121 3 105 a and,

J 1 1 2 b R 5 10,991 c d 773 473 mol

Q 5 nDh 5 21,981 3 J 4

(b) From the steam tables, Appendix B.2: h^ 1 1 at 1 MPa, 200°C 2 5 2827.9 3 kJ/kg 4 h^ 2 1 at 1 MPa, 500°C 2 5 3478.4 3 kJ/kg 4 So,

Dh^ 5 h^ 2 2 h^ 1 5 650.5 3 kJ/kg 4

Since the steam tables give us the specific enthalpy, we must multiply by mass. Thus, we must use the molecular weight for water, MWH2O 5 0.018 3 kg/mol 4 : Q 5 mDh^ 5 1 2 3 mol 4 2 1 0.018 3 kg/mol 4 2 1 650.5 3 kJ/kg 4 2 1 103 3 J/kJ 4 2 5 23,418 3 J 4 The answer for part (b) is approximately 6% higher than for part (a). At 1 MPa, water is not an ideal gas but rather has attractive intermolecular interactions. The extra energy needed in part (b) results from that needed to pull the water molecules apart. We will learn more of these things in Chapter 4.

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2.6 Thermochemical Data for U and H ◄ 73

EXAMPLE 2.8 Determination of Mean Heat Capacity for Air

Use the data available in Appendix A.2 to calculate the mean heat capacity, cP, for air between T1 5 298 K and T2 5 300 to 1000 K, in intervals of 100 K. SOLUTION Using the definition of cP from Equation (2.33): T

3 cPdT Dh 298 5 cP 5 1 T 2 298 2 T 2 298

(E2.8A)

The heat capacity can be integrated with respect to temperature using the parameters in

Appendix A.2: T2

T

T

22 3 cPdT 5 R 3 3 A 1 BT 1 DT 4 dT 5 RBAT 1 298

T1

B 2 D T 2 R 2 T 298

(E2.8B)

where for air: A 5 3.355,

B 5 0.575 3 1023,

and D 5 20.016 3 105

(E2.8C)

The solution to Equation (E2.8A), using (E2.8B) and (E2.8C), is presented at intervals of 100 K in Table E2.8. TABLE E2.8 Calculated Values for Mean Heat Capacity of Air at Different Temperatures. T [K]

EXAMPLE 2.9 Heat Calculation Using Mean Heat Capacity for Air

Dh 3 J/mol 4

T 2 298

cP 3 J/mol K 4

300

58.35

2

29.17

400

3003.93

102

29.45

500

6001.75

202

29.71

600

9049.59

302

29.97

700

12146.51

402

30.22

800

15292.02

502

30.46

900

18485.87

602

30.71

1000

21727.89

702

30.95

You need to preheat a stream of air flowing steadily at 10 mol/min from 600 K to 900 K. Determine the heat rate required using the mean heat capacity data from Example 2.8. SOLUTION This process occurs at steady-state with one stream in and one stream out; hence Equation (2.19) can be written as follows: 0 0 0 0 0 S S 2 2 # # V V # # 1 MWgzb 2 n2 ah 1 MW 1 MWgzb 1 Q 1 Ws 0 5 n1 ah 1 MW 2 2 1 2 (Continued)

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74 ► Chapter 2. The First Law of Thermodynamics

where we have set the bulk kinetic and potential energies and shaft work to zero. A mole balance yields:

n# 1 5 n# 2 5 n# air so the first law balance simplifies to: # Q 5 n# air 1 h900 2 h600 2 5 n# air 3 1 h900 2 h298 2 2 1 h600 2 h298 2 4

(E2.9A)

where h900 is the enthalpy of air at 900 K, h600 is the enthalpy at 600 K, and h298 is the enthalpy at 298 K. Equation (E2.9A) was rewritten to use the definition for mean heat capacity given by Equation (2.33): 1 h900 2 h298 2 5 cP,900 1 900 2 298 2 (E2.9B) 1 h600 2 h298 2 5 cP,600 1 600 2 298 2

(E2.9C)

Substituting Equations (E2.9B) and (E2.9C) into (E2.9A) and using the values from Table E2.8, we get: # Q 5 n# Dh 5 10 3 30.71 1 900 2 298 2 2 29.97 1 600 2 298 2 4 5 94.363 3 J/min 4

EXAMPLE 2.10 Spring-Assembled Piston–Cylinder

Air is contained within a piston–cylinder assembly, as shown in Figure E2.10A. The crosssectional area of the piston is 0.01 m2. Initially the piston is at 1 bar and 25ºC, 10 cm above the base of the cylinder. In this state, the spring exerts no force on the piston. The system is then reversibly heated to 100ºC. As the spring is compressed, it exerts a force on the piston according to: F 5 2kx

where k 5 50,000 3 N/m 4 and x is the displacement length from its uncompressed position. (a) Determine the work done. (b) Determine the heat transferred. SOLUTION (a) Since the process is reversible, the system pressure is always balanced by the external pressure and the work done is given by: v2

W 5 2 3 PdV

(E2.10A)

v1

We can draw a free-body diagram to determine how all the forces acting on the piston balance, as shown in Figure E2.10A. The displacement of the spring, x, can be written in terms of the change in volume: x5

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V 2 V1 DV 5 A A

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2.6 Thermochemical Data for U and H ◄ 75

F = −kx

Pext = 1 bar

A = 0.01 m2 Air

10 cm

P1 = 1 bar T1 = 25°C

Figure E2.10A Piston–cylinder assembly with a spring attached to the piston. The initial state of the system for Example 2.10 is shown.

Fext = PextA

|Fspring| = kx

Piston

Figure E2.10B Schematic of the forces acting on the piston in Figure E2.10A as the gas expands.

FAir = PAirA

where DV 5 V 2 V1. We can then equate the force per area acting on each side of the piston to get: kx kDV P 5 Pext 1 5 Pext 1 2 (E2.10B) A A Plugging Equation (E2.10B) into (E2.10A) and integrating gives: V2

W 5 2 3 PextdV 2 V1

DV5 1V2 2V1 2

3

k 1 V2 2 V1 2 2 kDV 1 2 1 2 d DV 5 2P 2 V V 2 ext 2 1 A2 2A2

(E2.10.C)

0

To solve Equation (E2.10C), we must find V2. Applying the ideal gas law gives: k 1 V2 2 V1 2 P1V1 P2V2 V2 5 5 ¢Pext 1 ≤ T1 T2 T2 A2 Solving this quadratic equation for V2 gives: V2 5 0.00116 3 m3 4 which can be plugged back into Equation (E2.10 C) to get: W 5 2166 3 J 4 (b) To find the heat transferred during this process, we can apply the first law to this closed system: DU 5 Q 1 W (E2.10D) (Continued)

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76 ► Chapter 2. The First Law of Thermodynamics

The change in internal energy is given by: T2

T2

T2

Du 5 3 cvdT 5 3 1 cP 2 R 2 dT 5 R 3 3 1 A 2 1 2 1 BT 1 DT22 4 dT T1

T1

T1

5 RB 1 A 2 1 2 T 1

T2

B 2 D T 2 R 2 T T1

The heat capacity parameters for air can be found in Appendix A.2: A 5 3.355,

B 5 0.575 3 1023,

and D 5 20.016 3 105

Thus, Du 5 1,580 3 J/mol 4 and,

DU 5 nDu 5 ¢

P1V1 ≤Du 5 638 3 J 4 RT1

We can now solve for the heat transfer from Equation (E2.10D): Q 5 DU 2 W 5 803 3 J 4

Latent Heats When a substance undergoes a phase change, there is a substantial change in internal energy associated with it (see Section 2.1). We need to be able to determine a value for this energy change if we want to apply the first law to a process involving a phase change. Like heat capacities, the energetics characteristic of a given phase change are reported based on accessible measured data. For example, consider the vaporization of a liquid. Liquids are held together by attractive forces between the molecules. To vaporize a liquid, we must supply enough energy to overcome the forces of attraction. A typical experimental setup is schematically shown in Figure 2.11. A given amount of liquid substance A is placed in a well-insulated closed system at constant pressure. A measured amount of heat is supplied until A becomes all vapor. We chose a system at constant pressure, as depicted in Figure 2.11, so that the temperature will stay constant during the phase change. A schematic for the data acquired in this experiment is presented on the right of Figure 2.11. The temperature of the subcooled liquid and the superheated vapor increases as energy is supplied via heat. It is only the heat input at constant temperature during the phase transition that is reported as the enthalpy of vaporization. Examination of Equation (2.28) shows that if we measure the heat absorbed as A changes phase; it is equal to the enthalpy of the vapor minus the enthalpy of the liquid. We term this difference the enthalpy of vaporization: Dhvap 5 hv 2 hl If we need to calculate the energetics of a vapor condensing to a liquid, we simply use the negative of Dhvap. Similarly, the change in enthalpy from the liquid phase to the solid phase is reported as the enthalpy of fusion, Dhfus: Dhfus 5 hs 2 hl

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2.6 Thermochemical Data for U and H ◄ 77 Experiment: State 1

Data: State 2

P = const

T AAA AA A

A

Liquid A AAA

A

q

A A

Vapor A

A A A A A

q = Δh

A

T

Δhvap

Tvap

q

T

Figure 2.11 Schematic of the experimental determination of enthalpy of vaporization.

And the change from the solid phase to the vapor phase is the enthalpy of sublimation, Dhsub:

Dhsub 5 hv 2 hs How would you find the internal energy of vaporization, Duvap, given Dhvap? The term to describe the change of enthalpy during a phase transition at constant pressure is latent heat. Latent means “hidden,” and it is called “latent” because we cannot “sense” the heat input by detecting the temperature change, as is the case with “sensible” heat, described previously. The latent heat changes with temperature. However, we usually only know its value at one state. For example, the enthalpy of vaporization is typically reported at 1 bar, the so-called normal boiling point, Tb. To find Dhvap at another pressure (and therefore another temperature), we need to construct an appropriate hypothetical path (see Section 2.2). Figure 2.12 illustrates a path for the calculation of Dhvap,T at any T based on the measured value to which we have access at Tb. It consists of three steps. In step 1, we calculate the change in enthalpy of the liquid from T to Tb, using heat capacity data. In step 2, we vaporize the liquid at the normal boiling point, since this is the value we have for Dhvap. In step 3, we calculate the change in enthalpy of the vapor from the normal boiling point to T. Adding together the three steps, we get: T

Tb

Dhvap, T 5 3

clP dT

1 Dhvap, Tb 1 3 Tb

T

Tb

Tb ∫ c |pdT T

T

Step 3

Step 1

Δhvap,Tb T ∫ cvpdT Tb

Temperature

5 D hvap, Tb 1 3 DcvlP dT

Step 2

Tb

Δhvap,Tb Liquid

Vapor Phase

c02.indd 77

T

cvP dT

Figure 2.12 Hypothetical path to calculate Dhvap at temperature T from data available at Tb and heat capacity data.

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78 ► Chapter 2. The First Law of Thermodynamics where we used the following definition: DcvlP 5 cvP 2 c lP This procedure can likewise be applied to determine the values of Dhfus and Dhsub at different temperatures than that at which they are reported.

EXAMPLE 2.11 Determination of Heat Required to Evaporate Hexane

10 mol/sec of liquid hexane flows into a steady-state boiler at 25ºC. It exits as a vapor at 100ºC. What is the required heat input to the heater? Take the enthalpy of vaporization at 68.8ºC to be: Dhvap, 68.8°C 5 28.88 3 kJ/mol 4

SOLUTION This process occurs in an open system with one stream in and one stream out. Can you draw a schematic? We label the inlet state as “state 1” and the outlet as “state 2.” In this case, Equation (2.20) can be written on a molar basis as: 0 # # # # 0 5 n 1h 1 2 n 2h 2 1 Q 1 Ws (E2.11A) where we have set the macroscopic kinetic and potential energy and the shaft work equal to zero. A mole balance gives: n# 1 5 n# 2 5 n#

(E2.11B)

Plugging (E2.11B) into (E2.11A) and solving for the heat-transfer rate, we get: # Q 5 n# 1 h2 2 h1 2

(E2.11C)

The enthalpy change can be divided into three parts: (1) the sensible heat required to raise liquid hexane to its boiling point; (2) the enthalpy of vaporization, that is, latent heat; and (3) the sensible heat required to raise hexane in the vapor state to 100ºC. Thus, the enthalpy difference becomes: h2 2 h1 5

Dh

l,25°C S 68.8°C

1 Dhvap, 68.8°C 1

Dh

v,68.8°C S 100°C

(E2.11D)

Values for the first and third terms can be found from the appropriate heat capacity data: 342

Dh

l.25°C S 68.8°C

5 3 216.3dT 5 9485 3 J/mol 4 5 9.49 3 kJ/mol 4 298.2 373.2

and,

Dh

v,68.8°C S 100°C

5 3 R 1 A 1 BT 1 CT 2 2 dT 342

5 8.314B3.025 1 373.2 2 342 2 1

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53.722 3 1023 1 373.22 2 3422 2 2

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2.6 Thermochemical Data for U and H ◄ 79

2

16.791 3 1026 1 373.23 2 3423 2 R 3 5 5.20 3 kJ/mol 4

Summing together values for enthalpy in Equation (E2.11D) and plugging the result into Equation (E2.11C) gives the rate at which heat must be supplied: # Q 5 a10 3 mol/s 4 b a 1 9.49 1 28.85 1 5.20 2 3 kJ/mol 4 b 5 435 3 kJ/s 4

EXAMPLE 2.12 Determination of Heat Required to Evaporate H2O

A rigid vessel contains 50.0 kg of saturated liquid water and 4.3 kg of saturated vapor. The system pressure is at 10 kPa. What is the minimum amount of heat needed to evaporate all the liquid? SOLUTION A schematic of the process is shown in Figure E2.12. We label the initial state as “state 1” and the final state as “state 2.” The left-hand side of the figure shows the physical process while the right-hand side represents it on a Pv diagram. This heating process occurs in a closed system at constant volume. As water boils, the pressure in the vapor phase will increase. The increase in pressure will require an increase in the temperature of the system for boiling to proceed. (Remember, a given pressure constrains the temperature of the two-phase region.) Thus, energy is required for both the evaporation of water (latent heat) and the increase in temperature (sensible heat). Since bulk kinetic and potential energy are negligible, we can write the first law according to Equation (2.13a): DU 5 Q 1 W

(E2.12A)

Since our system is at constant volume, there is no work done and the heat needed equals the

change in internal energy. This result contrasts with Figure 2.11, where enthalpy is used, since, in that case, there is Pv work. A mass balance gives: mv2 5 m1l 1 mv1 The internal energy in state 1 must account for that of both the saturated liquid and the saturated vapor. Hence, we can write Equation (E2.12A) as: P v P1 = 10 kPa

Liquid

v

process

Vapor P2 = ?

P2

l 10 kPa

Q State 1

2 Liquid-vapor 1 v1 = v2

v

State 2

Figure E2.12 Schematic of evaporation of saturated liquid H2O in a rigid, closed system. (Continued)

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80 ► Chapter 2. The First Law of Thermodynamics U2 2 U1 5 1 mv2u^ v2 2 2 1 ml1u^ l1 1 mv1u^ v1 2 5 Q

(E2.12B)

State 1 is completely constrained. Looking up values for internal energy in the steam tables for saturated water: pressure (Appendix B.2) gives: u^ l1 5 191.8 3 kJ/kg 4 u^ v1 5 2437.9 3 kJ/kg 4 We now need to constrain state 2. We know we have saturated vapor, but we need to find a value for a thermodynamic property. Since the container is rigid, this process occurs at constant volume, as illustrated on the Pv diagram in Figure E2.12. We can find the volume of the vessel as follows: V1 5 V2 5 ml1v^ l1 1 mv1v^ v1 5 1 50 2 1 0.001 2 1 1 4.3 2 1 14.67 2 5 63.1 m3 We can now solve for the specific volume of state 2: v^ v2 5

V2 63.1 m3 m3 5 1.16 B R v 5 m2 54.3 kg kg

Looking up the value, we find the specific volume of saturated vapor matches at: P2 5 0.15 MPa If this value did not match a table entry, we would have to interpolate. Looking up the internal energy of state 2 gives: u^ v2 5 2519.6 3 kJ/kg 4 Solving Equation (E2.12B) for heat gives: Q 5 1 mv2u^ v2 2 2 1 ml1u^ l1 1 mv1u^ v1 2 5 1 54.3 2 1 2519.6 2 2 3 1 50 2 1 191.8 2 2 1 4.3 2 1 2437.9 2 4 5 117 3 106 J

Enthalpy of Reactions A large amount of energy is “stored” in the chemical bonds within molecules. When the atoms in molecules rearrange by undergoing a chemical reaction, the energy stored within the bonds of the products is typically different from that of the reactants. Thus, significant amounts of energy can be absorbed or liberated during chemical reactions. The energy change upon reaction is an important component in applying the first law to reacting systems. It can be characterized by a change in internal energy, Durxn; however, it is more common to report the change in enthalpy of reaction, Dhrxn, since experiments are more conveniently executed at constant pressure.

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2.6 Thermochemical Data for U and H ◄ 81

For example, consider the reaction of two molecules of hydrogen gas with one molecule of oxygen to form gaseous water at 298 K and 1 bar. The reaction stoichiometry can be expressed as follows: 2H2 1 g 2 1 O2 1 g 2 h 2H2O 1 g 2

(2.34)

The reactants contain three bonds per molecule of oxygen reacting: single bonds between hydrogen atoms in each of the two H2 molecules and a double bond between the oxygen atoms in O2. The two product H2O molecules have four oxygen–hydrogen single bonds. These bonds are covalent in nature and vary in energy based on how the interacting valence electrons overlap. Bond energies for the three different types of bonds in this system are reported in Table 2.1. Let’s consider the energetics of this reaction by the following path: One way to do this calculation is to first pull apart the reactant molecules into their constituent atoms and then have the atoms recombine to form the product molecules. This path, in essence, defines a reference state as the atomic form of each element in the system. A schematic of the energetics in this reaction path is shown in Figure 2.13. The energy differences between states depicted by the arrows are based on the bond energies. It takes 14.1 eV of energy 1 1 eV 5 1.6 3 10219 J 2 to dissociate two molecules of H2 and one molecule of O2 into four H atoms and two O atoms, respectively. However, when these atoms reform into two water molecules, they release 17.1 eV of energy. The net energy change represents the internal energy of reaction and is found to be 23.5 eV for every two molecules of water produced by this reaction. The negative sign indicates that the products are more stable than the reactants, and, consequently, energy is released. Reactions that release energy are said to be exothermic, while reactions that absorb energy are termed endothermic. To generalize to any reaction, we introduce the stoichiometric coefficient, νi. The stoichiometric coefficient equals the proportion of a given species consumed or produced in a reaction relative to the other species. It can be found as the number before the corresponding species in a balanced chemical reaction. By convention, it is unitless and positive for products, νproducts . 0, negative for reactants, νreactants , 0, and zero for inerts, νinerts 5 0. For example, in Reaction (2.34): νH2 5 22,

νO2 5 21,

and νH2O 5 2

In the previous discussion, we used atoms for our reference state, since this choice made it straightforward to see how the energy of molecules changes with atomic

TABLE 2.1

Bond Dissociation Energies of H,O Bonds

Bond

Energy [eV/molecule]

HiH

4.50

O wO

5.13

O iH

4.41

Source: Derived from average bond enthalpies reported in G. C. Pimentel and R. D. Spratley, Understanding Chemical Thermodynamics (San Francisco: Holden-Day, 1969).

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82 ► Chapter 2. The First Law of Thermodynamics Figure 2.13 Schematic representation of the

Energy

H,H,H,H O,O .14.1 eV 17.6 eV H2,H2,O2

ΔUrxn = −3.5 eV H2O,H2O

energy needed to dissociate the reactants and products of Reaction (2.34) into their atoms. The resulting energy difference characterizes the internal energy change of reaction, DUrxn.

rearrangement. However, this reference state is inconvenient in practice, since, in nature, species seldom exist as atoms at 298 K and 1 bar. We are free to pick any reference state that we desire as long as we stick to it. A more convenient reference state can be defind as the pure elements in their most stable form at 298 K for the temperature of interest and 1 bar. For example, at 298 K and 1 bar, the most stable form of oxygen is O2 gas while the most stable form of carbon is solid graphite. The enthalpy difference between a given molecule and this reference state is defined as the enthalpy of formation, Dhf. The enthalpy of formation can be represented as: Dhf

elements S species i The enthalpy of formation of a species containing only one element, as it is found in nature, is identically zero. Enthalpies in the form of Dhf are the most common thermochemical data available to calculate the enthalpy of reaction; Appendix A.3 shows some representative values for 25ºC and 1 atm. For example, the enthalpy of formation of liquid water is defined by the reaction: H2 1 g 2 1

1 O2 1 g 2 S H2O 1 l 2 2

since the elements found in water, hydrogen and oxygen are found as diatomic gases at 25ºC and 1 atm. The value for the enthalpy of formation for this reaction is found in Appendix A.3 to be Dh°f,298 5 2285.83 3 kJ/mol 4 . Appendix A.3 also reports the enthalpy of reaction for gaseous water at 298 K and 1 atm. Although water cannot physically exist in this state, the enthalpy of formation is representative of a hypothetical (but important!) change of state that is often useful. For example, we may be interested in a system in which water is reacting at higher temperatures, where it is a vapor. The first step in obtaining the enthalpy of reaction at the system T would be finding it at 298 K. Example 2.14 illustrates such a calculation. With the enthalpies of formation available, it is straightforward to calculate the enthalpy of reaction. Such a calculation path for the enthalpy of reaction at 298 K is illustrated in Figure 2.14. In the dashed (calculation) path, the reactants are first decomposed into their constituent elements, as found in nature. This part of the path is labeled Dh1. The constituent elements are then allowed to react to form products, as given by Dh2. Note that the stoichiometric coefficients of the reactants are negative, making the

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2.6 Thermochemical Data for U and H ◄ 83

Δh1 = Σνi (Δh°f,298)i

298.15 K, 1 bar

reactants

products

Δh2 = Σνi (Δh°f,298)i

Enthalpy

Elemental form found in nature

Reactants 298.15 K 1 bar

Δh°rxn,298

Products 298.15 K 1 bar

Figure 2.14 Calculation path of Dh°rxn from the standard enthalpies of formation, 1 Dh°f 2 i.

signs for Dh1 consistent with the definition of enthalpy of formation above. Equating the two paths yields: Dh°rxn,298 5 Dh1 1 Dh2 5 a νi aDh°f,298 b 1 a νi aDh°f,298 b reactants

5 a νi aDh°f,298 b

i

products

i

i

Thus, if enthalpies of formation are available for all the species in the chemical reaction of interest, the enthalpy of reaction can be determined by scaling each species’Dhf by its stoichiometric coefficient. In summary, Dh°rxn 5 a νih°i 5 a νi aDh°f b

(2.35) i

Often a reaction does not go to completion; that is, there remain some reactants in the outlet stream. For incomplete reactions, we must account only for the enthalpy of reaction for the species that did react in our energy balance. Example 2.16 illustrates such a case. In Chapter 9 we will learn how to quantify the extent to which a chemical reaction proceeds at equilibrium. EXAMPLE 2.13 Determination of Enthalpy of Reaction

Calculate the enthalpy of reaction at 298 K for the following reaction: H2O 1 g 2 1 CH3OH 1 g 2 S CO2 1 g 2 1 3H2 1 g 2 SOLUTION The enthalpy of reaction can be found from the enthalpy of formation data presented in Appendix A.3. In this case, Equation (2.35) can be written as follows: Dh°rxn 5 a νi 1 Dh°f, 298 2 i 5 1 Dh°f, 298 2 CO2 1 3 1 Dh°f, 298 2 H2 2 1 Dh°f, 298 2 H2O 2 1 Dh°f, 298 2 CH3OH The enthalpy of formation of H2 is zero by definition, since that is the form hydrogen takes at

298 K and 1 bar. Taking values for the others from Appendix A.3: Dh°298 5 1 2393.51 2 1 3 1 0 2 2 1 2241.82 2 2 1 2200.66 2 5 49.0 3 kJ/mol 4

The sign of the enthalpy of reaction is positive, indicating that this reaction is endothermic.

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84 ► Chapter 2. The First Law of Thermodynamics

Example 2.14 Land Areas for Alternative Energy Sources

Alternative energy sources that are renewable and that can reduce greenhouse gas emissions are actively being developed. In this example, you wish to explore the feasibility of two alternative energy sources to petroleum. The U.S. petroleum consumption is approximately 3 million m3 per day. You may assume that the fuel value of petroleum can be represented by octane. The density of octane is 0.70 g/cm3. Take the average power density for solar radiation over a 24-hour period to be 200 W/m2. (a) Assuming a solar cell efficiency of 10%, how much area would be needed to provide the equivalent energy to that used by petroleum in the United States? (b) Bioethanol provides another possible alternative energy source. It is a biofuel that can be generated biologically through fermentation of corn. The energy efficiency of photosynthesis, defined as the energy content of the biomass that can be harvested divided by the average energy of solar radiation impinging on the same area, is typically between 0.1 and 1% for crop plants (it goes up to around 3% for microalgae grown in bioreactors). How does the crop area needed for corn compare to the area calculated in part (a)? SOLUTION (a) First we need to calculate the amount of energy produced by the consumption of 3 million m3 of petroleum, as approximated by octane. This value can be found with the enthalpy of reaction. As you will recall, enthalpy is a “constructed” property that is useful to characterize the energy for processes in closed systems at constant pressure because it accounts for both the change in internal energy and the Pv work needed to move the system boundary.

Assuming complete combustion, we can write the balanced equation as: C8H18 1 l 2 1

25 O2 1 g 2 S 8CO2 1 g 2 1 9H2O 1 g 2 2

Using Equation (2.35) and finding the appropriate values in Appendix A.3 gives: kJ R Dh°rxn 5 a yi 1 Dh°f,298 2 i 5 8 1 Dh°f,298 2 CO2 1 9 1 Dh°f,298 2 H2O 2 1 Dh°f,298 2 C8H18 5 25,073 B mol

Why is the sign negative? To find the extensive value for energy, we need to determine the number of moles of octane consumed in a day:

n5

Vr 5 MW

3 3 106 3 m3 4 3 0.7B

g cm3

114.2B

g mol

R 3 106 B

cm3 R m3

5 1.8 3 1010 3 mol 4

R

Thus, H 5 nh 5 29.3 3 1013 3 kJ 4 5 29.3 3 1016 3 J 4

The energy density of the sun can be found from information given in the problem statement: Er 5 200B

J m 2s

R 3 0.10 3 24B

J h s R 3 3,600B R 5 1.7 3 106 B 2 R day h m

We can now find the area: A5

c02.indd 84

9.3 3 1016 3 J 4 H 5 5 5.4 3 1010 3 m2 4 J Er 6 1.7 3 10 B 2 R m

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2.6 Thermochemical Data for U and H ◄ 85

This represents solar cells filling a square parcel of around 150 miles on a side—approximately one-fifth the size of the state of Arizona. (b) The range of area for the biofuel can be found from the ratio of the efficiency to the solar cells. For 1% efficiency, this calculation gives A 5 5.4 3 1011 3 m2 4 and for 0.1% efficiency, A 5 5.4 3 1012 3 m2 4 . These represent square parcels with sides approximately 500 and 1,500 miles, respectively. The latter size is over half the land area of the United States. Comment: the preceding analysis examined the area requirements for two alternative energy sources. However, comparison of the two alternatives is more complex; we must assess the total cost of the systems, including water, capital, operations, and maintenance costs. We should also consider the risks from manufacturing and possible interactions with the food supply and climate change.

Example 2.15 Adiabatic Flame Temperature Calculations

Propane is placed in an adiabatic, constant pressure combustion chamber at 25°C and allowed to react as follows. What is the final temperature in each case? Assume complete combustion. (a) It is mixed with a stoichiometric amount of oxygen and reacts to form H2O and CO2. (b) It is mixed with a stoichiometric amount of air and reacts to form H2O and CO2. (c) It is mixed with a stoichiometric amount of air, and the carbon distribution in the product stream contains 90% CO2 and 10% CO. SOLUTION (a) For part (a), a balanced equation of the chemical reaction is written as

follows: C3H8 1 5O2 S 3CO2 1 4H2O

(E15.1)

A schematic of this process, assuming Reaction E15.1 goes to completion, is shown in Figure E2.15A. An energy balance on the closed system at constant pressure gives [see Equation (2.28)]: DH 5 a 1 nihi 2 2 2 a 1 nihi 2 1 5 Q 5 0

(E15.2)

We now need a path to calculate DH. A convenient choice is illustrated by the solid lines in Figure E2.15B. That figure also shows the overall energy balance constraint of Equation (E15.2) as a dashed line. Because enthalpy of reaction data are available at 298 K (Appendix A.3), we choose a hypothetical path where we first completely combust propane at 298 K, then

P = const.

Combustion process

Well insulated

C3H8, O2 T = 298 [K]

Initial state (1)

CO2 H2O T= ?

Final state (2)

Figure E2.15A Schematic of complete combustion of propane in a stoichiometric mixture of oxygen at constant pressure. The closed system is adiabatic. (Continued)

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86 ► Chapter 2. The First Law of Thermodynamics

2

1

298

298

=0

T

ΔH

T

Σi ni ∫ cP,i dT

2

T2

ΔHrxn,298

C3H8, O2

Figure E2.15B Hypothetical path for calculation of enthalpy in complete combustion of propane.

CO2, H2O

heat the products to a temperature T2, which makes the enthalpy change between states 1 and 2 zero. Using this hypothetical path, Equation (E15.2) becomes: T2

DHrxn,298 1 3 a 1 ni 2 2 1 cp 2 idT 5 0 298

The enthalpy of reaction can be found based on the reaction stoichiometry given by Reaction E15.1: Dhrxn,298 5 a νi 1 Dhof 2 i 5 νCO2 1 Dhof 2 CO2 1 νH2O 1 Dhof 2 H2O 1 νC3Hs 1 Dhof 2 C3Hs 1 νO2 1 Dhof 2 O2 Using values from Appendix A.3, we get: Dhrxn,298 5 3 1 2393.51 2 1 4 1 2241.82 2 2 1 1 2103.85 2 2 0 5 22.044 3 106 B

J mol

R

and using a basis of 1 mol C3H8, the extensive enthalpy of reaction is: DHrxn,298 5 nC3H8Dhrxn,298 5 22.044 3 106 3 J 4 The enthalpy of reaction must be counteracted by an equal but opposite increase in the sensible heat. This effect will allow us to calculate T2. The sensible heat can be found through a summation of heat capacities, as follows: T2

T2

T2

3 a 1 ni 2 2 1 cp 2 i dT 5 3 nCO2 1 cp 2 CO2 dT 1 3 nH2O 1 cp 2 H2O dT 298

298

(E15.3)

298

Values for heat capacity parameters can be found in Appendix A.3. They are summarized in Table E15.1: Table E2.15

c02.indd 86

Heat Capacity Parameters for Species in State 2 A

B

CO2

5.457

1.045 3 10

H 2O

3.470

1.45 3 1023

1.21 3 104

N2

3.280

5.93 3 1024

4.00 3 103

CO

3.376

5.57 3 1024

23.10 3 103

Species

D 3

21.157 3 105

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2.6 Thermochemical Data for U and H ◄ 87

Integrating each term on the right of Equation E15.3 gives: T2

3 nCO2 1 cp 2 CO2 dT 5 1 3 2 RBACO2 1 T2 2 298 2 1

BCO2 2

1 T22 2 1 298 2 2 2

298

2 DCO2 ¢

1 1 2 ≤R T2 298

T2

3 nH2O 1 cp 2 H2O dT 5 1 4 2 RBAH2O 1 T2 2 298 2 1 298

(E15.4A) BH2O 2

2

1 T2 2 1 298 2 2 2

1 1 2 DH 2O ¢ 2 ≤R T2 298

(E15.4B)

The only unknown, T2, can be solved for implicitly by the value at which the sum of the two preceding equations equal2DHrxn,298, 2.044 3 106 3 J 4 . T2 is found to be: T2 5 4,910 3 K 4 This value is known as the adiabatic flame temperature, and in this case is quite large. The adiabatic flame temperature indicates the maximum temperature a reactor can reach for a given fuel. If there is heat transfer out of the system, the temperature will be lower. If you used stoichiometric amount of air, do you think the adiabatic flame temperature would be higher, lower, or the same? How about excess air? How about if some CO formed as well? We will see the results in the cases that follow. (b) If we use air instead of pure oxygen, the reaction described in part (a) remains the same; thus, the enthalpy of reaction remains 22.044 3 106 3 J 4 . However, we must now account for the inert N2 which does not participate in the reaction, in the sensible heat term. This species will provide more “thermal mass,” and consequently, we expect the adiabatic flame temperature to be lower. We have 5 moles of oxygen, so we calculate: nN2 5 ¢

0.79 ≤5 5 18.8 3 mol 4 0.21

The value of nN2 is large compared to the two reaction products so we expect T2 will reduce significantly. To calculate the sensible heat, we must add a term for N2 as follows: T2

T2

T2

T2

3 a 1 ni 2 2 1 cp 2 i dT 5 3 nCO2 1 cp 2 CO2 dT 1 3 nH2O 1 cp 2 H2O dT 1 3 nN2 1 cp 2 N2 dT 298

298

298

298

Using the following equation in addition to Equations (E15.4A ) and (E15.4B): T2

BN2 2 2 3 nN2 1 cp 2 N2 dT 5 1 18.8 2 RBAN2 1 T2 2 298 2 1 2 1 T2 2 1 298 2 2

298

2 DN2 ¢

1 1 2 ≤R T2 298

(E15.4C)

(Continued)

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88 ► Chapter 2. The First Law of Thermodynamics

Again, values for heat capacity parameters for N2 can be found in Appendix A.3 and are reported in Table E15.1. The only unknown, T2, can similiarly be solved for implicitly by the value at which the sum of the three preceding equations equal2DHrxn,298, 2.044 3 106 3 J 4 . T2 is found to be: T2 5 2,370 3 K 4 Note T2 has dropped significantly from the value of 4,910 [K] found in part (a). (c) We must now account for CO production as well. In addition to Reaction E15.1, we need to consider the following reaction: 7 C3H8 1 O2 S 3CO 1 4H2O 2

(E15.5)

with, DhIIrxn,298 5 νCO 1 Dhof 2 CO 1 νH2O 1 Dhof 2 H2O 2 νC3H8 1 Dhof 2 C3H8 2 νO2 1 Dhof 2 O2 5 21.195 3 106 B

J mol

R

Because the carbon distribution in the product stream contains 90% CO2 and 10% CO, for a 1 mol basis of C3H8, we multiple the first reaction (E15.1) by nIC3H8 5 0.9 and the second reaction (E15.5) by nII C3H8 5 0.1. Thus, the total enthalpy of reaction is: II 6 DHrxn,298 5 nIC3H8DhIrxn,298 1 nII C3H8Dhrxn,298 5 21.96 3 10 3 J 4

and the following species concentrations are obtained: nCO2 5 2.7 3 mol 4 , nCO 5 0.3 3 mol 4 , 7 0.79 nH2O 5 4 3 mol 4 , and nN2 5 ¢ ≤ ¢0.9 3 5 1 0.1 3 ≤ 5 18.2 3 mol 4 . To calculate the 0.21 2 sensible heat, we must add a term for CO to the expression we used in part (b): T2

T2

T2

T2

T2

3 a 1 ni 2 2 1 cp 2 idT 5 3 nCO2 1 cp 2 CO2dT 1 3 nCO 1 cp 2 COdT 1 3 nH2O 1 cp 2 H2OdT 1 3 nN2 1 cp 2 N2dT 298

298

298

298

298

T2 is found to be: T2 5 2,350 3 K 4 Note T2 is similar to the value of 2,370 [K] found in part (b).

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2.6 Thermochemical Data for U and H ◄ 89

Example 2.16 Energy Balance for Multiple Incomplete Reactions

Consider an isobaric chemical reactor where the following two simultaneous chemical reactions occur: 1 CO 1 O2 S CO2 2

(1)

1 C 1 O2 S CO 2

(2)

Initially, the reactor contains 4 mol of CO, 4 mol of O2, and 2 mol of C. At the end of the reaction process, the reactor contains 2 mol of CO and 2 mol of O2. If the initial temperature is 25°C and the final temperature is 225°C, determine the amount of heat transferred during the process. SOLUTION An energy balance gives: 498

Q 5 DH 5 DHrxn 1 3 a 1 ni 2 2 1 cp 2 i dT 298

where the extensive enthalpy change of the system has been decomposed into a component for the reaction and a component for the sensible heat in a way similar to that shown in Figure E2.15B. To determine the enthalpy of reaction, we need thermochemical data for each of the two reactions listed above. For Reaction 1 we have: 1 o o o Dh1rxn,298 5 a νi 1 Dhf 2 i 5 1 Dhf 2 CO2 2 1 Dhof 2 CO 2 1 Dhf 2 O2 2 Dh1rxn,298 5 1 2393.51 2 2 1 1 2110.53 2 2 0 5 22.83 3 105 B

J mol

R

(E2.16A)

Similarly, for Reaction 2: 1 o o Dh2rxn,298 5 a νi 1 Dhof 2 i 5 1 Dhf 2 CO 2 1 Dhf 2 C 2 1 Dhof 2 O2 2 Dh1rxn,298 5 1 1 2110.53 2 2 0 2 0 5 21.11 3 105 B

J mol

R

(E2.16B)

The number of moles in the denominator of Equations (E2.16A) and (E2.16B) refer to the moles of CO or C, respectively, which have reacted. We can define this amount as the extent of reaction, j. We will learn more about the extent of reaction when we cover Chemical Reaction Equilibrium in Chapter 9. Thus, the number of moles of any species i can be related to the extent of k reactions through the stoichiometry of the reactions and their extent: ni 5 n0i 1 a νijk

(E2.16C)

k

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90 ► Chapter 2. The First Law of Thermodynamics

where n0i is the initial number of moles of i and ni is the final number of moles after reaction. Using Equation (E2.16C), the number of moles of CO can be written: nCO 5 n°CO 2 j1 1 j2 Substituting in values we get: 2 mol 5 4 mol 2 j1 1 j2 Similarly for O2, we have: nO2 5 n0O2 2 12 j1 2 12 j2 and, 2 mol 5 4 mol 2 12 j1 2 12 j2 Solving the simultaneous equations gives: j1 5 3 mol j2 5 1 mol To calculate the extensive enthalpy of reaction we can multiply the molar enthalpy of reaction for each reaction by its extent: DHrxn 5 j1Dhrxn,1 1 j2Dhrxn,2 5 29.59 3 105 3 J 4

(E2.16D)

Finally, to calculate the sensible heat, we need to determine the final species concentrations. Using Equation (E2.16C), we have: nCO,2 5 4 2 j1 1 j2 5 2 mol nO2,2 5 4 2 12 j1 2 12 j2 5 2 mol nCO2,2 5 0 1 j1 5 3 mol nC,2 5 2 2 j2 5 1 mol so, 498

3 a 1 ni 2 2 1 cp 2 i dT 5 52,000 3 J 4

(E2.16E)

298

where the data from Appendix A.3 were used, similarly to Example 2.15. Calculating the amount of energy transferred by heat: 498

Q 5 DH 5 DHrxn 1 3 a 1 ni 2 2 1 cp 2 i dT 5 29.1 3 105 3 J 4 298

where the values given by Equations (E2.16D) and (E.2.16E) were used. The negative sign indicates that heat must be removed from the reactor to the surroundings.

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2.6 Thermochemical Data for U and H ◄ 91

EXAMPLE 2.17 Enthalpy of Reaction at Different T

How would you calculate the enthalpy of reaction, Dhrxn, at any temperature T, given data for enthalpy of formation at 298 K and heat capacity parameters, available in Appendix A? SOLUTION Since enthalpy is a thermodynamic property, we can construct a hypothetical path that utilizes the available data. The enthalpy of reaction at any temperature T can then be found from the path illustrated in Figure E2.17. The reactants are first brought to 298 K. They are then allowed to react under standard conditions to make the desired products. The products are then brought back up to the system temperature, T. Adding these three steps gives the following integral: T

Dhrxn,T 5 Dhrxn,298 1 3 ¢ a nicP,i ≤ dT 298

i

Substituting in Equations (2.35) and (2.30) gives: T

Dhrxn,T 5

o a νi 1 Dh f , 298 2 i

Di 1 3 ¢R a νi ¢Ai 1 BiT 1 CiT 2 1 2 1 EiT 3≤ ≤dT T i

(E2.17)

298

When standard enthalpies of formation and heat capacity parameters are available, Equation (E2.17) can be solved explicitly for Dhrxn, T at any given T. Note the similarity between the paths in Figure 2.12 and Figure E2.17. In thermodynamics we can often apply concepts developed to solve one type of problem to many other cases.

T

Reactants

Products

Δhrxn,T

Δh1

Reactants

Δh3

Δh°rxn,298 298.15 K

Products

Δhrxn,T = Δh1 + Δh°rxn.298 + Δh3 298

Δh1 = − ∫R Σνi (Ai + BiT + CiT 2 + DiT −2 + EiT 3)dT T reactants T

Δh3 = − ∫R Σνi (Ai + BiT + CiT 2 + DiT −2 + EiT 3)dT 298 products

Figure E2.17 Calculation path of DhoT at temperature T from heat capacity data and

the enthalpy of reaction at 298 K.

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92 ► Chapter 2. The First Law of Thermodynamics

►2.7 REVERSIBLE PROCESSES IN CLOSED SYSTEMS One useful application of thermodynamics is in the calculation of work and heat effects for many different processes by applying the first law. This information allows engineers to use energy more efficiently, saving costs and resources. Since heat and work are path dependent, the specific process must be defined in order to perform the necessary calculations. In this section, we go through two such examples of these types of calculations using an ideal gas undergoing reversible processes. We will look at nonideal gases in Chapter 5. The intent is to gain some experience with applying the first law to get values for work and heat as well as to develop expressions that are useful in understanding the Carnot cycle (Section 2.9).

Reversible, Isothermal Expansion (Compression) Consider a reversible, isothermal expansion of an ideal gas. A schematic of a piston– cylinder assembly undergoing such a process is shown in Figure 2.15. The gas is kept at constant temperature by keeping it in contact with a thermal reservoir. A thermal reservoir contains enough mass so that its temperature does not noticeably change during the process. Can you predict the signs of DU, Q, and W ? Since the internal energy of an ideal gas is only a function of temperature, DU 5 0 For a reversible process, we can integrate over the system pressure (see Section 2.3): W 5 23 PdV

(2.36)

Applying the ideal gas relationship: V5

nRT P

Positive (+) Negative (−)

Zero (0)

ΔU Q W

Process Ideal gas Ideal gas Constant T reservoir

Constant T reservoir Initial state (1)

Final state (2)

Figure 2.15 An ideal gas in a piston–cylinder assembly undergoing a reversible, isothermal expansion. See if you can predict the signs of DU, Q, and W for this process in the table.

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2.7 Reversible Processes in Closed Systems ◄ 93

the differential in volume can be transformed into a differential in pressure (remembering, for this case, that T is constant): dV 5 2

nRT dP P2

(2.37)

Substituting Equation (2.37) into Equation (2.36) and integrating gives: 2

nRT P2 dP 5 nRT ln P1 1 P

(2.38)

P2 P1

(2.39)

W53 Now applying the first law, we get:

Q 5 DU 2 W 5 2nRT ln

Since P2 , P1, the sign for W is negative and for Q is positive. Did you get the sign right in the table in Figure 2.15? How do Equations (2.38) and (2.39) change if the gas undergoes a compression process instead of an expansion?

Adiabatic Expansion (Compression) with Constant Heat Capacity Consider when the same ideal gas undergoes an adiabatic, reversible expansion (as opposed to isothermal). We will assume that the heat capacity of this gas does not change with temperature, that is, constant heat capacity. This process is illustrated in Figure 2.16. Again, can you predict the signs of DU, Q, and W ? Neglecting macroscopic kinetic and potential energy, the first law for a closed system in differential form is obtained from Equation (2.14): 0

dU 5 dQ 1 dW

(2.40)

where the heat transfer was set to zero, since this process is adiabatic. From Equation (2.24) we get: dU 5 ncvdT

(2.41)

dW 5 2PdV

(2.42)

and for a reversible process:

Substituting Equations (2.41) and (2.42) into Equation (2.40) yields: ncvdT 5 2PdV

(2.43)

We can use the ideal gas law to relate the measured properties T, V, and P: d 1 nRT 2 5 d 1 PV 2 5 PdV 1 VdP

(2.44)

where we applied the product rule. Solving Equation (2.44) for dT and then plugging back into Equation (2.43) and rearranging gives: cvVdP 5 2 1 cv 1 R 2 PdV 5 2cPPdV

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(2.45)

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94 ► Chapter 2. The First Law of Thermodynamics Positive (+) Negative (−)

Zero (0)

ΔU Q W

Process Ideal gas Cv ≠ Cv (T )

Ideal gas Wellinsulated

Wellinsulated Initial state (1)

Final state (2)

Figure 2.16 An ideal gas in a piston–cylinder assembly undergoing a reversible, adiabatic expansion. In this example, cv is constant. See if you can predict the signs of DU, Q, and W for this process in the table.

Separating variables in Equation (2.45): 2

cP dV dP 5 cv V P

(2.46)

Now we integrate Equation (2.46) from the initial state 1 to the final state 2, 2k ln¢

V2 P2 ≤ 5 ln ¢ ≤ V1 P1

(2.47)

where k 5 cP/cv. Applying mathematical relationships of the natural logarithm, we can rewrite the left-hand side of Equation (2.47) as: 2k ln¢

V2 V2 ≤ 5 ln ¢ ≤ V1 V1

2k

5 ln ¢

V1 k ≤ V2

so, ln 1 P1V1k 2 5 ln 1 P2V2k 2 or, PVk 5 const

(2.48)

Now integrating for work: W 5 23 PdV 5 23 const V2kdV 5 5

const 1 1 B k21 2 k21 R k 2 1 V2 V1

1 nR 3 P2V2 2 P1V1 4 5 3 T2 2 T1 4 k21 k21

From the first law: DU 5 W 5

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1 nR 3 P2V2 2 P1V1 4 5 3 T2 2 T1 4 k21 k21

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2.8 Open-System Energy Balances on Process Equipment ◄ 95

►Summary A summary of the two cases presented in this section is shown in Table 2.2. In both cases, expansion of a piston provides useful energy to the surroundings in the form of work. However, each case represents a limit. In the isothermal process, all the energy delivered as work is provided by the surroundings in the form of heat. On the other hand, for the adiabatic case, the energy for work is provided by the internal energy of the gas in the system. An intermediate case exists where there is some heat adsorbed from the surroundings as well as some “cooling” of the gas in the system. A process is defined as polytropic if it follows the relation: PVg 5 const

(2.49)

Both the processes in this section can be considered polytropic. The isothermal expansion of an ideal gas follows Equation (2.49) with g 5 1 while the reversible, adiabatic expansion of an ideal gas with constant heat capacity has g 5 k 5 cP/cv. Can you think of another example of a polytropic process?

TABLE 2.2 Summary of Expressions for Change in Internal Energy, Heat, and Work for an Ideal Gas Undergoing a Reversible Process

Isothermal

Adiabatic, cv 2 cv 1 T 2

DU

0

Q

nR 3 T2 2 T1 4 k21

2nRT ln

W

nRT ln

P2 P1

P2 P1

0 nR 3 T2 2 T1 4 k21

► 2.8 OPEN-SYSTEM ENERGY BALANCES ON PROCESS EQUIPMENT In this section, we will examine examples of how to apply the first law to common types of process equipment. These systems will be analyzed at steady-state, when the properties at any place in the system do not change with time. Most cases will consist of one stream in and one stream out, which will be labeled streams 1 and 2, respectively. For these cases the mass balance becomes: m# 1 5 m# 2 And the energy balance, Equation (2.19) becomes: S S 0 5 m# 1 1 h^ 1 12V2 1 gz 2 1 1 m# 2 1 h^ 1 12 V2 1 gz 2 2

(2.50)

Equation (2.50) can be rewritten in molar terms as: S

2

S

2

V V 1 MWgz ≤ 1 1 n# 2 ¢ h 1 MW 1 MWgz ≤ 2 0 5 n# 1 ¢ h 1 MW 2 2

(2.50 molar)

where the molecular weight is used in the macroscopic kinetic and potential energy terms to convert from a mass basis to a molar basis.

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96 ► Chapter 2. The First Law of Thermodynamics It is important to remember that the examples in this section are restricted to cases when steady-state can be applied. If we are interested in start-up or shutdown of these processes, or the case where there are fluctuations in feed or operating conditions, we must use the unsteady form of the energy balance.

Nozzles and Diffusers These process devices convert between internal energy and kinetic energy by changing the cross-sectional area through which a fluid flows. In a nozzle the flow is constricted, increasing eK. A diffuser increases the cross-sectional area to decrease the bulk flow velocity. An example of a process calculation through a diffuser follows.

EXAMPLE 2.18 Diffuser Final Temperature Calcualtion Calculation

The intake to the engine of a jet airliner consists of a diffuser that must reduce the air velocity to zero so that it can enter the compressor. Consider a jet flying at a cruising speed of 350 m/s at an altitude of 10,000 m where the temperature is 10ºC. What is the temperature of the air upon exiting the diffuser and entering the compressor? SOLUTION A schematic diagram of the system, including the information that we know, is shown in Figure E2.15.

T2 = ?

T1 = 10° C → V1

→

V2 ≈ 0 m/s

= 350 m/s Diffuser

Figure E2.18 Schematic of the diffuser in Example 2.18.

This steady-state process occurs in an open system with one stream in and one stream out. In this case, we can write the first law using Equation (2.50): 0 0 0 0 0 S S 2 2 # # V V # # 0 5 n1 (h 1 MW 1 MWgz)1 2 n2(h 1 MW 1 MWgz)2 1 Q 1 Ws (E2.18A) 2 2 where the negligible terms have been set to zero. Note that the reference state for potential energy is set at 10,000 m. A mole balance gives: n# 1 5 n# 2 so that Equation (E2.18A) can be simplified to: T2 S2 1 eK 5 1 MW 2 V1 5 1 h2 2 h1 2 5 3 cP,air dT 2

(E2.18B)

T1

Looking up the value for heat capacity for air in Appendix A.3, we get: A 5 3.355,

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B 5 0.575 3 1023,

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2.8 Open-System Energy Balances on Process Equipment ◄ 97

Using the definition of heat capacity, we get the following integral expression: T2

T2

B 2 D T2 22 3 cPdT 5 R 3 3 A 1 BT 1 DT 4 dT 5 R c AT 1 2 T 2 T d 283

(E2.18C)

283

T1

Using Equation (E2.18C), Equation (E2.18B) becomes: S 1 B 1 1 1 MW 2 V12 5 RBA 1 T2 2 283 2 1 1 T22 2 2832 2 2 D ¢ 2 ≤R 2 2 T2 283

We now have one equation with one unknown, T2, which can be solved implicitly to give: T2 5 344 3 K 4 The temperature of the air increases because the kinetic energy of the inlet stream is being converted to internal energy.

Turbines and Pumps (or Compressors) These processes involve the transfer of energy via shaft work. A turbine serves to generate power as a result of a fluid passing through a set of rotating blades. They are commonly found in power plants and used to produce energy locally as part of chemical plants. This process was described in Section 2.1 and illustrated in Example 2.4. Pumps and compressors use shaft work to achieve a desired outcome. The term compressor is reserved for gases, since they are compressible. Typically they are used to raise the pressure of a fluid. However, they can also be used to increase its potential energy, as illustrated in Example 2.19.

EXAMPLE 2.19 Pump Power Calculation

You wish to pump 0.001 m3 /s of water from a well to your house on a mountain, 250 m above. Calculate the minimum power needed by the pump, neglecting the friction between the flowing water and the pipe. SOLUTION Can you draw a schematic of this process? We need to write the energy balance. This system is at steady-state, with one stream in and one stream out. When working with macroscopic potential energy, it is often convenient to write the balance on a mass (rather than mole) basis. We will neglect the bulk kinetic energy of the water at the inlet and outlet and the heat loss through the pipe. Since there are no frictional losses, the exit temperature is the same as the inlet; therefore, their enthalpy is equal. Thus, the first law simplifies to: 0

0

0

0

0

0

1 S 1 S 2 # # . 0 5 m1 (h 1 V 2 1 gz)1 2 m2 (h 1 V 1 gz)2 1 Q 1 Ws 2 2 .

or, rearranging, # # V2 # Ws 5 m2 1 gz 2 2 5 gz2 v^ 2 (Continued)

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98 ► Chapter 2. The First Law of Thermodynamics # where V is the volumetric flow rate. Solving for shaft work gives: 1 0.001 3 m3 /s 4 2 # Ws 5 B R 1 9.8 3 m/s2 4 2 1 250 3 m 4 2 5 2.5 3 kW 4 1 0.001 3 m3 /kg 4 2

Note that the sign for work is positive. Why? The actual work needed would be greater due to frictional losses.

Heat Exchangers These processes are designed to “heat up” or “cool down” fluids through thermal contact with another fluid at a different temperature. The radiator in your automobile is an example of a heat exchanger. In this application, energy is removed from the engine block to keep it from overheating during combustion. The most common design is when the two streams are separated from each other by a wall through which energy, but not mass, can pass. A calculation on a system employing this design is given in Example 2.20. An alternative design allows the fluids to be mixed directly. An example of such an open feedwater heater is given in Example 2.21.

EXAMPLE 2.20 Heat Exchanger Flow Rate Calculation

You plan to use a heat exchanger to bring a stream of saturated liquid CO2 at 0ºC to a superheated vapor state at 10ºC. The flow rate of CO2 is 10 mol/min. The hot stream available to the heat exchanger is air at 50ºC. The air must leave no cooler than 20ºC. The enthalpy of vaporization for CO2 at 0ºC is given by:

Dh^ vap, CO2 5 236 3 kJ/kg 4 at 0° C What is the required flow rate of air? SOLUTION First, let’s draw a diagram of the system including the information that we know, shown in Figure E2.20A.

nCO2 = 10

mol min

T1 = 0°C T4 = 20°C

Boundary 1 Sat. liq. CO2 Air

Vapor CO2 Q

Air

T2 = 10°C T3 = 50°C

Heat exchanger

Figure E2.20A Schematic of heat exchanger with boundary 1 depicted. There are several possible choices for our system boundary. We will choose a boundary around the CO2 stream, labeled “boundary 1” in Figure E2.20A. In this case, the heat transferred # from the air stream to evaporate and warm the CO2 stream is labeled Q. We next need to

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2.8 Open-System Energy Balances on Process Equipment ◄ 99

perform a first-law balance around boundary 1. The appropriate energy balance is for an open system at steady-state with one stream in and one stream out is: 0 0 0 0 0 S

S

V2 V2 # # # # 0 5 n1 (h 1 MW 1 MWgz)1 2 n2(h 1 MW 1 MWgz)2 1 Q 1 Ws 2 2 where we have set the bulk kinetic and potential energies and shaft work to zero. A mole balance yields: n# 1 5 n# 2 5 n# CO2 so the first-law balance on boundary 1 simplifies to: # Q 5 n# CO2 1 h2 2 h1 2 To determine the change in enthalpy, we must account for the latent heat (vaporization) and the sensible heat of the CO2 stream, that is: T2

1 h2 2 h1 2 5 Dhvap,CO2 1 3 cP, CO2dT T1

These can be found, in [J/mol], as follows. The latent heat is given by: Dhvap,CO2 5 ¢236 c

kJ kg

d ≤ ¢44 c

kg kmol

d ≤ 5 10,400 c

J mol

d

and the sensible heat is given by: T2

3 cP,CO2dT 5 R c A 1 T2 2 T1 2 1 T1

J 1 B 2 1 1 T 2 T 122 2 D ¢ 2 b R 5 353 c d 2 2 T2 T1 mol

where the numerical values for the heat capacity parameters, A, B, and D, are given in Appendix A.3. Thus, the energy transferred via heat to boundary 1 is: # Q 5 100,753 3 J/min 4 Now that we know the rate at which energy must be supplied to the CO2 stream, we can find the flow required for the air. We do this by choosing a different system boundary in the heat exchanger, which is labeled boundary 2 in Figure E2.20B. nCO2 = 10

mol min

T1 = 0°C T4 = 20°C

Sat. liq. CO2 Air

Boundary 2

Figure E2.20B

Vapor CO2 Q

Air

T2 = 10°C T3 = 50°C

Heat exchanger

Schematic of heat exchanger with boundary 2 depicted.

(Continued)

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100 ► Chapter 2. The First Law of Thermodynamics

A balance similar to that above yields: # 2Q 5 n# air 1 h4 2 h3 2

(E2.20A)

# Note that we must be careful about signs! We have included a negative sign on Q since the heat that enters boundary 1 must leave boundary 2. Rearranging Equation (E2.20A) gives: n# air 5 2

# Q 1 h4 2 h3 2

5 2 T4

# Q

# Q

5

3 cP,air dT

RBA 1 T4 2 T3 2 1

B 2 1 1 1 T4 2 T 23 2 2 D ¢ 2 ≤ R 2 T4 T3

T3

Looking up values for the heat capacity parameters in Appendix A.3, we get: 1 100,753 3 J/min 4 2 5 123 3 mol/min 4 n# air 5 877 3 J/mol 4

Alternatively, this problem could have been solved with a system boundary around the entire heat exchanger. In that case, a first-law balance would give: 0 5 n# CO2 1 h2 2 h1 2 1 n# air 1 h4 2 h3 2 which could then be solved for n# air.

EXAMPLE 2.21 Open Feedwater Heater Calculation

Superheated water vapor at a pressure of 200 bar, a temperature of 500ºC, and a flow rate of 10 kg/s is to be brought to a saturated vapor state at 100 bar in an open feedwater heater. This process is accomplished by mixing this stream with a stream of liquid water at 20ºC and 100 bar. What flow rate is needed for the liquid stream? SOLUTION The first step is to draw a diagram of the system with the known information, as shown in Figure E2.21. This example has two inlet streams in, so Equation (2.50) does not apply. If we assume that the rate of heat transfer and the bulk kinetic energy of the streams are negligible and the bulk potential energy and shaft work are set to zero, an energy balance reduces to: # # # 0 5 m1h^ 1 1 m2h^ 2 2 m3h^ 3

(E2.21A)

Similarly, a mass balance at steady-state gives: 0 5 m# 1 1 m# 2 2 m# 3

(E2.21B)

Rearranging Equation (E2.21B) and substituting into (E2.21A) gives: 0 5 m# 1h^ 1 1 m# 2h^ 2 2 1 m# 1 1 m# 2 2 h^ 3

(E2.21C)

We can look up values for the enthalpies from the steam tables (Appendix B). For state 1, the superheated steam is at 500ºC and 200 bar 1 5 20 MPa 2 , so: h^ 1 5 3238.2 3 kJ/kg 4

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2.8 Open-System Energy Balances on Process Equipment ◄ 101

Subcooled T2 = 20°C liquid P2 = 100 bar

Superheated vapor T1 = 500°C P1 = 200 bar kg m1 = 10 s

Open feedwater heater

Saturated vapor P3 = 100 bar

Figure E2.21 Schematic of the open feedwater heater in Example 2.21. For state 2, we use subcooled liquid at 20ºC and 100 bar: h^ 2 5 93.3 3 kJ/kg 4 and the saturated vapor at 100 bar (10 MPa) for state 3 is: h^ 3 5 2724.7 3 kJ/kg 4 Finally, rearranging Equation (E2.21C) and plugging in values gives: kg m# 1 1 h^ 1 2 h^ 3 2 5 1.95 c d m# 2 5 s 1 h^ 3 2 h^ 2 2

Throttling Devices These components are used to reduce the pressure of flowing streams. The pressure reduction can be accomplished by simply placing a restriction in the flow line, such as a partially opened valve or a porous plug. Since these devices occupy a relatively small volume, the residence time of the passing fluid is small. Hence, there is little energy loss by the transfer of heat. Consequently, we can neglect heat transfer. Since there is also no shaft work, the energy balance reduces to a very simple equation, as the next example illustrates.

EXAMPLE 2.22 Throttling Device Calculation

Water at 350ºC flows into a porous plug from a 10-MPa line. It exits at 1 bar. What is the exit temperature? SOLUTION First, let’s draw a diagram of the system, as shown in see Figure E2.22. A steady-state energy balance with one stream in and one stream out is appropriate for this system. We will assume that the bulk kinetic energy of the stream is negligible and that the porous plug is sufficiently small as not to allow a significant rate of heat transfer. Rewriting Equation (2.50) on a mass basis, we get: 0 0 0 0 0 0 S S V2 V2 # # # # ^ ^ 0 5 m1 1 h 1 1 gz 2 1 2 m2 1 h 1 1 gz 2 2 1 Q 1 Ws

2

2

(Continued)

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102 ► Chapter 2. The First Law of Thermodynamics T1 = 350°C P1 = 10 MPa

P2 = 1 bar Porous plug

Figure E2.22 Schematic of the throttling device in Example 2.22. Hence, the energy balance reduces this system to an isenthalpic process:11 h^ 1 5 h^ 2 Looking up the value for the inlet stream from Appendix B.4, we find: h^ 1 5 2923.4 3 kJ/kg 4 Since the enthalpy of stream 2 equals that of stream 1, we have two intensive properties to constrain the exit state: h^ 2 and P2. To find the temperature of stream 2, we must use linear interpolation. Inspection of the steam tables shows that T2 is somewhere between 200 and 250ºC. The following are taken from the superheated steam table at 100 kPa. P 5 100 kPa T[ºC]

h^ 3 kJ/kg 4

200

2875.3

250

2974.3

Interpolation gives: h^ 2 2 h^ at 200 2923.4 2 2875.3 T2 5 200 1 3 DT 4 B ^ R 5 200 1 3 50 4 c d 5 224 3 °C 4 2974.3 2 2875.3 hat 250 2 h^ at 200 11

In general, the energy balance of throttling processes reduces to this simple form.

► 2.9 THERMODYNAMIC CYCLES AND THE CARNOT CYCLE A thermodynamic cycle describes a set of processes through which a system returns to the same state that it was in initially. Typically cycles are used to produce power or provide refrigeration. Since the system returns to its initial state after the cycle has been completed, all the properties have the same values they had originally. The advantage of executing a thermodynamic cycle is that by having the system return to its initial state, we can repeat the cycle continuously. There are many different examples of thermodynamic cycles; in this section, we examine one such cycle—the Carnot cycle.12 In Chapter 3, we will learn that a Carnot cycle represents the most efficient type of cycle we can possibly have.

12

This cycle was conceived in 1824 by Sadi Carnot, a French engineer, to explore the maximum possible efficiency that could be obtained by the steam engines of his time.

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2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 103

QH

Adiabatic compression

Constant TH Isothermal expansion

State 1 State 2 T1 T2 = T1 P1 P2

State 4 T 4 = T3 P4

State 1 State 4 T1 T4 P1 P4

Isothermal compression

State 3 T3 P3

Well insulated

Adiabatic expansion

State 2 State 3 T2 T3 P2 P3

QC Constant TC

Figure 2.17 An ideal gas undergoing a Carnot cycle. The Carnot cycle consists of four reversible processes by which the gas is returned to its original state.

Figure 2.17 shows an ideal gas in a piston–cylinder assembly undergoing a Carnot cycle. In this cycle, the gas goes through four reversible processes through which it returns to its initial state. Two processes occur isothermally, alternating with two adiabatic processes. These processes were analyzed, individually, in Section 2.7. Consider a gas that is initially in state 1 at a pressure P1 and a temperature T1 as shown at the top of Figure 2.17. The first step of a Carnot cycle is a reversible isothermal expansion, in which the gas is exposed to a hot reservoir at temperature, TH; it gains energy via heat, QH, as indicated on the diagram. During this process, which takes the system from state 1 to state 2 (at P2 and T2 ), the pressure decreases while the temperature stays the same. The piston–cylinder assembly is then transferred into an adiabatic (well-insulated) environment and expanded further to state 3. In this step, both T and P decrease. In both expansion processes, work is done by the system on the surroundings; that is, we get useful work out. The system then undergoes two reversible compression processes. First, it is isothermally compressed by being placed in contact with a cold thermal reservoir at temperature TC. The gas loses an amount of energy via heat, QC to the cold reservoir. This process takes the system to state 4 1 P4, T4 2 . The system returns to its initial state (state 1) through an adiabatic compression. The net work obtained in a Carnot cycle is given by the sum of the work obtained in all four processes: 2Wnet 5 0 W12 0 1 0 W23 0 2 0 W34 0 2 0 W41 0

(2.51)

Since the overall effect of the power cycle is to deliver work from the system to the surroundings, the sign of Wnet is negative. The subscript “ij” on the terms for work in

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104 ► Chapter 2. The First Law of Thermodynamics Equation (2.51) refers to the work obtained in going from state i to state j. Absolute values are used to explicitly distinguish the steps where we get work out from those where we must put work in. The net work obtained from a Carnot cycle can also be calculated by applying the first law to the entire cycle. Since the cycle returns the system to its original state, its internal energy must have the same value as at the start of the cycle. Thus, DUcycle 5 0 5 Wnet 1 Qnet Comparing Equations (2.51) and (2.52), we see that: QH QC 0

(2.52)

0

2Wnet 5 Qnet 5 Q12 1 Q23 1 Q34 1 Q41 5 0 QH 0 2 0 QC 0 We see that the net work obtained is the difference in heat absorbed from the hot reservoir, QH, and expelled to the cold reservoir, QC. An alternative way of schematically representing a Carnot cycle is shown in Figure 2.18a. This schematic gives an overview of the energy transferred between the Carnot engine and the surroundings. Inside the circle labeled “Carnot engine” are the four processes depicted in Figure 2.17.

Efficiency The efficiency, h, of the cycle is defined as the net work obtained divided by the heat absorbed from the hot reservoir: h;

Wnet net work 5 heat absorbed from the hot reservoir QH

Hot reservoir, TH

Hot reservoir, TH

QH

Carnot engine

(2.53)

QH

Wnet

Carnot refrigeration

Wnet

QC

QC

Cold reservoir, TC

Cold reservoir, TC

(a )

(b )

Figure 2.18 Alternative representation for the Carnot cycle. (a) Carnot “engine”; (b) Carnot refrigerator.

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2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 105

For a given amount of energy available from the hot reservoir via QH, the greater the efficiency, the more work we obtain. For example, say the high temperature in the hot reservoir is obtained from the combustion of coal. A high efficiency means we can reduce the amount of coal we need to combust to produce a given amount of work.13 A refrigeration cycle allows us to cool a system so that we can store some ice cream and so on. In this case we want to expel heat from a cold reservoir. It takes work from the surroundings to accomplish this task. For example, your freezer at home needs electricity to keep the ice cream cold. Figure 2.18b shows a schematic way of representing the energy transferred in a refrigeration cycle. We supply work to the cycle in order to absorb energy via heat QC from the cold reservoir. We then expel the energy via heat QH to the hot reservoir. Thus, the direction of heat transfer is opposite that of the power cycle depicted in Figure 2.18a. The effectiveness of a refrigeration cycle is measured by its coefficient of performance, COP, which is defined as follows: COP 5

QC Wnet

(2.54)

We can see from Equation (2.54) that the higher the COP, the less work it takes to produce a desired level of cooling. Can you draw the analogous cycle to Figure 2.17 that goes in the circle labeled “Carnot refrigerator”?

EXAMPLE 2.23 Carnot Cycle Efficiency

Consider 1 mole of an ideal gas in a piston–cylinder assembly. This gas undergoes a Carnot cycle, which is described below. The heat capacity is constant, cv 5 1 3/2 2 R. (i) A reversible, isothermal expansion from 10 bar to 0.1 bar. (ii) A reversible, adiabatic expansion from 0.1 bar and 1000 K to 300 K. (iii) A reversible, isothermal compression at 300 K. (iv) A reversible, adiabatic compression from 300 K to 1000 K and 10 bar. Perform the following analysis: (a) Calculate Q, W, and DU for each of the steps in the Carnot cycle. (b) Draw the cycle on a Pv diagram. (e) Calculate the efficiency of the cycle. (d) Compare h to 1 2 1 Tc /TH 2 . (e) If what is found in part (d) is true, in general, suggest two ways to make the above process more efficient. SOLUTION (a) We will analyze each of the steps separately, with a little help from the results of Section 2.7. We label each state in a manner consistent with Figure 2.17. (i) The first process is a reversible, isothermal expansion at 1000 K from state 1 at 10 bar to state 2 at 0.1 bar. By definition, the internal energy change for an ideal gas at constant temperature is: DU 5 0 We can calculate the work using a result from Section 2.7: W53

P2 nRT dP 5 nRT ln 5 238,287 3 J 4 P P1 (Continued)

13

The steam engine was patented by James Watt in 1765. These first steam engines had efficiencies of only about 1%. Indeed, we can see there was much engineering that remained to be done!

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106 ► Chapter 2. The First Law of Thermodynamics

The negative sign indicates that the system is performing work on the surroundings (we are getting useful work out). To find the heat, we apply the first law: QH 5 DU 2 W 5 38,287 3 J 4 (ii) The second process is a reversible, adiabatic expansion from 0.1 bar and 1000 K to state 3 at 300 K. The pressure decreases during this process. By the definition of an adiabatic process: Q50 At constant heat capacity, the change in internal energy becomes: DU 5 ncv 1 T3 2 T2 2 5 28730 3 J 4 Applying the first law gives: W 5 DU 5 28730 3 J 4 (iii) The third process is a reversible, isothermal compression at 300 K. Again: DU 5 0 and, W 5 23 PdV 5 3

P4 nRT dP 5 nRT ln P P3

(E2.23A)

However, we now need to find P3 and P4. From Section 2,7, we know PVk 5 const for the polytropic, adiabatic processes (ii) and (iv). We first find k: k5

cv 1 R cP 5 5 1.67 cv cv

Setting PVk equal for states 2 and 3 gives: 5 PVk 5 P2V 1.67 2

1 nRT2 2 1.67

5 7347 5

P20.67

1 nRT3 2 1.67 P0.67 3

Solving for P3, we get: P3 5 B

1 nRT3 2 1.67 7347

1.5

5 0.0049 bar

R

Similarly for P4: PVk 5 P1V1.67 5 1

1 nRT1 2 1.67 P10.67

5 341 5

1 nRT4 2 1.67 P40.67

and, P4 5 B

1 nRT4 2 1.67 341

1.5

R

5 0.49 bar

Thus, the work given by Equation (E2.23A) is:

W 5 nRT ln

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0.49 bar 5 11,486 3 J 4 0.0049 bar

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2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 107

The work is positive for this compression process. From the first law, QC 5 DU 2 W 5 211,486 3 J 4 (iv) The fourth process is a reversible, adiabatic compression from state 4 at 300 K and 0.52 bar back to state 1 at 1000 K and 10 bar (process 4 S 1). After this process, the gas can repeat steps (i), (ii) . . . Again, for this adiabatic compression: Q50 At constant heat capacity, the change in internal energy becomes: DU 5 ncv 1 T1 2 T4 2 5 8730 3 J 4 Applying the first law gives: W 5 DU 5 8730 3 J 4 TABLE E2.23A

Results of Calculations for Carnot Cycle in Example 2.23 DU 3 J 4

W [J]

Q [J]

(i) State 1 to 2

0

238,287

38,287

(ii) State 2 to 3

28,730

28,730

0

(iii) State 3 to 4

0

11,486

211,486

(iv) State 4 to 1

8,730

8,730

0

0

226,800

26,800

Process

Total TABLE E2.23B

T, P, and v for Carnot Cycle in Example 2.23

State

T [K]

P [bar]

v 3 m3 /mol 4

1

1000

10

0.0083

2

1000

0.1

0.8314

3

300

0.0049

5.10

4

300

0.49

0.051

A summary of DU, W , and Q for the four processes and the totals for the cycle are presented in Table E2.23A. We get a net work of 26.8 kJ after one cycle. (b) To sketch this process on a Pv diagram, we first calculate the molar volume at each state using the ideal gas law. The results are presented in Table E2.23B. A sketch (not to scale) of the Pv diagram is presented in Figure E2.23. The work for a reversible process is given by the area under the Pv curve; hence, the net work is given by the shaded area in the box in Figure E2.23. Isotherms, TH and TC, for processes (i) and (iii) are also labeled. (c) The efficiency is given by Equation (2.53) h;

26,800 net work 5 5 0.70 heat absorbed from the hot reservoir 38,287

(E2.23B)

In practice, electrical power plants have efficiencies around 40%. (Continued)

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108 ► Chapter 2. The First Law of Thermodynamics

P 1

Q=0

TH Isotherm Q H = −W 2 wnet

Q=0

4 −Q

C

=W

3

TC Isotherm

v

Figure E2.23 Pv diagram of a Carnot cycle. The shaded area represents the net work obtained from one cycle.

(d) Applying the relation in the problem statement, we get: 12

TC 300 5 12 5 0.7 TH 1000

(E2.23C)

Comparing the values of Equations (E2.23B) and (E2.23C), we get: h512

TC TH

(E2.23D)

(e) If Equation (E2.23D) holds, the process can be made more efficient by raising TH or lowering TC. Note that these options will push the isotherms depicted in Figure E2.23 up and down, respectively. Thus either raising TH or lowering TC will serve to make the shaded box, which represents net work, larger. We will learn in Chapter 3 that, indeed, Equation (E2.23D) is true in general. However, we can also reach this conclusion by realizing that the isotherms in Figure E2.23 are fixed on the Pv plane.

►2.10 SUMMARY The first law of thermodynamics states that the total energy in the universe is a constant. Energy balances have been developed for closed systems and for open systems. For example, the integral equation of the first law for a closed system, written in extensive form, is: DU 1 DEK 1 DEP 5 Q 1 W

(2.12a)

For open systems, it is convenient to write the first law on a rate bases. The integral equation, in extensive form, is: a

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dU dEK dEP 1 S 1 S # 1 1 b 5 a m# in ah^ 1 V 2 1gzb 2 a m out ah^ 1 V 2 1gzb 1Q# 1 W# s dt dt dt sys in 2 2 in out out (2.21)

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2.10 Summary ◄ 109

We have also developed these equations in intensive forms, on a mass and a molar basis, and for differential increments. Given a physical problem, we must determine which form to use and which terms in these equations are important and which terms are negligible or zero. We must also identify whether the ideal gas model or property tables are needed to solve the problem. For some processes, it is convenient to define a hypothetical path so that we can use available data to solve the problem. We applied the first law to many engineering systems. Examples of closed systems included rigid tanks and adiabatic or isothermal expansion/compression in a piston–cylinder assembly. Steady-state open systems were illustrated by nozzles, diffusers, turbines, pumps, heat exchangers, and throttling devices. Transient open-system problems included filling or emptying of a tank, while the Carnot power and refrigeration cycles provided examples of thermodynamic cycles. However, you should understand the concepts well enough so that you are not restricted to the systems discussed above but rather are able to apply the first law to any system of interest. A process is reversible if, after the process occurs, the system can be returned to its original state without any net effect on the surroundings. This result occurs only when the driving force is infinitesimally small. The reversible case represents the limit of what is possible in the real world—it gives us the most work we can get out or the least work we have to put in. Moreover, only in a reversible process can we substitute the system pressure for the external pressure in calculating Pv work. Real processes are irreversible. They have friction and are carried out with finite driving forces. In an irreversible process, if the system is returned to its original state, the surroundings must be altered. We can compare the amount of work required in an irreversible process to that of a reversible process by defining the efficiency factor, h. Our strategy for actual, irreversible processes is often solving a problem for the idealized, reversible process and then correcting for the irreversibilities using an assigned efficiency factor. The thermodynamic property enthalpy, h, is defined as: h ; u 1 Pv

(2.18)

We first recognized the usefulness of h in describing streams flowing into and out of open systems. In such cases, we must account for both the internal energy of the stream and the flow work associated with it entering or leaving the system. Enthalpy accounts for both these effects. Enthalpy also describes the combined effects of internal energy and Pv work for closed systems at constant P. Therefore, experiments conveniently done in closed systems at constant pressure are reported using enthalpy. For example, experimental data for the energetics associated with phase changes and chemical reaction are typically reported using enthalpy. On a molecular level, internal energy encompasses the kinetic and potential energies of the molecules. Changes in internal energy present themselves in several macroscopic manifestations, including changes in temperature, changes in phase, and changes in chemical structure, that is, chemical reaction. A change in internal energy that is manifested as a change in temperature is often termed sensible heat. We can associate the increase in molecular kinetic energy and, therefore, in u with temperature to three possible modes in which the molecules can obtain kinetic energy: translational, vibrational, and rotational. Data for the change in energy due to sensible heat can be obtained using heat capacity or using property tables. The heat capacity for a species is often fit to a polynomial in temperature of the form: cP 5 A 1 BT 1 CT 2 1 DT22 1 ET 3

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(2.30)

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110 ► Chapter 2. The First Law of Thermodynamics The parameters A, B, C, D, and E are reported for some ideal gases in Appendix A.2. Heat capacity parameters at constant pressure of some liquids and solids are also reported in this appendix. An ideal gas presents a special case: All its internal energy is realized as molecular kinetic energy—it is a function of T only. We refer to a change in energy that results in phase transformations as latent heat. This energy is associated with the different degrees of attraction between the molecules in the different phases. Data for latent heats are reported as enthalpies of vaporization, fusion, and sublimation. These values are typically reported at 1 bar; however, using these values, we can construct hypothetical paths to find the latent heats at any pressure. The internal energy changes associated with chemical reactions can be attributed to the energetic differences between the chemical bonds of the reactants and the products. The enthalpy of a given chemical reaction can be determined from reported enthalpies of formation.

►2.11 PROBLEMS Conceptual Problems 2.1 In the two processes shown in the following figure, the same amount of heat, q, is supplied to equal amounts (in moles) of different gases, gas A and gas B. Both gases are initially at room temperature. The heat capacity of gas A is greater than the heat capacity of gas B. These processes take place at constant volume. Which gas has the greater final temperature? Explain. Process 1

Process 2

T1

T2

V = const

V = const A

B

A

A A

A

A A

A

A

B

B A

B

B

B A

B

q

B

B

A

B

q

2.2 In the two processes shown in the following figure, the same amount of heat, q, is supplied to the same gas, gas A. In both processes, the initial temperature and initial number of moles are equal. Process 1 occurs at constant volume; process 2 occurs at constant pressure. (a) Which gas has the greater final temperature? Explain. (b) For which process, if any, is it convenient to use the property enthalpy? Explain. Process 1

Process 2

T1 P = const

V = const A A

A A

A A

A

A

q

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A

A

A

A A

A

A A

A

A

T2

A

A

A A

q

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2.11 Problems ◄ 111 2.3 In the steady-state process shown in the following figure, steam flows through a porous plug, and its pressure drops from 10 MPa to 1 bar. Does the temperature increase, stay the same, or decrease? Explain. Water

T2 P2 = 1 bar

T1 P1 = 10 MPa Porous plug

2.4 In the steady-state process shown in the following figure, an ideal gas flows through a porous plug, and its pressure drops from 10 MPa to 1 bar. Does the temperature increase, stay the same, or decrease? Explain. Ideal Gas

T2 P2 = 1 bar

T1 P1 = 10 MPa Porous plug

2.5 In the steady-state process shown in the following figure, air flows through a turbine. Does the temperature increase, stay the same, or decrease? Explain.

Air in P1

Turbine

T1 Air out

P2 T2

2.6 An ideal gas flows into a well-insulated tank that is initially at vacuum, as shown in the following figure. How does T2 compare to Tin (higher, the same, or lower)? Explain.

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Ideal Gas Tin

Ideal Gas Tin

Pin

Pin

Initially Vacuum

Ideal Gas P2 = Pin T2

State 1

State 2

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112 ► Chapter 2. The First Law of Thermodynamics 2.7 Four processes are shown in the following figure. Each process occurs in a well-insulated piston cylinder assembly and starts from the same initial state, state 1. Put the final temperatures in order from highest to lowest. Explain. In Process A, a large block of mass, M, is removed from the piston. In Process B, an equivalent mass, M, is removed in incremental amounts. In Process C, a large block of mass, M, is placed on the piston. In Process D, an equivalent mass, M, is added in incremental amounts. Weightless, frictionless piston

Weightless, frictionless piston Patm

Well Insulated Patm

M

Process A T1 P1

State 1

∂m

TA

Process C T1 P1

State A

M

Patm

M TC

State C

State 1 Weightless, frictionless piston

Patm

Well Insulated Patm

Patm

M

Weightless, frictionless piston mT=M

Well Insulated

M

Well Insulated

mT =M ∂m

Patm

M

Process B

T1 P1

TB

State 1

State B

T1 P1 State 1

Process D

Patm M TD State D

2.8 State the conditions under which the following equations apply (try to be as specific as you can with the limitations). # # (a) 0 5 m# 1h1 1 m# 2h2 2 m# 3h3 1 Q 2 Ws (b) h2 2 h1 5 cP 1 T2 2 T1 2 (c) u2 2 u1 5 cP 1 T2 2 T1 2 (d) cP 5 cv 1 R (e) h 5 u 1 Pv (f) w 5 23 Pdv (g) Du 5 q 1 w (h) q 5 0 2.9 In Example 2.4, we solved a problem where 10.0 kg of water was reversibly compressed in a piston–cylinder assembly from a pressure of 20 bar and a volume of 1.0 m3 to a pressure of 100 bar. kJ kJ In this example, the work was calculated to be w 5 285 B R, the heat was q 5 210 B R, and kg kg the final temperature was T2 5 525°C. Qualitatively answer the following questions. You do not have to do any calculations, but you must choose the right answer and explain your reasoning. (a) Consider an adiabatic process that occurs from the same initial state (State 1) to the same final state (State 2). Will the magnitude of work required for the compression be (greater than, less than, or equal to) the value calculated in Example 2.4? Explain. (b) Consider an isothermal process that occurs from the same initial state (State 1) to the same final pressure 1 P2 2 . Will the heat transfer be greater than, less than, or equal to, the value calculated in Example 2.4? Explain. (c) Consider an irreversible process that occurs from the same initial state (State 1) to the same final state (State 2). Will the magnitude of work required for the compression be greater than, less than, or equal to, the value calculated in Example 2.4? Explain.

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2.11 Problems ◄ 113 (d) Consider an irreversible process that occurs from the same initial state (State 1) to the same final state (State 2). Will the heat transfer be greater than, less than, or equal to, the value calculated in Example 2.4? Explain. 2.10 Consider a cup of cold water. Come up with and sketch as many ways as you can think of to raise the temperature of the water. 2.11 Consider the compression of a spring by placing a large mass on it. The degree to which the spring compresses is related to its spring constant, k. The force exerted by the spring on the mass is given by: F 5 2kx where x represents the distance the spring compresses from its relaxed position. Does the compression of a spring represent potential or internal energy? 2.12 Take a thick rubber band and expand it by stretching it. If you hold it to your lips, you will sense that it is hotter. However, we have seen that the temperature of a gas in a piston–cylinder assembly cools upon expansion. Explain these opposite results in the context of an energy balance. 2.13 If you sprinkle water on a very hot skillet, it will evaporate. However, you get the paradoxical result that at higher temperatures the water drops take longer to evaporate than at lower temperatures. Explain this result. 2.14 On a hot summer day, your roommate suggests that you open the refrigerator to cool off your apartment. Choosing the entire apartment as the system, perform a first-law analysis to decide whether this idea has merit. 2.15 You are making plans to stay warm in the winter. Due to your busy schedule, you are typically away from your house all day. You know it costs a lot to operate the electric heaters to keep your house warm. However, you have been told that it is more efficient to leave your house warm all day rather than turn off the heat during the day and reheat the house when you get home at night. You think that thermodynamics may be able to resolve this issue. Draw a schematic of the system, the surroundings, and the boundary. Illustrate the alternative processes. What is your choice to save power? Justify your answer. 2.16 Explain why ice often forms on the valve of a tank of compressed gas (high pressure) when it is opened to the atmosphere and the gas escapes. Where does the H2O come from? 2.17 (a) What requires more heat input: to raise the temperature of a gas in a constant-pressure cylinder or in a constant-volume bomb? Explain. (b) Explain why you feel less comfortable on a hot summer day when the (relative) humidity is higher even though the temperature is the same. 2.18 Consider that towel you used to dry yourself with after your last shower. After being washed it must be dried so it can be used again. Estimate how much energy it takes to dry the towel after it has been washed. State all assumptions that you make. Now consider the dryer in your dwelling (or at the laundromat). If the towel is the only item placed in the dryer, estimate how much energy is used to actually dry the towel. What is the efficiency of the process? Can you suggest any ways to make it more efficient?

Numerical Problems 2.19 A rigid container contains saturated water at a pressure of 2 bar and a quality of 0.42. The water undergoes a process in which it is heated to a final temperature of 540°C. Determine the final pressure of the container, and the energy transferred by work and by heat during this process. 2.20 The normal human body temperature is approximately 37°C. For normal metabolic processes, it is important to maintain the body at this temperature. Hypothermia is a condition when the core body temperature drops below 35°C. You may assume the body contains 60% water.

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114 ► Chapter 2. The First Law of Thermodynamics (a) Consider a 70 kg person, and estimate the amount of energy lost by heat for the onset of hypothermia. State any assumptions that you make. (b) What mechanisms does your body have to allow more heat to be lost than that estimated in part a before the onset of hypothermia? 2.21 Consider the piston–cylinder assembly containing 0.20 kg of pure water, as shown in the following figure. The cross-sectional area of the piston is 0.50 m2, and its initial height is 0.7172 m above the base of the cylinder. The initial pressure is 1.0 bar. A spring with a linear spring constant, N R, is attached to the piston. Initially the spring exerts no force on the very thin m piston. A block of mass m 5 20,408 kg, is then placed on the piston, causing the gas to compress, until the forces again balance. During the compression process, 6.17 3 105 3 J/kg 4 of work is done on the system. k 5 1.62 3 105 B

F = −kx

m = 20,408 Patm = 1 bar kg

A = 0.5 m2

H2O

h1 = 0.7172 m m1 = 0.2 kg State 1

(a) Precisely draw the process on a Pext-v diagram. Draw the area that represents the work. (b) How much heat, in J/kg, is transferred during the process? 2.22 Consider boiling water to make a pot of tea. Say it takes roughly 10 min to bring 1 L of H2#O taken from the tap at 25ºC to boil. What is the total heat input, Q? What is the rate of heat input, Q? 2.23 Consider a process that takes steam from an initial state where P 5 1 bar and T 5 400°C to a state where P 5 0.5 bar and T 5 200°C. Calculate the change in internal energy for this process using the following sources for data: (a) the steam tables; (b) ideal gas heat capacity. 2.24 Consider a piston–cylinder assembly that contains 2.5 L of an ideal gas at 30ºC and 8 bar. The gas reversibly expands to 5 bar. (a) Write an energy balance for this process (you may neglect changes in potential and kinetic energy). (b) Suppose that the process is done isothermally. What is the change in internal energy, DU, for the process? What is the work done, W, during the process? What is the heat transferred, Q? (c) If the process is done adiabatically (instead of isothermally), will the final temperature be greater than, equal to, or less than 30ºC? Explain. 2.25 Five moles of nitrogen are expanded from an initial state of 3 bar and 88ºC to a final state of 1 bar and 88ºC. You may consider N2 to behave as an ideal gas. Answer the following questions for each of the following reversible processes: (a) The first process is isothermal expansion. (i) Draw the path on a Pv diagram and label it path A. (ii) Calculate the following: w, q, Du, Dh. (b) The second process is heating at constant pressure followed by cooling at constant volume. (i) Draw the path on the same Pv diagram, label it path B. (ii) Calculate the following: w, q, Du, Dh. 2.26 A 5-kg aluminum block sits in your lab, which is at 21ºC. You wish to increase the temperature of the block to 50ºC. How much heat in [J] must be supplied?

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2.11 Problems ◄ 115 2.27 A piston–cylinder assembly contains 3 kg of steam at a pressure of 100 bar and a temperature of 400ºC. It undergoes a process whereby it expands against a constant pressure of 20 bar, until the forces balance. During the process, the piston generates 748,740 [J] of work. Water is not an ideal gas under these conditions. Determine the final temperature in K and the heat transferred (in [J]) during the process. 2.28 Consider a piston–cylinder assembly that contains 1 mole of ideal gas, A. The system is well insulated. Its initial volume is 10 L and initial pressure, 2 bar. The gas is allowed to expand against a constant external pressure of 1 bar until it reaches mechanical equilibrium. Is this a reversible process? What is the final temperature of the system? How much work was obtained? For gas A: cV 5 1 5/2 2 R. 2.29 For the well-insulated piston–cylinder assembly containing 1 mole of ideal gas described in Problem 2.28, describe the process by which you can obtain the maximum work from the system. Calculate the value for the work. What is the final temperature? Why is T lower than that calculated in Problem 2.28? 2.30 The insulated vessel shown below has two compartments separated by a membrane. On one side is 1 kg of steam at 400ºC and 200 bar. The other side is evacuated. The membrane ruptures, filling the entire volume. The final pressure is 100 bar. Determine the final temperature of the steam and the volume of the vessel.

H2O T1 = 400°C P1 = 200 bar

Vacuum

Insulation

2.31 A membrane divides a rigid, well-insulated 2 -m3 tank into two equal parts. The left side contains an ideal gas 3 cP 5 30 J/ 1 mol K 2 4 at 10 bar and 300 K. The right side contains nothing; it is a vacuum. A small hole forms in the membrane, gas slowly leaks out from the left side, and eventually the temperature in the tank equalizes. What is the final temperature? What is the final pressure? 2.32 A well-insulated rigid container contains two 10 L compartments initially at 300 K. Each compartment contains argon gas. The compartments are separated by a well-insulated piston that is held in place by a restraining pin. One compartment is initially at a pressure of 1 bar and the other is at 5 bar. After the pin is removed, the piston moves, but no heat is transferred through the piston. Determine the final temperature, pressure and volumes of each compartment. (Thanks to Prof. Frank Foulkes for providing the idea for this problem.) 2.33 You have 0.4 mol of gas A within a piston–cylinder assembly. The initial pressure is 20 bar, and the initial temperature is 675 K. The system then undergoes an adiabatic process where the piston expands against constant external pressure of 1 bar until the forces balance. (a) What is the final temperature of the system? (b) How much work is done during the process? You may assume gas A behaves as an ideal gas during this process. In checking the lab notes for data for gas A, you come across the following experiment that was performed on a closed system containing pure gas A within a rigid container. The container was connected to a heat source (in this case a resistance heater) and was otherwise well insulated. As a known amount of heat, Q, was provided through the resistive heater, the temperature, T, of the system was measured. The following data were collected for one mole of species A:

c02.indd 115

T [K]

Q [J]

293

0

300

30.3

350

735.8

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116 ► Chapter 2. The First Law of Thermodynamics 400

1240.4

450

1933.9

500

2505.5

550

3248.0

600

3915.6

650

4498.1

700

5155.7

2.34 One kg of liquid n-octane 1 C8H18 2 is placed in a closed rigid container with 50% excess air at 100°C and 1 bar. It undergoes an isothermal process in which it reacts and completely combusts to form CO2 (g) and H2O (g). State any assumptions that you make in solving this problem. Take the heat capacity of liquid n-octane to be: cP 5 250 B

J mol K

R

(a) What is the final system pressure? (b) How much heat must be removed to keep the chamber at 100°C? (c) Consider the same amount of C8H18 reacts with a stoichiometric amount of O2 and forms the same products, how do you think the value you calculated in part B will change? (More Q, Less Q, No change in Q). Conceptually explain your answer. No calculations! 2.35 Repeat Problem 2.34, except consider now the combustion products contain CO (g) as well as CO2 (g) and H2O (g) with the ratio of CO2 / CO produced being 4/1. All other conditions remain the same as in Problem 2.34. 2.36 Approximately half the dry mass of the human body consists of proteins. The backbone of a protein is a long polypeptide chain. In its natural or “native” state, the protein folds back on itself and is held together by a large number of intramolecular interactions, which include van der Waals interactions and hydrogen bonding (which we will discuss further in Chapter 4). The resulting structure formed by the protein is important for its function. Under conditions in which the interactions are overcome, the protein unfolds and becomes “denatured.” The heat capacity of a protein in solution can be measured using a system similar to the one schematically shown in Figure 2.10b. A plot of heat capacity versus temperature for a protein is shown in the following figure. 50

cp [kJ/(mol K)]

45 40 35 30 25 20 15 40

50

60

70

80 T °C

Answer the following questions: (a) Physically explain, where is the protein in its native state where is it denatured; what is the cause of the “hump”?

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2.11 Problems ◄ 117 (b) Estimate the enthalpy change associated with the denaturation process, Dhd. Explain what physical process this number represents. (c) What would cP versus T look like if denaturation occurred in a two-step process? 2.37 Fuel cells are a promising alternative energy technology. They are based on producing energy by the following reaction: 1 H 2 1 g 2 1 O2 1 g 2 S H 2O 1 g 2 2 One type of fuel cell, the solid oxide fuel cell, operates at high temperatures. A solid oxide fuel cell is fed 0.32 mol/s H2 and 0.16 mol/s O2, and the reaction goes to completion. The heat loss rate (in W) is given by: # Q 5 27.1 1 T 2 T0 2 2 4.2 3 1028 1 T4 2 T40 2 where T is in K and T0 is the ambient temperature, which can be taken to be 293 K. (Thanks to Prof. Jason Keith for providing the information for this problem.) (a) Explain physically what the first and second terms on the right-hand side of the preceding heat rate equation represent. (b) Calculate the temperature at which the rate of heat loss is equal to the rate of heat generation. 2.38 Consider a piston–cylinder assembly containing 10 kg of water. Initially the gas has a pressure of 20 bar and occupies a volume of 1.0 m3. Under these conditions, water does not behave as an ideal gas. (a) The system now undergoes a reversible process in which it is compressed to 100 bar. The pressure–volume relationship is given by: Pv1.5 5 constant What is the final temperature and internal energy of the system? Sketch this process on a Pv diagram. Draw the area that represents the work for this process. Calculate the work done during this process. How much heat was exchanged? (b) Consider a different process by which the system gets to the same final state as in part (a). In this case, a large block is placed on the piston, forcing it to compress. Sketch this process on a Pv diagram. Draw the area that represents the work for this process. Calculate the work done during this process. How much heat was exchanged? (c) Can you think of a process by which the system could go from the initial state to the final state with no net heat exchange with the surroundings? Describe such a process, putting in numerical values wherever possible, and sketch it on a Pv diagram. 2.39 Consider the piston–cylinder assembly containing a pure gas shown below. The initial volume of the gas is 0.05 m3, the initial pressure is 1 bar, and the cross-sectional area of the piston is 0.1 m2. Initially the spring exerts no force on the very thin piston. F = − kx m = 2040 kg

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Patm

m = 1020 kg m = 1020 kg

F = − kx

Patm

A = 0.1 m2

A = 0.1 m2

Process A

Process B

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118 ► Chapter 2. The First Law of Thermodynamics (a) A block of mass m 5 2040 kg is then placed on the piston. The final volume is 0.03 m3, and the final pressure is 2 3 105 Pa. You may assume that the force exerted by the spring on the piston varies linearly with x and that the spring is very “tight,” so that the volume of the gas is never less than the final volume. Precisely draw the process on a PV diagram, labeling it “process A.” Draw the area that represents the work for this process. What is the value of the work? [Hint: First find the value for the spring constant, k.] (b) Consider instead a process in which a block of mass 1020 kg is placed on the piston in the original initial state, and after the gas inside has been compressed another block of mass 1020 kg is placed on the piston. Draw the process on the same PV diagram, labeling it “process B.” Draw the area that represents the work for this process. What is the value of the work? (c) Describe the process in which it will take the least amount of work to compress the piston. Draw the process on the same PV diagram, labeling it “process C.” Draw the area that represents the work for this process. What is the value of the work? 2.40 A rigid tank of volume 0.5 m3 is connected to a piston–cylinder assembly by a valve, as shown below. Both vessels contain pure water. They are immersed in a constant-temperature bath at 200ºC and 600 kPa. Consider the tank and the piston–cylinder assembly as the system and the constanttemperature bath as the surroundings. Initially the valve is closed and both units are in equilibrium with the surroundings (the bath). The rigid tank contains saturated water with a quality of 95% (i.e., 95% of the mass of water is vapor). The piston–cylinder assembly initially has a volume of 0.1 m3. Temperature bath TB = 200°C P = 600 kPa

Surroundings

Pure H2O V = 0.5 m3 Quality = 95% H2O Vinitial = 0.1 m3

The valve is then opened. The water flows into the piston–cylinder until equilibrium is obtained. For this process, determine the work done, W, by the piston; the change in internal energy, DU, of the system; and the heat transferred, Q. 2.41 Consider the well-insulated, rigid container shown below. Two compartments, A and B, contain H2O and are separated by a thin metallic piston. Side A is 10 cm long. Side B is 50 cm long. The cross-sectional area is 0.1 m2. The left compartment is initially at 20 bar and 250ºC; the right compartment is initially at 10 bar and 700ºC. The piston is initially held in place by a latch. The latch is removed, and the piston moves until the pressure and temperature in the two compartments become equal. Determine the final temperature, the final pressure, and the distance that the piston moved. Latch Well-insulated wall

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Thin metallic piston

A P1,A = 20 bar T1,A = 250°C H2O

B P1,B = 10 bar T1,B = 700°C H2O

10 cm

50 cm

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2.11 Problems ◄ 119 2.42 When you open a can of soda (or beer), compressed CO2 expands irreversibly against the atmosphere as it bubbles up through the drink. Assume that the process is adiabatic and that the CO2 has an initial pressure of 3 bar. Take CO2 to be an ideal gas, with a constant heat capacity of cP 5 37 3 J/ 1 mol K 2 4 . What is the final temperature of the CO2 after it has reached atmospheric pressure? 2.43 Find the Pv work required to blow up a balloon to a diameter of 1 ft. Does the value you calculate account for all the work that is required? Explain. 2.44 You have a rigid container of volume 0.01 m3 that you wish to contain water at its critical point. To accomplish this task, you start with pure saturated water at 1 bar and heat it. (a) How much water do you need? (b) What is the quality of water that you need to begin this process? (c) How much heat in [J] will it take? 2.45 In an attempt to save money to compensate for a budget shortfall, it has been determined that the steam in ChE Hall will be shut down at 6:00 P.M. and turned back on at 6:00 A.M., much to the chagrin of a busy class of chemical engineers who have an outrageously long computer project due the following day. The circulation fans will stay on, however, leaving the entire building at approximately the same temperature. Well, things aren’t going as quickly as you might have hoped and it is getting cold in the computer lab. You look at your watch; it is already 10:00 P.M. and the temperature has already fallen halfway from the comfortable 22ºC it was maintained at during the day to the 2ºC of the outside temperature (i.e., the temperature is 12ºC at 10:00 P.M.). Seeing as you will probably be there all night and you need a diversion, you decide to estimate what the temperature will be at 6:00 A.M. You may assume the outside temperature stays constant at 2ºC from 6:00 P.M. to 6:00 A.M. You may take the heat transfer to be given by the following expression:

q 5 h 1 T 2 Tsurr 2 where h is a constant. (a) Plot what you think the temperature will be as a function of time. Explain. (b) Calculate the temperature at 6:00 A.M. 2.46 Steam at 6 MPa, 400ºC is flowing in a pipe. Connected to this pipe through a valve is a tank of volume 0.4 m3. This tank initially contains saturated water vapor at 1 MPa. The valve is opened and the tank fills with steam until the pressure is 6 MPa, and then the valve is closed. The process takes place adiabatically. Determine the temperature in the tank right as the valve is closed. 2.47 Consider filling a cylinder of compressed argon from a high-pressure supply line as shown below. Before filling, the cylinder contains 10 bar of argon at room temperature. The valve is then opened, exposing the tank to a 50 bar line at room temperature until the pressure of the cylinder Argon 50 bar

Pinitial = 10 bar

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120 ► Chapter 2. The First Law of Thermodynamics reaches 50 bar. The valve is then closed. For argon take cP 5 1 5/2 2 R and the molecular weight to be 40 kg/kmol. You may use the ideal gas model. (a) What is the temperature right after the valve is closed? (b) If the cylinder sits in storage for a long time, how much heat is transferred (in kJ/kg)? (c) What is the pressure of the cylinder when it is shipped (after it was stored for a long time)? 2.48 A well-insulated piston–cylinder assembly is connected to a CO2 supply line by a valve, as shown below. Initially there is no CO2 in the piston–cylinder assembly. The valve is then opened, and CO2 flows in. What is the temperature of the CO2 when the volume inside the piston–cylinder assembly reaches 0.1 m3? How much CO2 has entered the tank? Take CO2 to be an ideal gas, with a constant heat capacity of cP 5 37 3 J/ 1 mol K 2 4 .

PE = 1 bar

Valve

CO2

P = 3 bar, T = 283 K

2.49 A rigid tank has a volume of 0.01 m3. It initially contains saturated water at a temperature of 200ºC and a quality of 0.4. The top of the tank contains a pressure-regulating valve that maintains the vapor at constant pressure. This system undergoes a process whereby it is heated until all the liquid vaporizes. How much heat (in [kJ]) is required? You may assume that there is no pressure drop in the exit line.

v

Value maintains pressure in system constant

T1 = 200°C x1 = 0.4 V = 0.01 m3

Il

2.50 You wish to measure the temperature and pressure of steam flowing in a pipe. To do this task, you connect a well-insulated tank of volume 0.4 m3 to this pipe through a valve. This tank initially is at vacuum. The valve is opened, and the tank fills with steam until the pressure is 9 MPa. At this point the pressure of the pipe and tank are equal, and no more steam flows through the valve. The valve is then closed. The temperature right after the valve is closed is measured to be 800ºC. The process takes place adiabatically. Determine the temperature (in [K]) of the steam flowing in the pipe. You may assume the steam in the pipe stays at the same temperature and pressure throughout this process.

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2.11 Problems ◄ 121 Steam T=? P = 9 MPa

Steam T=? P = 9 MPa

H2 O

Initially vacuum

P2 = 9 MPa T2 = 800°C

State 2

State 1

2.51 An ideal gas undergoes an adiabatic, reversible expansion process in a closed system: (a) If cp is constant, show that: T2 P2 5¢ ≤ T1 P1

k/k21

(b) Determine the relationship between temperatures and pressures for an ideal gas if the heat capacity is given by: cP 5 A 1 BT 1 CT2 2.52 Methane vapor enters a valve at 3 bar and 25°C and leaves at 1 bar. If the methane undergoes a throttling process, what is the exit temperature, in °C? Under these conditions, you may assume methane is an ideal gas. 2.53 You wish to heat a stream of CO2 at pressure 1 bar, flowing at 10 mol/s, from 150°C to 300°C in a countercurrent heat exchanger. To do this task, you have been asked to use a stream of highpressure steam available at 40 bar and 400°C, as shown in the following figure. You may assume the pressure of each stream stays constant as it flows through the heat exchanger (i.e., neglect the pressure drop of the flowing streams). The entire system is well insulated, as shown. It is undesirable for the steam to condense in the heat exchanger tubes. What is the minimum volumetric flow rate, in m3 /s, required of the inlet steam to keep it from condensing at the exit? nco = 10 2

mol s

Tin = 150 [°C] Pin = 1 [bar]

Heat Exchanger CO2

Tout = 300 [°C]

CO2

Steam

Steam

Tin = 400 [°C] Pin = 40 [bar]

Insulation

2.54 At steady-state, an ideal gas enters a compressor with a mass flow rate of 10.0 [mol/s]. The inlet pressure is 1.00 bar and the inlet temperature is 25.0°C. The gas exits at 25.0 bar and 65.0°C. The ideal gas heat capacity is given by: cP 5 3.60 1 0.500 3 1023 T R where T is in [K].

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122 ► Chapter 2. The First Law of Thermodynamics (a) Assuming the compressor is adiabatic, calculate the power (in kW) required. (b) In the real process, there is a finite amount of heat transfer. If the compressor operates between the same initial state and final state as in Part A, will the actual power required be greater than, equal to, or less than that calculated in Part A? Explain. 2.55 Answer the following questions: (a) Argon gas enters a heat exchanger at a volumetric flow rate of 4.0 3 m3 /min 4 , a temperature of 100°C and a pressure of 2 bar. It leaves at 200°C and 1 bar. Compute the heat supplied, in W. You may assume ideal gas behavior. (b) Isobutane gas enters a heat exchanger at a flow rate of 3.0 3 m3 /min 4 , a temperature of 100°C and a pressure of 1.5 bar. It leaves at 200°C and 1 bar. Compute the heat supplied, in W. You may assume ideal gas behavior. (c) A stream of argon gas flowing at 4.0 3 m3 /min 4 , 100°C and 2 bar is mixed at steady-state with a second stream containing isobutane gas flowing at 3.0 3 m3 /min 4 , 100°C and 1.5 bar. The gas mixture exits at 200°C and 1 bar. Compute the heat supplied, in W. You may assume ideal gas behavior. 2.56 A stream of pure liquid benzene flowing at a rate of 10 mol/s at 40°C and 10 bar enters a vessel where it is flashed. At steady-state, 6 mol/s of the benzene leaves as vapor, and the rest leaves as liquid. The flash vessel operates at a pressure of 5 bar; determine the rate of heat input required (in W). At 1 atm, benzene boils at 80.1°C and has an enthalpy of vaporization of 30,765 J/mol. 2.57 In a biological system, N2 is often bubbled through a fermentor to maintain anaerobic conditions. As the N2 bubbles through the fermentor, the gas strips water from the liquid. Consider an isothermal continuous fermentation process operated at 30°C, where 0.5 g/min of water are evaporated. (a) Does heat need to be added of removed to compensate for the water evaporation? (b) What is the resulting heat load on the fermentor? 2.58 An electric generator coupled to a waterfall produces an average electric power output of 5 kW. The power is used to charge a storage battery. Heat transfer from the battery occurs at a constant rate of 1 kW. (a) Determine the total amount of energy stored in the battery, (in [kJ]) in 10 hours of operation. (b) If the water flow rate is 200 kg/s and conversion of kinetic energy to electric energy has an efficiency of 50%, what is the average velocity (in m/s) of the water? Where does the rest of the energy go? 2.59 Air enters a well-insulated turbine operating at steady, state with negligible velocity at 4 MPa, 300ºC. The air expands to an exit pressure of 100 kPa. The exit velocity and temperature are 90 m/s and 100ºC, respectively. The diameter of the exit is 0.6 m. Determine the power developed by the turbine (in kW). You may assume air behaves like an ideal gas throughout the process.

Air in V1 = 0 P1 = 4 MPa T1 = 300°C

Turbine

Air out

V2 = 90 m/s P2 = 100 kPa T2 = 100°C

2.60 Ethylene 1 C2H4 2 at 100ºC passes through a heater and emerges at 200ºC. Compute the heat supplied into the unit per mole of ethylene that passes through. You may assume ideal gas behavior.

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2.11 Problems ◄ 123 2.61 Propane at 350ºC and 600 cm3 / mol is expanded in a turbine. The exhaust temperature is 308ºC, and the exhaust pressure is atmospheric. What is the work obtained? You may assume ideal gas behavior. 2.62 Consider 20 mol/s of CO flowing through a heat exchanger at 100ºC and 0.5 bar. (a) At what rate must heat be added (in kW) to raise the stream to 500ºC? (b) Consider the same molar flow rate of n-hexane and the same rate of heat input calculated in part (a). Without doing any calculations, explain whether you expect the final temperature of n-hexane to be greater or less than 500ºC. 2.63 Steam enters a well-insulated nozzle at 10 bar and 200ºC. It exits as saturated vapor at 100 kPa. The mass flow rate is 1 kg/s. What is the steady-state exit velocity? What is the outlet cross-sectional area? 2.64 Propane enters a nozzle at 5 bar and 200ºC. It exits at a velocity of 500 m/s. At steady-state, what is the exit temperature? Assume ideal gas conditions. 2.65 Consider a diffuser operating at steady-state with an outlet twice the area of the inlet. Air flows in with a velocity of 300 m/s, a pressure of 1 bar, and a temperature of 70ºC. The outlet is at

1.5 bar. What is the exit temperature? What is the exit velocity? 2.66 A stream of air is compressed in an adiabatic, steady-state flow process at 50 mol/s. The inlet is at 300 K and 1 bar. The outlet is at 10 bar. Estimate the minimum power that the compressor uses. You may assume air behaves as an ideal gas. 2.67 Sulfur dioxide 1 SO2 2 with a volumetric flow rate of 5000 cm3 /s at 1 bar and 100ºC is mixed with a second SO2 stream flowing at 2500 cm3 /s at 2 bar and 20ºC. The process occurs at steadystate. You may assume ideal gas behavior. For SO2, take the heat capacity at constant pressure to be: cP 5 3.267 1 5.324 3 1023T R (a) What is the molar flow rate of the exit stream? (b) What is the temperature of the exit stream? 2.68 A mass flow controller (MFC) is used to accurately control the molar flow rate of gases into a system. A schematic of an MFC is shown below. It consists of a main tube and a sensor tube, to which a constant fraction of the flowing gas is diverted. In the sensor tube, a constant amount of heat is provided to the heating coil. The temperature difference is measured by upstream and downstream temperature sensors, as shown. A control valve can then be opened or closed to ensure the desired flow rate. Inlet

Sensor tube

Main tube

Valve Outlet

Gas in Upstream temperature sensor (T1) Downstream temperature sensor (T2)

Gas out Valve control Power supply

Electronics

(a) Flow rates are typically reported as standard cubic centimeters per minute (SCCM), which represents the volume the gas would have at a “standard” pressure of 1.0135 bar and a “standard” temperature of 0ºC. What molar flow rate (in [mol/s]), does 1 SCCM correspond to?

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124 ► Chapter 2. The First Law of Thermodynamics (b) Consider controlling the flow of N2. Develop an equation for the molar flow rate of N2 in SCCM in terms of the measured temperature difference, the heat input to the heating coil, and the fraction of gas diverted to the sensor tube. State the assumptions that you make. (c) Instead of recalibrating the MFC for any gas that is used, conversion factors allow you to correct the MFC readout for different gases. Consider controlling SiH4 instead of N2. What conversion factor must be applied? 2.69 Using data from the steam tables, come up with an expression for the ideal gas heat capacity of H2O in the form:

cv 5 A 1 BT Compare your answer to the values in Appendix A.2. 2.70 Steam at 8 MPa and 500ºC flows through a throttling device, where it exits at 100 kPa. What is the exit temperature? 2.71 A monatomic, ideal gas expands reversibly from 500 kPa and 300 K to 100 kPa in a piston– cylinder assembly. Calculate the work done if this process is (a) isothermal; (b) adiabatic. 2.72 Compare the change in internal energy for the following two processes: (a) Water is heated from its freezing point to its boiling point at 1 atm. (b) Saturated liquid water is vaporized at 1 atm. Repeat for the change in enthalpy. 2.73 Calculate the values of the heat capacity of Ar, O2, and NH3 at 300 K. Account for their relative magnitudes in terms of the three modes (translational, rotational, and vibrational) in which molecules can exhibit kinetic energy. 2.74 Two kilograms of water, initially saturated liquid at 10 kPa, are heated to saturated vapor while the pressure is maintained constant. Determine the work and the heat transfer for the process, each in kJ. 2.75 A rigid vessel contains 5 kg of saturated liquid water and 0.5 kg of saturated vapor at 10 kPa. What amount of heat must be transferred to the system to evaporate all the liquid? 2.76 Consider the cooling of a glass of tap water by the addition of ice. The glass contains 400 ml of tap water at room temperature, to which 100 gm of ice is added. Assume the glass is adiabatic and that thermal equilibrium is obtained. The ice is originally at 210°C when removed from the freezer and put in the glass. For ice, Dhfus 5 26.0 3 kJ/mol 4 . (a) What is the final temperature? (b) What % of the cooling is achieved by latent heat (the melting of ice)? 2.77 One mole of saturated liquid propane and 1 mole of saturated vapor are contained in a rigid container at 0ºC and 4.68 bar. How much heat must be supplied to evaporate all of the propane. At 0ºC, Dhvap 5 16.66 3 kJ/mol 4 You may treat propane as an ideal gas. 2.78 Calculate the enthalpy of reaction at 298 K for the following reactions: (a) CH4 1 g 2 1 2O2 1 g 2 S CO2 1 g 2 1 2H2O 1 g 2 (b) CH4 1 g 2 1 2O2 1 g 2 S CO2 1 g 2 1 2H2O 1 l 2 (c) CH4 1 g 2 1 H2O 1 g 2 S CO 1 g 2 1 3H2 1 g 2 (d) CO 1 g 2 1 H2O 1 g 2 S CO2 1 g 2 1 H2 1 g 2 (e) 4NH3 1 g 2 1 5O2 1 g 2 S 4NO 1 g 2 1 6H2O 1 g 2

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2.11 Problems ◄ 125 2.79 Calculate the adiabatic flame temperature of acetylene gas at a pressure of 1 bar under the following conditions. The reactants are initially at 298 K. Assume that the acetylene reacts completely to form CO2 and H2O: (a) It is combusted in a stoichiometric mixture of O2. (b) It is combusted in a stoichiometric mixture of air. (c) It is combusted with twice the amount of air as needed stoichiometrically. 2.80 Calculate the adiabatic flame temperature of the following species in a stoichiometric mixture of air at a pressure of 1 bar. The reactants are initially at 298 K. Assume that they react completely to form CO2 and H2O. Compare the answers and comment. (a) propane (b) butane (c) pentane 2.81 In an experiment, methane is burned with the theoretically required amount of oxygen for complete combustion. Because of faulty operation of the adiabatic burner, the reaction does not proceed to completion. You may assume all of the methane that does react forms H2O and CO2. If the reactants are fed into the reactor at 25ºC and 1 bar and the exiting gases leave at 1000ºC and 1 bar, determine the percentage of methane that passes through the reactor unburned. 2.82 The following data are from a system that undergoes a thermodynamic cycle among three states. Fill in the missing values for DU, Q, and W. Is this a power cycle or a refrigeration cycle?

DU 3 kJ 4

Process

W [kJ]

Q [kJ]

State 1 to 2

350

State 2 to 3

800

State 3 to 1

2750

800 2500

2.83 One mole of air undergoes a Carnot cycle. The hot reservoir is at 800ºC and the cold reservoir is at 25ºC. The pressure ranges between 0.2 bar and 60 bar. Determine the net work produced and the efficiency of the cycle. 2.83 One mole of air undergoes a Carnot refrigeration cycle. The hot reservoir is at 25ºC and the cold reservoir is at 215+C. The pressure ranges between 0.2 bar and 1 bar. Determine the coefficient of performance. 2.84 A Rankine cycle is shown below. This cycle is used to generate power with water as the working fluid. It consists of four unit processes in a thermodynamic cycle: a turbine, a condenser, a compressor, and a boiler. The state of each stream is labeled on the plot and defined in the table below. The mass flow rate of water is 100 kg/s. Kinetic and potential energy effects are negligible. Answer the following questions:

State

1

T[ºC]

520

P [bar]

100

Quality

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2

3

4 80

0.075

0.075

90% sat. vapor

sat. liquid

100

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126 ► Chapter 2. The First Law of Thermodynamics Rankine cycle WS Turbine 1

Fuel air

2

QH

QC

Boiler

Cooling water

4 Condenser 3 WC Compressor

(a) (b) (c) (d) (e)

Sketch all four processes on a Pv diagram. Include the vapor–liquid dome. From the plot above, explain why the net power is negative. # # Determine the heat-transfer rates in the boiler, QH, and the condenser, QC. Determine the net power developed in the cycle. What is the thermal efficiency, h of the cycle? h;

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net power delivered by the plant rate of heat tranfer in the boiler

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► CHAPTER

3 Entropy and the Second Law of Thermodynamics Learning Objectives To demonstrate mastery of the material in Chapter 3, you should be able to: ► State and illustrate by example the second law of thermodynamics—that is, entropy analysis—and its basic concepts, including directionality, reversibility and irreversibility, and efficiency. ► Write the integral and differential forms of the second law for (1) closed systems and (2) open systems, under steady-state and transient (uniformstate) conditions. Convert these equations between intensive and extensive forms and between mass-based and molar forms. ► Apply the second law of thermodynamics to identify, formulate, and solve engineering problems for isothermal and adiabatic processes in the following systems: rigid tank, expansion/compression in a piston—cylinder assembly, nozzle, diffuser, turbine, pump, heat exchanger, throttling device, filling or emptying of a tank, and vapor compression power and refrigeration cycles. ► State the assumptions used to develop the Bernoulli equation, and apply this equation when appropriate to solve engineering problems. ► Develop a hypothetical reversible path to calculate the entropy change between any two states. Develop the expression for the entropy change of an ideal gas, a liquid, or a solid when the heat capacity is known. ► Calculate the entropy difference between two states if you are given either heat capacity data or property tables. Calculate entropy changes for species undergoing phase change or chemical reaction. ► Describe how a vapor-compression power cycle produces power and a refrigeration cycle achieves refrigeration. Identify the key issues in selecting the working fluid. Solve for the net power obtained and the efficiency of a reversible power cycle and the coefficient of performance of a reversible refrigeration cycle. Correct those values for real systems by using isentropic efficiencies. ► Describe the context in which exergy analysis is commonly used. Calculate the exergy of a species in a system or the exthalpy of a flowing stream if you are given either heat capacity data or property tables. Calculate the ideal work, useful work, and lost work for a process. ► Describe the molecular basis for entropy, including its relation to spatial and energetic configurations. Relate macroscopic directional processes to molecular mixing.

127

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128 ► Chapter 3. Entropy and the Second Law of Thermodynamics

► 3.1 DIRECTIONALITY OF PROCESSES/SPONTANEITY Thermodynamics rests largely on the consolidation of many observations of nature into two fundamental postulates or laws. Chapter 2 addressed the first law—the energy of the universe is conserved. We cannot prove this statement, but based on over a hundred years of observation, we believe it to be true. In order to use this law quantitatively— that is, to make numerical predictions about a system—we cast it in terms of a thermodynamic property: internal energy, u. Likewise, the second law summarizes another set of observations about nature. We will see that to quantify the second law, we need to use a different thermodynamic property: entropy, s. Like internal energy, entropy is a conceptual property that allows us to quantify a “law” of nature and solve engineering problems. This chapter examines the observations on which the second law is based; explores how the property s quantifies these observations; illustrates ways we can use the second law to make numerical predictions about closed systems, open systems, and thermodynamic cycles; and discusses the molecular basis of entropy. We first examine several examples of the type of observation on which the second law is based—the directionality of processes. These examples are taken from scenarios with which you are probably familiar. First, consider the tank of compressed gas shown in Figure 3.1a as the system. It is initially in state 1. The surroundings are at atmospheric temperature and pressure. When the valve is opened, gas will spontaneously flow from the system to the surroundings until the pressure in the tank reaches 1 atm. With time, the system reaches state 2, where the pressure inside the cylinder is equal to the pressure outside. During this process, energy is conserved. Hence, if we consider the first law, the energy of the universe is identical in each of the two states. However, there is clearly a direction in which this process occurs spontaneously. It would be absurd to declare that the gas will spontaneously flow from the atmosphere (state 2) into the cylinder (state 1). Since the driving force that pushes the system from state 1 to state 2 is pressure, we label this as an example of mechanical directionality. Similarly, there is a clear directionality if we place a high-temperature block in a room at 25ºC, as illustrated in Figure 3.1b. With time, the block cools to room temperature (state 2). Again, the system will spontaneously go from state 1 to state 2, but it will not go spontaneously from state 2 to state 1. Again, energy is conserved. Since a temperature gradient provides the driving force for this process, we label this example “thermal directionality”. Figure 3.1c illustrates a system in which two different gases, gas A and gas B, are initially separated by a diaphragm in state 1. Upon removing the diaphragm, the gases will eventually mix completely, obtaining state 2. Again, we do not observe a mixed gas to spontaneously separate into pure species. This provides an example of chemical directionality. In all three examples, the first law says nothing about which direction the system spontaneously will go. However, there is a clear directionality associated with each process. Assigning a direction to these processes is easy, since we have experience with them. In other cases, the direction in which a process will go may not be so obvious. For example, consider the following scenario: An excess amount of zinc has been found in the groundwater near the former site of a metal-plating plant. You have been tasked with developing a process to clean up the groundwater. It has been suggested to form a zinc precipitate through reaction with lime, Ca 1 OH 2 2. Is this a reasonable approach? How much Zn can you expect to remove? Another way to phrase these questions is, “To what extent can zinc react with lime so that it does not violate my experience about the directionality of nature?” As we will soon see, the second law of thermodynamics provides a quantitative statement about the directionality of nature and allows us to predict which

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3.2 Reversible and Irreversible Processes (Revisited) and their Relationship to Directionality ◄ 129

High-pressure gas

Patm Tatm

Open valve Patm

State 1

(a ) Mechanical Directionality

State 2

Patm Tatm Block at High Temperature

Tatm

State 1

State 2 (b ) Thermal Directionality

Gas A

Remove membrane

A&B

A&B

Membrane

Gas B State 1

State 2 (c ) Chemical Directionality

Figure 3.1 Examples of directionality in common processes observed in engineering. (a) The expansion of a compressed gas illustrates mechanical directionality. (b) The cooling of a hot block exhibits thermal directionality. (c) The mixing of gas A and gas B represents chemical directionality.

way a process, such as the one described above, will spontaneously go. Thus, with knowledge of the second law of thermodynamics, you can evaluate whether the proposed solution to clean up groundwater is possible.

► EXERCISE

In 5 minutes, come up with as many examples as you can that illustrate the directionality of nature.

► 3.2 REVERSIBLE AND IRREVERSIBLE PROCESSES (REVISITED) AND THEIR RELATIONSHIP TO DIRECTIONALITY From one perspective, the second law of thermodynamics addresses directionality. From another, it is about the reversibility and irreversibility of processes. In this section, we review examples of when mechanically and thermally driven processes are reversible and when they are irreversible.

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130 ► Chapter 3. Entropy and the Second Law of Thermodynamics Process I: Mechanical Process Let’s review the mechanical process from which we learned about reversibility and irreversibility. In Section 2.3, we considered a piston–cylinder assembly that underwent an isothermal expansion/compression, as shown in Figure 3.2a. When we remove the 1020-kg block, as depicted on the left, the piston expands irreversibly. Likewise, when we replace the block on the piston, it compresses irreversibly. In this case, the driving force for change is a pressure difference. The processes are illustrated on the Pv curve at the bottom. Notice that the irreversible processes have a definite directionality. The arrows that describe the expansion process do not overlap with those that describe the compression process. As we saw, the work needed to compress the assembly was greater than the work obtained from expanding it and is represented by a very different directional process (with different arrows and different shaded area on the Pv curve). The irreversible expansion and compression processes are distinct and different. The reversible process is executed by changing the force that acts on the piston by differential amounts, as shown to the right. In this case, the expansion and the compression curves on the Pv diagram meet. The reversible process can be reversed at any point in the process and, therefore, does not have a directionality like those real processes illustrated in Section 3.1. Remember, a reversible process is an idealization and represents the limiting case where a process is perfectly executed. In terms of work, a reversible process represents the upper limit of the work we get out of the system for expansion mT = 1020 kg ∂m Hot reservoir, TH Patm

vs.

QH

A = 0.1 m2 0.4 m

Carnot engine

Wnet

Reversible

QC Cold reservoir, TC

Irreversible

Reversible

QH .04 .08 Molar volume (m3/mol)

(a ) Process I: Isothermal Expansion/Compression

Isothermal expansion

State 3 State 4

State 3

e bl si

Carnot cycle:

State 2

er ev R

Cold reservoir, TC

State State 1 2 Well-insulated

2 Irreversible expansion

"external" pressure (bar)

Q

0.4 m

1 mol of pure, ideal gas

Irreversible

Fexternal/area or

vs.

A = 0.1 m2

1 mol of pure, ideal gas

1

Hot reservoir, TH

State State 1 4

Patm

Irreversible compression

m = 1020 kg

QC

Adiabatic expansion

Isothermal expansion

Adiabatic expansion

(b ) Process II: Thermal Heat Engine

Figure 3.2 Illustration of irreversible and reversible processes. (a) Mechanical process of isothermal expansion/compression. (b) Thermal process in which work can be obtained from a Carnot engine.

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3.3 Entropy, the Thermodynamic Property ◄ 131

and the lower limit for the work we must put in for compression. The reversible process represents the best we can do and serves as a useful reference to which we can compare real, irreversible processes. Additionally, for a reversible process the work obtained in expansion is exactly the same as that required in compression. In short, the reversible expansion and compression processes follow equivalent paths. Process II: Thermal Heat Engine In Section 2.9, we learned how to execute a Carnot cycle. This represents the thermal analogy to the mechanical process described above. In this case, the driving force for change is a temperature difference. Consider the two thermal reservoirs shown in Figure 3.2b. We know that energy will spontaneously transfer from the hot body to the cold body in the form of heat. In the irreversible process illustrated on the left-hand side, an amount of heat, Q, flows spontaneously from the hot reservoir to the cold reservoir, and we get no work out. Again, this process is directional in that energy will not flow spontaneously from the cold reservoir to the hot reservoir. If instead we insert a Carnot engine between these two reservoirs, we can use a reversible transport of energy via heat to obtain work. Again, the reversible process (the Carnot heat engine) represents the maximum work we can get out of the system. The process can also be reversed. By reversing the work and putting it into the Carnot cycle, we transfer the heat from the cold body to the hot body in the form of a Carnot refrigerator. In an intermediate case to those discussed above, we place an irreversible (real) heat engine between these two reservoirs. In this case, the work obtained would be less than that provided by the reversible Carnot cycle. Similarly, a real refrigeration cycle would not represent the reverse of the real heat engine but rather would require more work to get a desired level of refrigeration than the corresponding Carnot refrigerator. Let’s summarize this discussion: • •

Irreversible processes are distinct and show directionality. Reversible processes do not show directionality and represent the best that we can do (e.g., the maximum work we get out or minimum work we put in).

In each example in this section, we can see the driving force for an irreversible process and the direction that each process wants to go. In each example, we see how to make the process reversible so it produces the maximum work (or consumes the minimum). In more complex systems, these effects may not be as obvious. In such cases, we look for answers using the second law of thermodynamics and the related property, entropy.

► 3.3 ENTROPY, THE THERMODYNAMIC PROPERTY We would like to generalize our experience with the directionality of nature (and the limits of reversibility) into a quantitative statement that allows us to do calculations and draw conclusions about what is possible, what is not possible, and whether we are close to or far away from the idealization represented by a reversible process. Indeed, it would be nice if we had a thermodynamic property (i.e., a state function) which would help us to quantify directionality, just as internal energy, u, was central in quantifying the conservation of energy (the first law of thermodynamics). It turns out the thermodynamic property entropy, s, allows us to accomplish this goal. Three historical milestones have established three corresponding distinct contextual paradigms for entropy. First, the property entropy was conceived by Rudolph Clausius in 1865, based largely on Sadi Carnot’s work on maximizing the efficiency of cyclic

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132 ► Chapter 3. Entropy and the Second Law of Thermodynamics processes. He coined the term entropy from the Greek word meaning “transformation,” deliberately choosing a word that sounded like energy to emphasize its equal importance. Clausius related entropy to reversible heat transfer and temperature. This definition is the basis for entropy in classical thermodynamics and will be presented below. In 1877, Ludwig Boltzmann conceptualized entropy in terms of the behavior of molecules. This formulation is the basis of statistical mechanics. In this context, entropy is related, in the most general sense, to molecular probability and statistics.1 States that can exhibit a larger number of different molecular configurations are more probable and have greater entropy. Since macroscopic systems contain such a large number of atoms, knowledge of the probable behavior of the molecules in a system leads to knowledge about how the system will behave as a whole. Based on this view, entropy is often interpreted as the degree of disorder in a system, or, as J. Willard Gibbs prefers, the “mixedup-ness.”2 We will learn more about the molecular origins of entropy in Section 3.10. According to the molecular concept of entropy, there are more molecular configurations accessible when a system has high entropy and is disordered than when it has low entropy and is more ordered. We can view this axiom in terms of information; since there are more possibilities from which to choose in the disordered state, we have less of a chance of guessing the precise molecular configuration of the disordered state as compared to its more ordered counterpart. In analogy, Claude Shannon in 1948 conceived of “entropy” associated with missing information and thus gave birth to the field of information theory. In information theory, “entropy” is seen as a measure of the uncertainty of the true content of a message. In fact, Shannon mathematically defined information “entropy” for bits of information using the identical formula that Boltzmann applied to molecular configurations. Similarly, “entropy”-based arguments have expanded into such diverse fields as economics, theology, sociology, art, and philosophy. We can gain some insight about how entropy is defined for the macroscopic systems in classical thermodynamics by borrowing from Boltzmann’s molecular views, where entropy can be seen as the degree of disorder. Consider a closed system where energy transfer can occur by work or by heat. Energy transfer by work occurs in a very directed way. For example, in the rotation of a shaft or the movement of a piston, the interaction of the system with the surroundings occurs via a boundary that moves in a specific and welldefined direction; that is, all the molecules in the shaft or the piston have the same (angular) speed and are moving in the same direction. Similarly, electrical work is achieved by directed flow of electrons in a wire. On the other hand, energy transfer via heat is driven by temperature, which can be related to the random motion of molecules and therefore can be related to a “disordered” form of energy transfer. In Boltzmann’s terms, the effect of energy transfer by work is directed and ordered and should not affect the entropy. Conversely, we should be able to relate entropy to the disordered energy transfer by heat. Correspondingly, the thermodynamic property entropy, s, is defined in terms of heat absorbed during a reversible process. In differential form, the change in entropy of a substance undergoing a reversible process is equal to the incremental heat it absorbs divided by the temperature: ds ;

dqrev T

(3.1)

1 In fact, the equation for his statistical based entropy, S 5 k log V, serves as Boltzmann’s epitaph and is engraved on his tombstone. In the equation above, V is the number of distinct, different molecular configurations to which a macroscopic state has access. 2 We will use the common term disordered to represent what may be more objectively viewed as greater spread or dispersion.

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3.3 Entropy, the Thermodynamic Property ◄ 133

We can integrate Equation (3.1) between the initial and final states to give: final

dqrev

initial

T

Ds 5 3

(3.2)

For entropy to be considered a thermodynamic property, the entropy change from the initial state to the final state must have the same value no matter what path is taken. Since the definition in Equation (3.2) is written in terms of the path-dependent property qrev, the path independence of s is not obvious. However, we can logically show that entropy is indeed a thermodynamic property, independent of path, and that it is defined by Equation (3.1). Thus, any process that goes between the initial state and the final state has the entropy defined by Equation (3.2), be it reversible or irreversible. The proof can be demonstrated either by using arbitrarily small Carnot cycles3 or, more formally, through the general examination of reversible and adiabatic surfaces using the principle of Caratheodory.4 Those interested in the specific details of these arguments are referred to the sources cited. The property s quantitatively tells us about the directionality of nature, reversibility vs. irreversibility, and the maximum work that we can get out of a process (or the minimum work we need to put in). We cannot deduce this supposition directly from the definition in Equation (3.1), but rather, as we shall see, it just works out that way! Specifically, we will need to determine the entropy change of the universe for a particular process of interest. Recall that the universe is comprised of the system together with the surroundings. In this section, we will illustrate how the entropy change of the universe relates to directionality and reversibility through two cases: adiabatic expansion/compression (case I), and thermodynamic cycles (case II). For each case, we will look both at the reversible processes that represent the best we can do and at the corresponding irreversible processes. These cases provide an interesting juxtaposition. In case I, the entropy change to the surroundings is always zero; therefore, we can determine the entropy change of the universe solely by examining the entropy change of the system. Conversely, in case II, the entropy change to the system is always zero, allowing us to examine the entropy change of the universe through the entropy change of the surroundings. The conclusions from these two cases will be generalized in Section 3.4 to form the second law of thermodynamics. In Example 3.1, we verify the second law for a set of isothermal expansion/ compression processes where both the entropy change of the system and that of the surroundings need to be taken into account.

Case I: Adiabatic Expansion/Compression To illustrate how entropy tells us about the directionality of nature, we first pick a set of four mechanical processes similar to those described in Figure 3.2a: (1) reversible expansion, (2) irreversible expansion, (3) reversible compression, and (4) irreversible compression. In this case, however, we choose adiabatic rather than isothermal processes. The adiabatic process represents the limit of no heat transfer between the system and the

3

See, e.g., Kenneth Denbigh, The Principles of Chemical Equilibrium, 3rd ed. (New York: Cambridge University Press, 1971). 4 See, e.g., Adrian Bejan, Advanced Engineering Thermodynamics, 2nd ed. (New York: Wiley, 1997).

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134 ► Chapter 3. Entropy and the Second Law of Thermodynamics surroundings. If you look at Equation (3.1), you might induce why an adiabatic process was chosen for this first illustrative case. In Example 3.1, we show that identical conclusions are formed for isothermal expansion/compression processes, which represent the limit of rapid heat transfer. As in our discussion of Figure 3.2, the reversible expansion gives the most work we can get out of the system while the irreversible expansion has a definite directionality. Similarly for compression, the reversible process defines the least amount of work that we must put into the system while the irreversible compression process has a definite directionality. For each process, we will calculate three forms of entropy: Dssys, the entropy change of the system (note that we will often omit the subscript “sys” and write it as Ds); Dssurr, the entropy change of the surroundings; and Dsuniv, the entropy change of the universe. These three forms are related by: Dsuniv 5 Dssys 1 Dssurr

(3.3)

First consider a reversible, adiabatic expansion of a piston–cylinder assembly in which the system expands from state 1 to state 2. This process is marked “reversible” on the PT diagram shown in Figure 3.3. The entropy change for the system is given by: final

Ds 5 s2 2 s1 5 3

dqrev

50

T

initial

since there is no heat transfer. This process is termed isentropic. For a reversible, adiabatic process, the entropy of the system remains constant. The entropy change for the surroundings is zero, since there is no heat transferred to the surroundings. (For closed systems, all adiabatic processes, reversible or irreversible, result in Dssurr 5 0, since qsurr 5 0). Inspection of Equation (3.3) shows that the entropy change of the universe is also zero. The changes in entropy for the reversible, adiabatic expansion are summarized in the left column of Table 3.1. An energy balance gives: Du 5 u2 2 u1 5 wrev Since energy from the system is required in the form of Pv work to execute the expansion, wrev is negative and state 2 has a lower internal energy and, consequently, a lower temperature. P

Adiabatic expansion

ve r

r ve Re

sib le

sib le

Initial (1)

e Irr

Final (2) State (3) Energy needs to be added via heat

T

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Figure 3.3 PT diagrams of adiabatic reversible and irreversible expansions. Note that after an irreversible expansion, heat must be expelled to bring the system to the same state as after the reversible expansion.

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3.3 Entropy, the Thermodynamic Property ◄ 135 TABLE 3.1 Summary of Entropy Change for Reversible and Irreversible Expansion (or Compression) of the Adiabatic Piston–Cylinder Assembly in Case I

Reversible Process

Irreversible Process

s2 5 s1 q50 Ds 5 0 Dssurr 5 0 Dsuniv 5 0

s3 . s1 q50 Ds . 0 Dssurr 5 0 Dsuniv . 0

How about an irreversible process? The irreversible process brings the system to a new state that we will label “3” on the PT diagram. We wish to ask, “Does the entropy of the system go up, go down, or remain the same?” Well, an energy balance on the irreversible expansion gives: Du 5 u3 2 u1 5 wirrev We know that 0 wirrev 0 , 0 wrev 0 because a reversible process gives us the maximum possible work out from an expansion. If we now compare the previous two equations, we can see that: u3 . u2 So, for an ideal gas, T3 . T2 as shown in Figure 3.3. How about the entropy change? Our definition of entropy requires that we have a reversible process. So to calculate entropy, we must construct a reversible process to go from state 1 to state 3.5 In this case, we can first go reversibly and adiabatically from state 1 to state 2. Second, we can reversibly and isobarically transfer heat into the system to bring the gas from state 2 to state 3. From state 2 to state 3, qrev is positive, so: final

Ds 5 s3 2 s2 5 3 initial

dqrev

T3

cP dT . 0 5 3 T T T2

From the previous equation, we see that s3 . s2 5 s1. Therefore, for the irreversible expansion, Ds 5 s3 2 s1 . 0 Thus, for an irreversible, adiabatic expansion, the entropy of the system increased. Again, the entropy change to the surroundings is zero, since qsurr 5 0. Summing together the entropy changes of the system and surroundings, we find the entropy change of the universe increases for the irreversible process. The entropy changes for the irreversible case are summarized in the right column of Table 3.1. 5

We will frequently calculate entropy changes of irreversible processes by constructing an alternative reversible process between the initial state and the final state.

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136 ► Chapter 3. Entropy and the Second Law of Thermodynamics We will now apply the same analysis to an adiabatic compression. The PT diagrams for this process are shown in Figure 3.4. The reversible compression brings the system from state 1 to state 2. The entropy change for the system is given by: final

D s 5 s2 2 s1 5 3

dqrev T

50

initial

Again, for a reversible, adiabatic process the entropy of the system remains constant. Likewise, the entropy changes of the surroundings and universe are zero. We see that the results for reversible, adiabatic compression are identical to the results presented in Table 3.1 for reversible, adiabatic expansion. An energy balance gives: Du 5 u2 2 u1 5 wrev In this case, state 2 is at a higher temperature, since we are adding energy to the system via work to compress the gas. How about an irreversible process? An energy balance on an irreversible compression gives: Du 5 u3 2 u1 5 wirrev . wrev where the inequality results because we know a reversible process takes the least amount of work for a compression. If we now compare the previous two equations, we can see that: u3 . u2 So, T3 . T2 Thus state 3 is shown at a higher temperature in Figure 3.4. To find the entropy change of the system, again let’s go from state 1 to state 3 by a reversible process. As with the expansion, we can first go adiabatically from state 1 to state 2. Second, we must isobarically transfer heat into the system to bring the gas from state 2 to state 3. From state 2 to state 3, qrev is positive and: final

Ds 5 s3 2 s2 5 3

dqrev T

.0

initial

P

Adiabatic compression

R

rs ib

ev ers

le

ible

Energy needs to be added via heat Final (2) State (3)

Irre

ve

Initial (1) T

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Figure 3.4 PT diagrams of adiabatic reversible and irreversible compression. Note that after an irreversible compression, heat must be expelled to bring the system to the same state as after the reversible compression.

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3.3 Entropy, the Thermodynamic Property ◄ 137

The results show that for the irreversible compression, s3 . s2 5 s1 Thus, for an irreversible, adiabatic compression the entropy of the system also increased! The entropy change to the surroundings is again zero while the entropy change to the universe has increased. In summary, the results for the irreversible, adiabatic compression are identical to the irreversible expansion presented in Table 3.1. If we generalize, we see that: For an adiabatic reversible process (whether compression or expansion), the entropy of the system remains unchanged while for an irreversible process (whether compression or expansion), the entropy of the system increases. In both cases, the entropy changes for the surroundings are zero. Therefore, the entropy change of the universe remains unchanged for a reversible process and increases for an irreversible process.

The mechanical expansion/compression process described above is convenient since it is adiabatic, so the entropy change of the surroundings is, by definition, zero. We have seen from Equation (3.3) that the entropy change of the universe is the sum of the change of the system and the surroundings, so our next question is, “What happens if the entropy change of the surroundings is not zero?” In our next case, we will consider a set of cyclic processes in which there are only entropy changes in the surroundings and see that we come to a similar generalization as above. Case II: Carnot Cycle We wish to calculate the entropy change for the Carnot cycle, the reversible process by which we converted heat into work as illustrated in Figure 2.17. We then analyze an irreversible cyclic process that has definite directionality. Since this is a cyclic process in which the system returns to its initial state, all properties must return to their initial value, that is, for the system Ducycle 5 0, and similarly Dscycle 5 0. The entropy change to the surroundings for the entire cycle equals the entropy change of each of the four steps: 2

entropy change to surroundings = for the Carnot cycle

final

3

qH

0

TH

dqrev T

initial

Process 1 2 isothermal expansion

final

1

3

dqrev T

2 final

1

initial

Process 2 3 adiabatic expansion

3

qC

0

TC

dqrev T

initial

final

1

3

dqrev T

initial

Process 3 4 isothermal compression

Process 4 1 adiabatic compression

where negative signs are used for q because the heat transfer to the surroundings is the negative of the heat transfer to the system; that is, if energy enters the system, it must leave the surroundings. The previous equation reduces to: Dssurr 5 Dssurr, H 1 Dssurr, C 5 2

qH TH

2

qC TC

(3.4)

We will now come up with alternative expressions for the right-hand side of Equation (3.4) based on our first-law analysis of the reversible processes presented in Section 2.7.

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138 ► Chapter 3. Entropy and the Second Law of Thermodynamics We can rewrite Equation (2.39) as follows: qH 5 2RTH ln

P2 P1

(3.5)

qC 5 2RTC ln

P4 P3

(3.6)

and,

For the two adiabatic processes 1 2 S 3 and 4 S 1 2 , Equation (2.48) can be applied: PVk 5 const Hence, for the adiabatic process 2 S 3: k

k

nRTC V3 P2 P2 5¢ ≤ 5¢ ≤ ¢ ≤ P3 V2 nRTH P3

k

Rearranging the previous equation gives: TC P2 5¢ ≤ P3 TH

k/12k

Similarly, we find the pressures of states 1 and 4 can be related by: TC P1 5¢ ≤ P4 TH

k/12k

Equating the previous two equations, we get: P1 P2 5 P3 P4 or rearranging: P3 P2 5 P3 P4 Dividing Equation (3.5) by (3.6) and using the previous equation gives: P2 qH TH TH P1 52 D T 52 qC P3 TC TC ln P4 ln

or, 2

qH TH

2

qC TC

50

(3.7)

For both Equations (3.7) and (3.4) to be true, the total entropy change to the surroundings for the four reversible processes in the Carnot cycle must be zero: DsH, surr 1 DsC, surr 5 0

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3.3 Entropy, the Thermodynamic Property ◄ 139

As an aside, if we look at the efficiency of the Carnot power cycle (see Section 2.9), we get: h;

0 qH 0 2 0 qC 0 0 qC 0 0 wnet 0 net work 5 512 5 0 qH 0 0 qH 0 0 qH 0 heat absorbed from the hot reservoir

(3.8)

We can relate the heat transferred to the temperature by Equation (3.7), giving: h512

TC TH

(3.9)

We first encountered this relationship in Example 2.23. Equation (3.9) represents the highest efficiency that we can possibly have in operating between a hot reservoir at TH and a cold reservoir at TC. To improve the efficiency further requires a hotter energy source or a colder energy sink. Similarly, we can use Equation (3.7) to write the coefficient of performance of a Carnot refrigeration cycle as: COP 5

0 qC 0 wnet

5

0 qC 0 0 qH 0 2 0 qC 0

5

TC TH 2 TC

Now let’s consider an irreversible cycle. We know that an irreversible process produces less work than a reversible one. If the heat absorbed from the hot bath is identical, less net work means more heat must be discarded to the cold bath (since wnet 5 qH 1 qC ): 1 qC 2 irrev, surr . 1 qC 2 rev, surr So for an irreversible cycle, 1 DsC 2 surr 5

1 qC 2 irrev, surr TC

is greater than for the Carnot cycle; thus,

Dssurr 5 1 DsH 2 surr 1 1 DsC 2 surr . 0 So we find that the entropy change to the surroundings for an irreversible cycle is greater than zero. The changes in entropy for the reversible Carnot cycle and the irreversible power cycle are given in Table 3.2. We can summarize this analysis as follows: For the set of reversible processes described in the Carnot cycle, the entropy of the surroundings remains unchanged, while for an equivalent irreversible cycle the entropy of the surroundings increases. In both cases, the entropy changes for the system are zero. Therefore, the entropy change of the universe remains unchanged for a reversible process and increases for an irreversible process.

Comparison of Tables 3.1 and 3.2 shows a common theme. The entropy change for the universe is zero for the reversible process and greater than zero for the irreversible process. TABLE 3.2 Summary of Entropy Change for the Reversible Carnot Cycle and the Irreversible Cycle in Case II

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Reversible Cycle

Irreversible Cycle

Dscycle 5 0

Dscycle 5 0

Dssurr 5 0

Dssurr . 0

Dsuniv 5 0

Dsuniv . 0

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140 ► Chapter 3. Entropy and the Second Law of Thermodynamics

► 3.4 THE SECOND LAW OF THERMODYNAMICS In case I of Section 3.3, we looked at the adiabatic expansion and compression of a piston–cylinder assembly. In these processes, the entropy change to the surroundings is zero. We saw that the entropy change of the system is zero for a reversible process and that the entropy change for the system is greater than zero for an irreversible process. In case II, we looked at the entropy change associated with a thermodynamic cycle, where the entropy change to the system is, by definition, zero. The entropy change of the surroundings is zero for a reversible process, and the entropy change for the surroundings is greater than zero for an irreversible process. It turns out that we can generalize our observations above for all processes by considering the entropy change of the universe. Equation (3.3) indicates that the entropy change of the universe is the sum of the entropy change to the system and the entropy change to the surroundings. We call this generalization the second law of thermodynamics: For any reversible process, the entropy of the universe remains unchanged, while for any irreversible process, the entropy of the universe increases.

or, in other words, Entropy is time’s arrow—the larger the entropy of the universe, the more recent the event.

Just as with the first law, we believe these statements because all the observations we have made and used to test this statement are consistent with it (just as in the two specific cases above). If we quantify the statement above (so that we can solve problems such as determining whether we can remove zinc with lime), we have, for the second law of thermodynamics, Dsuniv $ 0

(3.10)

Dsuniv 5 0

(3.10r)

Dsuniv . 0

(3.10i)

For a reversible process,

and for an irreversible process,

Equation (3.10r) sets the limit of reversibility. This case represents the best we can possibly do for a given design. Equation (3.10i) tells us about directionality. If you can determine the entropy of the universe for two states, the one with the higher entropy associated with it happened more recently. If the entropy of the universe is the same, the process is reversible and can occur in any direction.

EXAMPLE 3.1 Calculation of Dsuniv for Processes with Contributions from Both Dssys and Dssurr

c03.indd 140

In Section 2.3, we learned about reversible and irreversible processes in the context of a piston–cylinder assembly undergoing isothermal expansion and compression processes. Four of these processes are summarized in Figure 3.2: (i) Irreversible expansion, process A (ii) Irreversible compression, process B (iii) Reversible expansion, processes E (iv) Reversible compression, processes F

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3.4 The Second Law of Thermodynamics ◄ 141

Calculate Dsuniv for each of these isothermal processes and show that the result is consistent with the second law of thermodynamics. SOLUTION In this example, we must consider both the entropy change of the system and the entropy change of the surroundings. We outline the solution methodology with process A, the isothermal, irreversible expansion from state 1 at 2 bar and 0.04 m3 /mol to state 2 at 1 bar and 0.08 m3 /mol. Since the process is isothermal, we assume the temperature of the surroundings and that of the system are identical. Their values can be found using the ideal gas law: T 5 Tsurr 5

Pv 5 962 3 K 4 R

To calculate the entropy change of the system, we must devise a reversible process between the same states, 1 and 2 so that we can apply Equation (3.2). Such a process is depicted in Figure 2.7 by process E. Since the property entropy is independent of path, the entropy change of the system for process E and process A must be identical. Applying the definition provided by Equation (3.2) to the isothermal process, we get: qrev dqrev Dssys 5 3 5 T T

(E3.1A)

We can find the heat transfer by applying the first law to the reversible process: Du 5 qrev 1 wrev The internal energy change for an ideal gas undergoing an isothermal process is zero; thus, qrev 5 2wrev 5 5545 3 J/mol 4

(E3.1B)

where we have used the value reported in the analysis in Chapter 2. Substituting Equation (E3.1B) into (E3.1A) gives: Dssys 5

5545 5 5.76 3 J/mol K 4 962

(E3.1C)

This value is reported for Dssys, for both process A and process E in Table E3.1. We next calculate the entropy change for the surroundings. To compute this value, a conceptual argument is useful. The change in entropy of the surroundings is identical for reversible heat transfer and for irreversible heat transfer, as long as the magnitude of q is the same. Macroscopically, we can envision the same effect on the surroundings for a given amount Table E3.1 Summary of Entropy Change for the Isothermal Expansion/Compression Processes Described in Section 2.3

Irreversible Processes Expansion (Process A) Dssys 3 J/ 1 mol K 2 4

Compression (Process B)

Reversible Processes Expansion (Process E)

Compression (Process F)

5.76

25.76

5.76

25.76

Dssurr 3 J/ 1 mol K 2 4

24.14

8.32

25.76

5.76

Dsuniv 3 J/ 1 mol K 2 4

1.61

2.56

0

0 (Continued)

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142 ► Chapter 3. Entropy and the Second Law of Thermodynamics of heat transfer. Heat represents the transfer of energy by a temperature gradient. From a molecular perspective, temperature is representative of the random motion of molecules. Thus, energy transfer via heat represents the greatest possible increase in disorder, that is, increase in entropy—whether it occurs reversibly or irreversibly. In contrast, when energy is transferred via work, molecules are specifically directed. Therefore, the entropy does not increase. Hence, for this calculation we use the actual magnitude of q found for process A. Since the heat transfer to the surroundings is the negative of the heat transfer to the system— that is, heat that leaves the surroundings enters the system—we get: Dssurr 5

2qA Tsurr

5

wA 4000 52 5 24.14 3 J/mol K 4 Tsurr 962

(E3.1D)

where the value of work for process A is taken from the analysis in Chapter 2. The entropy change of the surroundings is recorded in Table E3.1. Finally, the entropy change of the universe is obtained by adding the values in Equations (E3.1C) and (E3.1D): Dsuniv 5 Dssys 1 Dssurr 5 1.61 3 J/ 1 mol K 2 4 We see that the entropy change for the universe is greater than zero for this irreversible process, as the second law requires. Values for Dssys, Dssurr, and Dsuniv for processes B, E, and F can be found in a similar manner. The values that are obtained are reported in Table E3.1. For both irreversible processes, the entropy change for the universe is greater than zero; on the other hand, the entropy change of the universe equals zero for the reversible processes. In the compression processes, the entropy change of the system is negative. This result is possible as long as the entropy change of the surroundings is suitably positive.

►3.5 OTHER COMMON STATEMENTS OF THE SECOND LAW OF THERMODYNAMICS Historically, the second law was construed in the context of producing power (work) from cyclic processes in which heat was absorbed from a hot reservoir and rejected to a cold reservoir. Thus, the second law is often stated in terms of work produced and heat rejected. For example, the second law is often defined as follows: Heat cannot be caused to flow from a cooler body to a hotter body without producing some other effect. Clausius

We can see that the Clausius statement is consistent with the general definition presented in Section 3.4. If a positive quantity of heat, q, flows reversibly from a cool body at TC to a hot body at TH, with no other effect, the associated entropy change is equal to: Dsuniv 5 Dssurr 5 2

q TC

1

q TH

,0

The quantity on the right-hand side is less than zero, since TC , TH. Since Dsuniv is negative, it violates the second law of thermodynamics, unless there is “some other effect” that has a positive Dsuniv large enough to offset this transfer of energy via heat. Another common form of the second law is given by: It is impossible to build an engine which operating in a cycle can convert all the heat it absorbs into work. Kelvin and Planck

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3.6 The Second Law of Thermodynamics for Closed and Open Systems ◄ 143

Again, we can see that this statement is consistent with the general definition presented in Section 3.4. Let’s consider the most efficient engine possible, the Carnot engine. If qC 5 0. Equation (3.4) gives: qH Dsuniv 5 Dssurr 5 2 ,0 TH which also violates the second law of thermodynamics. There are many other specific forms of the second law. It is clear that the statements above can be viewed in terms of the generalizations presented in Section 3.4, but the reverse is not so clear.

► 3.6 THE SECOND LAW OF THERMODYNAMICS FOR CLOSED AND OPEN SYSTEMS In general, there are two ways to apply the second law of thermodynamics: DSuniv $ 0

(3.11)

1. We may determine whether a process is possible (and estimate how efficient it is). This applies to real, irreversible processes for which the entropy change of the universe is greater than 0: DSuniv . 0

(3.11i)

2. The second law can also be used to provide an additional constraint to solve a problem; that is, it provides us another equation to use. To apply the second law in this way, we must assume the process is reversible. In this case, we apply: DSuniv 5 0

(3.11r)

Calculation of Ds for Closed Systems In a closed system, mass cannot transfer across the system boundary. We can write Equation (3.11) as: DSsys 1 DSsurr $ 0

(3.12)

The change in entropy of the surroundings is identical for reversible heat transfer and for irreversible heat transfer, as long as the magnitude of Q is the same (see the discussion in Example 3.1). If the surroundings are at constant temperature, Tsurr , we can write the entropy change to the surroundings as: final

dQsurr

DSsurr 5 3 Tsurr initial

final

Qsurr 1 5 dQsurr 5 3 Tsurr Tsurr initial

We must be careful about sign conventions. If heat flows into the system, it must flow out of the surroundings, that is: Qsurr 5 2Q

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144 ► Chapter 3. Entropy and the Second Law of Thermodynamics Substitution of the previous two equations into Equation (3.11) gives: DSsys 1

Qsurr Tsurr

5 n 1 sfinal 2 sinitial 2 1 5 n 1 sfinal 2 sinitial 2 2

Qsurr Tsurr Q Tsurr

$0

(3.13)

$0

As with the first law, the second law can be written in differential form: nds 1

dQsurr Tsurr

$0

(3.14)

Where do we obtain values of s?

EXAMPLE 3.2 Calculation of Dsuniv by Selecting a Hypothetical Reversible Process

An insulated tank 1 V 5 1.6628 L 2 is divided into two equal parts by a thin partition. On the left is an ideal gas at 100 kPa and 500 K; on the right is a vacuum. The partition ruptures with a loud bang. (a) What is the final temperature in the tank? (b) What is Dsuniv for the process? SOLUTION A schematic of the system is shown on the left-hand side of Figure E3.2. (a) We can apply the first law of thermodynamics to find the final temperature. If we choose the entire tank as the system, there is no work or heat transfer across its boundaries during this process. Hence, DU 5 Q 1 W 5 0

(E3.2)

The internal energy in the final state is the same as it was initially. For an ideal gas u 5 u (T only); thus, to follow Equation (E3.2), the temperature does not change: T2 5 500 K ∂m

Reversible isothermal expansion

Well insulated

V = 0.8314 L P = 100 kPa T = 500 K

Process chosen to calculate entropy change for the system Vacuum

V = 0.8314 L P = 100 kPa T = 500 K

Figure E3.2 The irreversible process in the problem statement is on the left-hand side. The hypothetical reversible path between the same initial and final states shown on the right-hand side is used to calculate the entropy change of the system.

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3.6 The Second Law of Thermodynamics for Closed and Open Systems ◄ 145

(b) We must apply the second law of thermodynamics. Since s is a property of the system, it depends only on the final state and the initial state, not on the process (or path). To calculate Dssys, we choose a reversible path by which the system goes from the same initial state to the same final state as the real system. We can then use the definition for change in entropy, since it final

involves a reversible process, ¢ 3

dqrev T

; Ds≤. Such a hypothetical path is illustrated on the

initial

right-hand side of Figure E3.2. For this hypothetical reversible process, we have: DU 5 Qrev 1 Wrev 5 0 or, 1.6628 L

Qrev 5 2Wrev 5

3

PdV 5 3

nRT dV V

0.8314 L

Solving with numerical values, we get: qrev 5

Qrev n

5 RT ln ¢

J J V2 ≤ 5 ¢8.314 c d ≤ 1 500 3 K 4 2 ln 2 5 2881 c d V1 mol K mol

Plugging into the definition for Ds: Dssys 5 3

dqrev T

5

qrev T

5

2881 3 J/mol 4 500 3 K 4

5 5.76 B

J mol K

d

where we pulled T out of the integral since the hypothetical process is isothermal. Since there is no heat transferred to the surroundings in the real irreversible process, Dssurr 5 0 Finally, the entropy change of the universe can be found by adding that of the system to the surroundings according to Equation (3.3): Dsuniv 5 Dssys 1 Dssurr 5 5.76 J/ 1 mol k 2 Note that this value is greater than zero, indicating an irreversible process, as we might expect for this very spontaneous process: Dsuniv . 0

EXAMPLE 3.3 Entropy Change in Obtaining Thermal Equilibrium

Consider an isolated system containing two blocks of copper with equal mass. One block is initially at 0ºC while the other is at 100ºC. They are brought into contact with each other and allowed to thermally equilibrate. What is the entropy change for the system during this process? Take the heat capacity for copper to be: cP 5 24.5 3 J/ 1 mol K 2 4 SOLUTION We label the cooler block “block I” and the hotter block “block II” as illustrated in Figure E3.3. The blocks go from state 1, in which they are at different temperatures, to state 2, in which their temperatures are equal. The pressure remains constant during the process. (Continued)

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146 ► Chapter 3. Entropy and the Second Law of Thermodynamics

We must first find the final temperature, T2. From the first law for an isolated system at constant pressure, we get: DH 5 nIDhI 1 nIIDhII 5 0

(E3.3A)

where the subscripts “I” and “II” refer to the appropriate block. For constant heat capacity, Equation (E3.3A) can be written: nIcP 1 T2 2 T1,I 2 1 nIIcP 1 T2 2 T1,II 2 5 0

(E3.3B)

Since the blocks have equal mass, nI 5 nII 5 n

(E3.3C)

Solving Equation (E3.3B) subject to (E3.3C) gives: T2 5

T1,I 1 T1,II 2

5 323 3 K 4

To get the total entropy change for the system, we can add the changes of each block individually: DS 5 nIDsI 1 nIIDsII

(E3.3D)

We can apply Equation (3.1) to solve for DsI and DsII. We first need to find dqrev. We do that by constructing a reversible path at constant pressure from T1 to T2. Such a process can be conceived by placing each block in thermal contact with a series of reservoirs that are each just differentially hotter (or cooler) than the block. Then qrev is given by: T2

qrev 5 Dh 5 3 cPdT T1

or, dqrev 5 cPdT Thus, the entropy change is described by: Ds 5 3

dqrev

T2

5 3 T

cPdT T

(E3.3E)

T1

State 1

Block I

T1,I = 0°C

State 2

Block II

T1,II = 100°C

Block I

T2,I = T2

Block II

T2,II = T2

Figure E3.3 Initial (state 1) and final (state 2) states of two copper blocks that obtain thermal equilibrium.

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3.6 The Second Law of Thermodynamics for Closed and Open Systems ◄ 147

Applying Equation (E3.3E) to the system of two blocks with constant heat capacity, we get: DS 5 n1cP ln ¢

T2 T2 ≤ 1 nIIcP ln ¢ ≤ T1,I T1,II

Rearranging and plugging in numerical values yields: Ds 5

cP cP T2 T2 T22 DS ≤ 1 ln ¢ ≤ R 5 ln ¢ ≤ 5 Bln ¢ 2n 2 T1,I T1,II 2 T1,I 3 T1,II

5¢

J J 24.5 3232 5 0.30 c c d ≤ ln d 1 273 2 1 373 2 2 mol K mol K

Since this is an isolated system, the entropy change to the surroundings is zero. The positive value we obtained confirms that energy spontaneously flows from a hot block to a cold block.

EXAMPLE 3.4 Entrophy Calculation for a Phase Change

Calculate the change in entropy when 1 mole of saturated ethanol vapor condenses at its normal boiling point. SOLUTION For ethanol, we can find that the enthalpy of vaporization and normal boiling point are: Dhvap,C2H6O 5 38.56 3 kJ/mol 4 and Tb 5 78.2 3 °C 4 The boiling point is, by definition, the temperature at which the phase change occurs reversibly. Hence, we can apply it directly to the definition of entropy, Equation (3.2). From an energy balance at constant P, we get: Dh 5 hl 2 hv 5 qrev 5 2Dhvap C2H2O

(E3.4)

We then substitute Equation (E3.4) into the definition for entropy. Realizing that this process occurs at constant temperature Tb, we get: liquid

Ds 5 3

dqrev T

vapor

5

2Dhvap,C2H6O Tb

Plugging in the values above gives: Ds 5

238.56 3 kJ/mol 4 351.4 K

5 20.1098 c

kJ mol K

d

The entropy change for the system has a negative value since the liquid is in a more ordered state than the vapor. However, this process does not violate the second law, since it is the entropy change of the universe that must be greater than zero. Therefore, we can also say that the entropy change of the surroundings must be at least 0.1098 3 kJ/ 1 mol K 2 4 for this process to occur.

Calculation of Ds for Open Systems In open systems, mass can flow into and out of the system. Thus, the entropy from flowing streams transfers from the surroundings to the system. An example of an open system with two streams in and two streams out is shown in Figure 3.5. As was the case with the first law, it is often convenient to discuss rates of flow [mol/sec] and energy transfer [J/s] or [W].

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148 ► Chapter 3. Entropy and the Second Law of Thermodynamics Stream 2 in

n2 s2

in n 1 s1

Ws

Stream 3 out

System

n 3 s3

Stream 1

n4

Qsys

s4

Tsurr

out

Stream 4

Figure 3.5 Schematic of open system with two streams in and two streams out.

Dividing Equation (3.12) by Dt and taking the limit as the time step becomes zero, the second law becomes: a

dS dS dS 5a b 1a b $0 b dt univ dt sys dt surr

(3.15)

At steady-state, the entropy change of the system is zero: a

dS b 50 dt sys

(3.16)

In addition to the heat exchanged with the surroundings, the entropy change in the surroundings is affected by the mass flow out of or into the system. Each mole of an outlet stream contains a quantity of entropy, sout, that it adds to the surroundings, while each inlet stream takes a quantity of entropy, sin, away from the surroundings. Therefore, at constant temperature, Tsurr, the rate of entropy change with the surroundings can be written: # Qsurr dS # # a b 5 a noutsout 2 a ninsin 1 dt surr Tsurr out in # Q # # 5 a noutsout 2 a ninsin 2 Tsurr out in Applying Equations (3.16) and (3.17), at steady-state Equation (3.15) becomes: # #n s 2 #n s 2 Q $ 0 a in in T a out out out in surr

(3.17)

(3.18)

In the case of one stream in and one stream out, Equation (3.18) can be written: # #n 1 s 2 s 2 2 Q $ 0 (3.19) out in Tsurr where the outlet and inlet molar flow rates are identically n# . In differential form, the steady-state entropy balance becomes: dQ n# ds 2 $0 Tsurr

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(3.20)

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3.6 The Second Law of Thermodynamics for Closed and Open Systems ◄ 149

For unsteady-state problems, the entropy change of the system, 1 dS/dt 2 sys, must be included. Adding this term to Equation (3.18) gives: # Q dS # # 1 a noutsout 2 a ninsin 2 $0 (3.21) dt Tsurr out in

EXAMPLE 3.5 Using Entropy to Help Calculate the Exit Velocity from a Nozzle

Steam enters a nozzle at 300 kPa and 700ºC with a velocity of 20 m/s. The nozzle exit pressure is 200 kPa. Assuming this process is reversible and adiabatic determine (a) the exit temperature and (b) the exit velocity. SOLUTION First we draw a schematic of the process including all the information we know. We label the inlet state “1” and the outlet state “2,” as shown in Figure E3.5. A first-law balance for an open system at steady-state gives: 0 0 0 0 S S 2 2 V V # # 1 gz 2 2 1 Q 1 Ws 0 5 m1 1 h^ 1 1 gz 2 1 2 m2 1 h^ 1 2 2 We have written this equation on a mass basis in anticipation of using the steam tables for thermodynamic property data. Using a mass balance, m# 1 5 m# 2, and the definition for kinetic energy, we get: S

2

S

2

V1 V2 h^ 1 1 5 h^ 2 1 2 2

(E3.5A)

Equation (E3.5A) has two unknowns, h^ 2 and V2; therefore, we need another equation. For a reversible adiabatic process, the second law gives Ds 5 0, or: S

s^ 1 5 s^ 2

(E3.5B)

It is worthwhile, at this point, to reflect on the overall solution methodology. State 1 is thermodynamically determined, since we know T1 and P2. Thus, we can find any other property, including s^ 1. Once the specific entropy of state 1 is determined, we know s^ 2 from Equation (E3.5B). This value, along with P2, constrains state 2, so we can, in principle, find any other property, including temperature and enthalpy. From the steam tables, for state 1, we get: h^ 1 5 3927.1 3 kJ/kg 4 and,

s^ 1 5 8.8319 3 kJ/ 1 kg K 2 4 5 s^ 2

T1 = 700°C P1 = 300 kPa

T2 = ? P2 = 200 kPa

V1 = 20 m/s

V2 = ?

Figure E3.5 Nozzle of Example 3.5 with known and unknown properties of steam delineated. (Continued)

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150 ► Chapter 3. Entropy and the Second Law of Thermodynamics We now go to the table for P 5 200 kPa and determine (by interpolation) that steam has S s^ 2 5 8.8319 3 kJ/ 1 kg K 2 4 when: T2 5 623 3 °C 4 h^ 2 5 3754.7 3 kJ/kg 4

and,

Now we can solve Equation (E3.5A) for the outlet velocity: V2 5 "2 1 h^ 1 2 h^ 2 2 1 V21 5 "2 1 172.4 3 103 2 3 J/kg 4 1 400 3 m2 /s2 4 5 587.5 3 m/s 4 S

S

The steam lost enough internal energy in cooling from 700ºC to 623ºC to manifest itself in a very high exit velocity (587.5 m/s). Again, this example illustrates the large amounts of energy manifest in u.

EXAMPLE 3.6 Entropy Generated by an Open Feedwater Heater

In Example 2.21, we analyzed an open feedwater heater. Superheated water vapor at a pressure of 200 bar, a temperature of 500ºC, and a flow rate of 10 kg/s was brought to a saturated vapor state at 100 bar by mixing this stream with a stream of liquid water at 20ºC and 100 bar. The flow rate for the liquid stream was found to be 1.95 kg/s. What is the entropy generated during this process? SOLUTION First let’s draw a diagram of the system, as shown in Figure E3.6. If we assume the rate of heat transfer is negligible, the entropy change reduces to the sum of the outlet stream minus the inlet streams: a

dS 5 m# 3s^ 3 2 1 m# 1s^ 1 1 m# 2s^ 2 2 b dt univ

(E3.6A)

A mass balance at steady-state gives: m# 3 5 m# 1 1 m# 2 5 11.95 3 kg/s 4

(E3.6B)

We can look up values for the entropies from the steam tables (Appendix B). For state 1 [superheated steam is at 500ºC and 200 bar 1 5 20 MPa 2 ], s^ 1 5 6.1400 c

kJ kg K

d

For state 2, we use subcooled liquid at 20ºC and 100 bar: s^ 2 5 0.2945 c

kJ kg K

d

and saturated vapor at 100 bar (10 MPa) for state 3: s^ 3 5 5.6140 c

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d

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3.7 Calculation of Ds for an Ideal Gas ◄ 151

Subcooled T2 = 20°C liquid P2 = 100 bar kg m2 = 1.95 s

Superheated vapor

Open feedwater heater

T1 = 500°C P1 = 200 bar kg m1 = 10 s

Saturated vapor P3 = 100 bar

Figure E3.6 Schematic of the open feedwater heater. Finally, rearranging Equation (E3.6A) and plugging in values gives: a

kg kJ dS 5 b ¢11.95 c d ≤ ¢5.6140 c b d≤r s dt univ kg K

2b ¢10.0 c

1¢1.95 c

kg

kg s

s

d ≤ ¢6.1400 c

d ≤ ¢0.2945 c

kJ kg K kJ

kg K

d≤

d ≤ r 5 5.11 c

kW d K

The entropy increases, as we would expect for this spontaneous process.

►3.7 CALCULATION OF Ds FOR AN IDEAL GAS This section illustrates how to calculate the change in entropy of an ideal gas between two states if P and T for each state are known. We will define the initial state as state 1, at P1 and T1, and the final state as state 2, at P2 and T2. Since entropy is a state function, we can construct any path that is convenient between state 1 and state 2 to calculate Ds. Figure 3.6 illustrates such a hypothetical path. We choose a reversible process for our hypothetical path so that we can apply the definition of entropy. The first step consists of isothermal expansion, while the second step is isobaric heating. To find Ds, we will calculate the entropy change for each step and add them together. Details of the analysis for each step follow. Step 1: Reversible, isothermal expansion The change in internal energy for an ideal gas is, by definition, zero; thus, the differential energy balance is: du 5 dqrev 1 dwrev 5 0 Solving for the differential heat transfer, we get: dqrev 5 2dwrev 5 Pdv

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152 ► Chapter 3. Entropy and the Second Law of Thermodynamics

State 1 (T1, P1) Real path

Step 1

Pressure (Pa)

P1

P2

Δsreal

Δshypothetical (reversible)

Step 2

State 2 (T2, P2)

T1

Figure 3.6 Plot of a process in which a system goes from state 1 to state 2 in TP space. The change in entropy is calculated along a reversible hypothetical path.

T2 Temperature (K)

Applying this expression for dqrev to the definition of entropy, Equation (3.2), gives: dqrev P 5 3 a bdv Dsstep1 5 3 T T Using the ideal gas law 1 P/T 5 R/v 2 and recognizing that the volume at the end of step 1 is RT1 /P2 gives: RT1/P2

RT1 /P2 R P2 P2 Dsstep1 5 3 a bdv 5 R lnB R 5 R ln¢ ≤ 5 2R ln¢ ≤ v v1 P1 P1 v1

Step 2: Reversible, isobaric heating Again, we want to solve for the reversible heat transfer, qrev, and use it in the definition of entropy. Applying the first law gives: du 5 dqrev 1 dwrev 5 dqrev 2 Pdv Solving for dqrev: dqrev 5 du 1 Pdu 1 vdP 5 d 1 u 1 Pv 2 5 dh where the term vdP was added to the right-hand side since d P 5 0 for a constant-pressure process. Thus the entropy change is equal to:

Dsstep2

dqrev

T2

cP dh 53 53 5 3 dT T T T T1

where the definition of heat capacity, Equation (2.29), is used.

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3.7 Calculation of Ds for an Ideal Gas ◄ 153

Adding together the two steps of our hypothetical reversible path gives: T2

cP P2 Ds 5 Dsstep2 1 Dsstep1 5 3 dT 2 R ln¢ ≤ T P1

(3.22)

T1

Equation (3.22) is true, in general, for the entropy change associated with an ideal gas between state 1 and state 2. In this expression, the property, Ds just depends on other properties—that is, cP, T, P—so it is independent of path. Therefore, Equation (3.22) can be applied to any process, be it reversible or irreversible. It is not limited only to the reversible processes for which it was developed. If cP is constant, Equation (3.22) becomes: Ds 5 cP ln¢

T2 P2 ≤ 2 R ln¢ ≤ 1 cP is constant 2 T1 P1

(3.23)

In a similar manner, it can be shown that for an ideal gas, the entropy change between 1 T1, v1 2 and 1 T2, v2 2 is given by (see Problem 3.17): T2

cv v2 Ds 5 3 dT 1 R ln¢ ≤ v1 T

(3.24)

T1

With cv constant, Equation (3.24) becomes: Ds 5 cv ln¢

EXAMPLE 3.7 Calculation of Entropy Change for an Irreversible, Isothermal Compression

T2 v2 ≤ 1 R ln¢ ≤ 1 cv is constant 2 v T1 1

(3.25)

A piston–cylinder device initially contains 0.50 m3 of an ideal gas at 150 kPa and 20ºC. The gas is subjected to a constant external pressure of 400 kPa and compressed in an isothermal process. Assume the surroundings are at 20ºC. Take cP 5 25R and assume the ideal gas model holds. (a) Determine the heat transfer (in kJ) during the process. (b) What is the entropy change of the system, surroundings, and universe? (c) Is the process reversible, irreversible, or impossible? SOLUTION (a) To determine the heat transfer during the process, we can apply the first law. The change in internal energy of an ideal gas undergoing an isothermal process is zero: DU 5 Q 1 W 5 0 Solving for Q gives: Q 5 2W 5 Pext 1 V2 2 V1 2 5 P2 ¢

P1V1 2 V1 ≤ 5 V1 1 P1 2 P2 2 P2

Finally, plugging in numerical values: Q 5 1 0.50 3 m3 4 2 1 2250 3 kPa 4 2 5 2125 3 kJ 4 (Continued)

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154 ► Chapter 3. Entropy and the Second Law of Thermodynamics

(b) To find the entropy change of the system, we can apply Equation (3.23): Dssys 5 cP ln

T2 P2 2 R ln T1 P1

The first term is zero, since the process is at constant temperature. Thus, DSsys 5 nDssys 5 a

P2 PV b ¢2ln ≤ T 1 P1

Plugging in numerical values gives: DSsys 5 ¢

kJ 150 3 kPa 4 0.5 3 m3 4 400 ≤ a2ln b 5 20.25 c d 3 4 150 K 293 K

The entropy change of the surroundings is given by: DSsurr 5

Qsurr T

5

2Q T

5

125 3 kJ 4 293 3 K 4

5 0.43 B

kJ K

R

The entropy change of the universe is equal to that of the system plus the surroundings: DSuniv 5 DSsys 1 DSsurr 5 0.18 3 kJ/K 4 (c) Since entropy of the universe increases, this process is irreversible. The irreversibility arises from the finite pressure difference.

EXAMPLE 3.8 Calculation of Entropy Change Between Two States

Calculate the entropy change when 1 mole of air is heated and expanded from 25ºC and 1 bar to 100ºC and 0.5 bar. SOLUTION At these low pressures, we can assume that air is an ideal gas. The entropy change between the initial state (1) and final state (2) is given by Equation (3.22): T2

cP P2 Ds 5 3 dT 2 R ln ¢ ≤ T P1

(E3.8A)

T1

We can integrate the first term on the right-hand side as follows: T2

373

373

cP D 21 23 3 T dT 5 R 3 3 AT 1 B 1 DT 4 dT 5 RBA ln T 1 BT 2 2T 2 R 5 0.793R 298

T1

(E3.8B)

298

where the following parameters were used for air: A 5 3.355, B 5 0.575 3 1023, and D 5 20.016 3 105 Substituting Equation (E3.8B) into (E3.8A) gives: Ds 5 R 1 0.793 2 ln2 2 5 0.83 3 J/ 1 mol K 2 4

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3.7 Calculation of Ds for an Ideal Gas ◄ 155

An alternative approach is to use the mean heat capacity values in Example 2.8: Ds 5 cP ln ¢

J T2 P2 1 373 ≤ 2 R ln ¢ ≤ 5 29.37 ln a b 2 8.314 ln a b 5 0.83 c d T1 P1 298 0.5 mol K

We have interpolated between 300 K and 400 K to get cP 5 29.37 3 J/ 1 mol K 2 4 . In this case, these two approaches give identical values. Do you think that is always true?

One mole of pure N2 and 1 mole of pure O2 are contained in separate compartments of a rigid container at 1 bar and 298 K. The gases are then allowed to mix. Calculate the entropy change of the mixing process.

EXAMPLE 3.9 Calculation of Entropy Change of Mixing

SOLUTION A schematic of the mixing process represented by DSmix is shown in Figure E3.9A. We are free to choose any path to calculate the change in entropy of the mixing process. One possible solution path is shown in Figure 3.9B. In the first step (step I), each of the pure species, N2 and O2, is reversibly and isothermally expanded to the size of the container of the mixture. During this process, the pressure drops to pO2 and pN2 respectively. The entropy change for each step is given by Equation (3.22): DSIO2 5 nO2 1 sO2 1 at pO2, T 2 2 sO2 1 at P, T 2 2 5 2nO2R¢ln

pO2 P

≤

and, P,T

O2

N2

O2

O2 O 2

P,T

O2

O2

N2

N2

O2

N2

O2

Mixing process

N2 N2

O2

P,T

N2

N2

N2

N2

O2

N2

N2

O2

N2

N2

O2

N2

O2

O2

O2

N2

O2

Figure E3.9A Mixing of N2 and O2 in a rigid container at constant P and T. pO2,T

P,T O2

O2

O2 O2 O2 O2 O2 O2

ΔSIO2 = nO2[sO2(at pO2,T ) − sO2 (at P,T)]

O2

O2 O2

O2

O2 P,T

O2

O2

pN ,T 2

P,T N2 N2 N2

N2 N2 N2 N2

N2

ΔSIN2 = nN2[sN2(at pN2,T ) − sN2 (at P,T)]

N2

N2

N2 N2

ΔSII = 0

N2

O2 O2

N2 O2 N2

N2

N2

O2

O2

N2 O2

N2 O2 N2 N2 O2 N2

O2

N2

N2

Figure E3.9B Hypothetical solution path to calculate DS for the mixing process (continued)

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156 ► Chapter 3. Entropy and the Second Law of Thermodynamics

DSN2 5 nN2 1 sN2 1 at pN2, T 2 2 sN2 1 at P, T 2 2 5 2nN2R¢ln I

pN2 P

≤

where the partial pressures PO2 and PN2 equal 0.5 bar. The next step (step II) is to superimpose both these expanded systems. Since O2 behaves as an ideal gas, it does not know N2 is there and vice versa; thus, the properties of each individual species do not change. Therefore, II DS II 5 nO2DsII O2 1 nN2DsN2 5 0

so, DS 5 DSIO2 1 DSIN2 1 DSII 5 2 nO2R¢ln

pO2 P

≤ 2 nN2R¢ln

pN2 P

≤ 5 11.53 c

J K

d

The path shown in Figure E3.9B has interesting implications in terms of the molecular viewpoint of the entropy increase for each species. Increased entropy dose not directly result because N2 mixes in O2, but rather because O2 has more room to move around in, so that there is more uncertainty where it is; therefore, information is lost, and the entropy is higher.

EXAMPLE 3.10 Reformulation of Polytropic Process in Terms of Entropy

In Section 2.7, we came up with the following expression for a reversible, adiabatic process on an ideal gas with constant heat capacity, based on first-law analysis: PVk 5 const Come up with the same equation based on second-law analysis, starting from Equation (3.23). SOLUTION Equation (3.23) can be set to zero for a reversible, adiabatic process: Ds 5 cP ln ¢

T2 P2 ≤ 2 R ln ¢ ≤ 5 0 T1 P1

Rearranging, we get: cP

ln ¢

P2 T2 ≤ 5 ln ¢ ≤ T1 P1

R

(E3.10)

Applying the ideal gas law and the relation R 5 cP 2 cv, Equation (E3.10) becomes: ¢

cP cP 2cv c 1 PV P2 T2 P nR 2 2 ≤ 5 ¢ PV ≤ 5 ¢ ≤ T1 P1 1 nR 2 1

However, since n1 5 n2, we get: cv

¢ or, with,

V1 P2 ≤ 5¢ ≤ P1 V2

cP

PVk 5 const cP k5 cv

The analysis is much easier using what we have learned about entropy.

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3.7 Calculation of Ds for an Ideal Gas ◄ 157

EXAMPLE 3.11 Entropy Change for the Expansion of Argon From a Compressed Cylinder

Consider a cylinder containing 4 moles of compressed argon at 10 bar and 298 K. The cylinder is housed in a big lab maintained at 1 bar and 298 K. The valve develops a leak and Ar escapes to the atmosphere until the pressure and temperature equilibrate. After a sufficient amount of time, the argon in the room reaches its background mole fraction of 0.01. Estimate, as closely as possible, the entropy change for the universe. You may assume that argon behaves as an ideal gas. Do you think this expansion process as a whole can be treated as reversible? SOLUTION A schematic of the process is shown in Figure E3.11A. We label the initial state “state 1” and the final state “state 4.” We are tempted to solve this problem as an open system where our boundary is contained by the walls of the cylinder; however, instead we will choose a system containing all the gas in the cylinder. Our problem is then reduced to that of a fixed amount of gas expanding in a closed system, with some of the gas then mixing with the environment. Our solution path is shown in Figure E3.11B. In step 1, we isothermally expand the gas to 1 bar. Our calculation for this part will be similar to the approach in Example 3.7.6 In step 2, we divide the gas into two parts: that which remains in the cylinder and that which is in the air-filled room. The entropy change for this step is zero. Step 3 involves calculating the entropy change of the mixing of the Ar with the other gases in the atmosphere (such as we

Real system

PE = 1 bar P = 10 bar T = 298 K

P = 1 bar T = 298 K Process

Tsurr = 298 K

Q State 1

State 4

Figure E3.11A Real system of a gas undergoing an expansion from a cylinder. Calculation Path

Ar outside the cylinder

PE = 1 bar Similar to Example 3.7

P = 1 bar T = 298 K

P = 1 bar T = 298 K

Tsurr = 298 K

State 1

State 2

Q

Similar to Example 3.9

PAr = 1 bar T = 298 K

Step 3

Step 2

Step 1 P = 10 bar T = 298 K

Δs = 0

P = 1 bar T = 298 K

Ar mixed with air

State 3 Ar remaining in the cylinder

P = 1 bar T = 298 K State 4

Figure E3.11B Calculation path for Ds from state 1 to state 4. 6

Alternatively, the open-system approach to step 1 will be presented at the end of this example.

(Continued)

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158 ► Chapter 3. Entropy and the Second Law of Thermodynamics

calculated in Example 3.9). We will assume that the Ar mixes until it reaches its composition in air (about 1%) and that no air diffuses back into the cylinder; that is, the cylinder remains pure Ar. Step I: State 1 to state 2 To solve for the entropy change of the universe in step I, we must determine the entropy change of the system as well as the entropy change of the surroundings. Since the initial and final states are constrained, the entropy change of the system can be calculated from Equation (3.23): 0 Dssys 5 cP ln ¢

T2 P2 ≤ 2 R ln ¢ ≤ T1 P1

Thus, the value for the entropy change of the system is given by: Dssys 5 2R ln ¢

J J P2 1 ≤ 5 2¢8.314 c d ≤ ln a b 5 17.1 c d P1 mol K 10 mol K

As the argon does work on the surroundings during the expansion process, it will lose energy. An equal amount of energy must be transferred in via heat to keep the temperature constant at 298 K. To find the entropy change to the surroundings, we need to determine the amount of heat transferred. A first-law balance gives: Du 5 q 2 Pext 1 v2 2 v1 2 5 q 2 P2 1 v2 2 v1 2 5 0 where we have set the change in internal energy equal to zero, since we are treating Ar as an ideal gas at constant temperature. Solving for q, we get the entropy change of the surroundings to be: Dssurr 5

qsurr Tsurr

52

q T2

5

2P2 1 v2 2 v1 2 P2 5 2RB1 2 R T2 P1

5 2 1 8.314 2 1 0.9 2 5 27.5 c

J mol K

d

Thus, the entropy change of the universe is then given by: DsIuniv 5 Dssys 1 Dssurr 5 17.1 2 7.5 5 9.6 3 J/ 1 mol K 2 4 or, for the extensive value, DSIuniv 5 n 1 Dsuniv 2 5 38.4 3 J/K 4

(E3.11A)

Step II: State 2 to state 3 The entropy change for step II is zero: DSII univ 5 0 The number of moles remaining in the cylinder can be found using the ideal gas law (with T and V constant): 1 ncyl 2 3 5 1 ncyl 2 1

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3.7 Calculation of Ds for an Ideal Gas ◄ 159

Step III: State 3 to state 4 The entropy change for the system from state 3 to state 4 is similar to that of the gases in Example 3.9. The entropy of mixing of argon outside the cylinder is given by: DSIII sys 5 noutside 1 s 1 at pAr, T 2 2 s 1 at P, T 2 2 5 2noutsideR¢ln

pAr P

≤

Plugging in numbers, we get: III 5 2 1 3.6 2 1 8.314 2 1 ln 1 0.01 2 2 5 137.8 3 J/K 4 DSsys

Assuming the composition of the air in the room does not noticeably change by the dilute addition of argon, the entropy change to the surroundings is zero and:7 DSIII univ 5 137.8 3 J/K 4 Adding together the three steps in the solution path, we get: III DSuniv 5 DSIuniv 1 DSII univ 1 DSuniv 5 176.2 3 J/K 4

The value is greater than zero, as we would expect for this very irreversible process. It should be noted that solutions to this kind of problem are often presented where reversibility is assumed if the leak is “slow enough.” However, this approach is wrong. No matter how slow the leak, the driving force for the expansion is finite (as opposed to differential), so this process is not reversible. It would be just as absurd as to invoke reversibility in Example 3.9 by saying that the mixing of the gases in the two compartments occurred through a vary small hole and, therefore, very slowly. [Step I] alternative: Open-system analysis We begin with the unsteady-state open-system entropy equation, Equation (3.21), with no inlet and one exit stream: a

# # Qsurr Q dS dS dS 5 a b 1 n# ese 1 5 a b 1 n# ese 2 b dt univ dt sys Tsurr dt sys Tsurr

(E3.11B)

If the exit state is assumed to be uniform, we can integrate Equation (E3.11B) to give: t

Q DSuniv 5 n2s2 2 n1s1 1 se 3 n# edt 2 Tsurr

(E3.11C)

0

However, since the exit state and state 2 are identical, se 5 s2

(E3.11D)

# 3 nedt 5 n1 2 n2

(E3.11E)

and by a mole balance: 1

0 7

Alternatively, this problem can be solved by mixing the argon with an infinite amount of air, with the same result.

(Continued)

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160 ► Chapter 3. Entropy and the Second Law of Thermodynamics

Plugging Equations (E3.11D) and (E3.11E) into Equation (E3.11C) gives: DSuniv 5 n1 1 s2 2 s1 2 5

Q

(E3.11F)

Tsurr

We can find the entropy difference between states 1 and 2 by using Equation (3.23): T2

J cP P2 1 s2 2 s1 5 3 dT 2 R ln ¢ ≤ 5 0 2 ¢8.314 c d ≤ ln a b T P1 mol K 10 T1

5 17.1 c

J mol K

d

(E3.11G)

We must now solve for the heat transfer Q, by an energy balance. Integrating Equation (2.22), we get: t

t

n2u2 2 n1u1 5 2B 3 n# edtRhe 1 Q 5 2B 3 n# edtR 1 ue 1 Peve 2 1 Q 0

(E3.11H)

0

Applying Equation (E3.11E) and solving Equation (E3.11H) for Q gives: 1

n2 P2 Q 5 B 3 n# edtRPve 5 1 n1 2 n2 2 Peve 5 n1 ¢1 2 ≤Peve 5 n1 ¢1 2 ≤RT2 n1 P1 0

Therefore, we get: Q Tsurr

5 n1RB1 2

J P2 R 5 4 1 8.314 2 1 0.9 2 5 29.93 c d P1 K

(E3.11I)

Plugging the results from Equations (E3.11G) and (E3.11I) into Equation (E3.11F): DSuniv 5 4 1 17.1 2 2 29.9 5 38.5 3 J/K 4

(E3.11J)

The results for the closed- and open-system analysis, Equations (E3.11A) and (E3.11J), are identical to within round-off error.

►3.8 THE MECHANICAL ENERGY BALANCE AND THE BERNOULLI EQUATION We next consider a special case—flow processes that are at steady-state and reversible, with one stream in and one stream out. We wish to come up with an expression to evaluate the work in such a process. The first-law balance, in differential form, is given by: S

# # V2 1 MWgz b d 1 dQsys 1 dWs 0 5 2 n# c dah 1 MW 2

(3.26)

where MW is the molecular weight. The second law is: # #nds 1 dQsurr 5 0 T

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3.8 The Mechanical Energy Balance and the Bernoulli Equation ◄ 161

However, since: # # dQsurr 5 2dQsys Equation (3.27) can be substituted into (3.26) to give: S

2

V # 0 5 2 n# c dah 1 MW 1 MWgz b d 1 n# Tds 1 dW s 2 Rearranging to solve for work: S # dWs V2 5 dh 2 Tds 1 MWd c a 2 b 1 MWgdz § n#

(3.28)

We can simplify Equation (3.28) even further. For a reversible process, we can write: du 5 dqrev 1 dwrev 5 Tds 2 Pdv If we add d(Pv) to both sides, we get: dh 5 d 1 u 1 Pv 2 5 Tds 1 vdP Solving for 1 dh 2 Tds 2 and plugging into Equation (3.28) gives: S # dWs V2 5 c vdp 1 MWda b 1 MWgdz § 2 n#

(3.29)

Equation (3.29) is termed the differential mechanical energy balance. It is a useful form, since the work is written in terms of the measured properties P and v as well as bulk potential and kinetic energy. It is applicable to reversible, steady-state processes with one stream in and one stream out. If we integrate Equation (3.29), we get: 2

S S # 2 Ws /n# 5 3 vdP 1 MW c 1 V2 2 V21 2 /2d 1 MWg 1 z2 2 z1 2

(3.30)

1

There are two frequent cases where Equation (3.30)is applied: Case 1: No Work (Nozzles, Diffusers) 2

S

S

V2 2 V12 0 5 3 vdP 1 MW ¢ 2 ≤ 1 MWg 1 z2 2 z1 2 2

(3.31)

1

Equation (3.31) is the celebrated Bernoulli equation. Case 2: No eK, eP (Turbines, Pumps) 2 # Ws 5 3 vdP n#

(3.32)

1

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162 ► Chapter 3. Entropy and the Second Law of Thermodynamics

EXAMPLE 3.12 Power Generated by a Turbine

An ideal gas enters a turbine with a flow rate of 250 mol/s at a pressure of 125 bar and a specific volume of 500 cm3 /mol. The gas exits at 8 bar. The process operates at steady-state. Assume the process is reversible and polytropic with: Pv1.5 5 const Find the power generated by the turbine. SOLUTION

Since this process is at steady-state and is reversible, we can use Equation (3.32): P2 # Ws 5 3 vdP n#

(E3.12A)

P1

Since both v and P vary during the process, we must write v in terms of P to perform the integral in Equation (E3.12A). This polytropic process is described by: 1.5

Pv1.5 5 const 5 P1v1

where we have written the constant in terms of state 1, since we know both v and P. Solving for v gives: v5 ¢

P1v1.5 1 ≤ P

2/3

(E3.12B)

Substituting Equation (E3.12B) into (E3.12A) and integrating leaves: P2 # Ws P 5 1 P1 2 0.6667v1 3 P212/32dP 5 3 1 P1 2 0.6667v1 3 P0.3333 4 P21 n# P1

Plugging in numerical values, we get: # Ws 3 3 5 3 1 1.25 3 107 3 Pa 4 2 0.6667 a5 3 1024 3 m3 /mol 4 b a"8 3 105 3 Pa 4 2 "1.25 3 107 3 Pa 4 b n# 5 211,250 3 J/mol 4 Finally, solving for power gives: # Ws 5 1 250 3 mol/s 4 2 1 211,250 3 J/mol 4 2 5 22.8 3 MW 4 The sign is negative, since we get useful work out of the turbine. Note, for comparison, that a coal-fired power plant generates on the order of 1000 MW.

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3.8 The Mechanical Energy Balance and the Bernoulli Equation ◄ 163

EXAMPLE 3.13 Correction for Efficiency in a Real Turbine

In an actual expansion through the turbine of Example 3.12, 22.1 3 MW 4 of power is obtained. What is the isentropic efficiency, hturbine, for the process? The isentropic efficiency is given by: hturbine 5

# 1 Ws 2 real # 1 Ws 2 rev

# where 1 Ws 2 rev represents the power obtained in the reversible process from the same inlet # state and outlet pressure and 1 Ws 2 real is the power of the actual process. SOLUTION We run into several different definitions of efficiency, depending on the context. Isentropic efficiencies compare the actual performance of a process operation with the performance it would obtain if it operated reversibly; that is, we compare the given process to the best it could do. For a turbine, the same inlet state and same exit pressure are used in the calculation. Figure E3.13 shows both the actual process (solid line) and the ideal, reversible process (dashed line) on a Ts diagram. Both processes start at the same state and end at the same pressure. The final temperature of the actual process is higher than the reversible process, since less energy is removed via work. The isentropic efficiency is calculated to be: hturbine 5

22.1 3 MW 4 5 0.75 22.8 3 MW 4

We obtain an isentropic efficiency of 75%. We can use a similar approach to determine isentropic efficiencies of other unit processes, such as pumps or nozzles. The isentropic efficiency of a pump compares the minimum work needed from the same inlet state and an outlet at the same pressure to the actual work: hpump 5

# 1 Ws 2 rev # 1 Ws 2 real

Can you draw a figure analogous to Figure E3.13 for a pump? The isentropic efficiency of a nozzle compares the actual exit kinetic energy to that the fluid would obtain in a reversible process. Values between 70% and 90% are typical for turbines and pumps, while nozzles typically obtain 95% or better. T 1 P=

T2,real T2,rev

ar 125 b

Actual process

Δs = 0

T1

Isobars P=

r

8 ba

2rev

2real

Inaccessible States s1 = s2,rev s2,real

c03.indd 163

s

Figure E3.13 A Ts diagram llustrating the states between which the isentropic efficiency is calculated. The actual process is shown by the solid line, between states 1 and 2real, while the ideal, reversible process is shown by the dashed line between states 1 and 2rev.

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164 ► Chapter 3. Entropy and the Second Law of Thermodynamics

► 3.9 VAPOR-COMPRESSION POWER AND REFRIGERATION CYCLES In this section, we will examine the basic elements of common industrial power and refrigeration systems. These systems employ a thermodynamic cycle in which a working fluid is alternatively vaporized and condensed as it flows through a set of four processes. Recall from Section 2.9 that, after completing a cycle, the working fluid returns to the same state it was in initially so that the cycle can be repeated. We will use the principles of energy conservation and entropy to analyze the performance of power and refrigeration cycles. We will first examine the Rankine cycle, which is used to convert a fuel source to electrical power. We will then look at a vapor-compression cycle operated in “reverse” to expel heat from a cold reservoir and produce refrigeration.

The Rankine Cycle Say we wish to convert a fossil-fuel, nuclear, or solar energy source into net electrical power. To accomplish this task, we can use a Rankine cycle. The Rankine cycle is an idealized vapor power system that contains the major components found in more detailed, practical steam power plants. While hydroelectric and wind are possible alternative sources, the steam power plant is presently the dominant producer of electrical power. A schematic of the Rankine cycle is shown in Figure 3.7. The left-hand side shows the four unit processes in order: a turbine, a condenser, a compressor, and a boiler. States 1, 2, 3, and 4 are labeled. The right-hand side identifies each of these states on a Ts diagram. Each of the four individual processes operates as an open system at steadystate, such as those modeled in Section 2.8. Moreover, these processes are assumed to be reversible; hence, the efficiency we calculate will be the best possible for a given design scenario. The working fluid that flows through these processes is usually water. We will formulate our analysis on a mass basis, in anticipation of using the steam tables for thermodynamic data. Electrical power is generated by the turbine, while the energy from combustion of the fuel is input via heat transfer in a boiler. Energy transfer between the surroundings and the system is further needed to return the system to its initial state and complete the cycle. This energy transfer occurs via heat expulsion in the condenser and via work input to the compressor. A more detailed analysis of the four processes in the Rankine cycle follows. Rankine cycle T

Ws Turbine

1

1 QH Fuel air

2

QH Boiler

Ws QC

Cooling water

4 Condenser 3 Wc

4 Wc

3

QC

2

s

Compressor

Figure 3.7 The ideal Rankine cycle used to convert fuel into electrical power. The four unit processes are sketched on the left, while the path on a Ts diagram is shown on the right. The working fluid is typically water.

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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 165

We start from state 1 in the diagram in Figure 3.7, where the working fluid enters the turbine as a superheated vapor. As it goes through the turbine, it expands and cools while producing work. It exits the turbine at state 2. The rate of work produced can be determined by applying the first law [Equation (2.50)]. Assuming that bulk kinetic and potential energy and heat transfer are negligible, the power produced by the turbine becomes: # Ws 5 m# 1 h^ 2 2 h^ 1 2 where m# is the flow rate of the working fluid. Since the process is reversible with negligible heat transfer, the entropy remains constant, as depicted by the vertical line in the Ts diagram: s^ 1 5 s^ 2 The steam enters as a superheated vapor, and does not condense significantly in the turbine. If the steam were saturated when it entered the turbine, a significant fraction of liquid would be formed when the temperature dropped isentropically. The dashed line on the Ts diagram illustrates this case. This option is impractical, since too much liquid causes erosion and wear of the turbine blades. The steam next enters a condenser; it exits in state 3 as saturated liquid water. The change of phase occurs at constant pressure and requires that energy be removed from the flowing stream via heat. Thus, a low-temperature reservoir is needed. A first-law balance around the condenser gives: # QC 5 m# 1 h^ 3 2 h^ 2 2 Next it is desired to raise the pressure of the liquid, which is accomplished using a compressor. High-pressure water exits the compressor in state 4. The work required to compress the liquid is given by: # Wc 5 m# 1 h^ 4 2 h^ 3 2 5 m# v^ 1 1 P4 2 P3 2

(3.33)

where Equation (3.32) was integrated assuming v^ l is constant. Since the molar volume of the liquid is significantly less than that of the vapor, the work required by the compressor is much less than that produced by the turbine. Typically, a small fraction of the power produced by the turbine is used to compress the liquid, and the remaining power is the net power obtained by the cycle. The liquid that enters the compressor is saturated, by design, since most compressors cannot handle a two-phase mixture. Finally, the high pressure liquid is brought back to a superheated vapor state in the boiler. It is in this step that energy released from the combustion of fuel is transferred to the working fluid. The fuel provides the high-temperature reservoir for the boiler. The boiler isobarically heats the liquid to saturation, vaporizes it, and then superheats the vapor. The rate of heat transfer in the boiler is given by: # QH 5 m# 1 h^ 1 2 h^ 4 2 The vapor exits the boiler in state 1 and the cycle is repeated.

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166 ► Chapter 3. Entropy and the Second Law of Thermodynamics We define the efficiency of the cycle as the ratio of the net work obtained divided by the heat absorbed from the boiler: # # 0 Ws 0 2 Wc 0 1 h^ 2 2 h^ 1 2 0 2 1 h^ 4 2 h^ 3 2 # 5 hRankine 5 (3.34) QH 1 h^ 1 2 h^ 4 2 In this definition, it is assumed that the heat absorbed is proportional to the amount of fuel consumed. The Ts diagram provides a useful graphical aid in interpreting the Rankine cycle. From the definition of entropy, qrev 5 3 Tds Since we are assuming reversibility for the cycle, the heat absorbed by the water in the boiler, qH, and the heat expelled in the condenser, qC, are equal to the respective area under the Ts curve. These graphical depictions are illustrated in the first two diagrams of Figure 3.8. The net work produced by the cycle is given by the difference of these two quantities: qH 2 0 qC 0 5 0 ws 0 2 wc 5 wnet Thus, the net work is equal to the area of the box in the third diagram. If we can make this box bigger relative to qH, we increase the efficiency. Can you think of ways to accomplish this? q H − qC = wnet T

T

T

1

1

1

wnet

qH 4

4 3

2

4 3

s

qC

2

3

2

s

s

Figure 3.8 Graphical representation of the heat absorbed in the boiler, qH, the heat expelled in the condenser, qC, and the net work, wnet, in an ideal Rankine cycle.

EXAMPLE 3.14 Calculation of the Power and Efficiency of a Rankine Power Cycle

c03.indd 166

Steam enters the turbine in a power plant at 600ºC and 10 MPa and is condensed at a pressure of 100 kPa. Assume the plant can be treated as an ideal Rankine cycle. Determine the power produced per kg of steam and the efficiency of the cycle. How does the efficiency compare to a Carnot cycle operating between these two temperatures? SOLUTION We can refer to Figure 3.7 to identify the states of water as it goes through the cycle. It is useful to refer to this figure as we are solving the problem. Examining Equation (3.34), we see that we need to determine the enthalpies in the four states to solve for the

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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 167

efficiency. Steam enters the turbine at 600ºC and 10 MPa. Looking up values from the steam tables (Appendix B), we get: h^ 1 5 3625.3 3 kJ/kg 4 and,

s^ 1 5 6.9028 3 kJ/ 1 kg K 2 4 5 s^ 2

Since the entropy at state 1 is equal to the entropy at state 2. We also know the pressure at state 2, P2 5 100 kPa. Thus, state 2 is completely constrained, and we can determine its enthalpy from the steam tables. Since it exists as a liquid–vapor mixture, we must determine the quality, x, of the steam as follows: s^ 2 5 6.9028 3 kJ/ 1 kg K 2 4 5 1 1 2 x 2 s^ l 1 xs^ v 5 1 1 2 x 2 1 1.3025 3 kJ/kg K 4 2 1 x 1 7.3593 3 kJ/kg K 4 2 Solving for x gives:

x 5 0.925

Therefore, the enthalpy in state 2 is given by: h^ 2 5 1 1 2 x 2 h^ l 1 xh^ v 5 1 0.075 2 1 417.44 3 kJ/kg 4 2 1 0.925 1 2675.5 3 kJ/kg 4 2 5 2505.3 3 kJ/kg 4 The power generated by the turbine is found by the enthalpy difference between state 2 and state 1: w^ s 5 h^ 2 2 h^ 1 5 2505.3 2 3625.3 5 21120.0 3 kJ/kg 4 The enthalpy of state 3 is a saturated liquid at 100 kPa: h^ 3 5 h^ 1 5 417.44 3 kJ/kg 4 The enthalpy at state 4 can be determined from Equation (3.33): h^ 4 5 h^ 3 1 v^ l 1 P4 2 P3 2 5 427.34 3 kJ/kg 4 Note that since we are increasing the pressure of water in the liquid state, the work required is only 9.9 kJ/kg. This value is less than 1% of the work produced by expansion of vapor through the turbine (1120 kJ/kg). The net work is given by: w^ net 5 w^ s 1 w^ c 5 21120.0 1 9.9 5 21110.1 3 kJ/kg 4 Solving for the efficiency from Equation (3.34), we get: # # # hRankine 5 1 0 Ws 0 2 Wc 2 /QH 5 3 0 1 h^ 2 2 h^ 1 2 0 2 1 h^ 4 2 h^ 3 2 4 / 1 h^ 1 2 h^ 4 2 5 0.347 The 34.7% efficiency is the best-case scenario for this cycle, since we assumed reversible processes. In reality, we would not even achieve this value! The Carnot efficiency is given by Equation (3.9): hCarnot 5 1 2

TC 373 512 5 0.573 TH 873

The Rankine efficiency is lower than the Carnot efficiency. We can see the basis if we compare the net work graphically, as we did in Figure 3.8. The box representing net work for each cycle is shown in Figure E3.14. Recall that the steam that enters the turbine of the Rankine cycle is superheated to eliminate wear and corrosion on the turbine blades. In modifying the cycle in this way, we “crop off” a significant portion of the rectangle that represents the Carnot cycle.

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168 ► Chapter 3. Entropy and the Second Law of Thermodynamics

EXAMPLE 3.15 Modification of Rankine Analysis for Nonisentropic Steps

Redo the analysis of the Rankine cycle of Example 3.14 but include isentropic efficiencies of 85% in the pump and turbine. Determine the net power and the overall efficiency of the power cycle. SOLUTION Recall the discussion of isentropic efficiencies from Example 3.13. The isentropic efficiency of the turbine is given by: hturbine 5

# 1 Ws 2 actual 1 w^ s 2 actual 5 # 1 w^ s 2 rev 1 Ws 2 rev

The reversible work was found in Example 3.14. Solving for actual work gives: 1 w^ s 2 actual 5 hturbine 1 w^ s 2 rev 5 0.85 1 21120.0 2 5 2952.0 3 kJ/kg 4 As we suspect, the work we get out of the turbines in the real, irreversible process is less than that for the reversible process. Solving the energy balance around the turbine gives the enthalpy in state 2 as: 1 h^ 2 2 actual 5 1 w^ s 2 actual 1 h^ 1 5 2952.0 1 3625.3 5 2673.3 3 kJ/kg 4 This value is higher than that found in Example 3.14, indicating that the temperature entering the condenser for the actual process is higher than it would be in the reversible case. State 3 remains the same: h^ 3 5 h^ 1 5 417.44 3 kJ/kg 4 Since the reversible work represents the best we can possibly do, the actual work needed in the compressor must be greater than the reversible value. Hence, the isentropic efficiency in the compressor is given by: 1 w^ c 2 actual 5

1 w^ c 2 rev 9.9 5 11.6 3 kJ/kg 4 5 hcompressor 0.85

Solving for the enthalpy at the exit of the compressor gives: 1 h^ 4 2 actual 5 1 w^ c 2 actual 1 h^ 3 5 11.6 1 417.44 5 429.1 3 kJ/kg 4

Carnot Cycle

Rankine Cycle

T

T 4

1

1

wnet

wnet 4

3

3

2 s

2 s

Figure E3.14 Graphical depiction of the net work in a Carnot cycle and a Rankine cycle.

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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 169

The net work can be found by adding the actual work generated in the turbine to that consumed in the compressor: w^ net 5 w^ s 1 w^ c 5 2952.0 1 11.6 5 2940.4 3 kJ/kg 4 Similarly, the actual efficiency must be calculated using these values: # # # hRankine 5 1 0 Ws 0 2 Wc 2 /QH 5 3 0 1 h^ 2 2 h^ 1 2 0 2 1 h^ 4 2 h^ 3 2 4 / 1 h^ 1 2 h^ 4 2 5 0.294 Introducing irreversibilities in turbine and pump reduced efficiency from 34.7% to 29.4%.

The Vapor-Compression Refrigeration Cycle Refrigeration systems are important in industrial and home use when temperatures less than the ambient environment are required. Of the several types of refrigeration systems, the most widely used is the vapor-compression refrigeration cycle. It is essentially a Rankine cycle operated in “reverse,” where heat is absorbed from a cold reservoir and rejected to a hot reservoir. Due to the constraints of the second law, this process can be accomplished only with a concomitant consumption of power. A schematic of the ideal vapor-compression cycle is shown in Figure 3.9. The lefthand side shows the four unit processes in order: an evaporator, a compressor, a condenser, and a valve. Each of the four individual processes operates as an open system at steady-state. States 1, 2, 3, and 4 are labeled. The right-hand side identifies each of these states on a Ts diagram. Unlike in the Rankine cycle, the work required for refrigeration is not represented by the area enclosed on the Ts diagram because the expansion through the valve is irreversible. The working fluid is termed the refrigerant. In choosing a refrigerant, we must realize that both the evaporation and condensation processes contain phase transformations. Thus, in each of these processes, T and P are not independent. Specifying the temperature at which these processes occur, restricts the pressure for a given choice of refrigerant. For example, the evaporator temperature is determined by the temperature, TC, required from our refrigeration system. For a given working fluid, constraining TC also constrains the evaporator P. We typically want a species that boils at lower temperatures than water. Ideally, we choose a species that provides the desired refrigeration temperature at a pressure slightly above atmospheric. In that way there is a positive pressure against the environment. Common choices are CCl2F2 (refrigerant 12), CCl3F (refrigerant 11), CH2FCF3 (refrigerant 134a), and NH3. The first two species, the chlorofluorocarbons, are very stable if released to the environment. They have mostly been phased out of use because they lead to depletion of the ozone layer and also contribute to the greenhouse effect, which leads to global climate change. An analysis of the four processes in the vapor-compression refrigeration cycle follows. We start from state 1 in the diagram in Figure 3.9, where the working fluid enters the evaporator. In the evaporator, heat is transferred from the refrigerated unit to # the working fluid. This occurs at temperature TC. The working fluid absorbs QC as it

changes phase.

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170 ► Chapter 3. Entropy and the Second Law of Thermodynamics Refrigeration Cycle T

Valve

3

1

QH

Evaporator Refrigerated unit at low T

4

4 QC QH

High T reservoir

2 Condenser 3

1

Qc

2

s

Wc Compressor

Figure 3.9 The ideal vapor-compression refrigeration cycle. The four unit processes are sketched on the left, while the path on a Ts diagram is shown on the right.

It emerges in state 2, where it is a vapor. The heat transferred is given by: # QC 5 n# 1 h2 2 h1 2

(3.35)

where n# is the molar flow rate of the refrigerant. We must then compress the refrigerant to a high enough pressure in state 3 that it will condense at the temperature of the hot reservoir available to us, TH.The choice of refrigerant determines the outlet pressure of the compressor required at state 3. Since we are performing the compression in the vapor phase, where molar volumes are large, a significant amount of work is needed. The higher the pressure, the more work is required for a given refrigeration effect. The power of compression is given by: # Wc 5 n# 1 h3 2 h2 2 If the compression is assumed to be reversible, s3 5 s2 # The high-pressure vapor is then condensed at TH, expelling heat QH to the hot reservoir. This process occurs in the condensor where the fluid exits at state 4. # QH 5 n# 1 h4 2 h3 2 The high-pressure liquid is then expanded in a valve back to state 1 so the cycle can be repeated. A valve is used instead of the turbine that was used in the Rankine cycle. The amount of work that would be produced by a turbine is small, so we replace it with a valve to reduce the complexity. This step is represented by a throttling process, where: h4 5 h1

(3.36)

Since the pressure decreases as the refrigerant passes through the valve, its entropy increases, as shown in Figure 3.9. Can you locate the evaporator and condenser on the refrigerator you have at home?

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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 171

The coefficient of performance, COP, of a refrigeration cycle measures its performance. It is defined as the ratio of the heat absorbed from the cold reservoir (the refrigeration effect) to the work required: # QC h2 2 h1 COP 5 # 5 h3 2 h2 Wc

(3.37)

In real refrigeration systems, a finite temperature difference is needed to get practical heat-transfer rates in the evaporator and the condenser. Thus, the evaporator must operate at a lower temperature than the desired refrigeration temperature, while the condenser must operate at a higher temperature than the ambient heat reservoir. Thus, more work is required to obtain a given refrigeration effect. Moreover, irreversibilities in the compressor must be considered, also adding to required work load and further decreasing the COP. COPs of well-designed real refrigeration systems typically fall between 2 and 5.

EXAMPLE 3.16 Estimation of the COP of a VaporCompression Refrigeration Cycle

It is desired to produce 10 kW of refrigeration from a vapor-compression refrigeration cycle. The working fluid is refrigerant 134a. The cycle operates between 120 kPa and 900 kPa. Assuming an ideal cycle, determine the COP and the mass flow rate of refrigerant needed. Properties of refrigerant 134a can be found at http://webbook.nist.gov/chemistry/fluid/. Data can be viewed in an HTML table. SOLUTION A sketch of the process on a Ts diagram is shown in Figure E3.16. The following saturated data for refrigerant 134a are obtained from the NIST site (see citation Pg. 27): P [MPa]

T [K]

hl 3 kJ/mol 4

hv 3 kJ/mol 4

sl 3 J/ 1 mol K 2 4

sv 3 J/ 1 mol K 2 4

0.12

250.84

17.412

39.295 5 h2

90.649

177.89 5 s2

0.90

308.68

25.486 5 h4

42.591

119.32 5 s4

174.74

P = 0.9 MPa 317.6

308.7

3 QH

4

T [K]

WC P = 0.12 MPa

250.8

1

QC

2

s

Figure E3.16 A Ts diagram of the ideal vapor-compression refrigeration cycle of Example 3.16. (Continued)

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172 ► Chapter 3. Entropy and the Second Law of Thermodynamics

To get the amount of refrigeration, we can combine Equations (3.35) and (3.36): qC 5 1 h2 2 h1 2 5 1 h2 2 h4 2 5 39.295 2 25.486 5 13.809 3 kJ/mol 4 To find the state at the exit of the isentropic compressor, we recognize: s3 5 s2 At a pressure of 0.90 MPa, from the NIST website, we get the entropies to match state 3 at: P [MPa]

T [K]

hv 3 kJ/mol 4

sv 3 J/ 1 mol K 2 4

0.90

317.62

43,578

177.89

The work required by the compressor is: wc 5 1 h3 2 h2 2 5 43.578 2 39.295 5 4.283 3 kJ/mol 4 Solving for the coefficient of performance gives: # QC h2 2 h1 5 3.22 COP 5 # 5 h3 2 h2 Wc To get the desired 10 kW of refrigeration, we need: # QC 10 3 kW 4 5 5 0.73 3 mol/s 4 n# 5 qC 13.809 3 kJ/mol 4 The temperature of the evaporator is around 250 K, below the freezing point of water. On the other hand, the condenser operates between 308 K and 317 K. This temperature is warm enough to expel heat to the ambient environment.

►3.10 EXERGY (AVAILABILITY) ANALYSIS As engineers, our goal is to design processes that best utilize the resources that are available. In terms of energy resources, we would want to maximize the work we obtain from a process or minimize the work we need to supply. We have learned that a reversible process, where the entropy change of the universe is zero, represents the “best that we can do.” However, an alternative perspective is sometimes useful—especially in the context of analyzing complex systems containing many unit processes. In this section, we introduce exergy analysis (which is also called availability analysis). This analysis looks at the most work we can get out of a process when considering the environment immediately adjacent in the surroundings. This approach is useful for a methodological analysis of complex processes. Specifically exergy (availability) analysis allows us to look at each step of a multistep process and allows us to determine the relative magnitude of the irreversibilities and lost work. This approach allows us to focus design efforts. The intent here is to provide a brief introduction to exergy analysis within the context of the thermodynamics that you are learning. This topic may be developed further in your design class, where you examine chemical and biological plants with many unit processes.

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3.10 Exergy (Availability) Analysis ◄ 173

We define the environment as the part of the surroundings that are immediately adjacent to the system. The environment exists at a state labeled “0,” constrained by properties P0, T0, and so on. We assume that the environment is stable and that its temperature, pressure and composition are uniform and unchanging (i.e., the environment does not change as a result of the processes we are analyzing). When the system reaches the state of the environment (P0, T0, etc.), there is no longer any driving force for mechanical, thermal, or chemical change. At this point, we cannot get any more work from the system; hence, at this point, we say the system is in the dead state. We define the ideal work, wid, as the maximum amount of work we can obtain as a system undergoes a process from its initial state to the dead state.

Exergy Let’s consider the adiabatic expansion of a piston-cylinder that was discussed as case I in Section 3.3. The reversible process takes the gas from State 1 to State 2, as shown in Figure 3.10. In this figure, the temperature and pressure of the environment are labeled T0 and P0, respectively. Because the process reaches mechanical equilibrium, the pressure of state 2, P2, equals P0. However, the temperature of the system does not equilibrate with the environment because the cylinder is well insulated. A first law balance around the piston-cylinder assembly can be written as: 0 Du 5 q 1 w 5 3 wid,Pv 2 Po 1 v2 2 v1 2 4

We label the ideal work for this expansion process, wid,Pv. Its value indicates the maximum useful work we can obtain from this process. We do not count the work required to expand against the environment at constant pressure 3 5 Po 1 v2 2 v1 2 4 as part of the ideal work because we cannot make use of it. Thus, wid,Pv is lower in magnitude than Du. The reversible process shown in Figure 3.10 shows us the “best that we can do” in expanding the gas from state 1 to state 2; however, if we are clever, we can obtain even more work! Because the temperature in the system has not yet reached that of the surroundings, energy will flow spontaneously from the system to the surroundings if we remove the insulation. If we treat the gas in the system as a “hot reservoir” and the environment as a “cold reservoir,” we can place a Carnot engine between them and get more work! Such a process is shown in Figure 3.11. We next calculate the ideal work, wid, for the system shown in Figure 3.11 where a gas is first adiabatically and reversibly expanded in a piston-cylinder assembly and then heat from the piston-cylinder assembly is expelled to a Carnot engine to obtain even T0 P0

T0 P0 Wid,Pv Reversible Process

Wellinsulated

T1

gas P1

State 1

Adiabatic Expansion

gas P2 = P0 T2 > T0 State 2

Figure 3.10 Process by which to calculate the Pv component of ideal work.

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174 ► Chapter 3. Entropy and the Second Law of Thermodynamics T0 P0

T0 P0

Carnot engine Wid, Pv

Reversible Process

Wellinsulated

Ideal gas T1 P 1

Adiabatic Expansion

qH Wid, cycle

Ideal gas

q0 State 1

State 2 Dead State Cold reservoir, T0

Figure 3.11 Process by which to calculate the ideal work. The labeling for heat and work in the Carnot engine has been modified to match with the nomenclature for exergy analysis.

more work. As before, we denote the ideal work coming from the expanding cylinder as wid,Pv; the ideal work coming from the Carnot cycle is denoted as wid,cycle. A first law balance around the piston-cylinder assembly for both steps can now be written: Du 5 q 1 w 5 2qH 1 3 wid,Pv 2 Po 1 v0 2 v1 2 4

(3.38)

where q is the heat expelled to the Carnot engine. We must be careful with signs when we write our first law balance around the Carnot engine. As shown in Figure 3.11, the value for heat that enters the cycle 1 qH 2 is the negative of what leaves the piston cylinder (q). For the Carnot engine: Du 5 0 5 qH 1 q0 1 wid,cycle where the nomenclature in Figure 3.11 is used. We keep the sign for qH and q0 consistent with the definition of the first law, with a negative value indicating heat transfer from the system to the environment. As we shall see, the signs will soon take care of themselves. Rewriting the previous equation, we get: wid,cycle 5 2 1 qH 1 q0 2 5 2¢ 1 1

q0 T0 ≤ qH 5 2¢ 1 1 ≤ qH qH T

where we have used Equation (3.7) for a Carnot cycle to convert from heat to temperature. Because the temperature of the hot reservoir is changing as we execute the process, we rewrite the previous equation for a differential increment. Thus,

dwid,cycle 5 2¢ dqH 1 T0d¢

qH T

≤ ≤ 5 2dqH 2 T0ds

where the definition of entropy was used. Integrating from the initial state (1) to the dead state (0), we get:

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3.10 Exergy (Availability) Analysis ◄ 175 0

0

s0

3 dwid,cycle 5 2 3 dqH 2 3 T0ds 1

1

s1

or, wid,cycle 5 2qH 2 T0 1 s0 2 s1 2 Note the value for q0 given by the second term ends up being opposite in sign because the final (cooler) state 1 s0 2 has lower entropy than the initial (hot) state 1 s1 2 . We next solve the previous equation for heat: 2qH 5 q 5 wid,cycle 1 T0 1 s0 2 s1 2 Substituting this expression for q back into Equation (3.38) gives: Du 5 1 u0 2 u1 2 5 wid,cycle 1 T0 1 s0 2 s1 2 1 3 wid,Pv 2 Po 1 v0 2 v1 2 4 Solving for the negative of ideal work, 2wid, (which gives us a positive number if we get work out of the system), we have the magnitude of the maximum useful work that is available for a process starting at state 1 and ending at the dead state of the environment, state 0. Although we have framed this development in terms of the most work we can obtain, we could alternatively view wid, as the minimum amount of work we need to put into process to take a system from the dead state to state 1. We define the exergy of state 1, b1, as the magnitude of the ideal work from state 1 to the dead state, that is, 2wid or 2 1 wid,cycle 1 wid,Pv 2 : b1 ; 1 u1 2 u0 2 1 Po 1 v1 2 v0 2 2 T0 1 s1 2 s0 2

(3.39)

For systems where macroscopic kinetic energy and macroscopic potential energy are important, we can generalize to: S

2

S

V20 V1 b1 5 1 u1 2 u0 2 1 MW ¢ 2 ≤ 1 MWg 1 z1 2 z0 2 1 Po 1 v1 2 v0 2 2 T0 1 s1 2 s0 2 2 2 S

2

V1 5 1 u1 2 u0 2 1 MW 1 MWgz1 1 Po 1 v1 2 v0 2 2 T0 1 s1 2 s0 2 2

(3.40)

because the kinetic energy, eK,0, and potential energy, eP,0, at the dead state are zero, by definition. The exergy provides the maximum useful work we can obtain from a system in a given state. It is also useful to consider two other forms of work, the useful work and the lost work. The useful work, wu, is the amount of work that we can obtain from a system that we can actually use to do something. To find the useful work, we must not include the work it “costs” us to either expand against the environment or to flow into a system. The lost work, wl, is the difference between the ideal work (i.e., the maximum useful work) and the useful work we actually obtain: wl 5 wid 2 wu In Example 3.17, we will calculate each of these types of work for a specific process.

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176 ► Chapter 3. Entropy and the Second Law of Thermodynamics

Example 3.17 Closed System Exergy Analysis

Consider a piston-cylinder assembly containing an ideal gas, initially at 20.0 bar and 1000 K. The initial volume is 1.6 L. The system undergoes a reversible process in which it is expands to 1 bar. Take the environment to be at 1 bar and 20°C. Take cp 5 1 7/2 2 R. The pressure volume relationship during this process is given by: Pv1.5 5 const (a) Calculate the work obtained during this process. (b) Calculate the useful work that could be obtained by subtracting the energy needed to push back the ambient pressure. (c) Calculate the exergy and the ideal work. (d) Calculate the lost work. Solution (a) A schematic of the process is shown in Figure E3.17, where we define the initial state as state 1 and the final state as state 2. Because this is a reversible process, we can calculate work as follows: w 5 23 PEdv 5 23 Pdv For the limits on the integral, we need to know the molar volume of the initial and final state. For state 1, we apply the ideal gas law:

RT1 v1 5 5 P1

8.314 3 1000 B 2 3 106 B

J mol

J m3

R 5 0.0042 B

R

m3 R mol

We can calculate v2 from the equation in the problem statement: 1.5 Pv1.5 5 const 5 P1v1.5 1 5 P2v2

P0 = 1 bar T0 = 293 K

Process

Ideal gas

T1 = 1000 K P1 = 20 bar

P2 = 1 bar

State 1

State 2

Figure E3.17 Initial and final states of of the expansion process.

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3.10 Exergy (Availability) Analysis ◄ 177

Solving for v2 gives:

P1v1.5 1 v2 5 ¢ ≤ P2

1/1.5

5 1 20 3 0.00421.5 2

1/1.5

5 0.031 B

m3 R kg

Now we can solve for work: 0.031

w 5 23

0.0042

0.031

0.031

kJ P1v1.5 1 1 1 dv 5 2P1v1.5 B 0.5 2 0.5 R 5 210.6 B R 1 1.5 v2 v1 0.0042 mol 0.0042 v

Pdv^ 5 23

The sign of work is negative because we are transferring energy from the system to the surroundings to perform work. (b) To calculate the useful work, we need to find the molar volume, v0, at the dead state:

RT0 v0 5 5 P0

8.314 3 293 B 1 3 105 B

J mol

J m3

R 5 0.024 B

R

m3 R mol

As in Equation (3.38), the work needed to push back the envirounment is given by:

Po 1 n1 2 n0 2 5 2.0 B

kJ mol

R

So the amount of useful work is:

wu 5 w 1 Po 1 v1 2 v0 2 5 210.6 1 2.0 5 28.6 B

kJ mol

R

(c) To calculate exergy of state 1, we use Equation (3.39):

b1 5 1 u1 2 u0 2 1 Po 1 v1 2 v0 2 2 T0 1 s1 2 s0 2 5 cv 1 T1 2 T0 2 1 Po 1 v1 2 v0 2 2 T0 Bcv ln ¢

T1 v1 ≤ 1 R ln ¢ ≤ R v0 T0

1000 0.0042 5 b1 5 R 1 1000 2 293 2 1 1 3 105 1 0.0042 2 0.024 2 2 293R B2.5 ln ¢ ≤ 1 ln ¢ ≤R 2 293 0.024 5 9.5 B

kJ mol

R

Since the ideal work is the negative value of exergy, we get: wid 5 29.5B

kJ mol

R

(d) If we compare answers to part (c) and part (b), we see that the lost work, wl, is given by:

wl 5 wid 2 wu 5 29.5 1 8.6 5 20.9B

kJ mol

R

The magnitude of the lost work is about 10% of the useful work; results like this allow us to prioritize our engineering design and improvement efforts. In addition, we see that the value for exergy, which indicates the maximum useful work, calculated in part (c), is actually lower than the work calculated in part (a). Can you explain why?

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178 ► Chapter 3. Entropy and the Second Law of Thermodynamics In the preceding discussion, we learned how to calculate the maximum available work (i.e., the exergy) and lost work of a system that goes from an initial state, state 1, to the dead state, state 0. However, often we are interested a system that undergoes a process from state 1, to some final state, state 2, that is not yet at the dead state. In such a case, the magnitude of the lost work is given by the exergy difference between the states: wl 5 Db 5 b2 2 b1 Alternatively, the lost work can be represented by the increase in entropy of the universe due the irreversibilities in the process: 2 wl 5 T0Dsuniv

(3.41)

Example 3.18 will illustrate that calculation for lost work using Equation (3.41) gives us an equivalent answer to that calculated using the change in exergy between the final and initial states.

Exthalpy—Flow Exergy in Open Systems As we learned in Chapter 2, for open systems, we must account for the mass and the flow work of streams flowing into and out of the system. The flow work required to push a stream 1 into the system is P1v1. However, the environment provides a part of this work from the ambient pressure, P0v1. Hence, it does not “cost” us anything. Therefore, we add the difference in these two terms, 1 P1 2 P0 2 v1, to the exergy to account for flow work. This analysis leads to the definition of exthalpy of state 1 (also called flow exergy), defined as follows: bf,1 5 b1 1 1 P1 2 P0 2 n1 5 1 u1 2 u0 2 1 Po 1 v1 2 v0 2 2 T0 1 s1 2 s0 2 1 1 P1 2 P0 2 v1 Gathering like terms and using the definition for enthalpy, we get: bf,1 5 1 u1 1 P1v1 2 2 1 u0 1 P0v0 2 2 T0 1 s1 2 s0 2 5 1 h1 2 h0 2 2 T0 1 s1 2 s0 2

(3.42)

Like enthalpy, we use exthalpy, bf, for any stream flowing into or out of a system.

Example 3.18 Calculation of Internal Exergy Loss in a Heat Exchanger

c03.indd 178

A shell and tube heat exchanger is designed to warm air from the environment by condensing steam that is passed through the tubes on the other side. The maximum air flow is 30 kg/min, and the air is inlet at the temperature and pressure of the environment, 285 K and 1 bar, respectively. The steam flows through the system at a pressure of 10 bar and a flow rate of 3 kg/min. The quality of the steam at the inlet is 0.9, and at the exit is 0.2. (a) Calculate the lost work during this process using the “Exergy Method” by calculating the difference in exthalpy between the initial and final states. (b) Calculate the lost work during this process using the entropy change of the universe, i.e., Equation (3.41).

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3.10 Exergy (Availability) Analysis ◄ 179

Solution ENERGY CONSERVATION A schematic of the process is shown in Figure E3.18.

kg min T1 = 285 K P1 = 1 bar air mair = 30

T2 = ? P2 = 1 bar

water x4 = 0.2 P4 = 10 bar

x3 = 0.9 P3 = 10 bar

Heat Exchanger

KG mwater = 3 min

Figure E3.18 Heat exchanger in Example 3.18 with inlet and outlet stream properties labeled. First we need to find the outlet temperature of the air. To do this we perform an energy balance: # # 0 5 nair 1 h2 2 h1 2 1 mwater 1 h^ 4 2 h^ 3 2

(E3.18A)

kg R 30 B # min m mol air n# air 5 5 1034 B R 5 kg MWair min R 0.029 B mol

(E3.18B)

Where,

The enthalpy change for air is given by: T2

T2

h2 2 h1 5 3 cP,air dT 5 R 3 3 A 1 BT 1 DT22 4 dT T1

T1

5 RBA 1 T2 2 T1 2 1

1 B 2 1 1 T 2 T 21 2 2 D ¢ 2 ≤ R 2 2 T2 T1

(E3.18C)

Where the values for air can be found in Appendix A.2: A 5 3.355; B 5 0.575 3 1023;

and D 5 20.016 3 105

For water we look up the values of enthalpy for saturated steam at 10 bar (1 MPa) in the steam tables (Appendix B.2): kJ kJ h^ v 5 2778.1 B R and h^ l 5 762.79 B R kg kg (Continued)

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180 ► Chapter 3. Entropy and the Second Law of Thermodynamics

Using the definition of quality: kJ h^ 3 5 xh^ v 1 1 1 2 x 2 h^ l 5 0.9 3 2778.1 1 0.1 3 762.79 5 2576.6 B R kg Similarly, kJ h^ 4 5 xh^ v 1 1 1 2 x 2 h^ l 5 0.2 3 2778.1 1 0.8 3 762.79 5 1165.9 B R kg Substitution of these values for specific enthalpy gives: kJ # mwater 1 h^ 4 2 h^ 3 2 5 242,300 B R min

(E3.18D)

Substitution of Equations (E3.18B), (E3.18C), and (E3.18D) into Equation (E3.18A) and solving for T2 implicitly gives: T2 5 423.8 K (a) Calculation using the exergy method: First we find the change in exthalpy in the air stream. Here we use the extensive form of exthalpy, expressed in a rate form: # # Bf,2 2 Bf,1 5 n# air 3 1 h2 2 h1 2 2 T0 1 s2 2 s1 2 4 T2

T2

cP,air

P2 A s2 2 s1 5 3 dT 2 ln 5 R 3 B 1 B 1 DT23 RdT T P1 T T1

T1

5 RBA ln

423.8K J T2 1 1 1 1 B 1 T2 2 T1 2 2 ¢ 2 2 2 ≤ R 5 11.7 B R T1 2 T2 T 1 285K mol K

so, J J kJ # # mol R B4090B R 2 285 3 K 4 3 11.7 B R R 5 787 B R Bf,2 2 Bf,1 5 1034 B min mol mol K min For the exthalpy change of steam, we have: b^ f,4 2 b^ f,3 ; 1 h^ 4 2 h^ 3 2 2 T0 1 s^ 4 2 s^ 3 2 From the steam tables: s^ v 5 6.5864 B

kJ kg K

R and s^ l 5 2.1386 B

kJ kg K

R

So, s^ 3 5 xs^ v 1 1 1 2 x 2 s^ l 5 0.9 3 6.5864 1 0.1 3 2.1386 5 6.142B

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kJ kg K

R

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3.10 Exergy (Availability) Analysis ◄ 181

and, s^ 4 5 xs^ v 1 1 1 2 x 2 s^ l 5 0.2 3 6.5864 1 0.8 3 2.1386 5 3.028 B

kJ kg K

R

So, kJ b^ f,4 2 b^ f,3 5 1 1165.9 2 2576.6 2 2 T0 1 3.028 2 6.142 2 5 2523.4 B R kg And, kJ # # Bf,4 2 Bf,3 5 m# water 1 b^ 4 2 b^ 3 2 5 21570 B R min Finally, for the entire heat exchanger, we sum the total exthalpy change: kJ # # # # # # R Wl 5 DBf 5 1 Bf,2 2 Bf,1 2 1 1 Bf,4 2 Bf,3 2 5 2783 B min Thus, we lose 783 kJ of useful work for every minute the heat exchanger operates! This value represents the internal exergy loss in this process. Said another way, 783 kJ of work is lost because heat is transferred nonreversibly due to the temperature difference. If we could place a Carnot engine between the steam and air, we could generate 783 kJ/min of power. (b) Calculation using the change in entropy of the universe: Since there is no heat transfer to the surroundings, we can calculate the change in the entropy of the universe by the extensive entropy differences in each stream during the process: # DSuniv 5 n# air 1 s2 2 s1 2 1 m# water 1 s^ 4 2 s^ 3 2 Inserting values from above, we get: J kg J # mol R 3 11.7B R 1 3B R 3 1 3.028 2 6.142 2 B R DSuniv 5 1034B min mol K min mol K 5 2749B

J min K

R

We can then calculate the lost work based on Equation (3.41): # # Wl 5 2T0DSuniv 5 2285 3 K 4 3 2749B

J min K

R 5 2783B

kJ min

R

This value is identical to that calculated in part A.

Example 3.18 showed the mechanics of calculating internal exergy loss. This approach is particularly useful in evaluating complex processes, such as chemical or biological plants. Exergy analysis allows us to evaluate each unit process and allows us to identify the opportunities to reduce the largest inefficiencies. Thus, it is a useful tool to determine ways to become more energy efficient and apply toward sustainable engineering. We illustrate with a simple case next.

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182 ► Chapter 3. Entropy and the Second Law of Thermodynamics Exergy lost in flue gas

Useful work

Exergy lost in cooling water

Exergy lost in exhaust

Exergy In

Exergy lost internally

Boiler

Turbine

Condenser

Compressor

Figure 3.12 Graphical representation of exergy analysis for a vapor-compression power cycle.

Figure 3.12 shows a graphical representation of the results for an exergy analysis of a vapor compression power cycle like the one shown in Figure 3.7. The vertical scale represents how the exergy is divided among exergy lost (e.g., lost work) and useful work among the processes. An engineer can use such a diagram to compare the lost work for each of the unit processes (e.g., boiler, turbine, condenser, or compressor) of the power cycle and the useful work that is obtained in the turbine. Such a diagram is drawn to scale as a result of the thermodynamic analysis (not shown), and the width of the lines represents either the exergy lost (white) or the useful work (shaded). Inspection of Figure 3.12 shows that design improvement efforts should focus on the boiler and the turbine. Can you think of possible design modifications? This type of diagram becomes quite useful as the design becomes more complex and the number of unit processes increases. It allows recognition of the critical units needed for efficiency improvement. As the number of processes increases even further, the data can be reported in tabular format for convenience.

►3.11 MOLECULAR VIEW OF ENTROPY When we discussed internal energy, u, in the context of the first law, we gained insight through understanding it on a molecular level. To that end, we discussed the molecular components of kinetic and potential energy that atoms and molecules possess. In analogy, we may ask, what is the molecular basis of entropy, s? The molecular view of entropy relates, in the most general sense, to molecular probability. The more different molecular configurations a state exhibits, the more likely that state will exist and the greater its entropy. To examine the idea of molecular probability and configurations, consider a system with two different ideal gaseous species, which we will call atom A and atom B. Entropy is a measure of how many different ways we can configure these species. To illustrate this concept, let’s start with a system whose center is divided by a porous membrane. To simplify the problem, but still get our central idea across, we will assign roughly equal volume elements to each atom in the system. We can relax this constraint later if we wish. We will begin modestly, by considering there to be four atoms (two A atoms and two B atoms) in the system. We must realize, however, that we need to extend this concept to roughly 1023 atoms to understand the macroscopic systems that we encounter in our physical world. Initially, we will say that the system contains pure gas A on the left of the membrane and pure gas B on the right of the membrane, as shown in Figure 3.13a. In this case, we

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3.11 Molecular View of Entropy ◄ 183

know exactly where all the A atoms are located (and where all the B atoms are located). Next we allow the atoms to randomly distribute. Figure 3.13b illustrates that six possible distinguishable molecular configurations result. It seems logical to assume that all these configurations are equally likely;8 hence, there is a 1/6 chance the A atoms will be found in their initial state with two atoms on the left-hand side of the membrane, a 4/6 chance there will be 1 A to the left of the membrane, and a 1/6 chance there will be no A. The most probable state of the system is the one in which there is 1 A and 1 B on each side, or, in other words, when A and B are mixed! Moreover, in this mixed state, we are not certain exactly where the A atom is; we have lost some information about the system. This effect of mixing becomes even more dramatic as we increase the size of the system. We now wish to extend the size of the system toward macroscopic amounts. Figure 3.14 shows a stick diagram of the probabilities of the number of A atoms on the left-hand side of a porous membrane for system consisting of ideal gases A and B, as above. The upper-left-hand corner of this figure shows the case depicted in Figure 3.13, that is, a system consisting of 2 A atoms and 2 B atoms. The case with 1 A on the lefthand side is represented by a stick four times as great as the case of 0 A atoms or 2 A atoms, since its probability is four times higher. Let’s double the size of the system so that we have 4 A atoms and 4 B atoms. There are now five possibilities for A on the left-hand side: 0, 1, 2, 3, or 4. The probability for this system is presented in the next diagram to the right in Figure 3.14. If we consider each configuration equally likely, the number of configurations with 2 A atoms to the left-hand side represents 36 out of 70 possible configurations. Again, we find the case of complete mixing most probable. Again, in its most probable state, we become less certain about the exact positions of the A atoms. On the other hand, only one configuration in 70 leads to pure A on the left-hand side, where we know, with certainty, where all the A atoms are. Similar stick diagrams are shown for systems consisting of 8 A atoms and 8 B atoms, 16 A atoms and 16 B atoms, 32 A atoms and 32 B atoms, and 128 A atoms and 128 B atoms. As the number of species increases, the mixing becomes more pronounced; that is, the likelihood of having roughly identical numbers of A atoms and B atoms on each

Porous membrane Initial state

A

B

A

B

All possible configurations

A

B

B

A

A

A

B

B

A

B

A

B

A

B

B

B

A

A

B

A

(a) B

A

B

A (b)

Figure 3.13 System consisting of ideal gases A and B separated by a porous membrane. (a) Initial state of the system with all gas A on the left and all gas B on the right. (b) Possible configurations after the atoms are allowed to randomly redistribute. 8

This assumption forms the basis for the ergodic hypothesis upon which statistical thermodynamics is constructed.

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184 ► Chapter 3. Entropy and the Second Law of Thermodynamics System: 4 As & 4 Bs

System: 8 As & 8 Bs

Probability

System: 2 As & 2 Bs

0

1

2

0 1 2 3 4 System: 32 As & 32 Bs

Probability

System: 16 As & 16 Bs

0 2 4 6 8 System: 128 As & 128 Bs

NA = 76 0 4 8 12 16

0 8 16 24 32 Number As to the left of the membrane

0 32 64 96 128 Number As to the left of the membrane

Figure 3.14 Conditional probabilities of the number of A atoms on the left-hand side of a system consisting of ideal gases A and B separated by a porous membrane. Cases are shown for a system consisting of 2 A atoms and 2 B atoms (as depicted in Figure 3.13), 4 A atoms and 4 B atoms, 8 A atoms and 8 B atoms, 16 A atoms and 16 B atoms, 32 A atoms and 32 B atoms, and 128 A atoms and 128 B atoms. As the number of species increases, the mixing becomes more pronounced.

side of the membrane is great and the probability of finding all the A atoms on a given side becomes negligible. For example, in the case of 128 A atoms and 128 B atoms, only 1 in 5.8 3 1075 configurations leads to pure A on the left-hand side. Thus, if all molecular configurations are equally likely, finding pure A by chance is so improbable that it essentially will not happen. On the other hand, the configurations between 52 and 76 A atoms on the left-hand side of the membrane occurs 99.8% of the time. What does this argument about molecular probability say about macroscopic systems in which we have on the order of 1023 A atoms and 1023 B atoms? Regardless of the initial state, if all the molecular configurations in this system are allowed to occur with equal probability, the system will evolve to the point where, for all practical purposes, roughly the same number of A atoms and B atoms are on each side of the membrane. We can relate our understanding of molecular probability to our macroscopic observation of the second law if we say that the number of molecular configurations a system can take is proportional to its entropy.9 Thus, if we start with a macroscopic system with pure A on the left-hand side of a porous membrane and pure B on the right-hand side, it has only one possible molecular configuration, so its entropy is low. On the other hand, if it completely mixes, it has a large number of molecular configurations and its entropy is high. Thus, the spontaneous mixing of pure gases can be related on a molecular level to molecular probability and on a macroscopic level to entropy. In short, we can relate the irreversibility of macroscopic processes to mixing on a molecular level, where we become less certain of the exact molecular state of the system, because it can exist in many equivalent molecular configurations. We can relate the above discussion to the commonly expressed viewpoint that entropy relates to “disorder.” As entropy increases, we become less certain about the exact molecular state of a system; that is, there are more equivalent molecular configurations in which it can exist. Therefore, in a loose sense, we would view it as more 9

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It is actually mathematically proportional to es.

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3.11 Molecular View of Entropy ◄ 185

disordered. Thus, if we relate the degree of disorder to the number of configurations a system can have, higher entropy means more disorder. Several examples of directional processes were presented in Section 3.1. We know that the directionality of an irreversible process relates to an increase in entropy. Let’s consider a few examples of how it also corresponds to an increase in molecular configurations, that is, mixing.

Maximizing Molecular Configurations over Space Example 1: Chemical Directionality The example discussed above directly corresponds to chemical directionality with the mixing of molecules A and B. As Figure 3.14 illustrates, the more completely the two species mix, the greater the number of equivalent spatial configurations and, consequently, the greater the probability the system will be found in that state. This mixing occurs over space, as is illustrated in Figure 3.15. Example 2: Mechanical Directionality Another example of an increase in spatial configurations is given by a gas expanding from a compressed cylinder. In this case, the molecules increase in entropy because there are more configurations available in the larger volume after the gas has expanded; that is, we are less certain exactly where a particular molecule is located. This type of spreading or “mixing over space” is illustrated in Figure 3.16. In this figure, the box inset is an expanded molecular view partly within the cylinder and partly outside of it. We can see we have greater knowledge of where the species are in state 1 than in state 2; hence, state 1 has a lower value of entropy. In fact, the spatial mixing depicted in Figure 3.16 is very similar to the species mixing of Figure 3.15. If we examine the hypothetical solution path we used in Example 3.9 (Figure E3.9B), we see that the increase in entropy of a given species as ideal gases mix arises from the fact that it has more space in which to move, whether another species is there or not. Said another way, if we replace species B in Example 1 with empty space, as far as species A is concerned, the situation in Examples 1 and 2 are equivalent. Therefore, this spatial mixing is often referred to as “spreading.”

Gas A

A&B Remove membrane

Chemical directionality

Membrane

Mixing: Intermingling of molecules A and B

Gas B State 1

State 2

Figure 3.15 Example of chemical directionality as the mixing of molecules A and B over space.

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186 ► Chapter 3. Entropy and the Second Law of Thermodynamics Patm Tatm Open valve

Mechanical directionality

Highpressure gas

State 1

Spreading of molecule A: mixing over space

State 2

Figure 3.16 Example of mechanical directionality as the mixing of molecule A over space as the gas expands. The molecular projection in this illustration consists of a region within the cylinder and a region outside the cylinder.

Maximizing Molecular Configurations over Energy In determining the number of possible configurations of a set of molecules, we must consider not only where they are in space but also how their energy is distributed. This factor results from the quantized nature of energy on the molecular scale. The energy within a given system must be assigned to quantized energy levels in much the same way the mass in the system must be assigned to a specific space. The filling of energy levels is constrained by the total energy of the system. We can understand the role of how the molecules distribute over energy in analogy to the way we related the spatial configurations and probability in the discussion with Figures 3.15 and 3.16. The greater the number of equivalent configurations with which a set of molecules can distribute their energy, the greater the state’s entropy. Thus, an increase in entropy can be characterized by either “mixing” over space or “mixing” over energy. Mixing over energy decreases our knowledge of where the energy in the system is.10 Example 3: Thermal Directionality In addition to mixing spatially, irreversible processes are sometimes driven by a mixing of states at different energy levels. For example, consider the case shown in Figure 3.17 which shows an isolated system containing a hot solid, initially at T1,H, placed in contact with a cool solid, initially at T1,L . After time, the temperatures equilibrate at the intermediate temperature, T2. The thermal directionality of this process can be thought of as the “mixing” of states with different energy levels, which we will refer to as energetic mixing (and, as with spatial configurations, can just as appropriately be considered energetic spreading). The energy level diagrams at the bottom of the figure represent schematically how the energy of the electrons are distributed in the solids at the three different temperatures.11 In analogy to our discussion of spatial mixing, we can see that there are more configurations of energy states as the temperatures of the solids equilibrate, i.e., there are more ways to arrange the electrons in state 2 than state 1. Therefore, the 10

An alternative argument asserts that increases in spatial configurations are all essentially just manifestations of increases in energetic configurations, and that the molecular interpretation of entropy can be understood by only considering the latter. For more information, see F.L. Lambert, J. Chem. Ed, 84(9), 1548 (2007). 11 This representation is meant to be schematic. Actually, the energy levels in solids form bands of available states and are not discrete as is the case for individual atoms.

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3.11 Molecular View of Entropy ◄ 187 State 1

Solid at T1,H

Solid at T2

Energy

Solid at T1,L

State 2

Figure 3.17 Schematic representation of thermal directionality as the mixing of energy configurations between a cool solid initially at T1,L and a hot solid at T1,H.

probability the system will be in that state is greater than in state 1. Thus, state 2 has higher entropy and occurs later in time. Example 4: Kinetic Energy to Friction Other entropic processes can also be viewed of in terms of mixing of energy. Consider a block that is sliding across a surface. It slows down as macroscopic kinetic energy 1 EK 2 is converted into internal energy by friction. This irreversible process can be thought of as mixing of energy from a directed form to dissipation in all directions. Figure 3.18 illustrates this concept. There is only one configuration the block’s kinetic energy can be in—that of directed, forward motion. After the block slows down, the increase in temperature leads to greater lattice vibrations in which the energy can be distributed in many configurations. Hence, the conversion of kinetic energy to internal energy by friction represents an irreversible process in which entropy increases. Example 5: Chemical Reaction Let’s look at the reaction we studied when we learned about the energetics of reaction in Section 2.6. Two molecules of hydrogen gas will spontaneously react with one molecule of oxygen to give two molecules of water. The rearrangement of chemical bonds liberates 3.5 eV of energy. Suppose this reaction occurs adiabatically, so that the entropy of the surroundings does not change. The products will manifest the bond energy liberated by an increase in temperature. Since this process is spontaneous and irreversible, the second law tells us the products must have greater entropy than the reactants. We can understand this in terms of energy configurations. Recall that there are two types of molecular energy: potential energy, such as that stored within a chemical bond, and kinetic energy, which includes translational, vibrational, and rotational motion. In this discussion we will not consider vibrational and EK friction

Mixing of energy (by direction, from directed to random)

Figure 3.18 Example of energetically directional process being randomized as mixing of energy. Here the block’s kinetic energy is being dissipated into friction.

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188 ► Chapter 3. Entropy and the Second Law of Thermodynamics rotational motion; however, even with their inclusion, the essence of the argument does not change. Before reaction, the 3.5 eV of energy stored in the reactants can only exist in one configuration—that of bond energy between atoms in the H2 and O2 molecules. In other words, each bond has a specified and known strength. Once they react, the two molecules of water must share this exact amount of energy; however, they do not need to share it equally. In fact, there are many combinations of increased velocity these two molecules can have to sum up to the total 3.5 eV of energy. If we pick one of the water molecules, we are not sure of its exact speed. We have lost information. There are more energetic configurations and higher entropy. If we consider four molecules of H2 and two molecules of O2, the situation becomes even more drastic. Again, we can specify completely the energy configuration of the reactants. However, now the products are free to distribute 7.0 eV between four molecules. There are many configurations of molecular kinetic energy that can accomplish the feat. If we extend this concept to macroscopic sizes, we have on the order of 105 J of energy to distribute among 1 mole of water produced. The 1023 molecules will have many configurations of velocity (molecular kinetic energy) to distribute the energy of reaction. This state is much more probable than that of the reactants which distribute this energy in a specific way, that is, in the form of bond energy—so the entropy of reaction is positive and the reaction proceeds spontaneously. The above discussion focuses on the energy component of chemical reaction. However, there is also a spatial component, which counteracts the energy component. If the system completely reacts, it contains pure water. On the other hand, if some reactant remains, we have a mixture of three species. Those species have many more spatial configurations than pure water. Hence, as the reaction approaches completion, there is a trade-off between the entropy it gains by an increase in the energy configurations with an increase in temperature and the spatial configurations it loses due to “unmixing” in going to pure water.12 In general, the trade-off between these two components will determine how far a reaction will proceed. We will learn to quantify the extent to which species chemically react in Chapter 9. In summary, as a system evolves to states with more possible configurations (spatially or energetically) of its molecular states (more probable), its entropy increases.

EXAMPLE 3.19 Refrigeration by Adiabatic Demagnetization

Magnetic refrigeration cycles can be used to achieve supercold temperatures. They typically operate between a “hot” reservoir at liquid helium temperature (4.4 K) and a cold reservoir at very low temperature (as low as 0.0065 K). One configuration consists of a paramagnetic working material, such as gadolinium gallium garnet (GGG) or ferric ammonium alum (FAU), in the form of a rim of a wheel. The wheel is rotated between the high-temperature (4.4 K!) reservoir and the low-temperature reservoir. In the high-temperature reservoir, the working material expels heat as it is subjected to a large magnetic field (7 tesla). As the working material is rotated into the low-temperature reservoir, the field slowly becomes smaller and eventually zero. This demagnetization process may be assumed to be adiabatic and reversible. As the working material demagnetizes, it cools off. When the working material is then exposed to a low-temperature reservoir, it absorbs heat. From a molecular basis, explain how adiabatic demagnetization cools the working material. SOLUTION A paramagnetic material consists of many unpaired electrons, each of which has a definite spin state (spin-up or spin-down). In the absence of a magnetic field, the energies of

12

For every three molecules that react, only two molecules form; this effect also contributes to the decrease in the spatial component of entropy.

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3.11 Molecular View of Entropy ◄ 189

Magnetic Field

No Magnetic Field

Energy

State 1

State 2

Figure E3.19 Electronic spin states of a paramagnetic material in the presence of a magnetic field (state 1) and with no magnetic field present (state 2).

each of the electron states are identical. Thus, an equal number of electrons will occupy each state. When a magnetic field is applied to the material, the energies of the spin-up and spindown states are no longer identical. Consequently, more electrons occupy the lower energy state then the upper energy state. A schematic of the spin states with a magnetic field (state 1) and without a magnetic field (state 2) is illustrated in Figure E3.19. The entropy of this system can be thought of as having two components, that due to magnetization, smagnetization and that due to temperature, stemperature.We first consider smagnetization. Inspection of Figure E3.19 shows that there are fewer configurations possible (less disorder) in state 1, with the magnetic field applied, than in state 2. Thus, the entropy due to magnetization increases during the demagnetization process as the working material goes from state 1 to 2: Dsmagnetization . 0 However, since the overall process is reversible and adiabatic, the entropy change is zero: Ds 5 0 5 Dsmagnetization 1 Dstemperature Thus, Dstemperature , 0 The entropy of the temperature component is lowered as the temperature decreases, that is, T2 , T1 It is interesting to compare the two states in an adiabatic demagnetization cycle (Figure E3.19) with those in the conventional vapor-compression refrigeration cycle described in Section 3.9. Both cycles rely on the working substance absorbing heat from a cold reservoir as the working substance goes from a more ordered state to a less ordered state. In this example, the liquid-to-vapor transition of conventional refrigeration is replaced with the demagnetization of the paramagnetic material. Similarly, in both cycles, heat is rejected to a hot reservoir as the working substance transits to a more ordered state (liquid or magnetic). This example illustrates one approach to creatively developing engineering processes—applying the same fundamental approach in a new way that is better suited for the given application. Clearly, the super-low temperatures obtained by adiabatic demagnetization refrigeration systems could not be obtained by a conventional liquid–vapor transition, since vapors do not exist at these temperatures. However, in developing this process, the same fundamental type of process (i.e., ordered-to-disordered transition) is exploited. It is just accomplished in a way that is appropriate for the application, where the ordered-to-disordered transition occurs within a solid. Many clever engineering processes have been created by applying analogous fundamental mechanisms in this manner.

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190 ► Chapter 3. Entropy and the Second Law of Thermodynamics

►3.12 SUMMARY The second law of thermodynamics states that the total entropy of the universe increases or, at best, remains the same. It never decreases. The entropy of the universe remains unchanged for a reversible process, while it increases for an irreversible process. Entropy balances have been developed for closed systems and for open systems. In each case, we account for the entropy change of the universe by adding together the entropy change for the system and the entropy change for the surroundings. For example, the integral equation of the second law for a closed system, written in extensive form, is:

DSuniv 5 n 1 sfinal 2 sinitial 2 2

Q Tsurr

$0

(3.13)

In Equation (3.13), the surroundings are at constant temperature, Tsurr. For open systems, it is convenient to write the second law on a rate bases. The integral equation, in extensive form, is: # Q dS 1 a n# outsout 2 a n# insin 2 $0 dt Tsurr out in

(3.21)

The first term in Equation (3.21) describes the rate of entropy change of the system, while the next three terms describe the entropy change of the surroundings due to mass flow out of the system, mass flow into the system, and heat transfer. The first and second laws can be combined for flow processes that are at steady-state and reversible, with one stream in and one stream out, to give the Bernoulli equation: 2 # S2 S2 Ws V2 2 V1 ≤ 1 MWg 1 z2 2 z1 2 # 5 3 vdP 1 MW ¢ n 2

(3.30)

1

We have also used these equations in intensive forms, on a mass and a molar basis, and for differential increments. We applied the second law to many engineering systems. Given a physical problem, we must determine which terms in these equations are important and which terms are negligible or zero. Examples of closed systems included the rigid tank and adiabatic or isothermal expansion/ compression in a piston–cylinder assembly. Steady-state open systems include nozzles, diffusers, turbines, pumps, heat exchangers, and throttling devices. Transient open-system problems can entail filling or emptying of a tank. Finally, vapor-compression power and refrigeration cycles provided useful examples of thermodynamic cycles. In this case, the performance of the cycles is quantified by the efficiency and the coefficient of performance, respectively. You should understand the concepts well enough that you are not restricted to the systems discussed above but rather can apply the second law to any system that interests you. When we perform calculations on a reversible process, the second law provides an additional constraint that allows us to determine an unknown state or to calculate quantities such as heat or work. We can use these values to estimate the corresponding values of real processes. The isentropic efficiency compares the actual performance of a unit process with the performance it would obtain if it operated reversibly.

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3.13 Problems ◄ 191 The change in entropy between state 1 and state 2 is defined as: 2

dqrev Ds 5 3 5 T

(3.2)

1

Since entropy is defined in terms of heat absorbed during a reversible process, we can calculate the entropy between any two states by constructing a path that follows a reversible process from state 1 to state 2. In this way, we found that the entropy change of an ideal gas is given by: T2

cP P2 Ds 5 3 dT 2 R ln ¢ ≤ T P1

(3.22)

T1

Equation (3.22) is true, in general, for the entropy change associated with an ideal gas between state 1 and state 2; it is not limited only to the reversible process that we chose to develop it. For an ideal gas, the entropy change between 1 T1, v1 2 and 1 T2, v2 2 is analogously given by Equation (3.24). Alternatively, the entropy change between two states can be directly obtained from property tables, if they are available. Exergy can be used for a methodological analyses of complex processes. Exergy is related to the ideal work, the most work we can get out of a process that brings the system from its current state to the dead state. The dead state has the same properties as the environment, immediately adjacent in the surroundings. (Alternatively, we can conceive ideal work as least work we have to put in to go from the dead state to the state of interest.) Similarly exthalpy, or flow exergy, accounts for the flow work of inlet and outlet streams as well. Exergy analysis allows us to quantify the amount of useful work we are losing in the unit processes that comprise a system. We can then prioritize where to focus design and process improvement. The molecular view of entropy relates, in the most general sense, to molecular probability. The more different molecular configurations a state exhibits, the more likely that state will exist and the greater its entropy. In determining the number of possible configurations of a set of molecules, we must consider not only where they are in space but also how their energy is distributed. This factor results from the quantized nature of energy on the molecular scale. The greater the number of equivalent configurations with which a set of molecules can distribute in space or spread their energy, the greater the state’s entropy. Thus, an increase in entropy can be characterized by either “mixing” over space or “mixing” over energy. These molecular-based concepts have propogated to many other fields. Information theory mathematically defines information “entropy” for bits of information using the identical formula that Boltzmann applied to molecular configurations. Similarly, “entropy”-based arguments have expanded into such diverse fields as economics, theology, sociology, art, and philosophy.

►3.13 PROBLEMS Conceptual Problems

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3.1 A set of mixing processes is shown in the following figures. The volumes are represented by the size of the boxes. For each process, determine whether Ds is greater than zero, less than zero, or equal to zero. Explain your answer. You may assume O2 and N2 behave as ideal gases. Reflect on your results in parts (a) through (d), and comment on the common statement that entropy represents the “degree of disorder” in a system.

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192 ► Chapter 3. Entropy and the Second Law of Thermodynamics T

N2

T

N2

N2

N2

O2

O2

N2

N2

O2

O2

N2

T

O2

N2 Mixing process

O2 O2

O2

N2

O2 O2

N2

O2

N2

O2

N2

N2 O2

O2 N2 O2

O2

(a ) T

N2

T

N2

N2

N2

N2

N2

N2

N2

N2

N2

N2

T

N2

N2 Mixing process

N2 N2

N2

N2

N2 N2

N2

N2

N2

N2

N2

N2 N2 N2

N2 N2 N2

(b ) T

N2

T

N2

N2

N2

O2

O2

N2

N2

O2

O2

N2

T

O2

N2 Mixing process

O2 O2

O2

O2

O2 N2

O2

O2 N2

O2

O2

N2

N2 O2

N2 O2

N2

(c ) T

T T

N2

N2

O2

O2 O2

N2

O2

N2 N2 N2 N2

N2

O2 Mixing process

O2

N2

O2

O2

N2

O2

O2 O2

O2

O2 O2

N2 O2 N2

N2 N2 O2

O2

(d )

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3.13 Problems ◄ 193 3.2 State the conditions under which the following equations apply. Be as specific as you can with the limitations. (a) q 5 3 Tds (b) h2 2 h1 5 cp 1 T2 2 T1 2 (c) Dsuniv . 0 Qsurr (d) DSsurr 5 Tsurr QH QC (e) DSuniv 5 2 TH TC 2

2

(f) 0 5 3 vdP 1 ¢

V 22 2 V 1 ≤ 1 g 1 z2 2 z1 2 2

1

3.3 Consider two ideal gases, A and B, that are initially separated. For each of the following processes, the gases are allowed to mix. Determine if the entropy change of the system is positive, negative, or zero. Explain. (a) The process occurs isothermally, at constant P. (b) The process occurs adiabatically, at constant P. (c) The process occurs isothermally, at constant V. (d) The process occurs adiabatically, at constant V. 3.4 Consider an ideal gas. For each of the following processes, determine if the entropy change of the system is positive, negative, zero, or you cannot tell. Explain. (a) Isothermal compression (b) Adiabatic compression (c) Isobaric heating (d) Isochoric heating 3.5 Consider the following processes and determine if the entropy change of the system is positive, negative, zero, or you cannot tell. Explain. (a) Liquid water is frozen to form ice. (b) 1 mol of oxygen and 2 mol of hydrogen react isothermally and completely to form 1 mol of water vapor. (c) 1 mol of oxygen and 2 mol of hydrogen react adiabatically and completely to form 1 mol of water vapor. 3.6 Consider an ideal gas that undergoes two alternative processes from state 1 to state 2. The path for the first process (Path 1) is reversible, and the path of the second process (Path 2) is irreversible. Answer the following questions. (a) If we compare the entropy change of the system, we can say that the entropy change for Path 1 is (greater than, equal to, less than) the entropy change for Path 2. Explain. (b) If we compare the entropy change of the surroundings, we can say that the entropy change for Path 1 is (greater than, equal to, less than) the entropy change for Path 2. Explain.

Cable

Cable

3.7 A black box is bolted to a beam with a cable as shown in the following figure. Consider the box as the system. During the process, the box raises in height as the cable goes into the box by an unknown mechanism. Assume that there is some friction associated with this process.

During the process the cable goes into the black box

Black Box

Black Box

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194 ► Chapter 3. Entropy and the Second Law of Thermodynamics (a) During the process, is the change in total energy of the box: greater than zero, equal to zero, less than zero, or cannot tell? Explain. (b) During the process, is the change in entropy of the universe: greater than zero, equal to zero, less than zero, or cannot tell? Explain. (Thanks to Prof. Octave Lvenspiel for providing the idea for this problem.) 3.8 In Problem 2.12, you explained why a thick rubber band heats when stretched. Offer an alternative explanation in the context of entropy. You may consider this process isentropic. 3.9 Qualitatively sketch the Ts diagram corresponding to the real Rankine cycle given in Example 3.15. 3.10 Ts diagrams for two reversible thermodynamic power cycles are shown in the following figure. Both cycles operate between a high temperature reservoir at 500 K and a low temperature reservoir at 300 K. The process on the left is the Carnot cycle described in Section 2.9. The process on the right is a Stirling cycle, which is similar to a Carnot cycle, except that the two steps (state 4 to state 1) and (state 2 to state 3) are at constant volume. Which cycle, if either, has a greater efficiency? Explain. Carnot Cycle

T [K] 1

500

Stirling Cycle

T [K] 2

1

500

2

v = const v = const 300

4

300

3

4

3

s

s

3.11 Ts diagrams for two reversible thermodynamic power cycles are shown in the following figure. Both cycles operate between a high temperature reservoir at 500 K and a low temperature reservoir at 300 K. The process on the left is the Carnot cycle described in Section 2.9. The process on the right is a Brayton cycle, which is similar to a Carnot cycle, except that the two steps (state 1 to state 2) and (state 3 to state 4) are at constant pressure. Which cycle, if either, has a greater efficiency? Explain. Carnot Cycle

T [K] 1

500

Stirling Cycle 2 p = const

T [K] 2

500

1

3 300

4

300

3

4 s

p = const s

3.12 CdTe forms a II–VI compound semiconductor. This solid forms a well-ordered single crystal where Cd atoms and Te atoms sit in distinct sites adjacent to each other in a crystal lattice. Estimate the entropy of mixing from an initial state of 1 mole of pure solid Cd and 1 mole of pure solid Te to a final state of 1 mole of CdTe at constant temperature and pressure.

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3.13 Problems ◄ 195 3.13 In the 1959 Rede Lecture, The Two Cultures, C. P. Snow asserts: A good many times I have been present at gatherings of people who, by the standards of traditional culture, are thought highly educated and who have with considerable gusto been expressing their incredulity at the illiteracy of scientists. Once or twice I have been provoked and have asked the company how many of them could describe the Second Law of Thermodynamics. The response was cold: it was also negative. Yet I was asking something which is about the equivalent of: Have you read a work of Shakespeare’s? Do you think that this analogy (between the second law and Shakespeare) is appropriate? Write a paragraph defending your position. 3.14 In his book The Trouble Waters, Henry Morris proclaims: The Law of Increasing Entropy is an impenetrable barrier which no evolutionary mechanism yet suggested has ever been able to overcome. Evolution and entropy are opposing and mutually exclusive concepts. If the entropy principle is really a universal law, then evolution must be impossible. Is this argument scientifically sound? Explain. 3.15 “Four of a kind” is one of the best hands you can have in poker. Can you relate this statement to the concept of entropy? Would you say this hand has a high value of s? 3.16 The concept of entropy was developed in the nineteenth century, in order to study the efficiency of the steam engine, largely through the work of Sadi Carnot, Rudolph Clausius, and Lord Kelvin. However, it has had major implications well beyond the realm of engineering, including impacting the thought, philosophy, and theology of nineteenth-century Europe. Go to the library or the Web and find a nonengineering topic in which entropy plays a major role. Describe it in about a hundred words and cite your source(s).

Numerical Problems 3.17 Develop a general expression for Dssys for an ideal gas that goes from 1 v1, T1 2 to 1 v2, T2 2 based on the path below.

v1

Step 1 Step 2

Volume (m3/mol)

Δshypothetical State 1 (T1 ,v1)

Δsreal v2

Real path State 2 (T2 ,v2)

T1

Temperature (K)

T2

3.18 Develop a general expression for Dssys for an ideal gas that goes from 1 P1, T1 2 to 1 P2, T2 2 where heat capacity is given by: cP 5 A 1 BT 1 CT2

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196 ► Chapter 3. Entropy and the Second Law of Thermodynamics 3.19 A rigid vessel contains 10 kg of steam. The steam is initially at 10 bar and 300°C. After a period of time, the pressure in the vessel is reduced to 1 bar due to heat transfer with the surroundings. The surroundings are at a constant temperature of 20°C. Determine the change in entropy of the system, the surroundings, and the universe during this process. 3.20 A 10-kg block of copper is initially at 100°C. It is thrown in a very large lake that is at 280 K. What is the entropy change of the copper? What is the entropy change of the universe? 3.21 Calculate the change in entropy for the system for each of the following cases. Explain the sign that you obtain by a physical argument. (a) A gas undergoes a reversible, adiabatic expansion from an initial state at 500 K, 1 MPa, and 8.314 L to a final volume of 16.628 L. (b) One mole of methane vapor is condensed at its boiling point, 111 K; Dhvap 5 8.2 3 kJ/mol 4 (c) One mole of liquid water is cooled from 100°C to 0°C. Take the average heat capacity of water to be 4.2 JK21g21. (d) Two blocks of the same metal with equal mass are at different temperatures, 200°C and 100°C. These blocks are brought together and allowed to come to the same temperature. Assume that these blocks are isolated from their surroundings. The average heat capacity of the metal is 24 J K21 mol21. 3.22 Calculate the change in entropy of the universe for the process described in Problem 2.28. Repeat for Problem 2.29. 3.23 Determine the change in entropy of an ideal gas with constant heat capacity, cP 5 1 7/2 2 R, between the following states: (a) P1 5 1 bar, T1 5 300 K; P2 5 0.5 bar, T2 5 500 K (b) v1 5 0.05 m3 /mol, T1 5 300 K; v2 5 0.025 m3 /mol, T2 5 500 K (c) P1 5 1 bar, T1 5 300 K; v2 5 0.025 m3 /mol, T2 5 500 K 3.24 Compare the change in entropy (a) when water is heated from its freezing point to its boiling point at 1 atm and (b) when saturated liquid water is vaporized at 1 atm. 3.25 You have just cooled a glass of tap water at 20°C by adding ice, at 210°C. The glass originally contains 400 mL of tap water, to which 100 g of ice is added. Assume that the glass is adiabatic. Calculate the change in entropy of the universe after thermal equilibrium has been obtained. For ice, take Dhfus 5 26.0 3 kJ/mol 4 . 3.26 Consider a piston–cylinder assembly that initially contains 0.5 kg of steam at 400°C and 100 bar. For the isothermal expansion of the steam in this system to a final pressure of 1 bar, determine the following: (a) What is the maximum possible work (in [kJ]) that can be obtained during this process and the entropy change of the surroundings (in [kJ/K])? (b) Repeat part (a) using the ideal gas model for steam. Compare your answers. 3.27 Consider the piston–cylinder assembly shown at the top of page. It is well insulated and initially contains two 5000-kg blocks at rest on the 0.05-m2 piston. The initial temperature is 500 K. The ambient pressure is 5 bar. One mol of an ideal gas is contained in the cylinder. This gas is compressed in a process in which another 5000-kg block is added. The heat capacity of the gas at constant volume can be taken to have a constant value of (5/2) R, where R is the gas constant. (a) What are the initial and final pressures of the gas in the system? (b) Do you expect the temperature to rise or fall? Explain. (c) What is the final temperature? (This is not necessarily a polytropic process!) (d) Calculate Dssys and Dssurr. [Hint: You may want to refer to the Carnot cycle to get an idea of a possible set of reversible processes to pick.] (e) Does this process violate the second law of thermodynamics? Explain.

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3.13 Problems ◄ 197 m = 5000 Process consists of adding third kg 5000 kg block to compress piston m = 5000 kg m = 5000 kg

Psurr = 5 bar

A = 0.05 m

2

Well insulated

1 mol of pure, ideal gas T initial = 500 K 5 cv = R 2

3.28 Problem 3.27 consists of an irreversible process in which an ideal gas with constant heat capacity was compressed in a piston–cylinder assembly. As part of this problem, you were asked to calculate Dssys for this process. Entropy change is defined for a reversible process as: final

Ds 5 3

dqrev T

initial

Since entropy is a property, the change in entropy depends only on the final and initial states of the system, not on the path the system went through for a particular process. Therefore, we can pick any reversible path we want, as long as it takes us from the initial state to the final state. Calculate Dssys for the process depicted in Problem 3.27, using each of the following paths: (a) a reversible, adiabatic compression, followed by a reversible, isothermal expansion (two of the four steps in the Carnot cycle) (b) a reversible, isobaric heating followed by a reversible, isothermal compression (c) a reversible, isochoric (constant-volume) heating followed by a reversible, isothermal compression 3.29 Consider a well-insulated piston–cylinder assembly. O2, initially at 250 K and 1 bar, undergoes a reversible compression to 12.06 bar. You may assume oxygen is an ideal gas. Answer the following questions: (a) What is the entropy change for this process? (b) What is the final temperature of the oxygen? (c) What is the value of work for this process? (d) If the oxygen in this system had undergone an irreversible compression to 12.06 bar, would the final temperature be higher than or lower than that calculated in part (b)? Explain. 3.30 The insulated vessel shown below has two compartments separated by a membrane. On one side is 1 kg of steam at 400°C and 200 bar. The other side is evacuated. The membrane ruptures, filling the entire volume. The final pressure is 100 bar. Determine the entropy change for this process.

H2O T1 = 400°C P1 = 200 bar

Vacuum

Insulation

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198 ► Chapter 3. Entropy and the Second Law of Thermodynamics 3.31 A partition divides a rigid, well-insulated 1-m3 tank into two equal parts. The left side contains an ideal gas 3 cP 5 1 5/2 2 R 4 at 10 bar and 300 K. The right side contains nothing; it is a vacuum. A small hole forms in the partition, gas slowly leaks out from the left side, and eventually the temperature in the tank equalizes. What is the entropy change? 3.32 An insulated tank is divided by a thin partition. (a) On the left is 0.79 mole of N2 at 1 bar and 298 K; on the right is 0.21 mole of O2 at 1 bar and 298 K. The partition ruptures. What is DSuniv for the process? (b) On the left is 0.79 mole of N2 at 2 bar and 298 K; on the right is 0.21 mole of O2 at 1 bar and 298 K. The partition ruptures. What is DSuniv for the process? [Hint: Consider the entropy change of each gas separately and add them together. To do this, you will need to use the concept of partial pressure.] 3.33 Consider the well-insulated, rigid container shown in the following figure. Two compartments, A and B, contain H2O and are separated by a thin metallic piston. Side A is 50 cm long. Side B is 10 cm long. The cross-sectional area is 0.1 m2. Latch Well-insulated wall

A

B

P1,A = 10 bar T1,A = 700°C H2O

P1,B = 20 bar T1,B = 250°C H2O

50 cm

10 cm Thin metallic piston

The left compartment is initially at 10 bar and 700°C; the right compartment is initially at 20 bar and 250°C. The piston is initially held in place by a latch. The latch is removed, and the piston moves until the pressure and temperature in the two compartments become equal. Determine the entropy change of the universe for this process. 3.34 Consider the well-insulated container shown below. Two gases, gas A and gas B, are separated by a metallic piston. The piston is initially held in place by a latch 10 cm from the left of the container. Latch Well-insulated wall

Gas A P1,A = 10 bar T1,A = 500°C cv /R = 3/2

10 cm

Thin metallic piston Gas B P1,B = 1bar T1,B = 100°C cv /R = (5/2 + 1.5 × 10–3T )

20 cm

Gas A, which is located in the left compartment, is initially at 10 bar and 500°C. The heat capacity of gas A is constant: 1 cv,A/R 2 5 3/2. Gas B is located in the right compartment and is initially at 1 bar and 100°C. The heat capacity of gas B is given by 1 cv,B/R 2 5 5/2 2 1.5 3 1023 T where T is in Kelvin. You may use the ideal gas model for both gases. (a) The latch is removed and the piston moves until the pressure and temperature in the two compartments become equal. What are the final pressure and temperature? State any assumptions that you make. (b) Calculate the entropy change of the universe. Is this process possible?

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3.13 Problems ◄ 199 3.35 Steam at 8 MPa and 500°C flows through a throttling device, where it exits at 100 kPa. Determine the entropy change for this process. 3.36 A fast-talking salesperson comes to your doorstep and says she is down on her luck and is willing to sell you the patent rights to her most glorious invention. She brings out a mysterious black box and says it can take an inlet stream of ideal gas at 2 kg/s and 4 bar and cool part of it (0.5 kg/s) from 50°C to 210°C with no external parts, as shown below. 2 kg/s 4 bar 50°C

0.5 kg/s 1 bar –10°C

1.5 kg/s Mysterious black box

1 bar 70°C

You are feeling somewhat adventurous and are tempted by this offer but must ask the fundamental question: “Can it work?” Can it? Explain. 3.37 Steam enters a nozzle at 4 MPa and 640°C with a velocity of 20 m/s. This process may be considered reversible and adiabatic. The nozzle exit pressure is 0.1 MPa. (a) Draw a sketch of this process. Include all known information. (b) What is the entropy change of the steam? (c) What is the exit temperature? (d) What is the exit velocity? 3.38 Propane at 350°C and 600 cm3 /mol is expanded in a turbine. The exhaust pressure is atmospheric. What is the lowest possible exhaust temperature? How much work is obtained? You may assume ideal gas behavior and that heat transfer to the surroundings is negligible. 3.39 What is the minimum amount of work required to separate an inlet stream of air flowing at 20°C and 1 bar into exit streams of pure O2 and pure N2 at 20°C and 1 bar? 3.40 An adiabatic turbine is designed to take stream of steam flowing at 10 kg/s from an inlet at 10 bar and 500°C to an outlet at 1 bar. It is reported that at steady-state, this turbine can deliver 7,619 kW of power. Is this possible? Explain. 3.41 An ideal gas at a flow rate of 10.0 m3 /min enters a compressor at 25°C and 1 bar. It leaves at 1 MPa. During this process, heat is dissipated to the surroundings at a rate of 2100 W. You may take the surroundings to be at a constant temperature of 25°C. The heat capacity of the ideal gas is given by:

cP 5 2.00 1 0.0400T R where T is in [K]. Answer the following questions: (a) Assuming this process is reversible, calculate the temperature at the compressor outlet. (b) Calculate the minimum power required to compress the gas. (c) If the compressor efficiency is 70%, calculate the actual power needed. (d) Calculate the actual final temperature. 3.42 Consider a Carnot (reversible) power cycle operating between a hot reservoir of 727°C and a cold reservoir of 27°C. If 700 W of power are generated, calculate the total entropy change of the universe, the entropy change of the hot reservoir, and the entropy change of the cold reservoir.

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200 ► Chapter 3. Entropy and the Second Law of Thermodynamics 3.43 An ideal gas enters an adiabatic turbine with a molar flow rate of 10 [mol/s]. The inlet pressure is 100 bar, and the inlet temperature is 500°C. The gas exits at 1 bar. The ideal gas heat capacity is given by: cP 5 3.6 1 0.5 3 1023T R where T is in [K]. (a) At steady state, calculate the maximum power (in kW) generated by the turbine. (b) If the isentropic efficiency is 80%, calculate the actual power delivered. 3.44 A rigid tank of volume 0.1 m3 is filled from a supply line at 127°C and 2 bar, as drawn in the following figure. It is initially at vacuum. The valve is opened, and the tank fills until the pressures equilibrate. The valve is then shut. During this process, 6,000 [J] of heat transfers to the surroundings. The final temperature in the tank is 227°C. The surroundings are at 27°C. What is the 7 entropy change of the universe for this process? Take the heat capacity to be constant, cP 5 ¢ ≤R. 2 Ideal Gas Tin = 127°C p in = 2 bar

Ideal Gas Tin = 127°C p in = 2 bar

Tsurr = 27°C

Tsurr = 27°C

Initially Vacuum

Ideal Gas 3

V = 0.1 m

V = 0.1 m3

State 1

p2 = 2 bar T2 = 227°C

State 2

3.45 A rigid tank of volume 0.5 m3 is connected to a piston–cylinder assembly by a valve as shown below. Both vessels contain pure water. They are immersed in a constant-temperature bath at 200°C and 600 kPa. Consider the tank and the piston–cylinder assembly as the system and the constant-temperature bath as the surroundings. Initially the valve is closed, and both units are in equilibrium with the surroundings (the bath). The rigid tank contains saturated water with a quality of 95% (i.e., 95% of the mass of water is vapor). The piston–cylinder assembly initially has a volume of 0.1 m3. The valve is then opened. The water flows into the piston–cylinder assembly until equilibrium is obtained. For this process, calculate the change in entropy for the system, the surroundings, and the universe. Temperature bath TB = 200°C P = 600 kPa

Surroundings

Pure H2O V = 0.5 m3 Quality = 95% H2O Vinitial = 0.1 m3

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3.13 Problems ◄ 201 3.46 A rigid, well-insulated container is initially divided into three compartments. The top compartment contains a vacuum. It is separated from the middle compartment A by a frictionless mass of 1000 kg and area 0.098 m2. Compartment A contains 2 moles of ideal gas at 300 K and is separated from compartment B, on the bottom of the container, by a rigid partition. Compartment B initially contains 2 moles of the same ideal gas at 300 K and occupies a volume of 0.1 m3 A process is initiated by removing the partition. The mass then re-equilibrates in the container. What is the change in entropy? Take cP 5 1 5/2 2 R. Well insulated Vacuum

Vacuum

g

A = 0.098 m2 M = 1000 kg n1,a = 2 moles T1,a = 300 K V1,b = 0.1 m3 n1,b = 2 moles T1,b = 300 K

Ideal gas

remove partition M = 1000 kg Ideal gas partition T2 = ?

3.47 Consider the system shown below. Tank A has a volume of 0.3 m3 and initially contains an ideal diatomic gas at 700 kPa, 40°C. Cylinder B has a piston resting on the bottom, at which point the spring exerts no force on the piston. The piston–cylinder has a cross-sectional area of 0.065 m2, the piston has a mass of 40 kg, and the spring constant is 3500 N/m. Atmospheric pressure is 100 kPa. Tanks A and B are well insulated and do not transfer heat between each other. The valve is opened and gas flows into the cylinder until pressures in A and B become equal and the valve is closed. You may assume constant heat capacity. Determine the final pressure in the system. Assuming the gas in A has undergone a reversible, adiabatic expansion, find the final temperature in cylinder A. The temperatures in tank A and B are not necessarily equal. F = −kx

Patm

B

A

3.48 A steam turbine in a small electric power plant is designed to accept 4500 kg/hr of steam at 60 bar and 500°C and exhaust the steam at 10 bar. Heat transfers to the surroundings 1 Tsurr 5 300K 2 at a rate of 69.86 kW. Answer the following questions: # (a) Calculate the maximum power 1 Wmax 2 that the turbine can generate. (b) In this case, what is the exit temperature of the steam? (c) You know that the isentropic efficiency of the turbine is actually 66.5%. What is the actual power produced? (d) Do you expect the exit temperature to be higher or lower than that calculated in part (a). Explain. (Assume the heat transfer does not change.) (e) What is the actual exit temperature?

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202 ► Chapter 3. Entropy and the Second Law of Thermodynamics 3.49 Air flowing at 1 m3 /s enters an adiabatic compressor at 20°C and 1 bar. It exits at 200°C. The isentropic efficiency of the compressor is 80%. Calculate the exit pressure and the power required. 3.50 Steam enters a turbine at 10 MPa and 500°C and leaves at 100 kPa. The isentropic efficiency of the turbine is 85%. Calculate the exit temperature and the work generated per kg of steam flowing through. 3.51 Nitrogen gas at 27°C flows into a well-insulated device operating at steady-state. There is no shaft work. The device has two exit streams. Two-thirds of the nitrogen, by mass, exits at 127°C and 1 bar. The remainder exits at an unknown temperature and 1 bar. Find the exit temperature of the third stream. What is the minimum possible pressure of the inlet stream? Assume ideal gas behavior. 3.52 Consider a well-insulated piston–cylinder assembly containing 5 kg of water vapor, initially at 540°C and 60 bar, that undergoes a reversible expansion to 20 bar. The surroundings are at 1 bar and 25°C. Answer the following questions: (a) What is the entropy change (Dsuniv, Dssurr, and Dssys) for this process? (b) What is the final temperature of the water? (c) What is the value of work for this process? (d) What is the final volume of the system? 3.53 Consider a well-insulated, rigid tank containing 5 kg of water vapor in the same initial state as in Problem 3.52 (540°C, 60 bar). Again, the surroundings are at 1 bar and 25°C, as shown below. A tiny leak develops, and water slowly escapes until the pressure reaches 20 bar.

Very tiny leak hole

Tsurr = 25°C Psurr = 1 bar

Pure H2O P1 = 60 bar T1 = 540°C m = 5kg

Pure H2O P2 = 20 bar T2 = ?°C

Well insulated

Do you expect the final temperature to be higher than, lower than, or the same as that calculated in part (b) of Problem 3.52? Explain your answer. 3.54 Consider filling a “type A” gas cylinder with water from a high-pressure supply line as shown on next page. Before filling, the cylinder is empty (vacuum). The valve is then opened, exposing the tank to a 3-MPa line at 773 K until the pressure of the cylinder reaches 3 MPa. The valve is then closed. The volume of a “type A” cylinder is 50 L. (a) What is the change in entropy of the universe immediately after the valve is closed? (b) If the cylinder then sits in storage at 293 K for a long time, what is the entropy change of the universe? 3.55 A rigid tank has a volume of 0.01 m3. It initially contains saturated water at a temperature of 200°C and a quality of 0.4. The top of the tank contains a pressure-regulating valve that maintains the vapor at constant pressure. This system undergoes a process whereby it is heated until all the liquid vaporizes. You may assume there is no pressure drop or heat transfer in the exit line. The surroundings are at 200°C. What is the entropy change of the universe?

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3.13 Problems ◄ 203 Water 3 MPa; 773 K

Tsurr = 293 K Initially: vacuum

3.56 Consider the system sketched below in which a turbine is placed between two rigid tanks. Tank 1 initially contains an ideal gas at 10 bar and 1000 K. Its volume is 1 m3. Tank 2 is 9 m3 and is initially at vacuum. The heat transfer with the surroundings is negligible. Determine the maximum work (in [J]) obtainable by the turbine. You may take the heat capacity of the gas to be cP 5 1 5/2 2 R. You may neglect the volume in the turbine and assume the final temperature in the two tanks is equal. Ws V = 9 m3 V = 1 m3 P1 = 10 bar T1 = 1000 K

Ideal gas in Turbine

Ideal gas out

Initially vacuum

Rigid tank Rigid tank

3.57 A hot reservoir is available at 500°C and a cold reservoir at 25°C. Calculate the maximum possible efficiency of a power cycle that operates between these two reservoirs. 3.58 An ideal Rankine cycle operates with the following design: 100 kg/s of steam enters the turbine at 30 bar and 500°C and is condensed at 0.1 bar. Determine the power produced and the efficiency of the cycle. 3.59 Come up with four ways in which you can make the power cycle of Problem 3.58 more efficient. Illustrate how your ideas achieve increased efficiency using sketches like that in Figure 3.8. 3.60 An ideal Rankine cycle produces 100 MW of power. If steam enters the turbine at 100 bar and 500°C and is condensed at 1 bar, determine the mass flow rate of steam. Recalculate the mass flow rate assuming that the isentropic efficiency of the turbine and the compressor are 80%. 3.61 You are considering building a solar power plant which uses CCl2F2 as its working fluid. It enters the turbine as a saturated vapor at 1.7 MPa and leaves at 0.7 MPa. Based on the ideal Rankine cycle, determine the efficiency. Property data for CCl2F2 may be found at http://webbook.nist .gov/chemistry/fluid/. 3.62 Consider a refrigeration system based on an ideal vapor-compression cycle using R-134a as the refrigerant. It operates between 0.7 MPa and 0.12 MPa with a ﬂow rate of 0.5 mol/s. Calculate the following: (a) the rate of heat removal from the refrigerated unit (b) the power input needed to the compressor (c) the COP The properties of R-134a can be found at http://webbook.nist.gov/chemistry/fluid/.

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204 ► Chapter 3. Entropy and the Second Law of Thermodynamics 3.63 If the throttling valve in Problem 3.62 is replaced by an isentropic turbine, what is the COP? Is this modiﬁcation practical? Explain. 3.64 A two-stage cascade refrigeration system is shown below. The refrigerant is R134a. It consists of two ideal vapor-compression cycles with heat exchange between the condenser of the lowertemperature cycle and the evaporator of the higher-temperature cycle. The hotter cycle operates between 0.7 MPa and 0.35 MPa, while the cooler cycle operates between 0.35 MPa and 0.12 MPa. If the ﬂow rate in the hotter cycle is 0.5 mol/s, determine the following:

Valve

Valve

5 Refrigeration unit at low T

QC

1

P = 0.12 MPa Evaporator

P = 0.35 MPa

8 Condenser

6

Q

4

QH

Evaporator 2

Compressor 2

High T reservoir

Condenser

7 Wc

P = 0.7 MPa

3 Wc Compressor 1

(a) What is the ﬂow rate in the cooler cycle? (b) What is the rate of heat removal from the refrigerated unit? (c) What is the power input needed to the compressors? (d) What is the COP? (e) Compare the performance with the cycle in Problem 3.62. 3.65 Design a vapor-compression refrigeration system to cool a system to 25°C with the capability for up to 20 kW of cooling. You have a reservoir at 20°C to reject heat to. Refrigerants and their properties can he found at http://webbook.nist.gov/chemistry/ﬂuid/. 3.66 Modify the vapor-compression refrigeration system presented in Section 3.9 to apply to a refrigerator for home use. This system needs to provide cooling to two units: the freezer at 215°C and the main compartment at 5°C. Take the refrigeration capacity, QC, of each compartment to be equal. You are limited to one compressor and one condenser. Draw a schematic of the process and the associated Ts diagram. Select an appropriate refrigerant and deﬁne the states of the system. Refrigerants and their properties can be found at http://webbook.nist.gov/chemistry/ﬂuid/. 3.67 Consider an ideal, reversible magnetic refrigeration cycle shown in the ﬁgure on next page. A paramagnetic working material in the form of the rim of a wheel is rotated between a hightemperature reservoir and a low-temperature reservoir. In the high-temperature reservoir, the working material expels heat into the reservoir as it is subjected to a high magnetic ﬁeld. As the working material moves into the low-temperature reservoir, the ﬁeld becomes smaller and eventually zero. The demagnetization process causes the working ﬂuid to absorb heat from the low-temperature reservoir. The working material is gadolinium sulphate octahydrate. On the upper-left-hand side of the ﬁgure, helium enters the porous wheel at temperature 1.1 K and a magnetic ﬁeld of 0.9 tesla [T] and is forced to ﬂow in heat exchange with the moving wheel. The wheel absorbs heat from the helium as it is demagnetized. The temperature of the helium drops 0.2 K during this process. The helium then absorbs heat QC from the load and reenters the heat exchanger area at 1.1 K. Similarly, in the lower-right-hand side of the ﬁgure, helium enters the porous wheel at 8 K and a magnetic ﬁeld of 1.6 T. The wheel deposits heat into helium as the material is magnetized. The temperature rises to 9.5 K at 6.4 T.

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3.13 Problems ◄ 205 High-magnetic field region

T = 1.1 K B = 0.9 T

T = 9.5 K B = 6.4 T

QC

QH

Wheel T = 0.9 K B=0

T=8K B = 1.6 T

Low-magnetic field region

Approximately how much heat is being expelled by the cold reservoir? How much heat is being absorbed by the hot reservoir? Estimate the coefficient of performance for this process. How does the numerical value compare with a conventional refrigeration cycle? How is work being supplied to perform the refrigeration process? A Ts diagram for gadolinum sulphate octahydrate is provided. Gadolinium sulphate octa-hydrate Gd2 (SO4)3 8H2O

H, T = 0 Tesla 0.1 0.4 0.9 1.5 1.6

R

S tot

2.0

2.5 3.6

1.0

4.9

6.4

8.1 10 Tesla

0.5 0

1

2

3

4 5 6 7 Temperature, K

8

9

10

3.68 Consider the oxidation of cuprous oxide to form cupric oxide by the following reaction: 2Cu2O 1 s 2 1 O2 1 g 2 4 4CuO 1 s 2 Calculate Dsrxn. This task can be done in the same type of path described in Section 2.6 for Dhrxn. You can calculate the values of entropies of formation from the data in Appendix A.3 by applying the following relationship: o

o

Dsof

5

Dh f 2 Dg f T

Physically explain the sign of Dsrxn. Does the formation of CuO violate the second law of thermodynamics? Explain. 3.69 Determine the exergy of the following states. Take the environment to be at 25°C and 1 bar. (a) Argon at 500 K and 2 bar (b) Propane at 500 K and 2 bar (c) Water at 500 K and 2 bar

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206 ► Chapter 3. Entropy and the Second Law of Thermodynamics 3.70 Determine the exergy of a system of pure water containing 1 kg ice and 10 g water vapor at 220°C. Take the environment to be at 10°C and 1 bar. 3.71 1 kg of copper at 600°C and 1 bar is immersed in 20 kg of water at 20°C and 1 bar in a well-insulated container. Consider the system to be both the copper and the water. Calculate the change in internal energy, energy and exergy of the system for the process. Take the environment to be 25°C and 1 bar. 3.72 An ideal gas is contained in a piston–cylinder assembly. The pressure of the gas is initially balanced by two 2000 kg blocks plus the atmospheric pressure of 1 [bar]. The piston has a crosssectional area of 0.098 3 m2 4 and is initially 0.2 [m] from the base of the cylinder. The gas is allowed to expand in the two-step isothermal expansion process drawn in the following figure. One 2000 kg block is removed. The gas expands until the pressures are equal. The next block is then removed, and the gas expands until it reaches a pressure of 1 [bar]. The temperature in the system remains at 300 [K] throughout the process. The surroundings are at a temperature of 300 [K] and a pressure of 1 [bar]. The heat capacity is given by:

cP 5 3.6 1 0.5 3 1023T R where T is in [K]. Determine the following quantities: (a) The work obtained during this process (b) The useful work obtained (c) The exergy and the ideal work (d) The lost work

P0 = 1 [bar] T0 = 300 [K] m = 2000 Kg

m = 2000 Kg

Process: Isothermal Expansion

m = 2000 Kg

Process: Isothermal Expansion

A = 0.098 m2 Pure, ideal gas

Initial State

0.2 m

Intermediate State

Final State

3.73 Steam enters a turbine with a mass flow rate of 5 [kg/s]. The inlet pressure is 60 bar, and the inlet temperature is 500°C. The outlet contains saturated steam at 1 bar. The surroundings are at 20°C. At steady state, (a) Calculate the power generated by the turbine. (b) Calculate the isentropic efficiency of the turbine. (c) Calculate the change in exthalpy during the process. 3.74 1 mol of steam is initially at 10 bar and 200°C. The surroundings are at 20°C and 1 bar. (a) Calculate the exergy of the system. (b) Calculate the change in exergy for a process where the steam is heated at constant pressure until the volume doubles. (c) Calculate the change in exergy for a process where the steam isothermally expands until its volume doubles.

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3.13 Problems ◄ 207 3.75 Ethylene 1 C2H4 2 at 100°C and 1 bar passes through a heater and emerges at 200°C. Calculate the change in exthalpy per mole of ethylene that passes through. You may assume ideal gas behavior. The environment is at 20°C. 3.76 Consider a tank containing 100 kg of water initially at 70°C. Due to heat transfer, the temperature of water in the tank drops to 50°C. The surroundings are at 10°C. Calculate the lost work. 3.77 An open feedwater heater is used to take inlet stream of water vapor at 5 bar and 200°C and have it leave as saturated liquid at 5 bar. This is accomplished by mixing it with an appropriate amount of a second inlet stream at 5 bar and 20°C. Calculate the lost work. 3.78 You wish to heat a stream of CO2 at pressure 1 bar, flowing at 10 mol/s, from 150°C to 300°C in a countercurrent heat exchanger. To do this task, you are using a stream of high-pressure steam available at 40 bar and 400°C, as shown in the following figure. The steam exits the heat exchanger as a saturated vapor. You may assume the pressure of each stream stays constant as it flows through the heat exchanger (i.e., neglect the pressure drop of the flowing streams). The entire system is well insulated as shown. Calculate the change in exthalpy of each stream and the lost work during this process. nCO2 = 10

mol s

Tin = 150 [°C] Pin = 1 [bar]

Heat Exchanger CO2

Tout = 300 [°C]

CO2

Steam

Steam

Tin = 400 [°C] Pin = 40 [bar]

Insulation

3.79 An ideal gas at 6 MPa and 200°C is flowing in a pipe, as shown in the following figure. Connected to this pipe through a valve is a tank of volume 0.4 m3. This tank initially is at vacuum. The valve is opened, and the tank fills with the ideal gas until the pressure is 6 MPa, and then the valve is closed. Heat transfer occurs from the tank to the surroundings in the amount of 3.9325 kJ per each mol of gas that flows into the tank. The surroundings are at 25°C. The heat capacity is given by: cP 5 3.6 1 0.5 3 1023T R where T is in K. Ideal Gas 6MPa, 200°C

Tsurr = 25°C

Initially Vacuum

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208 ► Chapter 3. Entropy and the Second Law of Thermodynamics (a) Determine the final temperature of the gas in the tank immediately after the valve is closed. (b) Determine the entropy change of the universe for this process. 3.80 Steam at 6 MPa and 400°C is flowing in a pipe. Connected to this pipe through a valve is a tank of volume 0.4 m3. This tank initially is at vacuum. The valve is opened, and the tank fills with the steam until the pressure is 6 MPa, and then the valve is closed. Heat transfer occurs from the tank to the environment in the amount of 95 kJ per each kg of gas that flows into the tank. The environment is at 25°C. (a) Determine the final temperature of the steam in the tank immediately after the valve is closed. (b) Determine the entropy change of the universe for this process. (c) Determine the lost work during this process.

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► CHAPTER

4 Equations of State and Intermolecular Forces Learning Objectives To demonstrate mastery of the material in Chapter 4, you should be able to: ► Given a chemical species, identify which intermolecular interactions are significant. Given different species, qualitatively compare the magnitude of their dipole moments, polarizabilities, intermolecular interactions, LennardJones parameters ε and s, and van der Waals parameters a and b. ► Given two of the measured properties P, v, and T, calculate the value of the third using: cubic equations of state (e.g., van der Waals, Redlich–Kwong, Peng–Robinson), the virial equation, generalized compressibility charts, and ThermoSolver software. Apply the Rackett equation, the thermal expansion coefficient, and the isothermal compressibility to find molar volumes of liquids and solids. ► State the molecular components that contribute to internal energy. Describe and illustrate by example the following intermolecular interactions: point charges, dipoles, induced dipoles, dispersion (London) interactions, repulsive forces, and chemical effects. Define a van der Waals force, and relate it to the dipole moment and polarizability of a molecule. Ultimately, you want to be able to relate macroscopic thermodynamic behaviors to their molecular origins as much as possible. ► Define a potential function. Write equations for the ideal gas, hard sphere, Sutherland, and Lennard-Jones potentials and relate the terms to intermolecular interactions. ► State the molecular assumptions of an ideal gas. Describe how the terms in the van der Waals equation relax these assumptions. Identify how the general form of cubic equations of state accounts for attractive and repulsive interactions in a similar manner. ► State the principle of corresponding states on molecular and macroscopic levels. Apply this principle to develop expressions to solve for parameters in equations of state from the critical property data of a given species. Describe why the acentric factor was introduced and its role in constructing the generalized compressibility charts. ► Write the van der Waals mixing rules for coefficients a and b. Explain their functionality in terms of molecular interactions. Write the mixing rules for the virial coefficients and for pseudocritical properties using Kay’s rules.

209

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210 ► Chapter 4. Equations of State and Intermolecular Forces Apply these mixing rules to solve for P, v, or T of a mixture using equations of state or generalized compressibility charts.

►4.1 INTRODUCTION Motivation The intensive thermodynamic properties that can be experimentally measured are pressure, temperature, molar volume, and composition. For any pure species, only two intensive properties are independent; thus, we can graphically map a “surface” from experimental data using P, v, and T as coordinates and plot it as we did in Section 1.6. Alternatively, we could tabulate the data as was done for water in the steam tables (Section 1.7). However, in solving problems, it is often inconvenient to have to resort to graphs or tables for numerical values (as well as slopes for derivatives and areas for integrals). We therefore seek to come up with an equation that relates these measured variables by fitting experimental data. In the language of math, we want an equation of the form: f 1 P, v, T 2 5 0 Such an equation is fit to experimental data and is known as an equation of state (EOS) since it allows us to calculate the unknown measured property from the two that constrain the state. Equations of state can be explicit in pressure, that is, P 5 f 1 T, v 2

(4.1)

v 5 f 1 T, P 2

(4.2)

in molar volume,

or in terms of the dimensionless compressibility factor, z, z5

Pv 5 f 1 T, v 2 RT

z5

Pv 5 f 1 T, P 2 RT

or,

In developing an equation of state, the goal is to come up with an equation that fits experimental data as accurately as possible; there may or may not be a physical basis for its form. An equation of this type is called constitutive (as opposed to a fundamental equation such as the first law). What are other constitutive equations that you have encountered as a chemical engineer? In practice, there are hundreds of analytical equations of this form in the literature from which to choose! It would be quite burdensome (and very impractical) to examine every one to decide when to use a particular form. Consequently, we will take a different tack. We will start with the “friendliest” equation of state, the ideal gas model. After examining its limitations, we will explore how we can describe deviations from ideal gas behavior. We will investigate the generalities of these different forms as well as try to develop an intuition about what equations to use and when to use them. To this end, we

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4.2 Intermolecular Forces ◄ 211

will examine the molecular origins of macroscopic thermodynamic property behavior. Perilous straits indeed, but a journey with rich rewards!

The Ideal Gas As you know well, the most common equation of state is the ideal gas model. It can be written explicitly for pressure in terms of the intensive properties v and T as follows: P5

RT v

(4.3)

The ideal gas model is of the form presented in Equation (4.1) in that it relates the measured variables P, T, and v. The ideal gas equation can be derived directly from the kinetic theory of gases for a gas consisting of molecules that are infinitesimally small, hard round spheres that occupy negligible volume and exert forces upon each other only through collisions. Stated more concisely, the assumptions of the ideal gas model are that molecules: 1. Occupy no volume 2. Exert no intermolecular forces (except when they collide with each other or with the container’s walls) As we shall see in Section 4.2, the absence of intermolecular forces leads to the internal energy being independent of pressure. It depends only on temperature, that is, the molecular kinetic energy of the molecules. Hence, uideal gas 5 f 1 T only 2 As the pressure goes to zero, all gases approach ideal gas behavior.

►4.2 INTERMOLECULAR FORCES Internal (Molecular) Energy “Molecular” energy, or internal energy, u, can be divided into two parts: molecular kinetic energy and molecular potential energy. Kinetic energy results from the translational, rotational, and vibrational motion of the molecules; hence, kinetic energy manifests itself by the molecules’ velocities and is directly related to the measured variable temperature via Maxwell–Boltzmann statistics. The potential energy results from the position of one atom or molecule relative to others in the system. As we saw in Section 2.6, a significant component of molecular potential energy is related to the covalent bonds between atoms of the same molecule. We refer to this type of potential energy as intramolecular potential energy. When these bonds rearrange in a chemical reaction, there can be large changes in the molecular potential energy. In this section, we will examine another component of molecular potential energy— that related to interactions between different molecules (or atoms that are not covalently bonded), the intermolecular potential energy. As we will soon learn, the intermolecular potential energy between molecules depends on how close the molecules are relative to one another. Because the intermolecular potential energy depends on the molecules’ positions, we can relate it directly to the measured variable pressure. At constant temperature, as the pressure increases, the average distance between molecules decreases, and their positions relative to one another get closer. Therefore, as the pressure increases,

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212 ► Chapter 4. Equations of State and Intermolecular Forces the intermolecular interactions become more important, and we need to consider their effect on the thermodynamic properties. On the other hand, an ideal gas has no intermolecular forces, so the positions of the molecules relative to one another do not matter. Consequently, the internal energy of an ideal gas is independent of pressure and depends only on temperature. We wish to relax the ideal gas assumption so that we can develop more general equations of state. To accomplish this task, we need to establish the relationship for the internal energy of a system as a function of distance between the molecules and their orientations. We are specifically interested in the intermolecular potential energy component of internal energy.

The Electric Nature of Atoms and Molecules The intermolecular interactions between species arise from the electronic and quantum nature of atoms. An atom can be viewed as containing a fixed, positively charged nucleus surrounded by a relatively mobile, negatively charged electron cloud. When a molecule is in close enough proximity to another molecule, the electrically charged structure of its atoms can lead to attractive and repulsive forces. The attractive forces include electrostatic forces between point charges or permanent dipoles, induction forces, and dispersion forces. Since intermolecular interactions result from the electric nature of atoms, it is often useful to apply the concept of an electric field when discussing the effect of a given speS cies on the system. Recall the electric field intensity, E, is defined as the force per unit charge exerted on a positive test charge, Q, in the field. It is related to the negative gradient of the molecular potential energy, G, by: S

E5

F 2=G 5 Q Q

(4.4)

The electric field intensity from a given molecule is the same regardless of the species with which it interacts. The principle of superposition says the total electric field in the system is given by the vector sum of the individual electric fields of all the species in a system. Therefore, if we understand the behavior of a single molecule, we can quantify its contribution to theSenergy of the macroscopic system as a whole. Equation (4.4) relates the electric field, E, to the intermolecular forces, F, and to the intermolecular potential energy, G. In the discussion that follows, we will refer to all of these quantities, as appropriate, to characterize the intermolecular interactions that cause the system to deviate from ideal gas behavior.1 It is important to realize that there is presently no direct quantitative relationship between molecular physics and classical thermodynamics except in very simple systems. So why are we studying molecular physics? This topic warrants study for four reasons: 1. To strengthen our intuition about nonideal behavior and our judgment about what equations of state to use. 2. To understand why some mathematical relationships for f (P, v, T) work better than others.

1

Unit systems related to electrical quantities can be confusing. Depending on the unit system, these equations can take different forms. The equations below are written for CGS (Gaussian) units as opposed to SI. Equations that are denoted with “CGS units” are valid only in those units. See Appendix D for further discussion.

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4.2 Intermolecular Forces ◄ 213

3. While there is no quantitative connection between molecular physics and classical thermodynamics, with the increased computational power of modern computers, and the development of such techniques as density functional theory and Monte Carlo simulations, this quantitative connection may soon be realized. 4. To develop “mixing rules” for how thermodynamic properties of a mixture depend on composition. This ability allows us to extend data of pure systems to mixtures. Note that while thermodynamic data for pure species are readily available, the availability of data for the particular mixture you may be interested in is much less likely since there is an infinite number of permutations of mixtures.

Attractive Forces Electrostatic Forces Electrostatic interactions between molecules can result from the net charges of ions and also from permanent charge separation in neutral species. In this section, we will examine the nature of these interactions. Point charges The simplest electrostatic interaction results from the force of attraction between two species with nonzero charge. There are two types of electric charge, positive and negative. If two species have the same type of charge, they will repel each other, while species of unlike charge attract. Consider species i and j which have positive charges, Qi and Qj, respectively, and are separated by a distance r. The resulting repulsive force is illustrated in Figure 4.1. If the length r is much greater than the radius of i or j, these charges can be treated as point charges. Recall from basic physics that the force between point charges is inversely proportional to the square of the distance between the charges, that is: Fij 5

QiQj r2

CGS units

This equation is written for CGS units and is known as Coulomb’s law. See Appendix D for the SI units equivalent of Coulomb’s law. The potential energy between species i and j is found by rearranging Equation (4.4), and using the expression for force given by Coulomb’s law: Gij 5 23 Fij dr 5

QiQj r

CGS units

(4.5)

Examination of Equation (4.5) shows the potential energy is negative for unlike charges. A lower (more negative) energy indicates a more stable system, that is, an attractive force. For like charges, the potential is positive, indicating a repulsive interaction. At large separations 1 r S ` 2 , the potential goes to zero, indicating that the point charges do not interact.

F

Qi

Qj

+

+

F

r

Figure 4.1 The coulombic repulsion between two like point charges separated by a distance r.

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214 ► Chapter 4. Equations of State and Intermolecular Forces

+Q

→ E

−Q

→ E

Figure 4.2 Electric field lines from positive and negative point charges.

The effect of a given point charge on its neighbors can also be visualized in terms of electric fields. Electric field lines for positive and negative point charges are shown in Figure 4.2. The field lines represent “lines of force” and point in the direction a positive test charge would move if it were put in that field. The density of field lines is proportional to their strength; the closer the lines are spaced, the stronger the field at that point. Examination of Figure 4.2 shows that the positive test charge would move away from the positive point charge on the left and toward the negative point charge on the right and that the force exerted by the field falls off as the distance away from the point charge increases. These qualitative observations are consistent with the analytical relation given by Equation (4.5). If the “lines of force” given by the electric fields caused by the species in system sufficiently affect the system’s bulk behavior, intermolecular forces are significant and we can no longer treat the system using the ideal gas model. Point charges exert strong forces and fall off relatively slowly with distance. It is uncommon for an isolated net charge to exist in nature since it will typically find an oppositely charged species and combine. However, some molecular examples exist, including the following: 1. Ionic solids (e.g. NaCl crystals) Ionic solids are made up of positively charged and negatively charged ions within a crystal lattice. The bond energy of the solid results from attractive forces of the oppositely charged ions, as given by Equation (4.5). For example, in table salt, NaCl, a positively charged sodium ion, Na1, is electrostatically attracted to the negatively charged chloride ions, Cl2, that surround it. In Problem 4.19, you will calculate the bond strength of this ionic solid. 2. Electrolytes (e.g., 18M H2SO4) Net charges exist in the liquid phase in electrolyte solutions and molten salts. In an H2SO4 acid bath, for example, H1 and 22 SO4 exist in the liquid and exert Coulombic forces. The polar structure of water allows charged species to be stable. The water forms an electrostatic cloud that shields the ions from one another. The electrostatic interactions of the charged species within the electrolyte solution form a key component in the behavior of electrochemical systems.2 2

For a treatment of the behavior of ions in electrochemical solutions, see J. O. M. Bockris and A. K. N. Reddy, Modern Electrochemistry (New York: Plenum Press, 1970).

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4.2 Intermolecular Forces ◄ 215

3. Ionized gases or plasmas Point charges exist in the gas phase in the form of plasmas. Plasmas exist in many forms, from the Earth’s ionosphere to the glow discharge plasmas used in etching to the deposition of thin films in integrated circuit manufacturing. The thermodynamic properties of point charges requires particular attention to these strong electrical forces and will not be covered further in this text. Electric Dipoles While they have a net neutral charge, some molecules are configured so that there is an overall separation of charge. These molecules may be treated as an electric dipole (two poles). In a dipole, there is a region of positive charge 1 1Q 2 next to a region of negative charge 1 2Q 2 . The magnitudes of positive and negative charges are equal. A dipole is illustrated on the left of Figure 4.3. The electric field lines associated with the dipole are shown on the right. We can see from these field lines that dipoles can exert forces on other species in their vicinity. The strength of the dipole is characterized by the dipole moment, µ, a vector that points from the negative charge to the positive charge. The magnitude of the dipole moment is equal to the product of the magnitude of charge, Q, and the distance by which the positive charge and the negative charge are separated, l. The common unit of the dipole moment is the debye [D]: 1/2 c 1D 5 10218 1 erg cm3 2 d

Dipole moments are commonly found in nature at the molecular level. Molecules are formed by covalent bonds between the valence electrons of their atoms. When the sharing of electrons in a covalent bond is unequal, an atom can gain electron density at the expense of the atom to which it is bonded. For example, consider an HCl molecule. The Lewis dot structure is depicted on the left of Figure 4.4. The valence electrons of the chlorine atom are indicated by dots, while the valence electron of hydrogen is an “x.” The bond in this molecule is formed by the sharing of one electron from chlorine and one from hydrogen. However, the Cl is highly electronegative since it needs an electron to complete its outer shell. It pulls the electron from H much more strongly than hydrogen pulls the electron from Cl. Thus, the electrons are not shared equally; rather,

μ = QI

+Q

I −Q

+Q

→ E

μ −Q

Figure 4.3 Schematic of charge separation leading to an electric dipole. Electric field lines are depicted on the sketch to the right.

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216 ► Chapter 4. Equations of State and Intermolecular Forces

+++ + H + H × Cl

μ

–

–

Cl –

–

–

Figure 4.4 The Lewis dot structure of an HCl molecule and the electric field lines emanating from its dipole.

the lone hydrogen electron spends more time close to the Cl atom than the shared Cl electron does next to hydrogen. The result is a net charge separation with the Cl obtaining a net negative charge and the H a net positive charge, as depicted on the right of Figure 4.4. This separation of charge leads to a permanent dipole moment. In the case of HCl, the magnitude of the dipole moment is around 1.1 D. Diatomic molecules with electronegative (F, Cl) and electropositive (Li, Na) atoms have large dipole moments. Diatomic molecules with dissimilar atoms, such as HCl, will always exhibit some degree of charge separation; however, it may be small. Thus, these molecules will have permanent dipoles. On the other hand, for polyatomic molecules with more than two atoms, we must look at the molecular structure to see whether a dipole exists. Dipole moments in these molecules are caused by nonsymmetric distributions of the electron cloud in the molecule. Symmetric molecules have no dipole moment. The greater the molecular asymmetry, the greater the dipole moment. For example, the electron cloud in CH3Cl is pulled strongly to the electronegative Cl and exhibits a dipole of 1.87 D. On the other hand, CH4 is symmetric and does not have a permanent dipole moment. There are many other molecular examples of dipoles: H2O, HF, and so on. Can you think of some? Will CO2 exhibit a dipole? Dipole moments of several representative molecules are presented in Table 4.1. The interaction of one dipole with a neighboring dipole is termed a dipole–dipole interaction. Such an interaction can lead to deviations from ideal gas behavior. The dipole–dipole interaction depends on the relative orientation of the molecules. However, to relate this type of electrostatic force to macroscopic behavior, we must average over all orientations. For example, consider the case of two HCl molecules in proximity to each other. As the electric field from the dipole depicted in Figure 4.4 shows, each HCl can exert an electrical force on its neighbor. The bottom of Figure 4.5 illustrates dipoles in the two extreme limits of orientation. In the lowest energy configuration, the negative side of one dipole is aligned next to the positive side of the other. This leads to electrostatic attraction, as depicted in Figure 4.5. Conversely, in the highest energy configuration, like-charged sides of the dipole align, leading to electrostatic repulsion. Dipoles may take any orientation in between these two limits. If the orientation of the two dipoles were completely random, the average force would be zero, since attractive and repulsive orientations would occur equally. As the top of the Figure 4.5 shows, however, molecular dipoles in gases and liquids are free to rotate. This movement allows the energetically favored lower-energy attractive interactions to occur more frequently, as the dipoles tend to rotate to align. On the other hand, thermal energy leads to a randomization of orientation.

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4.2 Intermolecular Forces ◄ 217 TABLE 4.1

Dipole Moments, Polarizabilities, and Ionization Energies

Molecule

m [D]

a 3 cm3 3 1025 4

H2 He N2 O2 Ne

0 0 0 0 0

8.19 2.06 17.7 16 3.97

15.42 24.59 15.58 12.07 21.56

Cl2 Ar Kr Xe HF

0 0 0 0 1.91

46.1 16.6 25.3 41.1 5.1

11.5 15.76 14 12.13 16.03

HCl HBr HI H2O H2S

1.08 0.8 0.42 1.85 0.9

26.3 36.1 54.5 14.8 37.8

12.74 11.68 10.39 12.62 10.46

CH3OH NH3 NO N 2O SF6

1.7 1.47 0.2 0.2 0

32.3 22.2 17.4 30 44.7

10.84 10.07 9.26 12.89 15.32

SO2 CH4 CH3F CH3Cl CH3Br

1.63 0 1.85 1.87 1.81

38.9 26 26.1 45.3 55.5

12.35 12.61 12.5 11.26 10.54

CH2F2 CH2Cl2 CHF3 CHCl3 CF4

1.97 1.8 1.65 1.1 0

27.3 64.8 28 85 28.5

12.71 11.33 13.86 11.37 16

CHCl3 CCl4 CO CO2 CS2

0.45 0 0.12 0 0

82.4 105 19.8 26.3 87.4

11.68 11.47 14.01 13.78 10.07

C2H6 C2H4 C2H2 C3H8 HCN

0 0 0 0 3

44.7 42.2 34.9 62.9 25.9

11.52 10.51 11.4 10.94 13.6

1 CN 2 2 CH3OCH3

0.2 1.7

50.1 51.6

13.37 10.02

I [eV]

(contined)

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218 ► Chapter 4. Equations of State and Intermolecular Forces TABLE 4.1

Continued

Molecule

m [D]

a 3 cm3 3 1025 4

1 CH2 2 3 CH3 1 CO 2 CH3 C6H6

0.4 2.9 0

56.4 63.3 104

9.86 9.7 9.24

C6H5Cl C6H5NO2 o- C6H4Cl2 m- C6H4Cl2

1.69 4 2.5 1.72

122.5 129.2 141.7 142.3

9.07 9.94 9.06 9.1

−

Γ

−

Free rotation of a dipole

+

+

I [eV]

+

−

Γ1

+

− (1) vs.

Γ2

Lowest energy configuration

+

−

−

+

(2)

Highest energy configuration

Figure 4.5 Different possible orientations of a freely rotating electric dipole–dipole pair.

The trade-off between these two effects can be quantified according to the Boltzmann factor, e2G/1kT2. Thus, the ratio of the number of dipoles in any two states, N1 andN2 is related to their potential energy difference and the temperature according to: N1 5 e2c 1G1 2G22 / 1kT2 d N2 We can see that this equation implies that more dipoles will align in lower-energy orientations. Averaging the potential energy of the dipole–dipole interaction between species i and j over all possible orientations gives the following expression: Gij 5 2

2 m2i m2j CGS units 3 r6kT

(4.6)

where G is the average potential energy of the dipole–dipole interaction. The potential energy between dipoles has a 1/r6 dependence on position; it falls off much more quickly than the Coulomb interaction. It is also proportional to the square of the dipole moment of each species. The (kT) term in the denominator results from the averaging. At higher temperatures, the orientations are more randomly distributed and the attractive force decreases. In addition to permanent dipoles, species can exhibit higher-order terms in the multipolar expansion, such as quadrapoles (four poles), octapoles (eight poles), or higher-order multipoles. However, these higher-order terms do not measurably influence macroscopic property behavior.

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4.2 Intermolecular Forces ◄ 219

Induction Forces Induction results when the electric field from a dipole affects the electric structure of its neighbor. Since the negatively charged electrons are free to move about the atom, the electrons in molecule i can be displaced due to a neighboring dipole, molecule j. This displacement “induces” the separation of charge in molecule i, causing a dipole to form. Consequently i and j are attracted to each other. This phenomenon is termed induction. Dipoles can be induced in both polar and nonpolar species. As an example, let’s consider the effect of the electric field from the dipole HCl on an Ar atom in close proximity to it. Without any external influences, Ar has no net charge separation. However, as shown in Figure 4.6, the electrons in Ar will respond to the dipole field of HCl. Recall that the electric field lines show the direction in which a positive test charge will go. Hence, the negatively charged electrons will be attracted to the top of the Ar atom in Figure 4.6, leading to an induced dipole. The induced dipole in Ar will then be attracted to the permanent HCl dipole. Since the nature of induction is the same as a dipole-dipole attraction, we also expect a 1/r6 dependence. Again, the magnitude of induction depends on the orientation of dipole j. The average over all orientations of the potential energy between an induced dipole i and the permanent dipole j is given by: Gij 5 2

aim2j r6

CGS units

(4.7)

where a is the polarizability of molecule i. Polarizability is a parameter that characterizes the ease with which a molecule’s electron cloud can be displaced by the presence of an electric field. A larger displacement induces a stronger dipole. Electrons position themselves around the positively charged nucleus of an atom. Since the valence electrons of larger atoms are farther away from the nucleus, they are less rigidly “held” and the atom is more polarizable, Thus, in general, the larger the atom, the larger the value for a. Polarizability is also roughly additive; that is, the value of polarizability scales with the number of atoms. For example, ozone, O3, has roughly 1.5 times the polarizability of oxygen, O2, since the ratio of oxygen atoms is 3:2. A rough estimation of the polarizability of a molecule can be obtained by adding together the polarizability of all the atoms in the molecule.3 Hence molecules with more atoms have greater polarizabilities. Polarizabilities of several representative molecules are reported in Table 4.1.

+++ + H + −

−

Cl −

−

−

−−− Ar +++

Charge separation “induced” by the electric field from the HCl dipole

Figure 4.6 The induction of a dipole in argon from the electric field due to a HCl dipole.

3

A more accurate approach is to proportion polarizability to the type and number of covalent bonds in a molecule. Values for the contribution of several types of bonds (e.g., C2C, C5C, C2H, N2H, C2Cl, etc.) and molecular groups (e.g., C2O2H, C5O, C2O2C, etc) can be found in J. O. Hirshfelder, C. F. Curtiss, and R. B. Bird, Molecular Theory of Gases and Liquids, (New York: Wiley, 1954).

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220 ► Chapter 4. Equations of State and Intermolecular Forces Dispersion (London) Forces Nonpolar molecules, such as N2 and O2, show forces of attraction; otherwise they would not condense or freeze at low temperatures. Yet they do not have a dipole moment, and the pure species are not subject to dipole–dipole interactions and induction. The attractive interactions between nonpolar molecules result from a third type of interaction; dispersion (or London) forces. Dispersion is inherently a quantum-mechanical phenomenon; we would need to understand quantum electrodynamics to develop a rigorous model of dispersion. However, it can be viewed “classically” as follows: Nonpolar molecules are really only nonpolar when the electron cloud is averaged over time. In a given “snapshot” of time, the molecule has a temporary dipole moment. Dispersion forces result from the instantaneous nonsymmetry of the electron cloud surrounding a nucleus. The instantaneous dipole moment induces a dipole in a neighboring molecule, leading to an attractive force. Using quantum mechanics and perturbation theory, London developed the following expression for the energy of attraction of symmetric molecules i and j: Gij < 2

IiIj 3 aiaj ¢ ≤ CGS units 6 2 r Ii 1 Ij

(4.8)

where I is the first ionization potential, that is, the energy required for the following reaction: M S M1 1 e. Ionization potentials of common species are presented in Table 4.1. These “temporary dipole” interactions also have a 1/r6 dependence on position. They also depend on the polarizability of each species involved, since the extent of the instantaneous dipole is related to the looseness of the nucleus’s control of the valence electrons; similarly, the induction in the neighboring molecule depends on its polarizability. While Equation (4.8) was developed for nonpolar species, polar molecules are subject to dispersion interactions as well. The classical explanation of dispersion above would leave us to believe that dispersion is a relatively small force. However, as is often the case, our classical intuition is defied by quantum mechanics. It turns out that dispersion forces are surprisingly large in magnitude, as we will see in Examples 4.1 and 4.2. van der Waals Forces Dipole–dipole, induction, and dispersion forces are collectively referred to as van der Waals forces. The intermolecular potential energy for each of these interactions falls off as the sixth power of position. Thus, all three van der Waals interactions are of the form: Gij 5 2

C6 r6

(4.9)

The magnitude of the constant C6 is proportional to the strength of the attractive force. The subscript “6” indicates that the potential falls off as the sixth power of distance.

EXAMPLE 4.1 Comparison of van der Waals Forces for Pure Species

As best as you can, compare the strength of the dipole–dipole, induction, and dispersion interactions for each of the following pure species at 298 K: H2O, NH3, CH4, CH3Cl, CCl4. Discuss the results. SOLUTION The dipole moment, m, the polarizability, a, and the ionization energy, I, can be obtained from Table 4.1. They are summarized in Table E4.1. With these molecular parameters, we can determine the approximate dipole–dipole, induction, and dispersion potentials. Putting Equations (4.6), (4.7), and (4.8) in the form of Equation (4.9), for any pure species i, we get: dipole2dipole: 1 C6 2 dipole2dipole 5

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2 m4i CGS units 3 kT

(E4.1A)

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4.2 Intermolecular Forces ◄ 221

Table E4.1

Relative van der Waals Interactions for Species of Example 4.1

Molecule

m 3D 4

H 2O

1.85

14.8

12.62

NH3

1.47

22.2

CH4

0

CH3Cl

1.87

CCl4

0

a 3 cm3 3 1025 4

I [eV]

C6 3 1060 3 erg cm6 4

1 C6 2 dipole2dipole

1 C6 2 induction

233

190

10

33

10.07

145

76

10

60

26

12.61

102

0

0

102

45.3

11.26

507

198

32

277

11.47

1517

0

0

1517

105

1 C6 2 dispersion

induction: 1 C6 2 induction 5 2aim2i GCS units

(E4.1B)

3 dispersion: 1 C6 2 dispersion < a2i Ii CGS units 4

(E4.1C)

Note Equation (E4.1B) is multiplied by 2, since each species in a two-body interaction can induce a dipole in its neighbor. Equation (E4.1C) is rigorously valid only for the symmetric species CH4 and CCl4. However, we assume it provides a valid estimate for the other three species. The total van der Waals interaction is given by: C6 5 1 C6 2 dipole2dipole 1 1 C6 2 induction 1 1 C6 2 dispersion

(E4.1D)

The values for C6 calculated from Equations (E4.1D), (E4.1A), (E4.1B), and (E4.1C) for each of the five species are presented in Table E4.1. Even though it is nonpolar, CCl4 exhibits the largest intermolecular forces, approximately five times greater than the strongly polar CH3Cl. The magnitude results from the large polarizability associated with the four Cl atoms in CCl4. It is curious to note how large the dispersion forces can be! Although it is roughly the same size as ammonia or methane, water is more apt to be nonideal due to its more polar structure and concomitant dipole–dipole interactions. However, CH3Cl, which has a similar dipole moment to water, demonstrates much larger van der Waals interactions (over twice the value for C6) since this larger molecule is much more easily polarized and, consequently, has a larger dispersion interaction.

EXAMPLE 4.2 Magnitude of van der Waals Potentials in a Mixture

Consider a mixture of Ar and HCl. Predict the relative importance of the van der Waals interactions of the different “two-body” interactions in the mixture. Compare the unlike interactions to that which is obtained from prediction using the geometric mean of the like interactions. SOLUTION In a binary mixture, there are three possible interactions:Ar–Ar, HCl–HCl, and Ar–HCl.The like species interactions, Ar–Ar and HCl–HCl can be found using the same approach as in Example 4.1. Thus, Equations (E4.1A) through (E4.1D) are used. For the Ar –HCl. interaction, Equations (4.6), (4.7), and (4.8) can be used to get: 1 C6 2 dipole2dipole 5

2 m2Arm2HCl CGS units 3 kT 0

1 C6 2 induction 5

aArm2HCl

1 aHClm2Ar CGS units (Continued)

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222 ► Chapter 4. Equations of State and Intermolecular Forces

TABLE E4.2

Relative Magnitudes of Attractive Forces Between Different Molecules

Molecule–Molecule Ar–Ar HCl–HCl Ar–HCl

and,

C6 3 1060 3 erg cm6 4

1 C6 2 dipole2dipole

52

0

134 76

1 C6 2 induction

1 C6 2 dispersion

0

52

22

6

106

0

2

74

IArIHCl 3 1 C6 2 dispersion 5 aAraHCl ¢ ≤ CGS units 2 IAr 1 IHCl

Again, the total van der Waals interaction is given by: C6 5 1 C6 2 dipole2dipole 1 1 C6 2 induction 1 1 C6 2 dispersion The values of C6 obtained from these equations are shown in Table E4.2. In every case, the dispersion interaction is the largest. The magnitude of the unlike interaction, Ar–HCl, falls in between each of the like-species interactions. Often we estimate the magnitude of the unlike interaction as the geometric mean of the like interactions. In this case we would get: 1 C6 2 Ar2HCl 5 " 1 C6 2 Ar2Ar 1 C6 2 HCl2HCl 5 83 3 1060 3 erg/cm6 4 This value is different from that reported in Table E4.2 by 9%.

EXAMPLE 4.3 Prediction of the Relative Size of van der Waals Forces Based on Molecular Structure

Consider the following molecules: CCl4, CF4, SiCl4. List these species in order of their total van der Waals forces of attraction, C6, from the largest value to the smallest. Explain your choice based on molecular arguments. SOLUTION The molecular parameters for SiCl4 are not reported in Table 4.1. However, we can use qualitative molecular arguments to solve this problem. In general, attractive interactions include dispersion, dipole–dipole, and induction forces. The three species listed—CCl4, CF4, SiCl4—are all nonpolar and, therefore, exhibit only dispersion forces. The magnitude of these forces is related to the polarizability, a, of these species. Each atom in CF4 is from the second row of the periodic table. In SiCl4 each atom is in the third row. Thus, the valence electrons in CF4 are held in toward the nuclei the most tightly (least “sloshy”), so this molecule has the smallest dispersion forces. Conversely, the electrons in SiCl4 are the farthest away and most easily polarized. So we would expect that for these species: 1 C6 2 SiCl4 . 1 C6 2 CCl4 . 1 C6 2 CF4

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4.2 Intermolecular Forces ◄ 223

Intermolecular Potential Functions and Repulsive Forces To account for nonideal gas behavior, we want to describe how the intermolecular potential energy depends on the position between molecules. A function or plot of potential energy vs. molecular separation is called a potential function. The potential function is ultimately what determines how the internal energy of a gas depends on pressure. There are many models of potential functions that are used to approximate the relation between intermolecular energy and position. These models include both attractive and repulsive interactions. For net neutral species, the attractive forces can be described by the van der Waals forces discussed previously. We next examine two ways in which the repulsive forces are approximated and the potential functions which result. The Hard Sphere Model and the Sutherland Potential Repulsive forces among molecules result from their finite sizes. In the simplest model, the hard sphere model, we consider the molecules to be finite hard spheres of diameter s. Thus, molecules act like billiard balls; when they physically get close together, they run into each other and repel. Therefore, the potential between a pair of molecules is zero until the two molecules’ diameters touch, where the potential increases to infinity. Mathematically, the potential function is described by: G5 b

for r . s for r # s

0 `

A plot of the hard sphere potential function is presented in Figure 4.7a. The Sutherland model adds the van der Waals attractive term proportional to r26 to the hard sphere model. Therefore, the potential function is mathematically described by: 2 1 C6 2 G 5 c r6 `

for r . s for r # s

The Sutherland model is illustrated in Figure 4.7b. The Sutherland potential accounts for both attractive and repulsive forces; however, there is discontinuity right at r 5 s. What would the ideal gas model look like on the plots in Figure 4.7? The Lennard-Jones Potential In reality, molecules are not rigid but rather are bounded by diffuse electron clouds. Repulsive interactions occur when the molecules get so close that their electron clouds overlap, leading to coulombic repulsion as well as a possible violation of the Pauli exclusion principle. This effect leads to a violent repulsion of the two molecules. Quantum mechanics says the repulsion should have an exponential dependence on position, since atomic wavefunctions fall off exponentially at large distances. However, it is more convenient to represent the repulsive potential empirically in terms of an inverse power law expression, as follows: Gij 5

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1 Cn 2 rn

where 8 , n , 16

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224 ► Chapter 4. Equations of State and Intermolecular Forces Hard Sphere Model

1

Sutherland Model

1 σ

σ Γ

Γ

r (a)

r (C6) σ6

(b)

Figure 4.7 Potential functions. (a) Hard sphere model; (b) Sutherland model.

where Cn is a constant proportional to the magnitude of the repulsive force that falls off as the inverse of separation to the power n. If we consider both van der Waals attractive forces and quantum (repulsive) effects, we come up with an expression for the molecular potential energy of the form: Gij 5

1 Cn 2 rn

2

1 C6 2 r6

The attractive part is negative, since it lowers the energy, while the repulsive part is positive, raising the energy. Lennard-Jones recognized a mathematically convenient form of this equation came about if n 5 12, resulting in the Lennard-Jones potential function: s 12 s 6 G 5 4e B a b 2 a b R r r where,

C12 5 4es12

and

(4.10)

C6 5 4es6

A plot of the Lennard-Jones potential function is given in Figure 4.8. The parameters e and s can be physically interpreted as an energy parameter and a distance parameter, respectively. As illustrated in Figure 4.8, the energy parameter, e, is given by the depth of the potential well, while the distance parameter, s, is given by the distance at which attractive and repulsive potentials are equal and is characteristic of the molecular size. Some typical values of Lennard-Jones parameters are given in Table 4.2.4 The size parameter, s, increases with molecular size, and the energy parameter, e, scales with the magnitude of the van der Waals interaction. Figure 4.9a plots Lennard-Jones potential functions for O2, Cl2, and C6H6. These species are all nonpolar; the only van der Waals forces of attraction are from dispersion. Thus, their potential interactions depend only on the distance of separation between two

4

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From Hirshfelder et al., Molecular Theory of Gases and Liquids.

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4.2 Intermolecular Forces ◄ 225 Γ

σ ε

r

Figure 4.8 Plot of the Lennard-Jones potential as a function of distance between molecules.

molecules, not on their relative orientation. The relative contribution of attractive and repulsive interactions in these three species approximately scales with size; that is, as these species size get larger, their attractive forces get proportionately larger. We term these species simple molecules. For comparison, the Lennard-Jones potential functions for CH3OH and SO2 are included with the three simple species in Figure 4.9b. CH3OH has a much larger force of attraction in proportion to its size—presumably due to its polar structure. Moreover, its polar structure leads to a dependence on orientation of one CH3OH relative to another. Thus, the interactions with this species are inherently more “complex.” A more accurate potential function would include orientation as a variable. The potential function depicted in Figure 4.9b is an approximation of the average over all the orientations one methanol molecule can have relative to another. We say that CH3OH belongs to different class. SO2 has large repulsive interactions but relatively weak attractive forces.

TABLE 4.2

c04.indd 225

Lennard-Jones Parameters for Several Species

Gas

e/k 1 K 2

s 1 A+ 2

He H2 C2H4 C6H6 F2 Cl2 O2 N2 CCl4 CH4 Ne Ar Ke Xe CH3OH SO2

10.2 35.7 205 440 112 307 101 86 327 148.2 31.6 120 190 229 507 252

2.58 2.94 4.23 5.27 3.65 4.62 3.5 3.7 5.88 3.82 2.8 3.4 3.6 4.1 3.6 4.3

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600

600

400

400

200

200 Γ/k [K]

Γ/k [K]

226 ► Chapter 4. Equations of State and Intermolecular Forces

0 O2

−200

−600

O2

–200 Cl2

−400

0

Cl2

–400 C6H6

3

6

8

–600

12

3

SO2

C6H6 CH3OH 6

r [Å] (a)

8

12

r [Å] (b)

Figure 4.9 Comparison of the Lennard-Jones potential for (a) three different molecules and (b) five different molecules.

Principle of Corresponding States We now wish to generalize our treatment of the intermolecular interactions that lead to nonideal gas behavior. We first consider nonpolar molecules. Inspection of Figure 4.9a suggests that if we scale the intermolecular potential appropriately, we can come up with a universal expression that applies to all nonpolar molecules. The ability to scale intermolecular interactions in this way leads to the principle of corresponding states: The dimensionless potential energy is the same for all species.

In quantitative form, it says there exists a universal function that applies to all species if we scale the potential energy to the energy parameter and the distance between molecules to the size parameter. Therefore, we can write: F¢

Gii r , ≤ 50 Pi s i

(4.11)

Equation (4.11) is not restricted to the Lennard-Jones potential function but rather it says the dimensionless potential energy is some universal function of the dimensionless distance. We can extend the principle of corresponding states to macroscopic thermodynamic properties. In this form, we can write a general equation of state that applies to all species if we scale the measured properties P, v, and T appropriately. Van der Waals recognized that, for a given species, it was particularly suitable to scale the values to those at its critical point. The critical point represents a unique state, and that state is determined by the intermolecular interactions characteristic of a given species. Thus, we can construct a “reduced” coordinate system with the following three dimensionless groupings:5 Tr 5

T , Tc

Pr 5

P , Pc

and

vr 5

v vc

5

Pc, Tc, and vc are not all independent, since we need only two properties to constrain the state of the critical point. Thus both molecular and macroscopic versions of the principle of corresponding states have two independent scaling parameters.

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4.2 Intermolecular Forces ◄ 227

The principle of corresponding states says that there is some universal function that is the same (i.e., the same form and the same constants) for all substances: F¢

T P v , , ≤ 50 Tc Pc vc

(4.12a)

Alternatively, one of the dimensionless groupings might be the compressibility factor, z. Thus, z5

T P Pv 5F¢ , ≤ RT T c Pc

(4.12b)

Equation (4.12b) illustrates the macroscopic version of the principle of corresponding states: All fluids at the same reduced temperature and reduced pressure have the same compressibility factor.

We based the discussion above on nonpolar species. Figure 4.9b illustrates that there are different classes of molecules based on the particular nature of the intermolecular interactions involved. For example, CH3OH, with its strong dipole moment, behaves differently from the nonpolar species depicted in Figure 4.9a. We can improve the principle of corresponding states if we group molecules according to class and assert that within any one class intermolecular interactions scale similarly. To accomplish this objective, we introduce a third parameter characteristic of classes of molecules. There are many ways to introduce a parameter for classes of molecules; we will explore only one—the Pitzer acentric factor, v. It characterizes how “nonspherical” a molecule is, thereby assigning it to a class. The definition of v is somewhat arbitrary: v ; 21 2 log10 3 Psat 1 Tr 5 0.7 2 /Pc 4 , where Psat 1 Tr 5 0.7 2 is the saturation pressure at a reduced temperature of 0.7. This definition for a third parameter is convenient, since it gives a value of zero for the simple fluids Ar, Kr, and Xe. Moreover, other fluids have positive values less than 1. Since tabulated data for v are usually available, we seldom need to calculate v—we just know where to look v up. Appendix A.1 presents acentric factors for a number of common species. With the introduction of the acentric factor to categorize classes of molecules, the general macroscopic equation is of the form: F¢

T P v , , , v≤ 5 0 Tc Pc vc

(4.13)

Equation (4.13) is often written in the form: z 5 F0 1 Tr, Pr 2 1 vF1 1 Tr, Pr 2

(4.14)

where F0 and F1 depend only on the reduced pressure and temperature and the acentric factor is used to modulate the effect of the F1 term. Thus a perfectly “spherical” molecule (such as Ar) depends only on F0.

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228 ► Chapter 4. Equations of State and Intermolecular Forces

Chemical Forces The physical forces described above aptly account for most molecular interactions in the gas phase. We now direct our discussion toward the condensed phases. Solids and liquids form when the net attractive intermolecular forces are stronger than the thermal energy in the system and, consequently, hold the molecules together. While the force of attraction can sometimes be attributed to the electrostatic and van der Waals interactions described above, chemical forces also frequently play a role in condensed phases.6 Chemical forces are based on the nature of covalent electrons, the concept of the chemical bond, and the formation of new chemical species. The main difference between chemical and physical forces is that chemical forces saturate whereas physical forces do not, since chemical interactions are specific to the electronic wavefunctions of the chemical species involved. Indeed, a complete quantitative description of chemical interactions involves solution of the Schrödinger equation to describe the overlap of the molecular orbitals involved. We will consider chemical interactions only qualitatively. The goal of this discussion is to realize that there may be other important forces that govern the behavior of solids and liquids and to get a flavor of what these forces might be. The most prevalent chemical effects are due to hydrogen bonds and acid–base complexes. In both cases, there exists a sharing of valence electrons between different molecules. Hydrogen bonding is the “chemical bond” that results between an electronegative atom (usually F, O, or N) and a hydrogen atom bonded to another electronegative atom in a second molecule. Figure 4.10 illustrates the hydrogen bond that forms between the electronegative oxygen in water molecule 1 and the adjacent hydrogen in water molecule 2. Since the electronegative oxygen atom in water 2 pulls the hydrogen atom’s electron away from its nucleus, a partial charge separation results. The resulting positive charge on the hydrogen atom can be attracted to a partial negative charge of an electronegative atom in the adjacent oxygen atom in water 1, leading to an attractive force. So far, this sounds like the van der Waals interactions we just described, and we might be tempted to say that hydrogen bonds are caused by the dipole–dipole interactions. However, the mechanism of attraction is not purely electrostatic but rather has a significant amount of sharing of lone-pair electrons characteristic of a covalent bond. This leads to fundamental differences between hydrogen bonds and dipole–dipole forces. Hydrogen bonds form a relatively strong, highly directional interaction with characteristic saturation. Typically, this force is two orders of magnitude stronger than the van der Waals forces 1 1/r6 2 described above but one order of magnitude weaker than a covalent bond. In other words, if you plugged numbers into Equation (4.6) to try to account for the strength of a hydrogen bond, you would get a number whose magnitude is far below the strength experimentally observed. Moreover, the distance between the hydrogen-bonded atoms is considerably less than that predicted based on the hard sphere model. Thus, the hydrogen is actually Water “1” Hydrogen H bond + − O H Water “2” H + − O H Covalent bond

6

c04.indd 228

Figure 4.10 Hydrogen bonding in water.

In some unique gas systems, chemical forces also manifest themselves.

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4.2 Intermolecular Forces ◄ 229

penetrating into the electron cloud of the electronegative atom in the other molecule. The hydrogen bond leads to interesting thermodynamic phenomena. For example, the extensive network of hydrogen bonds leads to the open structure of ice. Thus, unlike most species, water expands when it freezes.7 To illustrate how chemical interactions can affect thermodynamic properties, consider two types of chemical behavior: solvation and association. Solvation is the tendency of unlike molecules to form chemical complexes. It is generically represented by the following reaction: A1B

h

AB

Association is the tendency of like molecules to form complexes (polymerize) and can be represented as follows: A1A

A2

h

Naturally, the ability of a molecule to solvate or associate is intimately linked to its electronic structure. Hydrogen bonding can lead to either of these behaviors. An example of solvation is given by a mixture of chloroform and acetone. Hydrogen bonding causes the unlike molecules to form a complex:

CH3

I

(B)

(A)

H-bond O II C

(AB)

I

CH3

CI I C H I CI

CH3

I

I

+

CI

I

O II C

I

I

CI I C H I CI

I

CI

CH3

Acid–base pairs also solvate. The dimerization of acetic acid illustrates association: I

HO

CH3

I

O II C

OH

(2A)

I

I

I

I

CH3

C II

O

2

O II C

CH3

OH (A2)

Can you think of other possible solvation and association reactions? To see how chemical effects can change the equilibrium behavior of a system, let’s examine Figure 4.11, which depicts species A and B in liquid–vapor equilibrium. We will assume ideal gas behavior in the vapor. We wish to compare the composition of the vapor phase for three scenarios occurring in the liquid: (i) A and B are the only species present in the system and Raoult’s law applies. (ii) Species A and B solvate in the liquid phase, but we are unaware of this chemistry. (iii) Species A associates in the liquid phase, but we are unaware of this chemistry. 7

Imagine what our world would look like if ice were denser than water. All the ice in the ocean would sink to the bottom and, consequently, be insulated from energy input from the sun. Thus, most of the ocean would be permanently frozen.

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230 ► Chapter 4. Equations of State and Intermolecular Forces Pideal i

Vapor

A A

Raoult's law: yAP ideal = xAPA sat

A B

A PPideal

iii

Vapor A

A B

Liquid AB

Solvation in liquid phase (chemical effect)

A

B

B

2A

Liquid A2

Association in liquid phase

Figure 4.11 Binary system where (i) system exhibits ideal behavior; (ii) species A and species B solvate in the liquid phase; (iii) species A associates in the liquid phase.

We wish to examine how solvation [scenario (ii)] or association [scenario (iii)] changes our perception of this system. In scenario (ii), species A and B solvate in the liquid phase. This reaction depletes the liquid of species A and B, since the complex AB is formed. Since AB is a different chemical species from A or B, some A and B in the vapor phase will then condense to compensate. This leads to lower total system pressure than in the ideal case [scenario (i)]; thus solvation effects lead to a “negative” deviation from Raoult’s law. Next we wish to examine what happens in scenario (iii), where species A associates in the liquid phase. The dimer is a different chemical species and assumed involatile. Let’s consider the point of view of molecule B. The association causes a higher mole fraction of B to be in the liquid than if A did not associate. Thus B will evaporate to compensate. This will lead to a higher system pressure than the ideal case. Association leads to “positive” deviations from Raoult’s law. The above arguments have provided a qualitative and intuitive feel for what we will spend a large percentage of the text developing: ways to predict equilibrium behavior and deviations from ideality in chemical systems.

EXAMPLE 4.4 Comparison of psat for H2O and CH3OH

c04.indd 230

Determine the saturation pressure, Psat, of water and methanol at 100°C, 50°C, and 25°C. Report values in [Pa]. Based on intermolecular forces explain (i) why the vapor pressure of methanol at a given temperature is greater than water and (ii) why vapor pressure increases with temperature.

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4.2 Intermolecular Forces ◄ 231

SOLUTION We can obtain the saturation pressure of water from the steam tables (Appendix B.1). The saturation pressure of methanol is given by the Antoine equation, ln 1 Psat 3 bar 4 2 5 A 2

B 3626.55 5 11.9673 2 3 4 T 2 34.29 TK 1C

where the constants were found consulting Appendix A.1. The resulting values of Psat are reported in Table E4.4: (i) Vapor pressure is a function of how easily molecules can “escape” from the liquid phase. This is dictated by the strength of the intermolecular interactions of the species involved given a certain thermal energy. The weaker the intermolecular forces, the larger Psat. Both methanol and water form hydrogen bonds in the liquid phase. Hydrogen bonds are much stronger than van der Waals interactions, so they will control the behavior of the vapor pressure for these two species. Water can form two hydrogen bonds/molecule, as shown in Figure E4.1, while CH3OH can form only one. Thus at a given temperature, water has stronger attractive forces in the liquid, and a lower vapor pressure. (ii) Explanation #1: Temperature is a measure of the molecular kinetic energy (part of the internal energy, u). While it is representative of the average molecular kinetic energy, species at thermal equilibrium have a distribution of energies. This distribution is given by the Maxwell-Boltzmann equation. A certain fraction of species (water or methanol in this case) will have enough kinetic energy to overcome the attractive forces (H-bonds) keeping them in the liquid phase. As this fraction increases, more molecules enter the vapor phase and Psat increases. Since the Maxwell-Boltzmann distribution depends exponentially on temperature, Psat also increases exponentially with temperature. Explanation #2: You may be tempted to use the following explanation for the temperature dependence of Psat. As T increases, more molecules would hit the walls of the container. This can be seen, for example, in the ideal gas equation P 5 RT/v. Thus Psat increases with T. This explanation is not wrong but it is incomplete! This would predict a linear relation between Psat and T not the exponential relation we observe experimentally. Viewed in another way, we may ask, do the number of molecules in the vapor phase increase as T increases? Explanation #1 asserts that they do. In explanation #2, however, you could get a higher Psat without adding any more species to the vapor. Viewing it the latter way is wrong. TABLE E4.4

Values of Psat for Water and Methanol Psat 3 Pa 4

T[°C]

Water

25

3.169 3

1.69 3 104

50

1.235 3 104

5.56 3 104

1.014 3

3.54 3 105

100

H O

Lone electron H pair

H

H

H

H

O Two H-bonds

H

O H

105

H O

c04.indd 231

Methanol 103

O H

Figure E4.1 Schematic of two H-bonds formed per H2O molecule.

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232 ► Chapter 4. Equations of State and Intermolecular Forces

►4.3 EQUATIONS OF STATE The van der Waals Equation of State Relation to Molecular Interaction We will now use our knowledge of intermolecular interactions to modify the ideal gas model for situations when potential interactions between the species are important. In this section, we will use the Sutherland potential function to describe the intermolecular interactions, that is, use the hard sphere model to account for repulsive forces and van der Waals interactions to describe attractive forces. This development leads to the van der Waals equation of state. This equation is particularly well suited for illustrating how the molecular concepts we learned about in Section 4.2 can be related to macroscopic property data. However, it should be emphasized that more accurate equations of state have been developed and will be covered next. First, let’s consider the “size” of the molecules based on the hard sphere model. The entire volume of the system will no longer be available to the molecules. We can account for this effect by replacing the volume term in the ideal gas model with one for available volume. Recall that in the hard sphere model, the molecules have a diameter s. Thus, the center of one molecule cannot approach another molecule closer than a distance s. The excluded volume of the two molecules is then 1 4/3 2 ps3. Dividing by 2 and multiplying by Avogadro’s number, NA, we get one mole of molecules occupying a volume b 5 1 2/3 2 ps3NA. To correct for size, we modify the ideal gas model to include only the unoccupied molar volume, 1 v 2 b 2 . Hence, we get: P5

RT v2b

since one molecule cannot occupy the space in which another molecule already sits. We still need to take into account attractive intermolecular forces. In the absence of net electric charge, the attractive forces in the gas phase can include dispersion, dipole– dipole, and induction, all of which have an r26 dependence. However, we do not have “distance” as a parameter in our equations, but rather volume, which is proportional to the cube of the distance 1 v < r3 2 . We can say, therefore, that all of these terms are proportional to v22. But how do we incorporate this into our equation of state? As we saw in Section 4.2, the variable most related to potential energy is pressure. So we correct the pressure by including a term that accounts for attractive forces. Attractive forces should decrease the pressure, since the molecules will not bang into the container as readily; hence, we subtract a correction term as follows: P5

RT a 2 2 v2b v

(4.15)

This equation was first proposed by the Dutch physicist van der Waals in 1873. Since it assumes a 1/r6 dependence for all attractive forces, any force with this functionality (be it dispersion, dipole–dipole, or induction) has been termed a “van der Waals force.” The parameter a in Equation (4.15) can be related to molecular constants by integrating the Sutherland potential function. This calculation gives a 5 1 2pN2AC6 2 / 1 3s3 2 . In practice, a and b are treated as empirical constants that account for the magnitude of the attractive and repulsive forces. Can you think of how we might find values for the constants a and b?

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4.3 Equations of State ◄ 233

We can rewrite Equation (4.15) as follows: Pv3 2 1 RT 1 Pb 2 v2 1 av 2 ab 5 0 van der Waals as a Cubic Equation of State This equation is termed a cubic equation of state since there are three roots for volume for fixed values of T and P given the values of the parameters a and b. We will consider other cubic equations of state shortly. Figure 4.12 illustrates the general characteristics of this equation for various isotherms. The three roots have different characteristics above the critical point than below it. Above the critical point there is one positive, real root and two roots containing negative or imaginary numbers. Only the positive, real root represents a physical value—the volume of the supercritical fluid. The other roots are mathematical artifacts with no physical basis. Below the critical point, however, there may be three real, positive roots. We can ascribe the lowest root to the molar volume of the liquid state and the highest root to the vapor state. We throw out the middle root where dP/dv . 0 on physical grounds. Can you think of a physical reason why this relationship is not possible at constant temperature? In reality, isotherms are horizontal in the two-phase envelope where vapor and liquid coexist. This discontinuity eludes description by cubic equations of state. One feature of the two-phase region can be determined by cubic equations. Maxwell’s “equal-area rule” (which will be verified in Chapter 6) provides a graphical means to determine Psat for a given T. It states that the saturation pressure is the pressure at which a horizontal line equally divides the area between the real isobar and the solution given by the cubic equation. Such a construct is illustrated in Figure 4.12, where the equal areas above and below the isobar fix the value for Psat. This procedure can be achieved by trial and error. If a higher saturation pressure were predicted, the upper area would be too small. Conversely, too low a value for Psat would make the upper area too large. van der Waals Parameters by Corresponding States To use the van der Waals equation, the parameters a and b must be determined for a species of interest. The most accurate values are obtained by fitting experimental PvT data. However, when these data are not available, we can use the principle of corresponding states (see Section 4.2).

T > Tc

Pressure

T = Tc

∂P ∂v

Tc

2 = ∂ P2 ∂v

=0

Tc

T < Tc

P sat Equal areas v sat(liq)

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v sat(vap)

Volume

Figure 4.12 Pv behavior of the van der Waals equation. This behavior is representative of other cubic equations of state.

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234 ► Chapter 4. Equations of State and Intermolecular Forces Recall that the principle of corresponding state scales property data to that at the critical point. We can relate the van der Waals parameters to the temperature and pressure at the critical point by noting that there is an inflection point on the critical isotherm, as shown in Figure 4.12. Mathematically, we can say: a

'2P 'P b 5 ¢ 2 ≤ Tc 5 0 'v Tc 'v

(4.16)

Thus, at the critical point we have: RTc a 2 2 vc 2 b vc

Pc 5

2RTc 'P 2a 2 3 b 505 2 1 vc 2 b 2 'v Tc vc

(4.18)

2RTc 6a '2P ≤ 505 2 4 2 3 1 vc 2 b 2 'v Tc vc

(4.19)

a

and,

¢

(4.17)

If we multiply Equation (4.18) by 2 and Equation (4.19) by 1 v 2 b 2 and add them together, we get: 05

6a 1 vc 2 b 2 4a 2 v3c v4c

Solving for vc gives: vc 5 3b

(4.20)

If we plug Equation (4.20) back into Equation (4.18) and solve for a, we get: a5

9 vcRTc 8

Finally, if we plug this equation back into Equation (4.17), we can solve for the van der Waals constants in terms of the critical temperature and the critical pressure: a5 and,

27 1 RTc 2 2 64 Pc

b5

1 RTc 2 8Pc

(4.21) (4.22)

Example 4.7 shows another approach to obtaining van der Waals parameters from properties at the critical point. We now have an equation of state for which we only need critical property data to solve for the parameters. This approach will not be as accurate as fitting PvT data to get the constants, a and b, but then again you do not have to go through the expense of laboratory measurements, since values of critical properties are usually known and readily available. We can use the results above to write the van der Waals equation in terms of the reduced variables Tr, Pr, and vr. We start with Equation (4.15): P5

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RT a 2 2 v2b v

(4.15)

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4.3 Equations of State ◄ 235

We can substitute for a and b and rearrange using Equations (4.20), (4.21), and (4.22) to give:

P ¢ ≤ 5 Pc

8¢ 3¢

T ≤ Tc

v ≤ 21 vc

2

3 v 2 ¢ ≤ vc

This equation can be written in reduced form as: Pr 5

8Tr 3 2 2 3vr 2 1 vr

(4.23)

If we compare Equation (4.23) to (4.12a), we see we have defined a universal function for the reduced pressure in terms of the reduced temperature and the reduced volume. Thus, we have a specific expression to delineate the corresponding states between species. We can calculate the compressibility factor at the critical point from Equations (4.20) and (4.22): zc 5

Pcvc 3 5 RTc 8

Thus, applying the principle of corresponding states to the van der Waals equation leads to a value of 0.375 for the compressibility factor at the critical point for all species. Experimental values for the compressibility factor at the critical point are around 0.29 for simple species and usually less for complex species. Thus, the value predicted by the van der Waals equation is considerably high—indicating its limitations in predicting PvT behavior. EXAMPLE 4.5 Prediction of the Relative Size of van der Waals Forces Based on Molecular Structure

Consider the following molecules: CCl4, CF4, SiCl4, SiCl3H. (a) List these species in order of their van der Waals constant a, from the largest value of a to the smallest. Explain your choice based on molecular arguments. (b) Repeat for the van der Waals constant b. Explain your choice based on molecular arguments. SOLUTION (a) The van der Waals a constant is representative of attractive interactions due to dispersion, dipole–dipole, and induction forces. The first three species listed—CCl4, CF4, SiCl4— are all nonpolar and exhibit only dispersion forces. The magnitude of these forces is related to the polarizability a, of these species. In CF4, the valence electrons are held in toward the nuclei the most tightly (least “sloshy”), so this has the smallest dispersion forces. Conversely, the electrons in SiCl4 are the farthest away and most easily polarized. So we would expect that for these species:

aSiCl4 . aCCl4 . aCF4 The fourth species, SiCl3H, has two forces, dispersion and dipole–dipole, which add together. So the question becomes where we place this species in the hierarchy above. This is a tough call. We expect a fairly strong dipole 1. 1 D 2 , as shown below: H

Cl Si Cl

Cl

+

−

(Continued)

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236 ► Chapter 4. Equations of State and Intermolecular Forces

However, polarizabilities are additive among atoms in a molecule, and this species replaces a very polarizable atom (Cl) with an almost nonpolarizable atom (H). We may say that the dipole wins out and:

aSiCl3H . aSiCl4 . aCCl4 . aCF4 however,

aSiCl4 . aSiCl3H . aCCl4 . aCF4 and even aSiCl4 . aCCl4 . aSiCl3H . aCF4 are possible. (b) The van der Waals constant b is representative of the volume a molecule occupies. SiCl4 is certainly the largest and CF4 the smallest, but how about SiCl3H vs.CCl4? Si is bigger than C, but Cl is bigger than H. If you imagine how the atoms stack—starting with a triangle of Cl and then either an Si in the middle with an H on top or a C in the middle with a Cl on top—you can see that CCl4 is larger. So for b:

bSiCl4 . bCCl4 . bSiCl3H . bCF4

EXAMPLE 4.6 Calculation of van der Waals Constants from Critical Properties

Calculate the van der Waals parameters from critical point data for the following gases: benzene, toluene, cyclohexane. Explain the relative magnitudes of a and b from a physical basis. SOLUTION The van der Waals parameters are calculated from the critical pressure and temperature as follows: 27 1 RTc 2 2 Jm3 B R 64 Pc mol2

1 RTc 2

m3 R mol

Pc 3 bar 4

Tc 3 K 4

Benzene

49.1

562

1.88

1.19 3 1024

Toluene

42.0

594

2.45

1.47 3 1024

Cyclohexane

40.4

553

2.21

1.42 3 1024

a5

b5

8Pc

B

The attractive interactions of all three compounds are dominated by dispersion interactions (parameter a), while size affects parameter b. Toluene has the highest values for a and b. Toluene has seven carbon atoms, whereas the other molecules have only six. This results in the largest polarizability as well as the largest size. Cyclohexane’s electrons are freer to move than the tight resonance structure exhibited by benzene. This leads to a greater polarizability than benzene. In fact the magnitude of dispersion forces is closer to toluene than benzene. Finally, cyclohexane has a three-dimensional structure, while the other two are planar and flat. Hence, the constant b, representative of size, is almost as large for cyclohexane as it is for benzene.

EXAMPLE 4.7 Alternative Determination of van der Waals Constants from Corresponding States

c04.indd 236

At the critical point, the three roots in volume to a cubic equation must converge. Thus,

1 v 2 vc 2 3 5 0

(E4.7A)

Use Equation (E4.7A) to write the van der Waals parameters a and b in terms of the critical pressure and temperature.

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4.3 Equations of State ◄ 237

SOLUTION We can rewrite Equation (4.15) as follows: v3 2 c

RT a ab 1 b d v2 1 v 2 50 P P P

or at the critical point: v3 2 B

RTc a ab 1 b R v2 1 v 2 50 Pc Pc Pc

(E4.7B)

v3 2 3v2vc 1 3vv2c 2 v3c

(E4.7C)

Expanding Equation (E4.7A) gives:

We can now set each of the terms in volume from Equation (E4.7B) equal to those from Equation (E4.7C). For the root of v0, we have: ab 5 v3c Pc

(E4.7D)

a Pc

(E4.7E)

For the root of v1, we have: 3v2c 5 and for the root of v2, we have: 3vc 5 B

RTc 1 bR Pc

(E4.7F)

We can solve Equation (E4.7E) for the parameter a: a 5 3vc2Pc

(E4.7G)

and then Equation (E4.7D) for b: b5

Pcv3c vc 5 a 3

(E4.7H)

Finally, solving Equation (E4.7F) for b and substituting the result of Equation (E4.7H) gives: vc 5

3RTc 8Pc

(E4.7I)

We can solve for the parameters a and b by substituting Equation (E4.7I) into Equations

(E4.7G) and (E4.7H), respectively: a5

27 1 RTc 2 2 64Pc

and, b5

RTc 8Pc

Note the expressions we obtained for the parameters a and b in this example match those given by Equations (4.21) and (4.22).

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238 ► Chapter 4. Equations of State and Intermolecular Forces

Cubic Equations of State (General) As we have just seen, the van der Waals equation is an example of a cubic equation of state because its highest term in volume is raised to the third power. The van der Waals equation is presented because of the clear way in which it incorporates the attractive and repulsive interactions we have discussed. We will see that many modern cubic equations have the same basic form as the van der Waals equation, but are considerably more accurate. In other words, if you need an accurate answer, there are better equations to use than the van der Waals equation. In fact, hundreds of different cubic equations of state exist. All these equations are approximate. They merely fit experimental data. Yet, in general, many can provide reasonable values for both the vapor and liquid regions of hydrocarbons and the vapor region for many other species. In this section, we will not attempt to go through a critical review of all the available cubic equations of state; rather, we will illustrate the scientific concepts and engineering application through a few commonly used cubic equation. The majority of other equations that have been proposed are variations of the forms we will study. The general form of a cubic equation is: v3 1 f1 1 T, P 2 v2 1 f2 1 T, P 2 v 1 f3 1 T, P 2 5 0 where fi 1 T, P 2 represent a function that can contain fitting parameters as well as the properties T and P. The three characteristic roots in volume follow the same trends as those we discussed with the van der Waals equation in relation to Figure 4.12. Table 4.3 illustrates some examples of the form: P5

RT 2 Attr v2b

All these equations use the same “repulsive” term as the van der Waals equation. The term indicated by “Attr” quantifies attractive interactions. In general, these terms are empirically established to best fit experimental data. The Redlich–Kwong equation, the Soave–Redlich–Kwong8 equation, and the Peng– Robinson equation are all commonly used. The Redlich–Kwong equation of state is given by: TABLE 4.3 Parameters for Some Popular Cubic Equations of State of the Form P 5 RT/ 1 v 2 b 2 2 Attr

8

c04.indd 238

Equation

Year

van der Waals

1873

Redlich–Kwong

1949

Soave–Redlich–Kwong

1972

Peng–Robinson

1976

Attr a v2 a/ "T 1 v v 1 b2 aa 1 T 2 1 v v 1 b2 aaT 1 2 v v 1 b 1 b1v 2 b2

This equation is also referred to as the Redlich–Kwong–Soave equation of state.

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4.3 Equations of State ◄ 239

P5

RT a 2 1/2 v2b T v1v 1 b2

(4.24)

The relationships for parameters a and b can be written in terms of critical temperature and critical pressure using the same methodology that we applied to the van der Waals equation. In this case, the alternative method illustrated in Example 4.7 is convenient to implement (Problem 4.35). After working through the math, we get: a5 ¢

1 3

9 1 "2 2 1 2

≤

R2Tc2.5 0.42748R2Tc2.5 5 Pc Pc

(4.24a)

and, 3

b5 ¢

0.08664RTc "2 2 1 RTc ≤ 5 3 Pc Pc

(4.24b)

Note the parameters a and b in the Redlich–Kwong equation are different from those for the van der Waals parameters and cannot be interchanged. The Redlich–Kwong equation works well over a wide range of conditions but departs significantly from measured values near the critical point. In reduced form, the Redlich–Kwong Equation gives: Pr 5

3Tr 1 2 vr 2 0.2599 0.2599"Trvr 1 vr 1 0.2599 2

and the compressibility factor at the critical point is found to be: zc 5

1 3

While this value is closer to experimental values than the van der Waals equation, it is still too high. The Peng–Robinson equation of state is given by: P5 with,

aa 1 T 2 RT 2 v2b v1v 1 b2 1 b1v 2 b2 a 5 0.45724

(4.25)

R2T2c Pc

b 5 0.07780

RTc Pc

a 1 T 2 5 3 1 1 k 1 1 2 "Tr 2 4 2 k 5 0.37464 1 1.54226v 2 0.26992v 2 The compressibility factor at the critical point is found to be: zc 5 0.307. The Peng– Robinson equation is an option in the equation of state menu of the ThermoSolver software that comes with the text. The Redlich–Kwong equation with critical constant estimation of parameters uses a “two-parameter” corresponding states expression represented, in general, by Equation (4.12a). On the other hand, the Peng–Robinson equation utilizes the third parameter, w;

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240 ► Chapter 4. Equations of State and Intermolecular Forces thus, we expect it to be better suited for different classes of molecules. The Soave–Redlich– Kwong equation of state is a three-parameter equation similar in form to the Peng–Robinson equation and is also commonly used.

The Virial Equation of State The virial equation of state has a sound theoretical foundation; it can be derived from first principles using statistical mechanics. This equation is given by a power series expansion for the compressibility factor in concentration (or the reciprocal of molar volume) about 1/v 5 0: z5

Pv B C D 511 1 21 31c v RT v v

(4.26)

Here B, C, ... are called the second, third, ... virial coefficients; these parameters depend only on temperature (and composition for mixtures). An alternative expression for the virial equation is a power series expansion in pressure: z5

Pv 5 1 1 BrP 1 CrP2 1 DrP3 1 c RT

(4.27)

By solving Equation (4.26) for P and substituting into Equation (4.27), it is straightforward to show the two sets of coefficients are related by: Br 5 Cr 5

B RT

C 2 B2 and so on 1 RT 2 2

(4.28)

A common question is: What power series expansion do I use? Well, Equation (4.26) is explicit in pressure and Equation (4.27) is explicit in volume, so if you need an expression that is explicit in one of these variables (so you can take a derivative, for example), use the appropriate form. The next issue is a question of accuracy. It turns out that at moderate pressures (up to about 15 bar) when you keep only the second virial coefficient, the power series expansion in pressure is better: z5

Pv BP 5 1 1 BrP 5 1 1 RT RT

From 15 to 50 bar, the virial equation should contain three terms, and the expansion in concentration is more accurate: z511

B C 1 2 v v

Using statistical mechanics, we can relate the virial coefficients to intermolecular potentials. We will leave the derivation to a physical chemistry course and merely present the results. The second virial coefficient, B, results from all the “two-body” interactions in the system, that is, all the interactions between two molecules; the third virial coefficient, C, results from all the “three-body” interactions in the system; and so on. From this point of view, can you see why you need to include more and more terms as the pressure increases? Additionally, if the pressure is so low that not even two-body

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4.3 Equations of State ◄ 241

interactions affect the system properties, we have an ideal gas. As an example, consider the case of spherically symmetric molecules. According to statistical mechanics, the second virial coefficient is given by the following expression: `

B 5 2pNA 3 1 1 2 e2G1r2/1kT2 2 r2 dr

(4.29)

0

The principle of corresponding states is often applied to the truncated virial equation. It can be put in the form given by Equation (4.14): Br 5 B102 1 vB112 Br 5

where,

BPc RTc

Several correlations of parameters B102 and B112 to reduced temperature have been proposed.9 For example, Abbott found that they can be calculated by: B102 5 0.083 2

0.422 T1.6 r

B112 5 0.139 2

0.172 T4.2 r

and,

respectively. The Beattie–Bridgeman equation of state is a specific version of the virial equation: z5

Pv B C D 511 1 21 3 v RT v v B 5 B0 2

where,

C 5 2B0b 1 D5

(4.30)

A0 c 2 3 RT T cB0 A0a 2 3 RT T

bcB0 T3

where A0, B0, a, b, and c are adjustable parameters. The Benedict–Webb–Rubin equation of state modifies the virial equation by adding an exponential term. The form is given by: z 5 1 1 ¢ B0 2

1

9

c04.indd 241

A0 C0 21 a aa 25 2 ≤ v 1 ab 2 v bv22 1 RT RT3 RT RT

b g g 3 2 ¢ 1 1 2 ≤ exp ¢ 2 2 ≤ RT v v v

(4.31)

S. M. Walas, Phase Equilibria in Chemical Engineering (Boston: Butterworth, 1985).

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242 ► Chapter 4. Equations of State and Intermolecular Forces It has been shown to model both liquid and vapor PvT behavior well even close to the critical point, where it is the most challenging to accurately model PvT behavior. However, you must have values for all eight coefficients, and you must have the computational muscle to do the calculations. The extension of the Benedict–Webb–Rubin equation by Lee and Kessler is presented in Appendix E and forms the basis for the generalized compressibility charts discussed in Section 4.4.

Example 4.8 Development of the Expression for the Virial Equation in Concentration

Show that a Taylor series expansion for the compressibility factor, z, about temperature, T, and concentration, c 5 1/v, gives the form of the virial equation shown in Equation (4.26). SOLUTION We are looking for an equation of the form: z 5 f 1 T, c 2 Because this function is continuous, we can write the Taylor series expansion about the point

c 5 0:

z 5 f 1 T, c 5 0 2 1 ¢

'f 1 T, c 2 'c

≤ c50 c 1 ¢

'2f 1 T, c 2 'c2

≤ c50

'3f 1 T, c 2 c2 c3 1¢ ≤ c50 1 c 2! 'c3 3!

Because all gases become ideal gases at low concentration 1 c0 2 : f 1 T, c 5 0 2 5 z 1 T, c 5 0 2 5

RT 51 Pv

So, z511 ¢

'f 1 T, c 2 'c

≤ c50 c 1 ¢

'2f 1 T, c 2 'c2

≤ c50

'3f 1 T, c 2 c2 c3 1¢ ≤ c50 1 c 3 2! 'c 3!

We can rewrite this expression as: z5

P 5 1 1 Bc 1 Cc2 1 Dc3 1 c cRT

where, B5 ¢

'f 1 T, c 2 'c

≤ c5c0

and, ¢ C5

'2f 1 T, c 2 'c2 2!

≤ c5c0 c

Finally, this can be rewritten substituting c 5 1/v to give Equation (4.26): z5

Pv B C D 511 1 21 31 c v RT v v

From this development, we see that the virial coefficients, B, C, D, … depend on T, but not on v because they are evaluated at a specified value of v (i.e., c 5 1/v 5 0).

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4.3 Equations of State ◄ 243

Example 4.9 Calculation of the Second Virial Coefficient from Hard Sphere Potential

Calculate an expression for the second virial coefficient, B, for the hard sphere potential model. SOLUTION We begin with Equation (4.29), the expression for the second virial coefficient in terms of intermolecular potential: `

B 5 2pNA 3 1 1 2 e2G1r2/kT 2 r2dr 0

To use the hard sphere model (Figure 4.7a), we divide the integral into two parts. For distances of 0 to r, the potential is infinite 1 ` 2 so , e2G1r2/kT 5 0. From r to `, the potential is zero. Thus, we get: s

B 5 2pNA B 3

`

r2dr

1 3 1 1 2 e20/kT 2 r2drR s

0

Evaluating the integrals gives: B 5 2pNA

r3 s 2 3 0

And finally, 2 B 5 pNAs 3 3

Example 4.10 Calculation of the Second Virial Coefficient from Lennard-Jones Parameters

(E4.9A)

Calculate the second virial coefficient, B, for CH4 over the temperature range 100–900 directly from the Lennard-Jones parameters. Compare the values to the following data reported in the literature. T [K]

B

3 cm3 /mol 4

T [K] B 3 cm3 /mol 4

110

120

130

140

150

160

180

200

2330

2273

2235

2207

2182

2161

2129

2105

225

250

275

300

350

400

500

600

283

266

253

242

226

215

20.5

8.5

SOLUTION We can solve this problem by substituting the Lennard-Jones potential (Equation 4.10) into the molecular formulation for the second virial coefficient, Equation (4.29), as follows: `

`

B 5 2pNA 3 1 1 2 e2G1r2/kT 2 r2dr 5 2pNA 3 D1 2 exp§ 0

s 12 s 6 24e B ¢ ≤ 2 ¢ ≤ R r r kT

¥Tr2dr (E4.10)

0

(Continued)

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244 ► Chapter 4. Equations of State and Intermolecular Forces

The integral in Equation (E4.10A) cannot be solved analytically, so we must perform it numerically. Varying length scales are in the integrand. Therefore, for numerical accuracy, it is convenient to put the variables in dimensionless form. In this way the scale for each of the variables is approximately one, and the likelihood of a numerical error is reduced. We can scale the dependent variable, B, according to the result for the hard sphere potential, from Example 4.9 [Equation (E4.9A)]. So the dimensionless variables become: B* 5

B 2 3 3 ps NA

; T* 5

kT * r ;r 5 e s

where the temperature, T, and the distance, r, are scaled by the Lennard-Jones parameters 1 e/k 2 and s, respectively. Substituting these expressions into Equation (E4.10A) gives: `

B*

24 3 1 r* 2 12 2 1 r* 2 6 4 B 5 2 3 5 3 3 B1 2 exp ¢ ≤ R 1 r* 2 2dr* T* 3 ps NA

(E4.10B)

0

Equation (E4.10B) can be integrated numerically. For example, in MATLAB, the commands might read: f = @(rstar)(1.-(exp(-4./Tstar*(rstar.^-12-rstar.^- ... 6))).*rstar.^2); Bstar = 3 * quad(f,jstart,jend) We then numerically integrate Equation (E4.10B) at different values of dimensionless temperature, and convert back to T and B, using the Lennard-Jones parameters in Table 4.2. The results are plotted as a solid curve in Figure E4.10A. The experimental data given earlier are also plotted for comparison. The second viral coefficient obtained from the Lennard-Jones potential reasonably represents the experimental data.

100 Lennard - Jones Experiment

B [cm3 /mol]

0

−100

−200

−300

−400 100

300

500 Temperature [K]:

700

900

Figure E4.10A Comparison of values for the second virial coefficient for CH4 obtained using the Lennard-Jones potential function with experimental data at different temperatures.

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4.3 Equations of State ◄ 245

Equations of State for Liquids and Solids Liquid and solid molar volumes are straightforward to measure in the lab. For example, there is data available for the molar volume of many liquids at room temperature or at their normal boiling point. Table 4.4 reports the volume of a sample set of liquids and solids at 20°C and 1 bar. The volumes of condensed phases are also much less sensitive to temperature and pressure than gases. The measured values can be adjusted for temperature or pressure changes by using a Taylor series expansion on density. For liquids significantly below the critical temperature and for solids, we can neglect all terms but the first (linear) term of the Taylor expansion. This approach leads to quantification of the temperature and pressure dependencies of volume with the thermal expansion coefficient,10 b, and the isothermal compressibility, k, respectively. These quantities are defined as: b;

1 'v ¢ ≤ v 'T P

(4.32)

and, 1 'v k ; 2 ¢ ≤T v 'P

(4.33)

From inspection of Equations (4.32) and (4.33), it can be deduced that b has SI units of 3 K21 4 and k has SI units of 3 Pa21 4 . Representative values of b and k are reported in Table 4.4. More extensive compilations are found in many engineering and materials handbooks. TABLE 4.4 Molar Volume, Thermal Expansion Coefficient, and Isothermal Compressibility of Some Liquid and Solid Species at 20°C and 1 bar

v 3 cm3 /mol 4

b 3 K21 4 3 103

k 3 Pa21 4 3 1010

73.33 86.89 39.56 58.24 130.77 14.75

1.49 1.24 1.12 1.12

12.7 9.4 12.1 11.1 15.5 0.40

Liquid Acetone Benzene Methanol Ethanol n-Hexane Mercury

0.181

Solid Aluminum Copper Iron Diamond

9.96 7.11 7.10 3.42

0.0672 0.0486 0.035 0.0036

0.145 0.091 0.048 0.010

Source: R. H. Perry, D. W. Green, and J. O. Maloney (eds.), Perry’s Chemical Engineers’ Handbook, 7th ed. (New York: McGraw-Hill, 1997); D. R. Lide, CRC Handbook of Chemistry and Physics, 83rd ed. (Boca Raton, FL: CRC Press, 2002–2003).

10

c04.indd 245

This quantity is also referred to as the volume expansivity.

05/11/12 6:51 PM

246 ► Chapter 4. Equations of State and Intermolecular Forces Some of the equations of state discussed above are applicable to liquids as well as gases. For example, the Benedict–Webb–Rubin equation of state provides reasonable estimates for most hydrocarbons. The generalized compressibility charts that will be discussed in the next section are based on an extension of this equation of state and can be used for both gas and liquid phases. Alternatively, correlations have been developed explicitly for the liquid phase. For example, the liquid volume at saturation is given by the Rackett equation: vl,sat 5

EXAMPLE 4.11 Temperature Correction for Molar Volume of Solid Cu

RTc 1 0.29056 2 0.08775v 2 311 112Tr22/74 Pc

(4.34)

Determine the molar volume of copper at 500°C from the data in Table 4.4. SOLUTION

We can rewrite Equation (4.32) as follows:

a

'v b 5 bv 'T P

Separation of variables leads to: dv 5 bdT v

(E4.11A)

Integration of Equation (E4.11A) from state 1 at 20°C to state 2 at 500° C gives: ln

v2 5 b 1 T2 2 T1 2 v1

Solving for the molar volume of solid in state 2, and plugging in values from Table 4.4, we get: v2 5 v1exp 3 b 1 T2 2 T1 2 4 5 7.28 3 cm3 /mol 4

(E4.11B)

Since values for the thermal expansion coefficient are usually small, Equation (E4.11B) is often rewritten using a series expansion for the exponential: v2 < v1 3 1 1 b 1 T2 2 T1 2 4

(E4.11C)

In this example, the use of the approximation given by Equation (E4.11C) results in an error of only 0.03% as compared to Equation (E4.11B).

►4.4 GENERALIZED COMPRESSIBILITY CHARTS The principle of corresponding states invokes a unique generalized relation between the compressibility factor and reduced temperature and pressure for a given class of molecules. It is sometimes convenient to have graphs or tables that quantify this relationship. In this section, we present charts and tabular data for the compressibility

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4.4 Generalized Compressibility Charts ◄ 247

factor, z, in terms of Pr, Tr, and v. To account for different classes of molecules, we use the form: z 5 z102 1 vz112

(4.35)

The first term on the right hand side of Equation (4.35), z102, accounts for simple molecules, while the second term, z112, is a correction factor for the “nonsphericity” of a species. Both z102 and z112 depend only on Tr and Pr. Values for z102 and z112 vs. Pr at different values of Tr are shown in Figures 4.13 and 4.14, respectively. These charts are developed based on the Lee-Kesler equation of state.11 The same data are reported in tabular form in Appendix C (Tables C.l and C.2).12 If you want to find the volume at a specific temperature and pressure, it is straightforward to use these the graphs or tables. First, determine the reduced temperature 1 T/Tr 2 and reduced pressure 1 P/Pr 2 and look up the acentric factor. Then go to Figures 4.13 and 4.14 or Tables C.1 and C.2, and determine the compressibility factor by Equation (4.35). You can then calculate the volume from z. When P or T is unknown, a trialand-error method must be used. The generalized compressibility factor using the Lee–Kesler equation is an option in the equation of state menu of the ThermoSolver software that comes with the text.

1.3 1.2 4

1.1

3

1 2

0.9 1.7

z (0)

0.8

Vapor-liquid dome

0.7

1.5 1.3

0.6

1.2 1.15

0.5 0.4

1.1

0.3

1.05

1.0

0.2 0.8 0.1 0 0.01

0.9

Tr = 0.7 0.1

1

10

Pr

Figure 4.13 Generalized compressibility factor—simple fluid term. Based on the Lee–Kesler equation of state.

11

See Appendix E to see how these were calculated. Lee and Kesler’s value for the critical compressibility factor (Pr 5 1 and Tr 5 1) is at the inflection point of the critical isotherm, while Tables C.1 and C.2 report the value obtained directly from the solution of their equation of state. 12

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248 ► Chapter 4. Equations of State and Intermolecular Forces 2

0.3

3

4

1.7

0.2 1.5 0.1

z (1)

1.3 1.2

0 1.05

1.1

1.15

1.0

−0.1

0.8

Tr = 0.7

0.9

−0.2 Tr = 0.7 −0.3 0.01

0.1

1.0 0.8 0.9

1

10

Pr

Figure 4.14 Generalized compressibility factor-correction term based on the Lee-Kesler equation of state.

EXAMPLE 4.12 Calculation of v by the Redlich–Kwong Equation of State and the Generalized Compressibility Charts

Calculate the volume occupied by 10 kg of butane at 50 bar and 60°C using the Redlich– Kwong equation and the generalized compressibility charts. SOLUTION Using the Redlich-Kwong equation of state We first find the Redlich–Kwong parameters a and b using critical properties: a5

JK1/2m3 0.42748R2Tc2.5 0.08664RTc m3 5 29.08 B R and b 5 5 8.09 3 1025 B R 2 Pc mol Pc mol

We can use these with the Redlich–Kwong equation: P5

a RT 2 1/2 v2b T v1v 1 b2

Solving by trial and error, we get:

v 5 1.2 3 1024 3 m3 /mol 4 and, V5

10 m 3v5 3 1.20 3 1024 5 0.021 3 m3 4 MW .05812

Using the compressibility charts We first find Pr, Tr, and v: Pr 5

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P 50 bar 5 5 1.32, Pc 37.9 bar

Tr 5

T 333.2K 5 5 0.78, Tc 425.2K

and v 5 0.193

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4.5 Determination of Parameters for Mixtures ◄ 249

From Tables C.1 and C.2, we get:

z1

z0 Pr

Pr

Tr

1.3

1.4

Tr

1.3

1.4

0.75

0.2142

0.2303

0.75

20.0934

0.78 (interpolated)

0.2116

0.2274

20.0871 20.0843

0.80

0.2099

0.2255

20.0825

20.0883

0.80

20.0903

By double linear interpolation, where the first interpolation is performed in the table, we get: z102 5 0.2116 1 z112 5 20.0843 1

0.02 1 0.2274 2 0.2116 2 5 0.2148 0.1

0.02 1 20.0903 2 1 20.0843 2 2 5 20.0855 0.1

Thus,

z 5 z102 1 vz112 5 0.198 The low value for the compressibility factor indicates that butane is a liquid. Now solving for volume: v5

zRT 0.198 3 8.314 3 333.15 m3 5 5 1.1 3 1024 B R 5 P 50 3 10 mol

and, V5

m 10 3 1.1 3 1024 5 0.019 3 m3 4 3v5 MW .05812

The compressibility charts and the Redlich–Kwong equation give similar values for liquid butane at 50 bar and 60°C.

►4.5 DETERMINATION OF PARAMETERS FOR MIXTURES Most chemical and biological engineering processes include systems in which the working fluid is a mixture of two or more components. In Chapter 6, we will learn how to carefully approach the thermodynamic properties of mixtures; however, it is useful to introduce the approach to apply equations of state to mixtures while intermolecular interactions are fresh in our mind. When approaching mixtures, it is useful to consider all the possible types of interactions between the molecules in the mixture. These include “like” interactions between two molecules of the same species and “unlike” interactions between molecules of different species. For example, a binary mixture of species 1 and 2 can have three types of interactions: like interactions between 1 and 1, like interactions between 2 and 2, and unlike interactions between 1 and 2. Similarly, a ternary mixture can have six types of interactions, a quaternary mixture 10 types, and so on. As we quickly can recognize, virtually an unlimited combination of mixtures is possible. Mixtures can vary not only by the

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250 ► Chapter 4. Equations of State and Intermolecular Forces choice of components, but also in the quantity of each component present. Therefore, the most practical approach to using equations of state is to form mixing rules whereby we develop equations for properties of the mixture based on pure component data and apply them to the types of interactions present. Only the virial equation provides a complete theoretical basis for mixing rules. Some mixing rules are ad hoc and generated as much by mathematical convenience as by any firm theory. Other mixing rules for equations of state, however, can be related to the physical origin of the terms involved. In this section, we first study the mixing rules proposed by van der Waals that can be applied to his equation and also to other cubic equations of state. We then study mixing rules for the virial equation, and finally, mixing rules for critical properties that we can use with the generalized compressibility charts.

Cubic Equations of State Let’s see how we develop mixing rules according to those originally proposed by van der Waals for his equation of state. As we have seen, the van der Waals a term is related to the attractive force between two molecules, while we can consider the van der Waals b term to be related to the volume that a species occupies. A schematic for a binary mixture of species 1 and 2 is shown in Figure 4.15. The van der Waals term a1 represents the attractive interaction between two molecules of species 1, as shown in the figure. These interactions are based on a so-called two body interaction; that is one “1” molecule must find another “1.” It will occur in proportion to the mole fraction of the first “1” times the mole fraction of the other “1,” that is, y21. Similarly the 2-2 interaction will occur in proportion to y 22. The unlike 1-2 interaction will occur in proportion to y1y2 since a “1” molecule must find a “2” molecule, while the 2-1 interaction is in proportion to y2y1. Summing these attractive interactions together to account for their relative proportions gives: amix 5 y21a1 1 y1y2a12 1 y2y1a21 1 y22a2 However, the 1-2 and 2-1 interactions are equivalent, so: a12 5 a21 so the mixing rule simplifies to: amix 5 y21a1 1 2y1y2a12 1 y22a2

(4.36)

The cross coefficient is often found from pure species data according to: a12 5 "a1a2

(4.37)

or, if data are available for the binary pair in the form of the binary interaction parameter, k12, the cross coefficient can be written: a12 5 "a1a2 1 1 2 k12 2

(4.38)

The van der Waals coefficient b represents excluded volume, so, on average, it is given by: bmix 5 y1b1 1 y2b2

c04.indd 250

(4.39)

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4.5 Determination of Parameters for Mixtures ◄ 251

b1 1

b2 2

1

a

1

2

2 a 21

1 2 1

a12 = a21

a2

a12

1

2

1 2

2

Figure 4.15 Van der Waals interactions in a binary mixture of species 1 and 2.

An extension of the above mixing rules to a multicomponent mixture gives: amix 5 a a yiyjaij

(4.40)

bmix 5 a yibi

(4.41)

i

j

where aii 5 ai, and,

i

The mixing rules defined by Equations (4.38), (4.40), and (4.41) can be applied to any van der Waals type cubic equation of state—such as the Redlich–Kwong equation, the Soave–Redlich–Kwong equation, or the Peng–Robinson equation. In the latter cases, Equation (4.40) is written as: amix 5 a a yiyj 3 aa 1 T 2 4 ij i

(4.42)

j

Virial Equation of State Let’s now consider application of mixing rules to the virial equation. Since there is a sound theoretical basis for the virial coefficients in terms of intermolecular interactions, we can relate the virial coefficients for mixtures in terms of intermolecular potentials via Equation (4.29) with no arbitrary assumptions; that is, these mixing rules are rigorous results from statistical mechanics. Consider first a binary mixture of 1 and 2. Again, there are three different types of “two-body” interactions characteristic of the second virial coefficient: 1-1 interactions characterized by G11 and therefore B11, 2-2 interactions characterized by G22 and B22, and 1-2 interactions characterized by G12 and B12. The three second virial coefficients characteristic of these interactions depend only on the intermolecular potential; they are independent of concentration and composition. Thus, the second virial coefficient for a binary mixture is proportional to the number of different possible binary interactions weighted by the amount of species present. It is given by: Bmix 5 y21B11 1 2y1y2B12 1 y22B22

c04.indd 251

(4.43)

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252 ► Chapter 4. Equations of State and Intermolecular Forces where B12 5 B21. Note that Bmix refers to the parameter for the entire mixture and is different from B12, which refers to a specific binary interaction. In general, for a mixture of n components, the second virial coefficient is given by: n

n

Bmix 5 a a yiyjBij

(4.44)

i51 j51

Similarly, the third virial coefficient, which depends on three-body interactions, can be written as: n

n

n

Cmix 5 a a a yiyjykCijk

(4.45)

i51 j51 k51

So for our binary system, for example: Cmix 5 y31C111 1 3y21y2C112 1 3y1y22C122 1 y32C222 where C112 5 C121 5 C211. Once again, it is the theoretical foundation of the virial equations from statistical mechanics that provides validity to this extension to mixtures and makes them such a powerful tool. In fact, the virial equation is the only equation of state for which rigorous mixing rules are available.

Corresponding States To apply corresponding states and generalized correlations to mixtures, we need the relationship of the pseudocritical properties, the critical properties of the mixture, to the pure component critical properties. Many mixing relationships have been proposed. The simplest and most commonly used approximation is known as Kay’s rules. The pseudocritical temperature, Tpc, is given by averaging the critical temperature’s of each species in proportion to the amount of that species present in the mixture: Tpc 5 a yiTc,i

(4.46)

Similarly, the pseudocritical pressure, Ppc, and acentric factor, v pc, become:

and,

Ppc 5 a yiPc, i

(4.47)

v pc 5 a yi, v c, i

(4.48)

respectively. There is absolutely no basis for these rules other than convenience. Alternatively, a geometric mean combining rule for critical temperature has been used: Tpc, ij 5 "Tc, jTc,j This has been extended to include an additional parameter the binary interaction parameter krij to better fit experimental data: Tpc,ij 5 "Tc,iTc,j 1 1 2 krij 2 The understanding of the relationship between mixing rules and the thermodynamic properties of mixtures is still incomplete and warrants study.

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4.5 Determination of Parameters for Mixtures ◄ 253

EXAMPLE 4.13 PvT Calculations for Pure Species and Mixtures

Calculate the following: (a) The volume occupied by 20 kg of propane at 100°C and 70 bar (b) The pressure needed to fill a 0.1 m3 -vessel at room temperature to store 50 mol of propane (c) The pressure needed to fill a 0.1 m3-vessel at room temperature to store a mixture of 20 mol of propane and 30 mol of ethane SOLUTION As a general strategy, first check whether conditions represent ideal gas behavior. If they do, Pv 5 RTcan be used. If they do not, we must use an approach that incorporates nonidealities. If P and T are given, we can use the compressibility charts directly. If T and v are given, an equation of state of the form P 5 f 1 T, v 2 is easier, since we can calculate P directly. An accurate equation of state is the Redlich–Kwong equation: P5

RT a 2 1/2 v2b T v1v 1 b2

where the relationships for parameters a and b can be found using the principle of corresponding states: a5

0.42748R2T2.5 c Pc

b5

and

0.08664RTc Pc

Why did we choose this equation instead of the van der Waals equation? (a) At 70 bar, propane is not an ideal gas. Since we are given T and P, we can use the compressibility charts directly. First, we need to find the reduced pressure and reduced temperature using the critical data available in Appendix A: Pr 5

P 70 bar 5 5 1.65 Pc 42.4 bar

and

Tr 5

T 373K 5 5 1.01 Tc 370K

We also have to look up the value of the acentric factor: v 5 0.153 Interpolating from Tables C.1 and C.2, z 5 z102 1 vz112 5 0.2822 1 0.153 3 1 20.0670 2 5 0.272 So, V 5 nv 5

m zRT 20 3 103 0.272 3 8.314 3 373 ¢ ≤5 ¢ ≤ 5 0.0548 m3 MW P 44 70 3 105

(b) Here we are given T and v, so we can use the Redlich–Kwong equation. Plugging in constants (Pc 5 42.24 bar and Tc 5 370 K):

a5

JK1/2m3 0.42748R2T2.5 c 5 18.35 Pc mol2

and

b5

0.08664RTc m3 5 6.29 3 1025 Pc mol

These parameters give (with room temperature 5 295 K): P5

RT a 2 1/2 5 1.01 MPa v2b T v1v 1 b2 (Continued)

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254 ► Chapter 4. Equations of State and Intermolecular Forces

(c) Now we have a mixture of propane (1) and ethane (2), so we must use mixing rules. We can use a1 and b1 for propane as above. For ethane (Pc 5 48.7 bar and Tc 5 305.5 K): a2 5

JK1/2m3 0.42748R2T2.5 c 5 9.90 Pc mol2

and

b2 5

0.08664RTc m3 5 4.52 3 1025 Pc mol

We will use the van der Waals mixing rules with y1 5 0.4 and y2 5 0.6: amix 5 y21a1 1 2y1y2"a1a2 1 y22a2

bmix 5 y1b1 1 y2b2

This gives: amix 5 9.73

JK1/2m3 mol2

and

bmix 5 5.23 3 1025

m3 mol

Plugging into the Redlich–Kwong equation: P5

amix RT 2 1/2 5 1.12 MPa v 2 bmix T v 1 v 1 bmix 2

Note that the pressures reported in parts (b) and (c) are too large for the ideal gas law to be accurate.

►4.6 SUMMARY

c04.indd 254

In this chapter, we studied equations of state, which relate the measured properties P, v, and T. Examples include cubic equations of state (e.g., van der Waals, Redlich–Kwong, Peng– Robinson), the virial equation (with several specific forms), and the generalized compressibility charts. The Rackett equation allows us to estimate the molar volume of liquids at saturation, while the thermal expansion coefficient and the isothermal compressibility allow us to determine how to correct for the volumes of liquids and solids with temperature and pressure, respectively. We developed an understanding of the underlying form of these equations by looking at the molecular behavior of chemical species. “Molecular” energy, or internal energy, u, can be divided into two parts: molecular kinetic energy and molecular potential energy. In Chapter 1, we saw that molecular kinetic energy is proportional to the macroscopic property temperature. In this chapter, we identified the basis for potential energy between molecules. Specifically, we related intermolecular interactions to: point charges, dipoles, induced dipoles, dispersion (London) interactions, repulsive forces, and chemical effects. Dipole–dipole, induced dipole, and dispersion interactions all demonstrate a r26 dependence on the distance between the molecules and are collectively referred to as van der Waals forces. The molecular parameters, dipole moment, and polarizability determine the magnitude of these interactions. The molecular assumptions of the ideal gas model were relaxed to develop the van der Waals equation of state, by including a r26 attractive term and a hard sphere repulsive term. This equation heuristically illustrates how molecular concepts can be applied to developing an equation of state. In fact, it was shown that the more accurate cubic equations that have been developed since van der Waals’s time have the same general form. Alternatively, the virial equation results from a power series expansion of the compressibility factor, either in concentration (1/v) or in pressure. The values for the parameters in a given equation of state must be determined before it can be applied. The best course is to fit these parameters with measured experimental data. When measured data are not available, we can use the principle of corresponding states. On a molecular scale, the principle of corresponding states asserts that the dimensionless potential energy is the same for all species. On a macroscopic scale, it translates to the statement that all fluids at the same reduced temperature and reduced pressure have the same compressibility factor. We applied the principle of corresponding states to relate the parameters of equations of state to the critical temperature and pressure by noting that there is an inflection point on the critical isotherm

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4.7 Problems ◄ 255 at the critical point. Relations were given for the following cubic equations: van der Waals [Equations (4.21) and (4.22)], Redlich–Kwong [Equations (4.24a) and (4.24b)], and Peng–Robinson. We can extend the principle of corresponding states to account for different classes of molecules, based on the particular nature of the intermolecular interactions involved. One way to accomplish this objective is by introducing a third parameter—the Pitzer acentric factor, v.We then write the compressibility factor in terms of z102, which accounts for simple molecules, and z112, a correction factor for the “nonsphericity”: z 5 z102 1 vz112

(4.35)

Both z102 and z112 depend only on Tr and Pr. Values for z102 and z112 vs. Pr at different values of Tr are presented as generalized compressibility charts and shown in Figures 4.13 and 4.14, respectively. They are also reported in Tables C.1 and C.2 in Appendix C. These charts are based on the Lee–Kesler equation of state (see Appendix E). We use mixing rules to extend equations of state to mixtures. The mixing rules allow us to extrapolate these equations to mixtures, from mostly pure component data. Mixing rules for van der Waals–type parameters a and b were developed based on a two-body attractive interaction and a hard sphere repulsion, respectively. The binary interaction parameter allows us to better describe the cross coefficient, a12; however, data from the mixture are needed. Mixing rules for the viral coefficients arise from a theoretical basis. Mixing rules for the second virial coefficient, B, are based on two-body interactions; for the third virial coefficient, C, on three-body interactions; and so on. Finally, Kay’s rules were presented, from which we can find the psuedocritical properties of the mixture from the pure component properties. These values allow us to apply the generalized compressibility charts to mixtures.

►4.7 PROBLEMS Conceptual Problems 4.1 Consider BClH2. In each of the following cases, when do you expect the compressibility factor to be closer to one. Explain. (a) At 300 K, 10 bar or at 300 K, 20 bar (b) At 300 K, 20 bar or 1000 K, 20 bar (c) Consider a mixture of BClH2 and H2 at 300 K, 10 bar. Qualitatively plot the compressibility factor vs. mole fraction BClH2. Point out any important features. 4.2 The Lennard-Jones potential function is often used to describe the molecular potential energy between two species. Rank each of the following sets of species, from largest to smallest, in terms of Lennard-Jones parameters s and e. If there is no noticeable difference, write that they are roughly the same. Explain your choice using molecular arguments. (a) O2, S2, I2. O H H H H H H H3C _C_C_C_OH, H3C _C_O_C_CH3 , H3C _C_C_CH3 H H H H H H =

(b)

n-butanol

diethylether

methyl ethyl ketone

4.3 Using your knowledge of intermolecular forces, explain the following observation: (a) At 300°C and 30 bar, the internal energy of water is less than at 300°C and 20 bar. (b) At 300 K and 30 bar, the compressibility factor of isopropanol 1 H3CCOHCH3 2 is less than that of n-pentane 1 C5H12 2 , but at 500 K and 30 bar, the compressibility factor of isopropanol (H3CCOHCH3) is greater than that of n-pentane 1 C5H12 2 .

_

H O H3C _ C _ CH3 H isopropanol

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256 ► Chapter 4. Equations of State and Intermolecular Forces 4.4 Consider comparing 1 mole of NH3 at 10 bar and 500 K behaving as a real gas (i.e., considering its intermolecular interactions) vs. 1 mole of NH3 at 10 bar and 500 K behaving as an ideal gas (i.e., hypothetically “turning off” the intermolecular interactions). Answer the following questions using molecular arguments. Explain your choice with diagrams and descriptions of the interactions involved. (a) In which case is the compressibility factor, z, higher? (b) In which case is the internal energy, u, higher? (c) In which case is the entropy, s, higher? You need consider only the “spatial” contribution to entropy. 4.5 Consider comparing 1 mole of NH3 at 10 bar and 500 K vs. 1 mole of Ne at 10 bar and 500 K. Answer the following questions using molecular arguments. Explain your choice with diagrams and descriptions of the interactions involved. (a) In which case is the compressibility factor, z, higher? (b) In which case is the entropy, s, higher? You need consider only the “spatial” contribution to entropy. 4.6 The normal boiling points of some halide silanes are reported below. Explain the order in terms of intermolecular forces. Species

SiClF3

SiBrF3

SiCl3F

SiBr3F

SiICl3

Boiling point [°C]

270.0

241.7

12.2

83.8

114

4.7 Three isomers of C3H6O2 have the following normal boiling points: propanoic acid 1 CH3CH2COOH 2 , 141°C; methyl acetate 1 CH3COOCH3 2 , 58°C; and ethyl formate 1 HCOOCH2CH3 2 , 53°C. Using your understanding of intermolecular interactions, explain why (i) propanoic acid boils at a much higher temperature than the other two and (ii) the boiling points of methyl acetate and ethyl formate are relatively close. 4.8 Normal boiling points are shown for sets of species in the following tables. Explain the order based on your understanding of intermolecular interactions: (a) Alkyl halides

Species

CH3CH3

CH3CH2Cl

CH3CH2Br

CH3CH2I

Boiling Point (°C)

288

12

38

71

(b) Alkanes Species

CH4

CH3CH3

CH3CH2CH3

CH3CH2CH2CH3

Boiling Point (°C)

2161

288

242

20.4

4.9 If the diatomic gas of Problem 3.47 were nonideal at the pressures in the problem and attractive forces dominate, qualitatively describe how the final temperature in tank A would change from the answer you obtained in that problem. 4.10 Consider a high-pressure tank at room temperature. It undergoes a process where a valve is opened and the gas escapes until the pressure reaches 1 bar. (a) The process is undertaken with an ideal gas, as shown as system A. Will the final temperature T2A be greater than, equal to, or less than 298 K? Explain. (b) Consider now a tank of propane, a real gas, at the same initial pressure as the tank in part A. This tank undergoes an identical process where it is opened and the gas escapes until it too reaches

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4.7 Problems ◄ 257 System A: Ideal gas

System B: Propane

PE = 1 bar

PE = 1 bar

P = high

P = 1 bar

P = high

T = 298 K

T2A = ?

T = 298 K

Ideal gas

Process

State 2A

State 1A

T2B = ?

Process

C3H8

Tsurr = 298 K

P = 1 bar

Tsurr = 298 K

State 1B

State 2B

a pressure of 1 bar. It obtains a final temperature T2B. This is shown as system B. Will the final temperature T2B be greater than, equal to or less than T2A? Explain. 4.11 The second virial coefficient for argon is reported versus temperature in the following table. Explain the trend with temperature in terms of dominant intermolecular interactions. What can you say about what is happening around 410 K? T [K] B

3 cm3 /mol 4

100

200

300

400

500

600

700

800

2183.5

247.4

215.5

21

7

12

15

17.7

4.12 Table 4.3 compares the van der Waals (1873), Redlich–Kwong (1949), and Peng–Robinson (1976) equations of state in similar forms. Based on intermolecular interactions, qualitatively analyze how the progression of equations may have given more accurate results.

Numerical Problems 4.13 At very high temperatures, a gas can be ionized and remain in thermodynamic equilibrium. Consider the case of gas containing only ions, A1. Your supervisor requests that you come up with a simple (one-parameter) equation of state for this gas. Your assistant leaves you a memo that she has fit the PvT data to an equation of state of the form: ¢PA1 1

a ≤vA1 5 RT vnA1

She tells you the data fit this equation well but, unfortunately, leaves you no numbers. Your meeting with your supervisor is in 10 minutes, and your assistant is nowhere to be found! In order to be ready for the meeting, you need to answer the following questions: (a) Is the form of this equation reasonable? Explain. (b) What sign would you expect for the constant, a? Will this be a small or large number? Explain. (c) What number will you use for n (it can be a fraction)? What are the units of a? Show your work. 4.14 Consider a mixture of O2 (a) and C3H8 (b): (a) Write expressions for the attractive interactions Gaa, Gbb, and Gab as a function of distance between the molecules, r. (b) How does Gab compare to "GaaGbb? (c) Write a general expression for the average attractive intermolecular interaction in a mixture as a function of mole fractions of O2 and C3H8 represented by ya and yb, respectively.

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258 ► Chapter 4. Equations of State and Intermolecular Forces 4.15 While returning to your dorm late last night with a hot cup of coffee, the heat overcomes you and, much to your chagrin, you drop the paper cup, spilling its entire contents. As you had just spent your laundry money, this is somewhat upsetting, especially since you still have a good deal of thermo left to study and your last clean pair of pants are now covered with coffee. You yearn for the old days of polystyrene (Styrofoam) cups, which never got hot. Being an ambitious student (and looking for a distraction), you decide to come up with a process to recycle polystyrene (Styrofoam) so that environmental concerns will no longer keep the coffee shop from using this very good insulating material. After several hours, you have come up with what you think is a very reasonable process (you cannot wait to call the patent attorney!) and have just a few final issues to resolve. In the purification process, you believe you have reduced the polystyrene to its monomer, styrene, shown below: C C

styrene

In this case, the reactor would consist of 100 moles of styrene in a volume of 30 L at a pressure of 10 bar. You are concerned that the temperature is beyond the limit for the decomposition of styrene, 289°C. Since you are studying for the thermo exam, and have just gotten to the van der Waals equation, you want to decide whether this would be a good equation of state to use. (a) What deviations from ideality would you expect at these reaction conditions? In order of importance, list the types of intermolecular forces you think contribute to nonideality. Is the van der Waals equation appropriate? Explain. (b) Your search for experimental values for the van der Waals constants, a and b, is futile; you do, however, find values for the critical constants for styrene: Pc 5 39 bar Tc 5 374°C Calculate the temperature of the reactor, using the van der Waals equation. Will the styrene decompose? (c) Your classmate, who’s taking a polymers class, says the polystyrene probably has not reduced to a monomer but still exists as a reduced polymer chain, perhaps five monomers long: C

C

C

C

C C

Using only the information above, what are reasonable values for the van der Waals constants, a and b, of this reduced polymer chain? Explain. (d) Calculate the temperature in the reactor at the same reactor volume and pressure and initial Styrofoam mass as for part (b), except where you have a five-unit polymer instead of the monomer. Explain the difference in value to that calculated in part (b). Will decomposition occur (assume around 289°C)?

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4.7 Problems ◄ 259 4.16 The London force is directly related to the polarizabilities of the corresponding molecules. Consider the following table of molecular polarizability, a: Species

a 1 1025 cm3 2

CH4 C2H6 C3H8 CH3Cl CH2Cl2 CHCl3 CCl4

26 44.7 62.9 45.6 64.8 82.3 105

From these data, come up with a model to account for the contribution of each atom to the polarizability of a molecule. Predict the polarizability of C4H10 and C2H5Cl. 4.17 Consider 2 neighboring Ar atoms in a system of pure Ar at 25 bar and 300 K: (a) What is the average distance between them (in Å)? (b) Calculate the potential energy due to gravity (between the two atoms). (c) Calculate the potential energy due to London interactions. (d) Compare the values obtained in parts (b) and (c). 4.18 As discussed in the text, the repulsive term in the Lennard-Jones potential should have an exponential dependence rather than r212. Graphically compare the features of the Lennard-Jones potential to one that has the same attractive term but whose repulsive term is given by: C1eC2/r C1 and C2 are constants that you need to choose so the term above fits as closely as possible to the Lennard-Jones potential (Figure 4.8). Comment on the differences between these potential functions. 4.19 Calculate the bond strength in [eV] of a sodium ion in a crystal of NaCl. For the salt lattice: (a) Consider only the six nearest-neighbor Cl2 ions. The Cl2 ions are at a distance r 5 2.76 A+ .

from the Na1 ion. (b) In addition to the six nearest-neighbor Cl2 ions, include the twelve next-nearest-neighbor Na1 ions at a distance "2r. (c) Now include the eight next-next-nearest-neighbor Cl2 ions at a distance "3r. (d) Finally, include the six next Na1 ions at a distance of 2r. 4.20 Using data from Table 4.2, estimate the equilibrium bond length that would exist in a molecule of Xe2. 4.21 Calculate the van der Waals parameters from critical point data for the following gases: He, CH4, NH3, and H2O. Explain the relative magnitudes of a and b from a physical basis. 4.22 Calculate the van der Waals parameter b for CH4, C6H6, and CH3OH. Based on these values, estimate the molecular diameter of each species. Compare the values obtained with those in Table 4.2. 4.23 Calculate the van der Waals parameter a for CH4, C6H6, and CH3OH. Based on these values, estimate the value of C6 for each species. Compare the values obtained with that calculated by Equation (4.8). 4.24 Consider a cylinder fitted with a piston that contains 2 mol of H2O in a container at 1000 K. Calculate how much work is required to isothermally and reversibly compress this gas from 10 L to 1 L, in each of the following cases: (a) Use the ideal gas model for water.

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260 ► Chapter 4. Equations of State and Intermolecular Forces (b) Use the Redlich–Kwong equation to relate P, v, and T: P5

where,

RT a 2 1/2 v2b T v1v 1 b2

a 5 14.24 3 1 JK1/2m3 2 /mol2 4 and b 5 2.11 3 1025 3 m3 /mol 4

(c) Use the Steam tables. Compare these three methods. 4.25 Determine the second and third virial coefficients using the van der Waals equation of state. Hint: Begin by writing the van der Waals equation in compressibility factor form and performing a power-series expansion. The following mathematical relation is useful: 1 5 1 1 x 1 x2 1 x3 1 c 12x 4.26 Determine the second and third virial coefficients using Redlich–Kwong equation of state. Hint: Begin by writing the Redlich–Kwong equation in compressibility factor form and performing a power-series expansion. The following mathematical relation is useful: 1 5 1 1 x 1 x2 1 x3 1 c 12x 4.27 The Dieterici equation of state is given by: RT exp ¢2 P5

a ≤ RTv

v2b

(a) Find an expression for the parameters a and b in terms of the critical properties Tc and Pc. (b) Find the compressibility factor at the critical point, zc, for a Dieterici gas. (c) Rewrite the Dieterici equation in virial form for molar volume. What are the expressions for the second virial coefficient, B, and the third virial coefficient, C? The following mathematical relations are useful: 1 5 1 1 x 1 x2 1 x3 1 c 12x and, ex 5 1 1 x 1

x2 x3 1 1 c 2! 3!

4.28 Verify Equations (4.28) by rewriting the expansion of the virial equation in pressure [Equation (4.27)] in terms of the virial expansion in the reciprocal of molar volume [Equation (4.26)]. 4.29 Consider the Berthelot equation of state given below. Show how to calculate the constants a and b using only critical point data. P5

RT a 2 2 v2b Tv

4.30 Find the reduced form of the Berthelot equation of state. See Problem 4.29.

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4.7 Problems ◄ 261 4.31 (a) Use the data in the steam tables to come up with an expression for the second virial coefficient for water vapor. (b) Calculate the value of BH2O using the principle of corresponding states. Compare the value to that obtained in part (a). A helpful value is the critical volume for water: vc 5 56 3 cm3 /mol 4 4.32 Calculate the saturation pressure of n-pentane at 90°C by applying the “equal area” rule to (a) the Redlich–Kwong equation; (b) the Peng–Robinson equation. Compare these results to the measured value of 5.7 bar. 4.33 At 230°C, the saturation pressure of ethane is 10.6 bar. Calculate the densities of the liquid and vapor phases using the Peng–Robinson equation. Compare to the reported values for the liquid and vapor densities of 0.468 and 0.0193 g/cm3. 4.34 Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases. (a) The total gross sales of N2 is 30,000 units. Take the volume of the cylinder to be 43 L, the pressure to be 12,400 kPa, and the cost to be $6.1/kg. Compare your result to that you would obtain using the ideal gas model. (b) Repeat for 30,000 units of O2 at 15,000 kPa and $9/kg. 4.35 For the Redlich–Kwong equation, develop expressions for the parameters a and b, the equation reduced form, and the value of the compressibility factor at the critical point as a function of the critical pressure and the critical pressure using an approach similar to Example 4.7. 4.36 Calculate the second virial coefficient, B, for C6H6 over the temperature range 100–900 K directly from the values of the Lennard-Jones parameters reported in Table 4.2. Compare the values to the following data reported in the literature. T [K] B

3 cm3 /mol 4

290

320

340

360

380

400

440

480

520

560

600

21590 21230 21050 2920 2810 2710 2570 2470 2390 2340 2290

4.37 The square-well potential function is given by: ` for r # s1 G 5 c2e for s1 , r , s2 0 for r $ s2 Answer the following questions: (a) Sketch a plot of square-well potential energy versus distance. (b) Using this potential function, develop an expression for the second virial coefficient. (c) For CH4, the parameters are reported to be s1 5 2.856, s2 5 4.678, and e/k is 132.2. Calculate the value of B for methane at 200 K and 400 K. Compare the results to the experimental values reported in Example 4.9. 4.38 In this problem we seek to develop an expression for the van der Waals constants a and b in terms of molecular parameters using the Sutherland model for potential energy. (a) Show that writing the van der Waals model in virial form gives an expression for the second a virial coefficient as: B 5 b 2 . The following mathematical relation is useful: RT 1 5 1 1 x 1 x2 1 x3 1 c 12x

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262 ► Chapter 4. Equations of State and Intermolecular Forces (b) Develop an expression for the second virial coefficient from the Sutherland model. In doing this, use the series expansion for the exponential, ex 5 1 1 x 1

x2 x3 1 1 c 2! 3!

and only keep the first two terms. (c) Relate the results from parts A and B to develop an expression for the van der Waals parameters, a and b. 4.39 Experimental values for the second virial coefficients for NH3 versus temperature are reported in the following table. Use these data to obtain the best estimates that you can for the Lennard-Jones parameters 1 e/k 2 and s. T [°C]

0

25

50

100

150

200

250

300

B 3 cm3 /mol 4

2345

2261

2209

2142

2101

275

258

244

4.40 Determine expressions for the thermal expansion coefficient, b, and the isothermal compressibility, k, for an ideal gas. 4.41 Using the steam tables, estimate the values for the thermal expansion coefficient, b,and the isothermal compressibility, k, of liquid water at 20°C and 100°C. 4.42 Use the Rackett equation to calculate the liquid-phase molar volume of each of the following species at the same temperature as the measured values reported. Which species had the greatest absolute percent error? The least? Can the trend be explained by molecular concepts? (a) methane 1 CH4 2 , vexp 5 37.7 cm3 /mol at 111 K (b) ethane 1 C2H6 2 , vexp 5 54.8 cm3 /mol at 183 K (c) n-octane 1 C8H18 2 , vexp 5 162.5 cm3 /mol at 293 K (d) water 1 H2O 2 , vexp 5 18.0 cm3 /mol at 293 K (e) acetic acid 1 C2H4O2 2 , vexp 5 57.2 cm3 /mol at 293 K 4.43 Calculate the following: (a) the volume occupied by 20 kg of ethane at 70°C and 30 bar (b) the pressure needed to fill a 0.1 m3-vessel at room temperature to store 40 kg of ethane 4.44 Calculate the volume occupied by 50 kg of propane at 35 bar and 50°C, using the following: (a) the ideal gas model (b) The Redlich–Kwong equation of state (c) The Peng–Robinson equation of state (d) The compressibility charts (e) The textbook software, ThermoSolver 4.45 For a lecture-demonstration experiment, it is desired to construct a sealed glass vial containing a pure substance that can be made to pass through the critical point by heating the vial in a person’s hand. Thus, at room temperature the vial should contain a liquid and its vapor. (a) From the list of critical properties, select a suitable substance to be sealed within the vial. (b) What magnitude of pressures must the vial withstand? (c) For a vial of 100 cm3, how much of the substance should be enclosed in the vial? (d) Describe the changes within the vial as it is heated if it contains an amount of substance that is less than that calculated in part (c). 4.46 Compare the compressibility factor of methane at Tr 5 1.1 and Pr 5 1.2 using the Peng– Robinson equation of state and the compressibility charts. Repeat the calculations for methanol.

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4.7 Problems ◄ 263 4.47 Using the generalized compressibility charts, calculate the molar volume of ammonia at 92°C and 306.5 bar. What phase is ammonia in? 4.48 Use the Redlich–Kwong equation to calculate the size of vessel you would need to contain 30 kg of acetylene mixed with 50 kg of n-butane at 30 bar and 450 K. The binary interaction coefficient is given by k12 5 0.092. 4.49 You wish to use the Redlich–Kwong equation of state to describe a mixture of carbon dioxide (1) and toluene (2). To be as accurate as possible with the mixing rules, you want to include the binary interaction parameter, k12. In the literature, you find reference to an experiment with the following conditions: n1 n2 V T P

2.0 mol 3.0 mol 10.0 L 400.0 K 1.353 MPa

Using the data above and critical point property data, estimate k12. 4.50 The process that you are designing requires that a vapor mixture of benzene and propane enter a holding tank at 480 K and 2 MPa, and a benzene mole fraction of 0.6. The design specifies that 100 kg of vapor are in the tank. What volume must the tank be? (a) Calculate using the ideal gas model. (b) Calculate using the Redlich–Kwong equation. (c) What is the percent error assuming an ideal gas? 4.51 Calculate the second virial coefficient, B11, for pure Ar, B22 for pure CH4, and the second virial coefficient for the unlike interaction in a mixture, B12 at 239.8 K directly from the values of the Lennard-Jones parameters reported in Table 4.2. Compare the values to the following data reported in the literature: B11 5 231.4 cm3 /mol, B22 5 273 cm3 /mol, and B12 5 248.1 cm3 /mol. Use the following “mixing” rules for the Lennard-Jones parameters:

s12 5

1 1 s1 1 s2 2 and e12 5 "e1e2 2

4.52 You are planning an experiment in which you have a mixture of 5 moles of hydrogen 1 H2 2 . 4 moles of water 1 H2O 2 , and 1 of mole ethane 1 C2H6 2 . You want to calculate the pressure of this mixture to determine which material to use to construct the vessel to contain these gases. The vessel needs to be able to hold 12.5 L 1 0.0125 m3 2 , and the maximum temperature in the laboratory is 27°C. You then go to the library and find the pure species parameters for the van der Waals equation, a and b. However, when you get back to the laboratory, and realize you forgot to label them. (a) Using only molecular arguments, match each species to its appropriate set of parameters. Explain your reasoning. a 3 Jm3 /mol 4

b 1 m3 /mol 2

0.564 0.025 0.561

6.38 3 1025 2.66 3 1025 3.05 3 1025

Species

(b) Calculate the van der Waals parameters, a and b, for the mixture. (c) Calculate the pressure of this mixture.

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264 ► Chapter 4. Equations of State and Intermolecular Forces 4.53 The following second virial coefficients have been reported for a mixture of n-butane (1) and carbon dioxide (2) at 313.2 K. B11 5 2625 3 cm3 /mol 4 B22 5 2110 3 cm3 /mol 4 B12 5 2153 3 cm3 /mol 4 From these data, do the following: (a) Predict the molar volume of a mixture of 25 mol % butane in carbon dioxide at 313.2 K and 10 bar. (b) Estimate the value of the binary interaction parameter, k12, at 313.2 K. 4.54 Re-solve Example 4.9 using the text software, ThermoSolver. Compare your answer to the answer that is given in the example. 4.55 Solve the following using ThermoSolver: (a) In Species Database, select Ethane. Report its critical temperature and pressure and Dhof,298. (b) In Saturation Pressure Calculator, find the saturation temperature of ethane at 40 bar. (c) In Equation of State Solver, find the volume and compressibility factor of ethane in the following states using the Lee–Kesler equation (generalized correlations) and the Peng–Robinson equation. Report the value of each and the percent difference between the two methods: (i) P 5 40 bar, T 5 290 K; (ii) P 5 40 bar; T 5 302 K.

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► CHAPTER

5 The Thermodynamic Web Learning Objectives To demonstrate mastery of the material in Chapter 5, you should be able to: ► Apply the thermodynamic web to relate measured, fundamental, and derived thermodynamic properties. In doing so, apply the fundamental property relations, Maxwell relations, the chain rule, derivative inversion, the cyclic relation, and Equations (5.22), (5.23), and (5.24). Use Figure 5.3 to rewrite partial derivatives with T, P, s, and v in more convenient forms. ► Develop hypothetical paths to calculate the change in a thermodynamic property between two states, using appropriate property data. Appropriate data may include heat capacity data, pressure or volume explicit equations of state, or thermal expansion coefficients and isothermal compressibilities. ► Write the exact differential for any intensive thermodynamic property in terms of partial derivatives of specified independent, intensive properties. For example, given h = h(T, P), write dh. Define what is meant by independent properties and dependent properties. ► Write Ds, Du, and Dh in terms of independent properties T and P or the independent properties T and v. Use these expressions to solve first- and second-law problems. ► Define a departure function. Use generalized enthalpy and entropy departure functions to solve first- and second-law problems for systems that exhibit nonideal behavior. ► Define Joule–Thomson expansion and the Joule–Thomson coefficient. Explain how Joule–Thomson expansion is used in liquefaction.

►5.1 TYPES OF THERMODYNAMIC PROPERTIES We have seen that the thermodynamic state of a system can be characterized by its properties. Our goal in this chapter is to develop mathematical expressions through which we can relate the properties of a system to one another and to forms in which data are typically reported. We begin by defining three distinct categories of thermodynamic properties: measured properties, fundamental properties, and derived properties.

Measured Properties As we explored in Chapter 1, the measured properties are: P, v, T, composition 265

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266 ► Chapter 5. The Thermodynamic Web Measured properties are those properties that are directly accessible from measurements in the laboratory. Can you think of a couple of ways in which each of the properties above can be measured?

Fundamental Properties Observations of nature led us to the two laws of thermodynamics presented in Chapters 2 and 3. In formulating these laws, we introduced two new properties:

u

(from conservation of energy)

s

(from directionality of nature)

Since internal energy and entropy come from the two fundamental postulates of thermodynamics—that energy is conserved (First law) and that entropy of the universe always increases (Second law)—we call them fundamental properties. These properties cannot be measured directly. In fact, it could be said that these are not real things (at least in the measurable sense) but rather constructs of our mind to generalize experimental observations.

Derived Thermodynamic Properties Finally, the most distant from direct experience are derived thermodynamic quantities. These cannot be measured in the lab, nor are they properties directly fundamental to the postulates that govern thermodynamics; they are merely some specific combination of the above two types of properties that are defined out of convenience. Consider, for example, enthalpy: h 5 u 1 Pv

(5.1)

In open systems, the mass that crosses the boundary between the surroundings and the system always contributes to two terms in the energy balance: internal energy and flow (Pv) work. Since these terms are always coupled, it is convenient to define a property that includes both terms. In this way we never need to explicitly account for flow work. Likewise, enthalpy is a convenient property for a closed system undergoing a process at constant pressure. In this case, we need to consider both the change in internal energy and the Pv work. Two other convenient properties are the Helmholz energy, a ; u 2 Ts

(5.2)

g ; h 2 Ts

(5.3)

and the Gibbs energy,

For the time being, we will not elucidate why a and g may be conveniently derived thermodynamic properties.1 However, you should realize that because they are combinations of state functions, they, too, must be properties that are independent of path.

1

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In Chapter 6, we will learn why Gibbs energy is so useful.

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5.2 Thermodynamic Property Relationships ◄ 267

►5.2 THERMODYNAMIC PROPERTY RELATIONSHIPS Dependent and Independent Properties In this section, we will develop a web of property relationships whereby we can relate the thermodynamic properties we need to solve problems to properties we can measure in the lab. We want to relate fundamental and derived thermodynamic properties, such as u, s, h, a, and g, to things we can measure, such as measured properties P, v, T or to quantities for which measured data are typically reported, for example, cv, cP, b, and k. We will exploit the rigor of mathematics to allow us to develop an intricate web of these relationships. As with searching for sites on the Internet, there is usually more than one way to obtain our final answer; some are quicker, while others are slower. We limit our present discussion to constant composition systems; we will learn about mixtures that can change in composition in Chapter 6. Recall that the state postulate says that for systems of constant composition, values of two independent, intensive properties completely constrain the state of the system. In mathematical terms, the change in any intensive thermodynamic property of interest, z, can be written in terms of partial derivatives of the two independent intensive properties, x and y, as follows: dz 5 a

'z 'z b dx 1 a b dy 'x y 'y x

(5.4)

The total differential dz is exact; that is, the integral of dz is independent of path. On the other hand, the partial differential, ∂, indicates we are specifying a constraint to the path. For example, consider the surface representing z as a function of x and y, as shown in Figure 5.1. The slopes of the dashed lines represent the two partial derivatives written in Equation (5.4). In taking the partial derivative, (∂z/∂x)y, we evaluate the change in the dependent property z with respect to the independent property x over a path where the independent property y is constant. However, in inspecting the figure, we can clearly see that the value of the partial derivative (the slope of the line tangent to the curve) will be different at different values of the constant y. Similarly, the partial derivative, (∂z/∂y)x, represents the change in z with respect to y over a path x is constant. Through the use of partial derivatives, we can isolate the effect of one independent property by holding the second independent property constant.

z

∂z ∂y

x

∂z ∂x

y

y x

Figure 5.1 The surface represents the value of the dependent property z at any given value of independent properties x and y. The slope of the two dashed lines gives the values of the partial derivatives at the point indicated.

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268 ► Chapter 5. The Thermodynamic Web The form of Equation (5.4) is general and we can use it to express any of the properties we examined in Section 5.1 in terms of two independent properties. For example, say we want to calculate the change in internal energy for a first-law analysis of a closed system. We may choose to relate the differential change in internal energy, du, to the measured properties temperature, T, and molar volume, v. In the form of Equation (5.4), we write: du 5 a

'u 'u b dT 1 a b dv 'T v 'v T

(5.5)

In Equation (5.5), the intensive properties T and v constrain the state of the system; we call these two the independent properties. For brevity, we will use the notation: u 5 u 1 T, v 2 as an equivalent form of Equation (5.5), that is, to indicate our choice of T and v as independent properties to constrain u. All the other properties in the system are dependent properties since they are all constrained by the two independent properties. In Equation (5.5) the change in internal energy is written as an exact differential, du. The exact differential is used since once changes to both T and v are specified, the internal energy u can change in only one way, as constrained by the state postulate. We are free to specify any two independent properties to constrain the exact differential du. For example, we can use any of the following forms: u 5 u 1 T, P 2 , u 5 u 1 T, s 2 , u 5 u 1 h, s 2 , and so on. However, it turns out that the measured properties T and v are particularly convenient choices for the independent properties when looking at changes in u.

Hypothetical Paths (revisited) Now let’s look at how we can use the construct of dependent and independent properties to help us solve thermodynamics problems. Say we are solving a first-law problem where we need to determine the change in internal energy of a gas between two states. In this case, we might formulate the problem in the form of Equation (5.5), where u is the dependent property that is constrained by our choice of independent properties T and v, that is, u 5 u(T,v). In other words, we frame the problem that the gas in the system undergoes a process from state 1 (T1 and v1) to state 2 (T2 and v2). Figure 5.2 presents a graphical representation of our problem on a Tv diagram. Such diagrams are very useful in formulating solutions to problems. We often use sketches like this one that delineate the process in which we are interested in terms of the independent properties we have chosen. Knowing that u is a state function, we recognize that we may choose any path that is convenient to calculate the change in this property for a given process. One possible path from state 1 to state 2 is to vary both T and v simultaneously, as occurs in the actual process. This path is depicted by the solid line in Figure 5.2. Another possibility is to develop a hypothetical path in which we change only one property at a time (see Section 2.2). In the hypothetical path depicted in Figure 5.2, we first perform an isothermal expansion to a volume large enough for the system to behave as an ideal gas. In step 2, we perform a constant volume heating to the final temperature T2. Finally, step 3 consists of an isothermal compression to v2. Because internal energy is a state function, it is independent of path; therefore, both paths will have the same value for Du. In practice, the hypothetical process is easier for our calculations. In that case, we can use available data for ideal gas heat capacity to calculate Du for step 2. It is not as easy to find heat capacity data for nonideal gases! Thus, the hypothetical path shown in Figure 5.2 has been constructed to make use of data available in the literature.

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5.2 Thermodynamic Property Relationships ◄ 269 Δureal = Δuhypothetical = Δu1 + Δu2 + Δu3 Step 2

Δuhypothetical State 1 (T1, v1)

Step 3

Volume

Step 1

Ideal Gas

Δureal Actual path

State 2 (T2, v2)

Temperature

Figure 5.2 Computational paths for the change in internal energy from state 1 to state 2.

This theme will commonly recur as we solve problems in thermodynamics. To solve for the hypothetical path depicted in Figure 5.2, we also need to calculate the changes in steps 1 and 3. This task will be made possible by developing a web of thermodynamic relations, as we will see in this chapter. The preceding example illustrates an important strategy in solving thermodynamic problems. We can construct an infinite number of possibe paths to connect one state to another. Solving a problem often reduces to picking the proper path. In considering which path to choose, two things should be considered. (1) What property data are available? (2) Within the constraints posed by consideration (1), what path yields the easiest calculation? As you become aware of the forms in which data are typically available, you will become more proficient at identifying which paths to choose. However, if the first path you choose does not work, you can always try another. Sometimes that requires you to select a different set of independent properties. We wrote Equation (5.5) in terms of 1 'u/'T 2 v. Is there a difference between 1 'u/'T 2 v and ('u/'T)P?

Fundamental Property Relations Let’s consider again the calculation of internal energy. This time we begin with the fundamental postulates of thermodynamics. For a closed system undergoing a reversible process with only Pv work, the relations developed in Sections 2.3 and 3.3 can be applied to the differential energy balance, Equation (2.14). Hence, the first law and second laws are combined to give: du 5 dqrev 1 dwrev 5 Tds 2 Pdv

(5.6)

We can apply the definition of the derived thermodynamic property h given by Equation (5.1) to get: dh 5 du 1 d 1 Pv 2 5 Tds 1 vdP

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(5.7)

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270 ► Chapter 5. The Thermodynamic Web Similarly, we can apply the definitions of the derived thermodynamic properties a and g given by Equations (5.2) and (5.3) to get: da 5 du 2 d 1 Ts 2 5 2sdT 2 Pdv

(5.8)

dg 5 dh 2 d 1 Ts 2

and,

5 2sdT 1 vdP

(5.9)

respectively. Equations (5.6) through (5.9) are known as the fundamental property relations. Although Equation (5.6) was derived for a reversible process, the ultimate expressions that define the fundamental property relations are only between properties. Therefore, these equations can be applied to any process: reversible or irreversible. We can make this statement since properties are independent of path and, therefore, independent of process. So even though these equations were derived for the specific case of a reversible process, we can apply these equations even if the physical process that leads from one state to another is irreversible! However, in the case of irreversibility, Tds is no longer equal to the differential heat transferred across the boundary and 2Pdv no longer gives the value of dw. If we apply the approach that we developed in the previous section, we can write the change in internal energy in terms of independent properties s and v, that is, u 5 u 1 s, v 2 : du 5 a

'u 'u b ds 1 a b du 's v 'v s

(5.10)

For both Equations (5.10) and (5.6) to be true, we must have: a

'u b 5T 's v

and

a

'u b 5 2P 'v s

(5.11)

The grouping represented by u 5 u 1 s, v 2 results in the partial derivatives of Equation (5.10) corresponding to thermodynamic properties as defined in Equation (5.11). While any two properties can be used to constrain u, no other grouping of independent properties x and y, for u 5 u 1 x, y 2 , allows us to write partial derivatives in terms of thermodynamic properties as we did in Equation (5.11). Therefore, we say that 5 u, s, v 6 form a fundamental grouping. From Equation (5.7), we can see that 5 h, s, P 6 also form a fundamental grouping. It follows that: a

'h b 5T 's P

and

a

'h b 5v 'P s

(5.12)

Likewise, the fundamental grouping 5 a, T, v 6 results in: a

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'a b 5 2s 'T v

and

a

'a b 5 2P 'v T

(5.13)

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5.2 Thermodynamic Property Relationships ◄ 271

and 5 g, T, P 6 gives: a

'g 'T

b 5 2s P

and

a

'g 'P

b 5v

(5.14)

T

The latter two fundamental groupings give us our first insight into the utility of g and a. These derived thermodynamic quantities are grouped with measured properties, T, P, and v. Thus, there is a direct link between the change of dependent properties g and a and what we can measure in the lab. Specifically, we may anticipate that the Gibbs energy will become useful when we approach phase equilibria. As we saw in Chapter 1, T and P form the criteria for thermal and mechanical equilibrium; thus, the fundamental grouping 5 g, T, P 6 is of special interest. We will learn more about such things in Chapter 6.

Maxwell Relations Additional relations between thermodynamic properties and their derivatives can be derived from the second derivatives of the fundamental property relationships. These relations are called Maxwell relations and can be obtained by noting that the order of partial differentiation of an exact differential does not matter. For example, we can equate the following two sets of partial derivatives of the exact differential du from the fundamental grouping 5 u, s, v 6 : B

' 'u ' 'u a b R 5B a b R 'v 's v s 's 'v s v

(5.15)

Substitution of Equation (5.11) into Equation (5.15) gives: a

'T 'P b 5 2a b 'v s 's v

(5.16)

Similarly, from the other three fundamental property relationships we get: a

'T 'v b 5a b 'P s 's P

(5.17)

a

's 'P b 5a b 'v T 'T v

(5.18)

2a

's 'v b 5a b 'P T 'T P

(5.19)

Can you verify these last three relations? In the Maxwell relations given by Equations (5.18) and (5.19), all the properties on the right-hand side are measured properties! These permit the calculation of change in entropy from the measured PvT data. The derivative relations of Equations (5.6), (5.7), and (5.9) then enable us to calculate changes in u, h, and g.

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272 ► Chapter 5. The Thermodynamic Web

Other Useful Mathematical Relations In this section, we present three other mathematical relations that will be of use in helping us surf the thermodynamic web. The first relationship is the chain rule, which can be written in general as follows: a

'y 'z 'z b 5a b a b 'x a 'y a 'x a

(5.20)

Derivative inversion allows us to flip partial derivatives as follows: a

'x b 5 'z y

1 'z a b 'y y

(5.21)

We are not through yet. We can derive an additional relation based on the mathematical behavior of state functions. We begin with Equation (5.4): dz 5 a

'z 'z b dx 1 a b dy 'x y 'y x

(5.4)

If we take the partial derivative of each term with respect to x at constant z, we get:2 a

'y 'z 'z 'x 'z b 5a b a b 1a b a b 'x z 'x y 'x z 'y x 'x z

(5.22)

Since we cannot simultaneously change z and keep it constant, a

'z b 50 'x z

(5.23)

Equation (5.23) is useful in its own right. It is easy to see that: a

'x b 51 'x z

(5.24)

Applying Equations (5.21), (5.23), and (5.24) to (5.22) and rearranging gives the cyclic relation:3 21 5 a

'y 'x 'z b a b a b 'z y 'x z 'y x

(5.25)

2

The mathematical development of Equation (5.22) from Equation (5.4) is actually more complex than this heuristic explanation. However, viewing it in this simplified manner is convenient and, in general, works. We present it here since we will use this method in other places. See Example 5.1 for an alternative development of Equation (5.25). 3 This relation is alternatively termed the triple product rule.

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5.2 Thermodynamic Property Relationships ◄ 273

This expression is easy to remember; each property appears in the numerator once, appears in the denominator once, and is held constant once. Hence, Equation (5.25) is termed the cyclic relation.

EXAMPLE 5.1 Alternative Derivation for the Cyclic Rule

Develop an expression for the cyclic relation by equating z 5 z 1 x, y 2 and y 5 y 1 x, z 2 SOLUTION We can write the change in dependent property z in terms of independent properties x and y according to Equation (5.4): dz 5 a

'z 'z b dx 1 a b dy 'x y 'y x

(E5.1A)

Alternatively, we can choose y as the dependent property and write it in terms of independent properties x and z: dy 5 a

'y 'x

b dx 1 a z

'y 'z

b dz

(E5.1B)

x

Solving Equation (E5.1A) for dy using derivative inversion gives: dy 5 2a

'y 'z

b a x

'y 'z b dx 1 a b dz 'x y 'z x

(E5.1C)

The first term on the right-hand sides of Equations (E5.1B) and (E5.1C) must be equal; therefore, a

'y 'x

b 5 2a z

'y 'z

b a x

'z b 'x y

(E5.1D)

Rearranging Equation (E5.1D) and applying derivative inversion, we get the cyclic rule: 21 5 a

'y 'x 'z b a b a b 'z y 'x z 'y x

(5.25)

This derivation of the cyclic rule does not depend on the heuristic mathematical argument presented earlier; yet it gives the equivalent result.

Using the Thermodynamic Web to Access Reported Data We have seen that problem solving in thermodynamics frequently involves construction of hypothetical paths to find the change in a given property between two states. In applying this procedure, we often come up with a partial derivative of one property with respect to another, holding a third constant. In this section, we will use the thermodynamic web to translate partial derivatives to forms in which experimental data are routinely reported, such as cv, cP, b, k, and derivatives of equations of state. Figure 5.3 presents a way to navigate the thermodynamic web when partial derivatives with T, P, s, and v are encountered. It provides 12 permutations of partial derivatives between these properties. Derivative inversion can also be applied to form 12 additional

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274 ► Chapter 5. The Thermodynamic Web −

∂P ∂v s Cyclic Relation

∂P ∂s v

–

T = ∂T ∂s cv

v

∂T ∂v

Relations without T

∂s ∂v

Maxwell Relation In the case of:

P

A ∂P ∂T

s

Relations without P

∂s ∂v

P explicit or β = ∂P ∂T κ EOS

−

T

Relations without s v

−

∂P ∂v

∂P ∂s

T

∂T ∂v

P

B A = BC C s

Relations without v

∂s ∂T

P

=

cP T

= 1 or v explicit βv EOS

T

Figure 5.3 Roadmap through the thermodynamic web. Partial derivatives of T, P, s, and v are related to each other and to reported properties cv, cP, b, and k.

relationships to make a complete set. Each of the terms in Figure 5.3 is related to other terms using either the cyclic rule or Maxwell relations and is delineated with the appropriate icon as defined in the upper right of the figure. In those cases where the cyclic rule is used, the term at the origin of the two arrows can be replaced by the product of the two terms at the other end of the arrows. For example, at the top of the figure, 2 1 'P/'v 2 s can be replaced by the product 1 'P/'s 2 v 1 's/'v 2 P. Any of the terms on one side of the double arrow of a Maxwell relation can be replaced by the term on the other side. For example, we can use 2 1 'T/'v 2 s for 1 'P/'s 2 v. Any derivative presented in Figure 5.3 can be rewritten by following a path in the diagram to ultimately lead to set of quantities for which measured data are conveniently reported. Three general sources for such data are (1) derivatives of equations of state; (2) the thermal expansion coefficient, b, and the isothermal compressibility, k; and (3) the heat capacities at constant pressure, cP, and at constant volume, cv. First, if we have a pressure explicit equation of state, P 5 f 1 T, v 2 , we can analytically assess the partial derivative 1 'P/'T 2 v. A volume explicit equation of state, v 5 f 1 T, P 2 , allows us to obtain 1 'v/'T 2 P. Second, we may use the thermal expansion coefficient, b, and the isothermal compressibility, k as an alternative to equations of state to assess 1 'P/'T 2 v and 1 'v/'T 2 P. Both alternatives are indicated in Figure 5.3. These properties were defined in Section 4.3 as: 1 'v b;2 a b v 'T P

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(4.32)

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5.2 Thermodynamic Property Relationships ◄ 275

and, 1 'v k;2 ¢ ≤ v 'P T

(4.33)

Using the cyclic rule (see Problem 5.20), we get: ¢

b 'P ≤ 5 k 'T v

(5.26)

whereas applying the definition for thermal expansion coefficient gives: ¢

'v ≤ 5 bv 'T P

(5.27)

Third, partial derivatives of s with T can be related to heat capacity by applying the fundamental property relations to the definition of heat capacity. Inserting Equation (5.6) into the definition of heat capacity at constant volume, we get: cv 5 ¢

'u 's 'v 's ≤ 5 T¢ ≤ 2 P¢ ≤ 5 T¢ ≤ 'T v 'T v 'T v 'T v

Thus, ¢

cv 's ≤ 5 'T v T

(5.28)

Similarly applying the fundamental property relation for h, Equation 5.7, to the definition of cP gives: cP 5 ¢

'h 's 'P 's ≤ 5 T ¢ ≤ 1 v¢ ≤ 5 T ¢ ≤ 'T P 'T P 'T P 'T P

or, ¢

cP 's ≤ 5 'T P T

(5.29)

Equations (5.28) and (5.29) are used in the appropriate places in Figure 5.3. Using the paths in the diagram, we can rewrite any derivative in Figure 5.3. For example, we can follow the diagram to represent as 2 1 'P/'v 2 s as 2 cP/vkcv. See if you can justify this relation (and the others indicated in Figure 5.3). This exercise will pay dividends when you encounter these forms in solving problems with the thermodynamic web.

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276 ► Chapter 5. The Thermodynamic Web

►5.3 CALCULATION OF FUNDAMENTAL AND DERIVED PROPERTIES USING EQUATIONS OF STATE AND OTHER MEASURED QUANTITIES Relation of ds in Terms of Independent Properties T and v and Independent Properties T and P The relations shown in Figure 5.3 allow us to express the dependent property s in terms of measured properties. We can write s in terms of independent properties T and v by applying the form of Equation (5.4): ds 5 a

's 's b dT 1 a b dv 'T v 'v T

Substitution of Equation (5.28) and the Maxwell relation (5.18) into this equation gives: ds 5

cv 'P dT 1 a b dv T 'T v

(5.30)

Equation (5.30) is generally applicable to a system with constant composition. Integration of this equation gives: cv 'P Ds 5 3 dT 1 3 a b dv T 'T v

(5.31)

If we choose the measured properties T and P as our independent properties, we get: ds 5 a

's 's b dT 1 a b dP 'T P 'P T

Substituting in Equation (5.29) and the Maxwell relation (5.19) into this equation gives: ds 5

cP 'v dT 2 a b dP T 'T P

(5.32)

Integration of Equation (5.32) gives: cP 'v Ds 5 3 dT 2 3 a b dP T 'T P

(5.33)

If we use the ideal gas model, Equation (5.33) simplifies to Equation (3.22). It is instructive to compare Equation (5.31) to Equation (5.33). Equation (5.31) was developed using T and v as the independent properties, while Equation (5.33) used T and P. We see that in the former case we get a partial derivative in P. Therefore, this form is amenable to a pressure-explicit equation of state. Conversely, Equation (5.33) gives a partial derivative for v and is more amenable to a volume-explicit equation of state. This observation holds generally; that is, T and v are convenient independent properties when we have a pressure-explicit equation of state, while T and P are convenient for a volume-explicit equation. This rule of thumb is reinforced in Example 5.4, where the choices of independent properties (T,v) and (T,P) are compared for a calculation using the pressure-explicit Redlich–Kwong equation of state.

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 277

Relation of du in Terms of Independent Properties T and v We can now go back to the calculation for Du illustrated by the hypothetical path in Figure 5.2. Recall that we can relate the differential change in internal energy to the independent properties T and v by Equation (5.5): du 5 a

'u 'u b dT 1 a b dv 'T v 'v T

(5.5)

The first term on the right-hand side of Equation (5.5) can be written using the definition of heat capacity at constant volume: a

'u b 5 cv 'T v

We usually want to evaluate this term under ideal gas conditions, since ideal gas heat capacity data are readily available for most gases. In Example 5.3, we will take another approach. The second term can be simplified using the fundamental property relation, Equation (5.6): a

T's 2 P'v 's 'u b 5a b 5 BTa b 2 P R 'v T 'v 'v T T

We can then use the Maxwell relation, Equation (5.13), to get: a

'u 'P b 5 BTa b 2 P R 'v T 'T v

(5.34)

In general, we can add together the effect of changes in both independent properties to get: du 1 T, v 2 5 cvdT 1 BTa

'P b 2 P Rdv 'T v

(5.35)

Integrating Equation (5.35) gives: 'P Du 1 T, v 2 5 3 cvdT 1 3 BTa b 2 P Rdv 'T

(5.36)

v

We can solve the second integral on the right-hand side of Equation (5.36) with the appropriate PvT data or with data for the thermal expansion coefficient and the isothermal compressibility. For example, if an equation of state is available of the form P 5 f 1 T, v 2 , we can take the partial derivative with respect to T at constant v, multiply by T, subtract P, and integrate. If no equation is available, we could solve graphically. The hypothetical path depicted in Figure 5.2 allows us to reduce Equation (5.36) to one term for each step. The first and third steps are isothermal, so the first term on the right-hand side goes to zero and we use only the second term. Conversely, step 2 is isochoric and we use only the first term. Furthermore, we constructed this path to carry out step 2 under ideal gas conditions, where heat capacity data are readily

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278 ► Chapter 5. The Thermodynamic Web available. The ideal gas heat capacity depends only on the individual molecular structure of the species themselves, not on the interactions between them.4 On the other hand, the second term relates to b and k or an equation of state and its derivatives, and it depends on how the molecules interact with one another. In constructing these hypothetical thermodynamic paths—be it for u, s, h, or other properties—we often choose T for one independent property and either P or v for the other independent property. In doing so, we relate the T dependent term to cP or cv, which can be reduced to the individual molecular structure of the species in the system. The P or v term then accounts for the interactions of the species with one another.

EXAMPLE 5.2 First-Law—ClosedSystem—Calculation Using the Thermodynamic Web

One mole of propane gas is to be expanded from 0.001 m3 to 0.040 m3 while in contact with a heating bath that keeps the temperature constant at 100ºC. The expansion is not reversible. The heat extracted from the bath is 10.4 KJ. Using the van der Waals equation of state, determine the work for the expansion. SOLUTION To find the work, we apply the first law: Du 5 q 1 w

(E5.2A)

Since we know q 5 10,400 J/mol, we just need to find Du. A schematic of the path is shown in Figure E5.2A. In this case, the actual path can be used for the computation. Writing du as a function of the independent properties T and v: du 5 a

'u 'u 'P b dT 1 a b dv 5 cvdT 1 BTa b 2 PR dv 'T v 'v T 'T v

(E5.2B)

where we have applied Equation (5.35). This process occurs at constant T, so integration of Equation (E5.2B) gives: 'P Du 5 3 BTa b 2 PR du 'T

(E5.2C)

v

State 2 (T2, v2)

0.040 m3

Volume

Δu

State 1 (T1, v1)

0.001 m3 Temperature

Figure E5.2A

4

c05.indd 278

100°C

Computational path to solve the problem in Example 5.2.

Recall the discussion after Equation (2.25).

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 279

From the van der Waals equation, we have: RT a 2 2 v2b v

(E5.2D)

'P R b 5 'T v v 2 b

(E5.2E)

P5

a

and differentiating:

Using Equations (E5.2D) and (E5.2E) in Equation (E5.2C) gives: Du 5 3 B

or,

Du 5

2a 2 v

0.040 m3

52 0.001 m3

5 2¢0.96B

Jm3 mol2

a Rdu v2

27 1 RTc 2 2 1 1 ¢ 2 ≤ v2 v1 64Pc

R≤a

J 1 mol 1 2 b B 3 R 5 936 0.04 0.001 m mol

Using this value of Du in Equation (E5.2A) gives: w 5 Du 2 q 5 2 9,460 3 J/mol 4 The value of the work is negative, indicating the gas is doing work on the surroundings.

EXAMPLE 5.3 Alternative Calculation Path for Du

Develop a methodology for calculating Du according to the path shown in Figure E5.3A, in which the change in T occurs when intermolecular interactions are important. SOLUTION There are many ways to get from one thermodynamic state to another using the thermodynamic web. We consider the path shown in Figure E5.3A as an alternative calculation path for Du to that presented in Figure 5.2. In this case, our hypothetical path consists of two steps: isochoric heating (step 1) followed by isothermal compression (step 2). However, for the temperature change in step 1, the gas no longer behaves as an ideal gas. Again, we can write the change in internal energy as a function of the independent properties T and v: du 5 a

'u 'u 'P b dT 1 a b dv 5 creal b 2 PRdv v dT 1 BTa 'T v 'v T 'T v

(E5.3A)

We have used the results given by Equation (5.35). In calculating the internal energy change in step 1, we must now recognize that the heat capacity is no longer under ideal gas conditions but rather in the region where intermolecular interactions are significant; therefore, cv now depends on both T and v, that is, creal 5 cv 1 T, v1 2 v (Continued)

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280 ► Chapter 5. The Thermodynamic Web

Alternative path to calculate Δu

Step 1

Volume

Step 2

ΔuStep 1

ΔuStep 2

State 1 (T1, v1)

State 2 (T2, v2)

Temperature

Figure E5.3A Alternative computational path to Figure 5.2 for the change in internal energy from state 1 to state 2.

In general, however, heat capacity data are available only at ideal gas conditions. Hence, we need to relate the real heat capacity at any given temperature along step 1 to the ideal gas heat capacity. Again, this task can be accomplished by utilizing the thermodynamic web. As illustrated in Figure E5.3B, we need to calculate 1 'cv /'v 2 T and then integrate from the volume of an ideal gas to the volume of the real gas, v1. We can develop this relationship by using the definition of heat capacity and then changing the order of differentiation (as we did in developing the Maxwell relations):5 ¢

'cv ' 'u ' 'u ≤ 5B a b R 5B ¢ b R 'v T 'v 'T v T 'T 'v T v

(E5.3B)

Using the value for 1 'u/'v 2 T obtained in Equation (5.34) and expanding, we get: ¢

'cv ' 'P 'P '2P 'P '2P ≤ 5 B ¢ BTa b 2 PR ≤ R 5 a b 1 T¢ 2 ≤ 2 a b 5 T¢ 2 ≤ 'v T 'T 'T v 'T v 'T v 'T v 'T v v

where we have used the product rule.

Path to calculate cvreal cvideal gas = cv (T only)

v∞

real

Volume

v v

v1

∫

ideal gas

∂cv dv ∂v T

cvreal = cvreal (T, v1)

Temperature

Figure E5.3B Computational paths to ideal gas. calculate creal v from cv

5 Again, we see a common theme of creative problem solving: being able to take what we learn in one place and apply it in a different context.

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 281

The differential change in heat capacity at any given temperature is therefore given by: dcv 5 BT¢

'2P ≤ Rdv 'T2 v

(E5.3C)

Integrating both sides of Equation (E5.3C) gives: real

vreal

ideal gas

videal gas

'2P 3 dcv 5 3 BT¢ 'T2 ≤ Rdv v

Solving for creal v , we get: v1

creal v

5 cv 1 T, v1 2 5

gas cideal v

'2P 1 3 BT¢ 2 ≤ Rdv 'T v

(E5.3D)

videal gas

Examining Equation (E5.3D), we see that if we have ideal gas heat capacity data and an appropriate equation of state, we can solve for the real gas heat capacity. We can then use this expression in Equation (E5.3A) to get: v1

du 5

gas bcideal v

'2P 'P 1 3 BT¢ 2 ≤ Rdv rdT 1 BTa b 2 PRdv 'T v 'T v videal gas

Relation of dh in Terms of Independent Properties T and P We can obtain an expression for the change in enthalpy in terms of the independent properties T and P by applying the thermodynamic web in a manner similar to that used for entropy and internal energy. We first write the differential expression in the form of Equation (5.4) using the independent properties T and P: dh 5 a

'h 'h b dT 1 a b dP 'T P 'P T

Applying the definition for heat capacity: a

'h b 5 cP 'T P

and the fundamental property relation for h, Equation (5.7), and the Maxwell relation (5.14): a

'h T's 1 v'P 's 'v b 5a b 5 Ta b 1 v 5 2Ta b 1 v 'P T 'P 'P 'T T T P

Substitution gives: dh 5 cPdT 1 B2Ta

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'v b 1 vRdP 'T P

(5.37)

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282 ► Chapter 5. The Thermodynamic Web Integrating this expression gives: Dh 5 3 cPdT 1 3 B2Ta

'v b 1 vRdP 'T P

(5.38)

In analogy to our discussion for Du, we usually construct a path whereby we take the system to a low enough pressure to apply the ideal gas heat capacity. Alternatively, we could calculate the real heat capacity at constant pressure using the same method we used to calculate creal v in Example 5.3. In Problem 5.25 you will verify that: Preal

creal P

5 cP 1 T, P 2 5

gas cideal P

'2v 2 3 BT ¢ 2 ≤ RdP 'T P

(5.39)

Pideal gas

EXAMPLE 5.4 First-law—OpenSystem—Calculation Using the Thermodynamic Web

The first step in manufacturing isobutene from isomerization of n-butane is to compress the feed stream of n-butane. It is fed into the compressor at 9.47 bar and 80ºC and optimally exits at 18.9 bar and 120ºC, so that it can be fed into the isomerization reactor. The work supplied to the compressor is 2100 J/mol. Calculate the heat that needs to be supplied into the unit per mole of n-butane that passes through. SOLUTION First, the process is sketched in Figure E5.4A. To find the heat in, we will apply the first law (i.e., do an energy balance). Assuming steady-state, the open-system energy balance with one stream in and one stream out can be written: S

S

# # V2 V2 0 5 n# 1 ah 1 MW 1 MWgzb 2 n# 2 ah 1 MW 1 MWgzb 1 Q 1 Ws 2 2 1 2 or, # # Q Ws q 5 # 5 h2 2 h1 2 # n n We know the power used by the compressor. Thus this problem reduces to finding the change in the thermodynamic property enthalpy from the inlet to the outlet. We know two intensive properties at both the inlet and outlet, so the values for the other properties (like enthalpy!) are already constrained. Since pressures on the order of 10 bar are being used, we do not expect the ideal gas law to hold. However, since enthalpy is a property and independent of path, we are free to choose whatever path is convenient.

wS

P1 = 9.47 bar T1 = 80°C

Compressor

q

c05.indd 282

P2 = 18.9 kPa T2 = 120°C

Figure E5.4A Schematic of the compressor for Example 5.4.

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 283

First, it may be helpful to find some data for n-butane. From Appendix A.2, we have an expression for the ideal gas heat capacity: cP 5 1.935 1 36.915 3 1023 T 2 11.402 3 1026 T2 R with T in [K]. Since this expression is limited to ideal gases, any change in temperature must be under ideal conditions. An equation of state may also be useful. Many equations are available; an accurate one is the Redlich–Kwong equation of state: P5

RT a 2 1/2 v2b T v1v 1 b2

We can find the parameters a and b using the principle of corresponding states. To do this, we first find critical parameters from Appendix A.1: 1 Tc 5 425.2 K; Pc 5 37.9 bar 2 . Applying Equations (4.24a) and (4.24b): a5

JK1/2m3 0.42748R2T2.5 0.08664RTc m3 c 25 B 5 29.08B R and b 5 5 8.09 3 10 R Pc mol2 Pc mol

Next we choose a path to solve this problem. We are constrained in that any change in temperature should be carried out when C4H10 behaves as an ideal gas. We first will solve this with T and v as independent properties, then show that it can be solved with T and P as independent properties. Solution with T and v The solution path is depicted in the Tv plane in Figure E5.4B. To execute this path, we need to know the molar volumes of states 1 and 2. The Redlich– Kwong equation is implicit in v; solving for the three roots and taking the largest gives: v1 5 2.59 3 1023 3 m3 /mol 4 and , v2 5 1.27 3 1023 3 m3 /mol 4

v

Step 2

v1, T1

Step 3

Step 1

Ideal gas

Δh =

q+

ws v2, T2

T

Figure E5.4B Solution path in the Tv plane for the compressor in Example 5.4. (Continued)

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284 ► Chapter 5. The Thermodynamic Web

Now we can write an expression for enthalpy in terms of our independent properties T and v: dh 5 a

'h 'h b dT 1 a b dv 'T v 'v T

For steps 1 and 3, T is constant: Dh1 or 3 5 3 a

'h b dv 'v T

(E5.4A)

From the fundamental property relation, Equation (5.7), Equation (E5.4A) becomes: Dh 5 3 a

T's 1 v'P 's 'P b dv 5 3 BTa b 1 va b Rdv 'v 'v T 'v T T

If we use the Maxwell relation (5.18), we get: Dh 5 3 BTa

'P 'P b 1 va b Rdv 'T v 'v T

(E5.4B)

We can now use the Redlich–Kwong equation, which is explicit in pressure: P5

RT a 2 1/2 v2b T v1v 1 b2

(E5.4C)

Differentiating Equation (E5.4C) gives: 'P R a 1 3/2 b 5 'T v v 2 b 2T v 1 v 1 b 2

(E5.4D)

a 1 2v 1 b 2 'P RT 1 1/2 2 b 52 2 1v 2 b2 'v T T 1 v 1 bv 2 2

(E5.4E)

a and, a

Plugging Equations (E5.4D) and (E5.4E) back in to Equation (E5.4B) gives: a 1 2v 1 b 2 R a RT Dh 5 3 BT c 1 3/2 1 1/2 2 R Rdv d 1 v B2 2 1v 2 b2 v2b 2T v 1 v 1 b 2 T 1 v 1 bv 2 2 Simplifying, Dh 5 3 B2

v 1 2v 1 b 2 bRT a 1 1 1 1/2 ¢ ≤ Rdv 1v 2 b22 3 v 1 v 1 b 2 42 T 2v 1 v 1 b 2

(E5.4F)

and integrating: Dh1 or 3 5

vf bRT a 3 v 1 1 1/2 ¢ lna b2 b2 1v 2 b2 1 v 1 b 2 vi T 2b v1b

We can now plug in numerical values. What should we use for vlarge? For step 1, we use T 5 353.15 K, getting: Dh1 5 1368 3 J/mol 4

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 285

For step 3, we use T 5 393.15 K, getting: Dh3 5 22542 3 J/mol 4 For step 2, we have an ideal gas undergoing an isochoric process: Dh2 5 3 a

' 1 u 1 Pv 2 'h b dT 5 3 a b dT 'T v 'T v

or, Dh2 5 3 B a

'u 'Pu b 1a b RdT 5 3 3 cv 1 R 4 dT 5 3 cPdT 'T v 'T v

where we used the ideal gas law to simplify. Plugging in numbers: 393

Dh2 5 R 3 3 1.935 1 36.915 3 1023 T 2 11.402 3 1026 T2 4 dT 353

and integrating: Dh2 5 8.314 3 1.935 1 393 2 353 2 1 18.458 3 1023 1 3932 2 3532 2 2 3.801 3 1026 1 3933 2 3533 2 4 Dh2 5 4696 3 J/mol 4 Finally, summing together gives the heat input as: Dh 5 Dh1 1 Dh2 1 Dh3 5 q 1 ws 5 3522 3 J/mol 4 q 5 Dh 2 ws 5 1422 3 J/mol 4 Solution with T and P The solution path is depicted in the TP plane in Figure E5.4C. In this case, we need a low pressure to obtain ideal gas behavior. If we write the change in enthalpy in terms of the independent properties T and P, we get: dh 5 a

'h 'h 'u b dT 1 a b dP 5 cPdT 1 B2Ta b 1 v RdP 'T P 'P T 'T P

(E5.4G)

P P2, T2 Δh =

q+

ws

Step 1

Step 3

P1, T1

Ideal gas Step 2

T

Figure E5.4C Solution path in the TP plane for the compressor in Example 5.4. (Continued)

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286 ► Chapter 5. The Thermodynamic Web where Equation (5.37) was used. The constant-pressure part, step 2, is equivalent to Dh2 above, so the result is identical. For steps 1 and 3, we must evaluate the second term on the righthand side of Equation (E5.4G) and integrate. We cannot write the Redlich–Kwong equation explicitly in v to solve. However, we can differentiate Equation (E5.4C) to get: dP 5 B2

a 1 2v 1 b 2 RT 1 1/2 2 Rdv 2 1v 2 b2 T v 1v 1 b22

(E5.4H)

At constant T, substituting Equation (E5.4H) into (E5.4G) gives: dh 5 B2Ta

a 1 2v 1 b 2 'v RT 1 1/2 2 Rdv b 1 v R B2 1v 2 b22 'T P T v 1v 1 b22

(E5.4I)

The partial derivative of volume with respect to temperature cannot be obtained directly. However, we can apply the cyclic relation as follows: 21 5 a

'v 'P 'T b a b a b 'T P 'v T 'P v

(E5.4J)

We can rewrite Equation (E5.4J) in terms of a form that allows us to use the P explicit Redlich– Kwong equation. Performing this manipulation and taking derivatives gives: a R 'P 2 3/2 2 a b v2b 2T v 1 v 1 b 2 'T v 'v 5 a b 52 'T P 'P a 1 2v 1 b 2 RT a b B2 1 1/2 2 R 2 'v T 1v 2 b2 T v 1v 1 b22

(E5.4K)

Substitution of Equation (E5.4K) into (E5.4I) and simplification gives: dh 5 B2

v 1 2v 1 b 2 bRT a 1 1 1/2 a 1 b Rdv 1v 2 b22 3 v 1 v 1 b 2 42 T 2v 1 v 1 b 2

(E5.4L)

If we integrate Equation (E5.4L), we get a result identical to Equation (E5.4F), so the rest of the problem is equivalent to the part above where we used T and v as independent properties. We come up with the same result applying the thermodynamic web to each path; the form h 5 h 1 T, v 2 yields an equivalent result to h 5 h 1 T, P 2 . However, the first choice of independent properties made the math easier. This result is not surprising in light of the discussion after Equation (5.33); that is, T and v are the convenient independent properties when we have a pressure-explicit equation of state. Solution Using Du A third alternative for calculating the enthalpy change is to apply the definition for h: Dh 5 Du 1 D 1 Pv 2 We have already calculated the volumes of states 1 and 2, so it is straightforward to obtain D 1 Pv 2 . We can calculate Du from Equation (5.36) in conjunction with the path shown in Figure (E5.4B). The result we obtain is equivalent to those presented above.

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 287

Alternative Formulation of the Web using T and P as Independent Properties As the name Thermodynamic Web implies, many approaches can be used to develop the useful relations between measured properties and the fundamental and derived properties that we need to solve problems. As we learned in Chapter 1, pressure and temperature are the properties that “drive” systems toward mechanical and thermal equilibrium, respectively. Because we will focus on equilibrium systems in the remainder of the text (phase equilibrium in Chapters 6–8 and chemical reaction equilibrium in Chapter 9), it is instructive to recast the thermodynamic web, exclusively in terms of independent properties, T and P. To achieve this objective, we will first write volume and entropy in terms of the independent properties, T and P, that is, v 5 v 1 T, P 2 and s 5 s 1 T, P 2 . We then use the fundamental property relations to apply this formulation to determine the differential changes in the “energy properties,” u, h, a, and g as a function of T and P. The development that follows is cast in terms of the measured quantities cP, b, and k; however, the latter two quantities can then be replaced by derivatives of equations of state if needed. Writing the intensive property volume in terms of independent properties T and P gives: dv 5 ¢

'v 'v ≤ dT 1 ¢ ≤ dP 'T P 'P T

Using Equations (5.26), (5.27) and the cyclic rule, we get: dv 5 bvdT 2 kvdP

(5.40)

Similarly writing entropy as a function of T and P gives: ds 5 ¢

's 's ≤ dT 1 ¢ ≤ dP 'T P 'P T

And applying Equation (5.29) and the Maxwell relation, Equation (5.19), we get: ds 5

cP 'v dT 2 ¢ ≤ dP T 'T P

Again using Equations (5.26), (5.27), and the cyclic rule, gives: ds 5

cP dT 2 bvdP T

(5.41)

We can use Equations (5.40) and (5.41) in the fundamental property relations to get values for the “energy parameters.” Substitution in Equation (5.6) for internal energy gives: du 5 Tds 2 Pdv 5 T B

cP dT 2 bvdP R 2 P 3 bvdT 2 kvdP 4 T

Rearranging like terms gives: du 5 1 cP 2 bPv 2 dT 1 1 kPv 2 bvT 2 dP

(5.42)

Similarly, using Equations (5.7), (5.8), and (5.9), we get:

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dh 5 1 cP 2 bPv 2 dT 1 vdP

(5.43)

da 5 2sdT 1 1 kPv 2 bvT 2 dP

(5.44)

dg 5 2sdT 1 vdP

(5.45)

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288 ► Chapter 5. The Thermodynamic Web

EXAMPLE 5.5 Developing Relations with Other Independent Variables

Use Equations (5.40) and (5.41) to develop an expression for s 5 s 1 T, v 2 in terms of cP, b, k, and v. From the relationship that is developed, determine a general relationship for c P 2 c v. SOLUTION We can write s 5 s 1 T, v 2 as: ds 5 ¢

's 's ≤ dT 1 ¢ ≤ dv 'T v 'v T

(E5.5A)

Substituting in Equation (5.40) gives: ds 5 ¢

's 's 's 's ≤ dT 1 ¢ ≤ 1 bvdT 2 kvdP 2 5 B ¢ ≤ 1 ¢ ≤ bv RdT 'T v 'v T 'T v 'v T 2¢

's ≤ kvdP 'v T

(E5.5B)

However, from Equation (5.41), we have: ds 5

cP dT 2 bvdP T

Because the second terms of the right-hand sides of Equations (E5.5A) and (E5.5B) must be equal: ¢

's ≤ kv 5 bv 'v T

or, ¢

b 's ≤ 5 k 'v T

(E5.5C)

Similarly, the first terms of the right-hand sides of Equations (E5.5A) and (E5.5B) must be equal: ¢

cP 's 's ≤ 1 ¢ ≤ bv 5 'T v 'v T T

So, ¢

b2v cP 's ≤ 5 2 k 'T v T

where Equation (E5.5C) was used. Substituting back into Equation (E5.5A) gives: ds 5 ¢

b2v b cP 2 ≤dT 1 dv k k T

Inspection of Equation (5.28) shows the first term on the right-hand side must also be equal to cv . Therefore, in general: T b2Tv cP 2 cv 5 k

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5.3 Calculation of Fundamental and Derived Properties Using Equations of State and Other Measured Quantities ◄ 289 Table 5.1 Relations of Properties in Terms of Independent Properties T and P. Both the General Case and the Ideal Gas Case are Shown. Dependent Property

General 5 f 1 T, P 2

ds 5

cP dT 2 bvdP T

dv 5

bvdT 2 kvdP

cP R dT 2 dP T P v v dT 2 dP T P

du 5

1 cP 2 bPv 2 dT 1 1 kPv 2 bvT 2 dP

1 cP 2 R 2 dT

dh 5

1 cP 2 bPv 2 dT 1 vdP

cPdT

da 5

2sdT 1 1 kPv 2 bvT 2 dP

2sdT

dg 5

2sdT 1 vdP

2sdT 1 vdP

1 1 Ideal gas ¢b 5 ; k 5 ≤ T P

If an equation of state is available, expressions can be calculated for b and k in Equations (5.40) to (5.45). We will illustrate this point using the ideal gas law, v5

RT P

Applying the definitions of b, and k gives: b;

1 R 1 'v 5 ¢ ≤ 5 v 'T P Pv T

1 ideal gas 2

and, 1 'v RT 1 k;2 ¢ ≤ 5 2 5 v 'P T Pv P

1 ideal gas 2

Substitution into Equations (5.40) to (5.45) and simplification gives the expressions listed in the ideal gas column in Table 5.1. This result illustrates that three of the “energy parameters,” u, h, and a only depend on T for an ideal gas. (We have seen this relation to be the case for u and h in Chapter 2.) On the other hand, the Gibbs energy g depends on both T and P. We will explore this aspect of the property Gibbs energy more in Chapter 6.

EXAMPLE 5.6 Change in Temperature for Compression of Iron

1 mole of iron is at 1000 K and 1 bar. It is reversibly compressed in a well-insulated system to 10,000 bar. What is the final temperature? SOLUTION Because this process is reversible and adiabatic, the entropy of the system does not change. Hence, we can solve this problem by writing s 5 s 1 T, P 2 and setting ds to 0. From Table 5.1: ds 5

cP dT 2 bvdP 5 0 T (Continued)

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290 ► Chapter 5. The Thermodynamic Web

Rearranging and substituting in the form for the heat capacity from Table A.2: 1 AR 1 BRT 2 cP dT 5 dT 5 bvdP T T Assuming iron is incompressible, we can integrate to get: AR ln ¢

T2 ≤ 1 BR 1 T2 2 T1 2 5 bv 1 P2 2 P1 2 T1

(E5.6A)

Values from Tables 4.4 and Appendix A give: A 5 2.104 B 5 2.98 3 1023 b 5 3.5 3 1025 3 K21 4 v 5 7.10 3 cm3 /mol 4 5 7.1 3 1026 3 m3 /mol 4 Substitution of these values into Equation (E5.6A) gives: 2.104 3 8.314 B

J mol K

R 3 ln ¢

J T2 1 ≤ 1 2.98 3 1023 B R 3 8.314 B R K mol K 1000 3 K 4 3 1 T2 2 1000 2 3 K 4

5 3.5 3 1025 B

J 1 m3 R 3 7.10 3 1026 B R 3 1 10,000 3 105 2 1 2 B 3 R K mol m

Finally, solving implicitly for T2 gives: T2 5 1006 3 K 4 The temperature does not rise very much!

►5.4

DEPARTURE FUNCTIONS

Enthalpy Departure Function Departure functions often provide us a convenient path for calculating the nonideal contribution to property changes for real gases (or liquids). The departure function of any thermodynamic property is the difference in that property between the real, physical state in which it exists and that of a hypothetical ideal gas at the same T and P. For example, the enthalpy departure is given by: dep

gas DhT, P 5 hT, P 2 hideal T, P

(5.46)

On a molecular level, we can consider this departure function to represent the change in enthalpy if we could “turn off” the intermolecular interactions in the real fluid. In this section, we will specifically explore how to calculate changes in enthalpy and entropy using departure functions; however, this methodology can be expanded to any property.6 6

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For example, in Problem 5.43, you will develop the departure function for internal energy.

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5.4 Departure Functions ◄ 291 Hypothetical ideal gas (T1,P1)

ideal gas

Δh T1→T2

(T2,P1) ideal gas

–Δh Tdep ,P

Δh P1→P2

1 1

State 1 (T1,P1)

(T2,P2)

Pressure

Real fluid

State 2 (T2,P2)

Δh dep T ,P

2 2

Temperature

Figure 5.4 Computational paths for the change in enthalpy from state 1 to state 2 using departure functions. The PT diagram on the left is for the real fluid while that on the right represents the hypothetical ideal gas in which all the intermolecular interactions are “turned off.”

We can use the enthalpy departure function to calculate the enthalpy difference between a species in an initial state 1 at temperature T1 and pressure P1 and final state 2 at T2 and P2. Figure 5.4 shows a calculation path constructed using departure functions. The PT diagram to the left side of the figure represents real, physical space, while the TP diagram on the right-hand side is the hypothetical, ideal gas state in which all the intermolecular interactions are turned off. We first “turn off” the intermolecular interactions the species exhibits at T1 and P1. We thus transform from a real fluid into a hypothetical ideal gas. The ideal gas then undergoes an isobaric temperature change to T2 followed by an isothermal change in pressure to P2. Finally, we “turn on” the intermolecular interactions, returning to the real, physical state 2. Adding together the four steps in Figure 5.4, we get: ideal gas ideal gas dep h2 2 h1 5 2Dhdep T1, P1 1 3 DhT1 h T2 1 DhP1 h P2 4 1 DhT2, P2

Using ideal gas heat capacity data for the temperature dependence and recognizing that the enthalpy of an ideal gas does not depend on pressure, we can simplify this equation to give: T2 dep h2 2 h1 5 2Dhdep T1, P1 1 B 3 cPdT 1 0R 1 DhT2, P2

(5.47)

T1

We now need to come up with an expression for the enthalpy departure function so that we can solve Equation (5.47). Since enthalpy departure at a given state is related to the intermolecular forces involved, we will need to use the PvT relation developed in Chapter 4 and then apply the relationships of the thermodynamic web to come up with an expression for the enthalpy departure function. In the development that follows, we will use the generalized compressibility charts and tables discussed in Section 4.4 to develop values for the generalized enthalpy departure function based on corresponding

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292 ► Chapter 5. The Thermodynamic Web states. If we wanted to apply departure functions using other property data, we might have to appropriately modify our approach. ideal gas

First, we add and subtract hT, P50 to Equation (5.46): dep

ideal gas

DhT,P 5 hT,P 2 hT,P

ideal gas ideal gas ideal gas ideal gas 5 1 hT,P 2 hT,P50 2 2 1 hT,P 2 hT,P50 2 5 hT,P 2 hT,P50

(5.48)

We have simplified this equation since the enthalpy of an ideal gas is independent of pressure, that is, ideal gas

hT,P

ideal gas

2 hT,P50 5 0

At constant T, Equation (5.37) becomes: dhT 5 B2Ta

'v b 1 vRdP 'T P

We wish to put PvT data in terms of the compressibility factor so that we can use the generalized compressibility charts. Thus, we want to put this equation in terms of z. By the definition of the compressibility factor, the molar volume is written as: v5

zRT P

Applying the product rule, the partial derivative with respect to temperature becomes: a

'v RT 'z zR R b 5B a b 1 'T P P 'T P P

Substitution gives: dhT 5 B2

RT2 'z zRT zRT RT2 'z 1 RdP 5 B2 a b 2 a b RdP P 'T P P P P 'T P

Since we are applying the corresponding states relation, we write this equation in reduced coordinates: dhTr T2r 'z 5 B2 a b RdPr RTc Pr 'Tr P Integrating between 0 and P, and plugging into Equation (5.48) gives: dep

DhTr,Pr RTc

ideal gas

5

hTr,Pr 2 hTr,Pr RTc

P

ideal gas

5

hTr,Pr 2 hTr,Pr 50 RTc

5 T2r 3 B2 0

1 'z ¢ ≤ RdPr Pr 'Tr P

(5.49)

If we have a relation for z from 0 to Pr, we can integrate Equation (5.49) to give the enthalpy departure. For example, we can use the generalized compressibility results of the form given in Section 4.4 to get simple fluid and correction enthalpy departure terms. The generalized enthalpy departure can then be determined according to: ideal gas

hTr,Pr 2 hTr,Pr RTc

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102

ideal gas

5B

hTr,Pr 2 hTr,Pr RTc

R

112

ideal gas

1 vB

hTr,Pr 2 hTr,Pr RTc

R

(5.50)

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5.4 Departure Functions ◄ 293

While the data presented in Section 4.4 can be numerically integrated, they were generated by Lee and Kesler according to their equation of state. This equation can be analytically integrated to find values for enthalpy departure that can be used in Equation (5.50). The results are shown in Appendix E. Plots of the simple fluid and correction terms that result are presented in Figures 5.5 and 5.6, respectively. Tables of their values are presented in Appendix C (Tables C.3 and C.4).

Entropy Departure Function Like the enthalpy departure function, the entropy departure function can be used to find the entropy change of a real fluid. It is defined as the difference in that property between the real, physical state and that of a hypothetical ideal gas at the same T and P: dep

ideal gas

DsT,P 5 sT,P 2 sT,P

(5.51)

The change in entropy from state 1 to state 2 can be written in analogy to Figure 5.4: dep

ideal gas

ideal gas

dep

s2 2 s1 5 2DsT1,P1 1 3 DsT1 h T2 1 DsP1 h P2 4 1 DsT2,P2 Substituting Equation (3.23) for the two ideal gas terms gives: T2

s2 2 s1 5

dep DsT2,P2

cP P2 dep 1 B 3 dT 2 R ln R 2 DsT1,P1 T P1

(5.52)

T1

0 4 0.5

3

–1.5

2

–2

1.7 1.15

–2.5 RTC

ideal gas (0) hT ,P – hT ,P r r r r

–1

1.1

–3

1.5 1.2 1.3

1.05 –3.5

1.0

–4 0.9

Vapor-liquid dome

–4.5

0.8 0.7 0.6

–5 –5.5

0.5 0.4 Tr= 0.3

–6 –6.5 0.01

1

0.1

10

Pr

Figure 5.5 Generalized enthalpy departure—simple fluid term. Based on the Lee–Kesler equation of state.

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294 ► Chapter 5. The Thermodynamic Web 2 4 1.7 1.5 1.3 1.1 1.15 1.2

0 Tr = 0.7

0.8

0.9

2

3 2

–4 0.9

RTC

ideal gas (1) hT ,P – hT ,P r r r r

1.05 1.0

0.8

–6 Vapor-liquid dome

0.7 0.6

–8

0.5 –10

0.4 Tr = 0.3

–12 0.01

0.1

1

Pr

10

Figure 5.6 Generalized enthalpy departure—correction term. Based on the Lee–Kesler equation of state.

As opposed to enthalpy, the entropy change with pressure of an ideal gas is nonzero. ideal gas

To calculate the entropy departure, we add and subtract sT,P50 to Equation (5.45): dep

ideal gas

DsT, P 5 sT,P 2 sT,P

ideal gas

ideal gas

5 1 sT,P 2 sT,P50 2 2 1 sT,P

ideal gas

2 sT,P50 2

(5.53)

Using the thermodynamic web, we can determine each difference on the right-hand side of Equation (5.53). At constant temperature, the entropy can be written in terms of the independent property P. According to Equation (5.32), dsT 5 2a

'v b dP 'T P

In the case of an ideal gas, the ideal gas law can be differentiated to give: ideal gas

dsT

5 2R

dP P

Integration gives: P ideal gas sT,P

2

ideal gas sT,P50

5 23R

dP P

0

For the real fluid, we get: dsT 5 2a

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'v zR RT 'z 1 b dP 5 2B a b RdP 'T P P P 'T P

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5.4 Departure Functions ◄ 295

Integrating from zero pressure to the pressure of the state of interest gives: P

sT,P 2

ideal gas sT,P50

zR RT 'z 5 3 2B 1 a b RdP P P 'T P 0

Finally, we get the entropy departure by inserting these results in Equation (5.47): P dep DsT,P

5 sT,P 2

ideal gas sT,P

5 R3 2 B 0

z21 T 'z 1 a b RdP P P 'T P

To use the generalized compressibility, we rewrite this equation in reduced coordinates: dep

DsTr,Pr R

ideal gas

5

sTr,Pr 2 sTr,Pr R

P

z21 Tr 'z 5 3 2B 1 ¢ ≤ RdPr Pr Pr 'Tr P

(5.54)

0

Again, Equation (5.54) can be integrated with the appropriate data or equation of state for z. Using the Lee–Kesler equation of state gives results that include a simple fluid term and a correction term. The form of entropy departure using this equation of state is given in Appendix E. The entropy departure can then be calculated by: ideal gas

sTr,Pr 2 sTr,Pr

5B

R

102

ideal gas

sTr,Pr 2 sTr,Pr R

R

112

ideal gas

1 vB

sTr,Pr 2 sTr,Pr R

R

(5.55)

Plots of the simple fluid and correction terms for entropy departure are given in Figures 5.7 and 5.8, respectively. Tables of values generated in the same way are presented in Appendix C (Tables C.5 and C.6). 0

1.1

–2

1.2 1.3

1.5

4 3 2 1.7

1.05 0.7

0.8

0.9

0.9

–4 0.8 0.7 R

ideal gas (0) ST ,P – ST ,P r r r r

0.6

1.0

–6

0.6

Vapor-liquid dome

0.5

–8

0.4 Tr = 0.3

–10

–12 0.01

0.1

1

10

Pr

Figure 5.7 Generalized entropy departure—simple fluid term. Based on the Lee–Kesler equation of state.

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296 ► Chapter 5. The Thermodynamic Web 3 2 1.7

0 –1 0.6

–2

0.8

0.7

1.05

0.9

–3 0.9

–5 0.8

–6 –7

0.7

Vapor-liquid dome

–8 R

1.2 1.3

1.0

–4

ideal gas (1) ST ,P – ST ,P r r r r

1.1

1.5

0.6

–9 –10 –11

0.5

–12 –13 –14

0.4

–15 Tr = 0.3

–16 –17 0.01

0.1

10

1 Pr

Figure 5.8 Generalized entropy departure—correction term. Based on the Lee–Kesler equation of state.

EXAMPLE 5.7 Alternative Solution to Example 5.4 Using Departure Functions

Repeat Example 5.4 using the Lee–Kesler generalized correlation data for enthalpy departure to account for nonideal behavior. SOLUTION The schematic of the process is drawn in Figure E5.4A. To find the heat, we need to calculate the enthalpy difference between the outlet (state 2) and the inlet (state 1). From Equation (5.47), we get: T2

q 1 ws 5 h2 2 h1 5

dep 2DhT1, P1

dep

1 B 3 cPdT 1 0R 1 DhT2, P2

(E5.7A)

T1

To find values of enthalpy departure, we can use the Lee–Kesler tables. They have the form: dep

DhTr,Pr RTc

ideal gas

5

hTr,Pr 2 hTr,Pr RTc

102

ideal gas

5B

hTr,Pr 2 hTr,Pr RTc

R

112

ideal gas

1 vB

hTr,Pr 2 hTr,Pr RTc

R

Looking up the critical properties and acentric factor for n-butane from Appendix A.1 gives: Tc 5 425.2 3 K 4 Pc 5 37.9 3 bar 4 v 5 0.199

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5.4 Departure Functions ◄ 297

Thus, the reduced coordinates for state 1 and state 2 are: T1,r 5

353.15 3 K 4 T1 5 5 0.83 Tc 425.2 3 K 4

and

P1,r 5

9.47 3 bar 4 P1 5 5 0.20 Pc 38.0 3 bar 4

T2,r 5

393.15 3 K 4 T2 5 5 0.925 Tc 425.2 3 K 4

and

P2,r 5

18.9 3 bar 4 P2 5 5 0.50 Pc 37.9 3 bar 4

We can find the values for the enthalpy departure terms in Tables C.3 and C.4 in Appendix C. For state 1, by interpolation: 102

ideal gas

B

hT1, r, P1, r 2 hT1, r, P1, r RTc

R

112

ideal gas

5 20.413 B

hT1, r, P1, r 2 hT1, r, P1, r RTc

R

5 20.622

so, ideal gas

hT1, r, P1, r 2 hT1, r, P1, r RTc

ideal gas

5B

hT1, r, P1, r 2 hT1, r, P1, r RTc

102

R 1vB

112

ideal gas

hT1, r, P1, r 2 hT1, r, P1, r RTc

R 5 20.536

(E5.7B)

and for state 2: 102

ideal gas

B

hT2, r, P2, r 2 hT2, r, P2, r RTc

R

112

ideal gas

5 20.771 and B

hT2, r, P2, r 2 hT2, r, P2, r RTc

R

5 20.994

so, ideal gas

hT2, r, P2, r 2 hT2, r, P2, r RTc

ideal gas

5B

hT2, r, P2, r 2 hT2, r, P2, r RTc

102

R 1vB

ideal gas

hT2, r, P2, r 2 hT2, r, P2, r RTc

112

R 5 20.969 (E5.7C)

The value for the integral of the heat capacity is the same as in Example 5.4: T2

393

ideal gas DhT1 h T2 5 3 cPdT5R T1

23 26 2 3 3 1.935136.915310 T211.402310 T 4 dT54,696 3 J/mol 4 (E5.7D)

353

Plugging in the values from Equations (E5.7B), (E5.7C), and (E5.7D) into Equation (E5.7A) gives: h2 2 h1 5 0.536RTc 1 4,696 2 0.969RTc 5 3,167 3 J/mol 4 The value calculated using the Redlich–Kwong equation of state differs from this value by 11.0%. What value do you think is more accurate? Solving for heat, we get: q 5 Dh 2 ws 5 1067 3 J/mol 4

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298 ► Chapter 5. The Thermodynamic Web

EXAMPLE 5.8 Enthalpy Departure for a van der Waals Gas

Develop an expression for the enthalpy departure function for a gas that obeys the van der Waals equation of state. Write it in terms of reduced coordinates. SOLUTION Since the van der Waals equation is explicit in pressure, it is convenient to choose T and v as the independent properties.7 Consequently, we use infinite volume as the limit of an ideal gas. In analogy to Equation (5.42), we write: dep

ideal gas

DhT,v 5 hT,v 2 hT,P

ideal gas

ideal gas

5 ahT,v 2 hT,v5 ` b 2 ahT,v

ideal gas

ideal gas

2 hT,v5 ` b 5 hT,v 2 hT,v5 `

Thus, we want to find the difference between the enthalpy at the volume of the state of interest and an infinite volume at constant temperature. We write the change in the dependent property h as: dh 5 a

'h 'h b dT 1 a b dv 'T v 'v T

At constant temperature, dhT 5 a

T's 1 v'P 's 'P 'P 'P b dv 5 BTa b 1 va b Rdv 5 BTa b 1 va b Rdv 'v 'v 'v 'T 'v T T T T v

where we have used the fundamental property relation for h, Equation (5.7), and a Maxwell relation, Equation (5.18). The derivatives can be found by differentiating the van der Waals RT a equation aP 5 2 2 b to give: v2b v dhT 5 BT

RT R 2a RTb 2a 1 va2 1 3 b Rdv 5 B2 1 2 Rdv 1v 2 b22 1v 2 b22 v2b v v

(E5.8)

Integrating Equation (E5.8) from the ideal gas state of infinite volume to volume v gives: v

Dhdep

2a RTb 2a RTb 2 5 3 B2 1 2 Rdv 5 1v 2 b22 1v 2 b2 v v v5 `

We can write the enthalpy departure in terms of reduced coordinates if we substitute in Equations (4.21) and (4.22) for the van der Waals constants a and b: Tr Dhdep 9 5 2 RTc 3vr 2 1 4vr 7 See the discussion after Equation (5.33); additionally, inspection of Figure 5.3 shows that derivatives of entropy with volume reduce to P explicit equations of state.

►5.5 JOULE-THOMSON EXPANSION AND LIQUEFACTION Joule-Thomson Expansion The pressure dependence of the thermodynamic property enthalpy leads to interesting phenomena in the unrestrained, free expansion of real gases. Figure 5.9 shows a schematic of a gas flowing through a porous plug. It enters the system in state 1 at P1 and T1 and it exits at a significantly lower pressure, P2. We wish to study the effect of this socalled Joule–Thomson expansion on the temperature of the gas at the exit, T2.

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5.5 Joule-Thomson Expansion and Liquefaction ◄ 299 P1

Gas in

T1

P2

T2

Gas out

Porous plug

Figure 5.9 Schematic of Joule–Thomson expansion through a porous plug.

Since the gas spends so little time in the plug, there is no opportunity for heat transfer; thus, we consider this process adiabatic. Additionally, the shaft work is zero. If kinetic energy effects are negligible, the first law for this steady-state, adiabatic throttling process reduces to: h2 2 h1 5 Dh 5 0 We call a process that occurs at constant enthalpy, such as this one, isenthalpic. We can determine the change in temperature that results as the pressure decreases in the isenthalpic throttling process if we know the derivative, 1 'T/'P 2 h. We call this relation the Joule–Thomson coefficient, mJT. mJT ; a

'T b 'P h

(5.56)

Figure 5.10 plots characteristic lines of constant enthalpy (isenthalps) on a TP diagram. In the shaded region, the slopes of the curves are positive; therefore, mJT . 0, as defined by Equation (5.56). In this region, the temperature will decrease as the pressure decreases during the throttling process. Since the decrease in temperature as the pressure drops corresponds to a decrease in molecular kinetic energy, the molecular potential energy must be increasing or else energy conservation would be violated. We can say the molecules are more stable when they are closer together at the higher pressure and, consequently, that attractive forces are dominant in this region. Conversely, in the nonshaded region, the slopes of the isenthalps are negative; therefore mJT , 0. The temperature will increase as pressure decreases, indicating that repulsive forces dominate the behavior in this region. These two regions are separated by the inversion line, where the slope of T vs. P is zero and where attractive and repulsive interactions exactly balance. For a given pressure, the temperature at which these interactions balance is known as the Boyle temperature. T Inversion line h = const μJT > 0

Net Attraction

μJT < 0

Net Repulsion

P

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Figure 5.10 Typical lines of constant enthalpy on a PT diagram. The inversion line separates the region of positive and negative Joule–Thomson coefficients.

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300 ► Chapter 5. The Thermodynamic Web We can use the thermodynamic web to develop an expression for mJT in terms of

PvT property relations and heat capacities. We begin with Equation (5.37): dh 5 cpdT 1 B2Ta

'v b 1 vRdP 'T P

We must use the real heat capacity given by Equation (5.39). During the Joule–Thomson expansion, dh is zero; thus, we can rewrite the previous equation as: 'T mJT 5 a b 5 'P h

BTa

'v b 2 vR 'T P 5 cp

BTa

'v b 2 vR 'T P

Preal

ideal gas

cP

(5.57)

'2v 2 3 BTa 2 b RdP 'T P Pideal gas

where we have substituted in Equation (5.39) for the real heat capacity, since intermolecular interactions are important. If we have an equation of state or other appropriate PvT data and the heat capacity of a fluid, we can evaluate Equation (5.57) for mJT. Alternatively, we could measure mJT experimentally using the porous plug in Figure 5.9 to find an unknown cP.

EXAMPLE 5.9 Joule–Thomson Coefficient from the Virial Equation

Develop an expression for the Joule–Thomson coefficient using the pressure-based expansion of the virial equation truncated at the second virial coefficient. Use the corresponding state relationships presented in Chapter 4 for the temperature dependence of B to develop a generalized correlation for mJT. SOLUTION To use Equation (5.57), we need to write the virial equation in a form that is explicit in volume. This can be done by rearranging Equation (4.27):

v5

RT RT 1 1 1 BrP 2 5 1 BrRT P P

(E5.9A)

We then take the first and second derivatives: 'v R dBr b 5 1 RBr 1 RTa b 'T P P dT

(E5.9B)

dBr '2v d2Br ≤ 5 Ra b 1 RT¢ 2 ≤ 2 'T P dT dT

(E5.9C)

a

¢

and,

Using Equations (E5.9A), (E5.9B), and (E5.9C) in Equation (5.57) gives: 'T mJT 5 a b 5 'P h

BTa

'v b 2 vR 'T P 5 cp

BTa

'v b 2 vR 'T P

Preal

ideal gas cP

'2v 2 3 BT¢ 2 ≤ P RdP 'T Pideal gas

RT2 a 5

dBr b dT

Preal ideal gas cP

(E5.9D)

d2Br dBr 2 3 bTBR a b 1 RT ¢ 2 ≤ R rdP dT dT Pideal gas

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5.5 Joule-Thomson Expansion and Liquefaction ◄ 301

To evaluate this expression for mJT, we can use the corresponding state relations given in Chapter 4. Using Equation (4.28) and the equations that follow Equation (4.29), we get: Br 5

BrTc Br B102 1 vB112 B 5 5 5 RT PcT PcTr PcTr

(E5.9E)

Substituting the generalized relations on page 241 into Equation (E5.9E) gives: Br 5

1 0.083 0.422 0.139 0.172 B¢ 2 2.6 ≤ 1 v ¢ 2 5.2 ≤ R Pc Tr Tr Tr Tr

(E5.9F)

We then take the first and second derivatives:

and,

a

1 0.083 1.097 0.139 0.894 dBr 1 ≤ 1 v ¢2 1 ≤R b 5 B ¢2 dT Pc TTr TT2.6 TTr TT5.2 r r

(E5.9G)

¢

d2Br 1 0.166 3.950 0.278 5.545 ≤ 5 B ¢ 2 2 2 2.6 ≤ 1 v ¢ 2 2 2 5.2 ≤ R dT2 Pc T Tr T Tr T Tr T Tr

(E5.9H)

Substituting Equations (E5.9G) and (E5.9H) into (E5.9D) and simplifying gives: 2 mJT 5

RT 0.083 1.097 0.139 0.994 B ¢2 1 2.6 ≤ 1 v ¢2 1 5.2 ≤ R Pc Tr Tr Tr Tr P

ideal gas cP

(E5.9I)

R 0.083 2.853 0.139 4.651 2 3 B B¢ 2 2.6 ≤ 1 v ¢ 2 5.2 ≤ R RdP Pc Tr Tr Tr Tr 0

where we have set Pideal to zero. Finally, integrating and substituting in T 5 TrTc, we get: 2 mJT 5

Tc 1.097 0.994 B ¢20.083 1 1.6 ≤ 1 v ¢20.139 1 4.2 ≤ R Pc Tr Tr

ideal gas

cP

R

0.083 2.853 0.139 4.651 2 Pr B ¢ 2 2.6 ≤ 1 v ¢ 2 5.2 ≤ R Tr Tr Tr Tr

(E5.9J)

Equation (E5.9J) presents a generalized relation for mJT. If the critical properties and acentric factor of a species are known, we can use Equation (E5.9J) to calculate the Joule–Thomson coefficient at a specified state at temperature T and pressure P.

Liquefaction Joule–Thomson expansion can be used to liquefy gases if it is performed in the region where mJT . 0 to the left of the inversion line in Figure 5.10. Liquefaction is an important process industrially; there is a significant market for liquefied gases. For example, liquid nitrogen, helium, and hydrogen are often used to remove energy from cryogenic systems. Additionally, separation of nitrogen and oxygen from air can be accomplished by liquefaction. However, the temperatures at which these gases condense are quite low. He condenses at 4.4 K while N2 condenses at 77 K. The liquefaction of these gases can require a significant amount of refrigeration. A schematic of such a liquefaction process is shown in Figure 5.11a. The gas is first compressed from state 1 to 2 to increase its pressure. However, during compression,

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302 ► Chapter 5. The Thermodynamic Web 5

2

1

3

Gas

Separator

4 Joule-Thomson expansion

Wc Compressor

6

Q cooler

Liquid (a)

8

Recycle

2

1

6

Q

3

4

5

Gas Wc

Heat exchanger

Joule-Thomson expansion

7 Liquid

(b)

Figure 5.11 Liquefaction of gases using Joule–Thomson expansion. (a) Basic liquefaction process using Joule–Thomson expansion and (b) Linde process.

the temperature of the gas also rises. It is then cooled from state 2 to state 3 to lower its temperature. These two processes are intended to bring it to the shaded region in Figure 5.10 and to put it in a state where a throttling process will bring it into the twophase region. It now goes through an isenthalpic Joule–Thomson expansion, from state 3 to state 4, where the temperature drops low enough to lead to condensation. The vapor and liquid streams at states 5 and 6, respectively, are then separated. An improvement to the liquefaction process is shown in Figure 5.11b. In this process, an additional heat exchanger is employed to recover the energy from the noncondensed gas. This gas is then recycled. The process depicted in Figure 5.11b is known as the Linde process.

EXAMPLE 5.10 Liquefaction of N2 by Joule–Thomson Throttling

Consider the liquefaction of N2 by Joule–Thomson throttling. If the inlet to the expansion valve is at T1 5 2122°C and P1 5 100 bar and the outlet is at 1 bar, determine the percentage of N2 that is liquefied. SOLUTION We can solve this problem using the generalized charts for enthalpy departure. Looking up properties for N2 from Appendix A.1, we get: Tc 5 126.2 3 K 4 Pc 5 33.8 bar

v 5 0.039 Since the acentric factor, v, is small, we will use only the simple fluid term in the generalized correlations. The reduced coordinates for state 1 are: T1, r 5

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151 3 K 4 T1 5 5 1.20 Tc 126.2 3 K 4

and

P1, r 5

100 3 bar 4 P1 5 5 3.0 Pc 33.8 3 bar 4

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5.5 Joule-Thomson Expansion and Liquefaction ◄ 303

We can then find the values for the enthalpy departure term in Table C.1 in Appendix C: ideal gas

hT1, r, P1, r 2 hT1, r, P1, r

B

RTc

R 5 22.81

(E5.10A)

For state 2, we have:

P2,r 5

1 3 bar 4 P2 5 0.03 5 Pc 33.8 3 bar 4

Looking at Figure 5.5, we see that the two-phase region is at: T2,r 5 0.61 T2 5 T2,rTc 5 0.61 3 126.2 3 K 4 5 77 3 K 4

so,

Applying Equation (5.47) for this isenthalpic process: dep

ideal gas

dep

2DhT1, P1 DhT1 h T2 DhT2, P2 h2 2 h1 505 1 1 RTc RTc RTc RTc

(E5.10B)

To find the ideal gas enthalpy change, we need to look up the heat capacity from Appendix A.2: cP 5 3.28 1 0.593 3 1023T R Therefore, ideal gas

DhT1 h T2 RTc

77

T2

1 cP 1 3 3.28 1 0.593 3 1023T 4 dT 5 21.28 (E5.10C) 5 3 dT 5 Tc R 126.2 3 151

T1

Rearranging Equation (E5.10B) and plugging in values from Equations (E5.10A) and (E5.10C), we get: dep

dep

DhT2, P2 RTc

5

DhT1, P1 RTc

ideal gas

2

DhT1 h T2 RTc

5 21.53

It is useful to view this process on an hP generalized enthalpy chart, as shown in Figure E5.10. The “lever rule” is illustrated. The quality can be found by: dep

dep

5 1 1 2 x 2 Dhl

dep

dep

Dh

dep

1 xDhv

Solving for x, we get: x5

Dh 2 Dhl 21.5 1 5.1 5 0.72 dep dep 5 Dhv 2 Dhl 20.1 1 5.1

Since the quality represents the fraction of vapor, we conclude that roughly 28% the inlet stream is liquefied. (Continued)

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304 ► Chapter 5. The Thermodynamic Web

0 4 0.5 –1 –1.5

3

Liquid 2

State 2

1.7 1.15

–2.5 RTC

ideal gas (0) hT ,P – hT ,P r r r r

–2

1.1

–3 –3.5

State 1 1.2

1.5

1.05

Vapor

1.0

–4 –4.5

0.9

Vapor–liquid dome

0.8 0.7 0.6

–5 –5.5 –6 –6.5 0.01

0.5 0.4 Tr = 0.3 1

0.1

10

Pr

Figure E5.10 Liquefaction process of N2 from state 1 to state 2.

►5.6 SUMMARY In this chapter, we developed a thermodynamic web to relate measured, fundamental, and derived thermodynamic properties. The web allows us to use available property data to solve first- and second-law problems for gases with nonideal behavior. Typically, we want relationships between fundamental and derived thermodynamic properties, such as u, s, and h, and things we can measure, such as measured properties P, v, T, or quantities in which measured data are typically reported, such as cv, cP, b, k, and equations of state. In solving these problems, it is often necessary to construct hypothetical paths to calculate the change in a desired property between two states. Similarly, the approach for developing solutions to the phase equilibria and chemical reaction equilibria problems in the second half of the text will rely on an ability to exploit property relations and form paths that allow us to use appropriate measured data. For a system with constant composition, the two properties that we choose to constrain the state of the system become the independent properties. We can write the differential change of any other property, the dependent property, in terms of these two properties, as illustrated by Equation (5.4). From a combined form of the first and second laws, we developed the fundamental property relations. We then used the rigor of mathematics to allow us to form this intricate web of thermodynamic relationships. Included in the web are the Maxwell relations, the chain rule, derivative inversion, the cyclic relation, and Equations (5.22) through (5.24). A set of useful relationships relating partial derivatives with T, P, s, and v is summarized in Figure 5.3. We use these relationships to solve first- and second-law problems similar to those in Chapters 2 and 3, but for real fluids. Departure functions often provide us a convenient path to calculate the nonideal contribution to property changes for real gases (or liquids). The departure function of any thermodynamic property is the difference in that property between the real, physical state in which the species exists and that of a hypothetical ideal gas at the same T and P. On a molecular level, we can consider the departure function to represent the change in the value of a property if we could “turn off” the intermolecular interactions in the real fluid. Plots of the simple fluid and correction terms for

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5.7 Problems ◄ 305 the enthalpy departure function are presented in Figures 5.5 and 5.6, respectively. Tables of their values are presented in Appendix C (Tables C.3 and C.4). Analogous data for entropy departure are presented in Figures 5.7 and 5.8 and in Appendix C (Tables C.5 and C.6). These values are obtained from the Lee–Kesler equation of state. Joule–Thomson expansion results from the unrestrained, free expansion of real gases. Such a process occurs at constant enthalpy and is termed isenthalpic. We can determine the change in temperature that results as the pressure decreases in the isenthalpic throttling process if we know the Joule–Thomson coefficient, mJT 5 1 'T/'P 2 h. Joule–Thomson expansion is the basis for liquefaction processes, such as those shown in Figure 5.11.

►5.7

PROBLEMS

Conceptual Problems 5.1 Consider the following partial derivatives. For an ideal gas, state whether the value is positive, negative, zero, close to zero, or you cannot tell. Explain your answer. ¢

'u 's 'u 's ≤ ,¢ ≤ ,¢ ≤ ,¢ ≤ 'T v 'T v 'v T 'v T

5.2 Consider the following partial derivatives. For an ideal gas, state whether the value is positive, negative, zero, close to zero, or you cannot tell. Explain your answer. ¢

'h 's 'h 's 'P ≤ ,¢ ≤ ,¢ ≤ ,¢ ≤ ,¢ ≤ 'T P 'T P 'P T 'P T 'T h

5.3 Consider the following partial derivatives. For a real gas dominated by attractive interactions, state whether the value is positive, negative, zero, close to zero, or you cannot tell. Explain your answer. ¢

'u 's 'u 's ≤ ,¢ ≤ ,¢ ≤ ,¢ ≤ 'T v 'T v 'v T 'v T

5.4 Consider the following partial derivatives. For a real gas dominated by attractive interactions, state whether the value is positive, negative, zero, close to zero, or you cannot tell. Explain your answer. ¢

'h 's 'h 's 'P ≤ ,¢ ≤ ,¢ ≤ ,¢ ≤ ,¢ ≤ 'T P 'T P 'P T 'P T 'T h

5.5 Consider the following partial derivatives. For a liquid, state whether the value is positive, negative, zero, close to zero, or you cannot tell. Explain your answer. ¢

'h 's 'h 's ≤ , ¢ ≤ , ¢ ≤ , ¢ ≤ , b, k 'T P 'T P 'P T 'P T

5.6 Consider the following partial derivatives. For an ideal gas, state whether the value is positive, negative, zero, close to zero, or you cannot tell. Explain your answer. ¢

'g 'T

≤ ,¢ P

'g 'P

≤ ,¢ T

'g 'P

≤ ,¢ P

'v ≤ 'P s

5.7 This question should be completed without doing any calculations. Consider the enthalpy departure function for the following cases. Rank them from the smallest magnitude to the largest. Explain. (a) Methane at 180°C and 1 bar (b) Methane at 240°C and 1 bar (c) Methane at 180°C and 10 bar (d) Water at 180°C and 1 bar (e) Water at 240°C and 1 bar (f) Water at 180°C and 10 bar

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306 ► Chapter 5. The Thermodynamic Web 5.8 Of the following mixture, which do you think has entropy departure function of larger magnitude (a) 50 mol% methane mixed with 50 mol% ethane (b) 50 mol% acetone mixed with 50 mol% chloroform? Explain. 5.9 You are using the Peng–Robinson equation of state to determine the entropy change of an ideal gas: P5

aa 1 T 2 RT 2 v2b v1v 1 b2 1 b1v 2 b2

Is it better to try s 5 s 1 T,v 2 or s 5 s 1 T,P 2 ? Explain. 5.10 Consider a gas that undergoes a process from state 1 to state 2. You know the ideal gas heat capacity and an equation of state. Which of the following hypothetical paths would be most appropriate to chose to calculate Du? Explain.

Path 1

Volume

Path 2

Path 3 State 1 (T1, v1)

State 2 (T2, v2) Δu

Path 4

Temperature

5.11 Consider a gas that undergoes a process from state 1 to state 2. You know the ideal gas heat capacity and an equation of state. Which of the following hypothetical paths would be most appropriate to chose to calculate Dh? Explain.

Pressure

Path 1

Path 2

Path 3 State 1 (T1, P1)

State 2 (T2, P2) Δh

Path 4

Temperature

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5.7 Problems ◄ 307 5.12 Consider the following property relation: a

'u b 'P s

(a) Come up with a physical process on a system which is described by the relation above. Sketch the process and describe it as completely as needed so that this relation holds. (b) Based on the process you chose in part (a), do you think the relation has a positive value, has a negative value, or is 0. Justify your answer.

Numerical Problems 5.13 Write equations analogous to Equation (5.5) for the exact differential of internal energy, du, in terms of each of the following sets of independent properties: (a) u 5 u 1 T, P 2 (b) u 5 u 1 T, s 2 (c) u 5 u 1 h, s 2 5.14 Using the thermodynamic web, show that for an ideal gas: u 5 u 1 T only 2 5.15 For an ideal gas, show that: cP 5 cv 1 R 5.16 Show that an ideal gas follows the cyclic relationship in P, v, and T. 5.17 Evaluate the derivative: a

'h b 'v T,P

for a pure species that follows the Peng–Robinson equation of state. The subscript T,P indicates that both temperature and pressure are held constant. 5.18 Using the van der Waals equation, find an expression for the derivative: a

'h b 'T s

in terms of a, b, cP, R, v, and T. 5.19 Consider the following equation of state: Pv 5 1 1 BrP 1 CrP2 RT where Br and Cr are constant parameters with no temperature dependence. In terms of Br, Cr, R, P, T, and cP, find the following expressions: a

'h 'h 'h 'h b ,a b,a b ,a b 'P T 'P s 'T P 'T s

5.20 Verify that: b 'P (a) a b 5 k 'T v (b) cP 2 cv 5

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vTb2 k

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308 ► Chapter 5. The Thermodynamic Web 5.21 Use the result of Problem 5.20(b) to calculate the difference, cP 2 cv, for liquid acetone at 20°C and 1 bar. Data can be found in Table 4.4. Repeat for benzene and copper. How do the values you obtain for cv compare to the value for cP? Explain. 5.22 Verify that: b 's b 5 k 'v T 's a b 5 2bv 'P T a

and,

5.23 Before the proliferation of personal computers, it was often convenient to summarize thermodynamic property data in the form of graphic diagrams. The Mollier diagram presents h (y-axis) vs. s (x-axis). Obtain an expression for the slope of an isochor (constant-volume line) on a Mollier diagram for (a) an ideal gas (in terms of T, v, cv, cP, and R); (b) a van der Waals gas (in terms of T, v, cv, cP, a, b and R). 5.24 Develop a general relationship for the change in temperature with respect to pressure at constant entropy: a

'T b 'P s

(a) Evaluate the expression for an ideal gas. (b) From the result in part (a), show that for an ideal gas with constant cP, an isentropic expansion from state 1 and state 2 yields Equation (2.49). (c) Evaluate the expression for a gas that obeys the van der Waals equation of state. 5.25 Derive Equation (5.39). 5.26 Your company has just developed a new refrigeration process. This process uses a secret gas, called Gas A. You are told that you need to come up with thermodynamic property data for this gas. The following data have already been obtained for the superheated vapor: P 5 10 bar v 3 m3 /kg 4

80

0.16270

100

0.17389

T [°C]

P 5 12 bar

s [kJ/(kg K)]

v 3 m3 /kg 4

s [kJ/(kg K)]

5.4960

0.13387

?

0.14347

As accurately as you can, come up with a value for s in the table above. Clearly indicate your approach and state any assumptions that you make. Do not assume ideal gas behavior. 's 5.27 Consider water at 250°C and 800 kPa. Determine a value for the partial derivative ¢ ≤ 'v T by the following methods: (a) Use the Redlich–Kwong equation of state. (b) Use the steam tables directly. (c) Use the steam tables together with the Maxwell relation given in Equation (5.18). Compare your answers and explain which one you think is the most accurate. 5.28 A gas can be described by the following equation of state: P5

RT v2b

ideal gas

5 35 J/ 1 mol K 2 . where b 5 5 3 1024 m3 /mol and cP real (a) Develop an expression of cP for this gas. (b) One mole of this gas flows through an adiabatic porous plug. It enters at 10 bar, 300ºC and it exits at 1 bar. What is the exit temperature?

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5.7 Problems ◄ 309 (c) Determine whether the Joule–Thompson coefficient, mTP 5 ¢

'T ≤ is positive, zero, or 'P h,

negative for this process. Is this result consistent with your result in part b? (d) What is the entropy change of the universe? Is this a reversible process? 5.29 One mole of methane in a piston-cylinder assembly undergoes an adiabatic compression from an initial state 1 P1 5 0.5 bar, v1 5 0.05 m3 /mol 2 to a final state 1 P2 5 10 bar, v2 5 3 3 1023 m3 /mol 2 . Use the van der Waals equation of state for nonideal gas behavior. (a) How much work is done on the system? (b) Do you expect the magnitude of work to be larger or smaller than what would be needed if methane behaved like an ideal gas? Explain based on molecular potential energy. 5.30 One mole of n-butane in a piston-cylinder assembly undergoes an irreversible isothermal expansion against a constant external pressure until the forces balance. The initial pressure is 10 bar, and the initial molar volume is 3 3 1023 m3 /mol, and the final volume is 0.05 m3 /mol. Take the temperature of the surroundings to be 298 K. Use the van der Waals equation of state and answer the following questions: (a) What is the initial temperature? (b) What is the final pressure? (c) How much work is done during this process? (d) How much heat is transferred this process? (e) What is the entropy change of the universe for this process? 5.31 Two moles of ethane in a piston-cylinder assembly undergo a reversible adiabatic compression. The initial pressure is 0.5 bar, and the initial volume is 0.1 m3. The final volume is 0.002 m3. Use the van der Waals equation of state to account for intermolecular interactions. Answer the following questions: (a) What is the initial temperature? (b) What is the entropy change of the system for this process? (c) What is the final pressure? What is the final temperature? (d) How much work in [J] is done during this process? (e) How much heat is transferred this process? (f) Do you expecting the magnitude of work to be larger or smaller than what you would obtain if ethane behaved like an ideal gas? Please comment on your answer based on molecular interactions. 5.32 Propane at 350°C and 600 cm3 /mol is expanded in an isentropic turbine. The exhaust pressure is atmospheric. What is the exhaust temperature? PvT behavior has been fit to the van der Waals equation with: a 5 92 3 105 3 1 atm cm6 2 /mol2 4 b 5 91 3 cm3 /mol 4 (a) Solve this problem using T and v as the independent properties, that is, s 5 s 1 T, v 2 (b) Solve this problem using T and P as the independent properties. 5.33 You need to design a heater to preheat a gas flowing into a chemical reactor. The inlet temperature is 27°C and the inlet pressure is 50 bar. You desire to heat the gas to 227°C and 50 bar. You are provided with an equation of state for the gas: z511

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aP "T

with

a 5 20.070 3 K1/2 /bar 4

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310 ► Chapter 5. The Thermodynamic Web and with ideal gas heat capacity data: cP 5 3.58 1 3.02 3 1023T R

where

T is in [K]

(a) Under these conditions, do attractive forces or repulsive forces dominate the behavior of this gas? Explain. (b) As accurately as you can, calculate, in [J/mol], the amount of heat required. 5.34 Consider the piston-cylinder assembly shown below; 250 moles of gas expand isothermally after the removal of a 10,000-kg block. (a) What is the internal energy change for the expansion process? (b) What is the entropy change of the universe for this process? Assume that the PvT behavior can be described by the van der Waals equation with a 5 0.5 3 Jm3 /mol2 4 ; b 5 4 3 1025 3 m3 /mol 4 ; and that the ideal gas heat capacity has a constant ideal gas 5 35J/ 1 mol K 2 . value of cP

Weightless, frictionless piston Patm A = 0.1 m2 m = 10,000 kg

Patm

Isothermal expansion

A = 0.1 m2 250 mol of van der Waals gas

0.4 m

State 1

State 2

5.35 Consider the piston-cylinder assembly shown below. It is well insulated and initially contains two 10,000-kg blocks at rest on the 0.05-m2 piston. The initial temperature is 500 K. The ambient pressure is 10 bar. Two moles of gas A are contained in the cylinder. This gas is compressed in a process where another 10,000-kg block is added. The following data are available for gas A: (i) Ideal gas heat capacity of gas A at constant pressure: cP 5 20 1 0.05T where cp is in [J/(mol K)] and T is in [K]. (ii) Gas A is can be described by the following equation of state: P5

aP RT 2 v2b T

with constants: a 5 25 K

and

b 5 3.2 3 1025 m3 /mol

Determine the temperature of gas A after this process. Note: This compression process is not isentropic. What is the entropy change of the universe for this process?

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5.7 Problems ◄ 311 Process consists of adding third

m = 10,000 10,000 kg block to compress piston kg

Psurr = 10 bar m = 10,000 kg

m = 10,000 kg

Well insulated

A = 0.05 m2 2 moles of gas A Tinitial = 500 K

5.36 One mole of CO is initially contained on one-half of a well-insulated, rigid tank. Its temperature is 500 K. The other half of the tank is initially at vacuum. A diaphragm separates the two compartments. Each compartment has a volume of 1 L. Suddenly, the diaphragm ruptures. Use the van der Waals equation for any nonideal behavior. Answer the following questions: (a) What is cv at the initial state? (b) Do you expect the temperature to increase, decrease, or remain constant. Justify your answer with molecular arguments. Be specific about the nature of the forces involved. (c) What is the temperature of the final state? (d) What is the entropy change of the universe for this process? 5.37 A well-insulated, rigid vessel is divided into two compartments by a partition. The volume of each compartment is 0.1 m3. One compartment initially contains 400 moles of gas A at 300 K, and the other compartment is initially evacuated. The partition is then removed and the gas is allowed to equilibrate. Gas A is not ideal under these circumstances but can be described well by the following equation of state: P5

RT a 2 2 v2b Tv

with constants: a 5 42 3 1 J K m3 2 /mol2 4 and

b 5 3.2 3 1025 3 m3 /mol 4 .

You may take the ideal gas heat capacity of gas A to be: cv 5 1 3/2 2 R Calculate the final temperature. 5.38 Consider filling a gas cylinder with ethane from a high-pressure supply line. Before filling, the cylinder is empty (vacuum). The valve is then opened, exposing the tank to a 3-MPa line at 500 K until the pressure of the cylinder reaches 3 MPa. The valve is then closed. The volume of the cylinder is 50 L. For ethane, use the truncated virial equation of state, in pressure: z5

with,

Pv 5 1 1 BrP RT

Br 5 22.8 3 1028 3 m3 /J 4

(a) What is the temperature immediately after the valve is closed? (b) If the cylinder then sits in storage at 293 K for a long time, what is the entropy change of the universe (from the original, unfilled state)?

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312 ► Chapter 5. The Thermodynamic Web 5.39 One mole per second of gas at 10 bar and 500°C flows through an isentropic, adiabatic turRT bine, where it exits at 1 bar. This gas can be described by the equation of state, P 5 . v2b 3 Where b 5 5 3 1024 m3 /mol. The ideal gas heat capacity is given by cv 5 R. Answer the follow2 ing questions: (a) What is the exit temperature? 'T (b) Is the coefficient m 5 ¢ ≤ positive, zero, or negative for this process? Is this consistent with 'P s your result in part A? (c) What is the work obtained by the turbine, in [J/mol]? 5.40 Following the process used in Example 5.5, develop an expression for the change in the dependent variable s in terms of the independent properties P and v, that is, s 5 s 1 P,v 2 . Write your answer using cp, b, k, and v. 5.41 Following the process used in Example 5.5, develop an expression for the change in the dependent variable h in terms of the independent properties T and v, that is, h 5 h 1 T,v 2 . Write your answer using cp, b, k, and v. 5.42 Following the process used in Example 5.5, develop an expression for the change in the dependent variable h in terms of the independent properties T and s, that is, h 5 h 1 T,s 2 . Write your answer using cp, b, k, and v. Consider an isentropic expansion process. Determine how h changes with T. 5.43 In analogy to Equation (5.43), develop an expression for the internal energy departure function in dimensionless coordinates: dep

DuTr, vr RTc

ideal gas

5

uTr, vr 2 uTr, vr RTc

5?

5.44 Develop expressions for the enthalpy and entropy departure functions for a gas that follows the Redlich–Kwong equation of state. 5.45 Calculate the enthalpy and entropy departure for water at 400°C and 30 MPa using generalized correlations. Compare these values to those in the steam tables. The ideal heat capacity of steam is useful in this calculation. 5.46 Calculate the enthalpy and entropy change of C2H6 from a state at 300 K and 30 bar to a state at 400 K and 50 bar using departure functions. 5.47 Repeat Problem 5.32 using the entropy departure function. 5.48 Methane flowing at 2 mol/min is adiabatically compressed from 300 K and 1 bar to 10 bar. What is the minimum work required? 5.49 Develop a relationship for the Joule–Thomson coefficient in terms of only the thermal expansion coefficient, the heat capacity at constant pressure, and measured thermodynamic properties. 5.50 What is mJT for an ideal gas? 5.51 Determine expressions for the thermal expansion coefficient, the isothermal compressibility, and the Joule–Thomson coefficient for a gas that obeys the van der Waals equation of state, in terms of T, v, cv, a, b, and R. 5.52 Determine mJT for steam at 1 MPa and 300°C using data from the steam tables. 5.53 Use the van der Waals equation of state to plot the inversion line for N2 on a PT diagram, as schematically shown in Figure 5.10. 5.54 Ethylene is liquefied by a Joule–Thomson expansion. It enters the throttling process at 50 bar and 0°C and leaves at 10 bar. What is the fraction of the inlet stream that is liquefied?

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5.7 Problems ◄ 313 5.55 The speed of sound, Vsound 3 m/s 4 , is formally equal to the partial derivative of pressure with respect to density at constant entropy: V2sound 5 a

'P b 'r s

Show that, a

'P 'P v2 ≤ b 5 2a b ¢ 'r s 'v s MW

where MW is the molecular weight. 5.56 Based on the definition in Problem 5.55, use the thermodynamic web to come up with an expression for 3 Vsound 4 in air. What is the value of 3 Vsound 4 in air at 20°C? You may consider air to be an ideal gas with cP 5 1 7/2 2 R. Based on this result, how far away is a bolt of lightning if you hear the thunder four seconds after you see the lightning. 5.57 Based on the definition in Problem 5.55, use the thermodynamic web to come up with an expression and a value for 3 Vsound 4 in water at 20°C. Use the steam tables for thermodynamic property data of liquid water. 5.58 We are interested in the thermodynamic properties of a strip of rubber as it is stretched (see below). Consider n moles of pure ethylene propylene rubber (EPR) that has an unstretched length z0. If it is stretched by applying a force F, it will obtain an equilibrium length z, given by: F 5 kT 1 z 2 z0 2

z

EPR

z0

EPR

where k is a positive constant.

F = mg = kT(z − z0)

Mass Unstretched state

Stretched state

The heat capacity of unstretched EPR is given by: cz 5 a

'u b 5 a 1 bT 'T z

where

a and b are constants

(a) Come up with fundamental property relations for dU and dA for this system, where A is the Helmholtz energy 1 A 5 U 2 TS 2 . Recall from mechanics that the work required for a reversible elastic extension is given by: dWrev 5 Fdz (b) Develop an expression that relates the change in entropy to the changes in temperature and length, that is, the independent properties z and T (and constants a, b, k, n and z0). In other words, for S 5 S 1 T, z 2 , find dS. Hint: You will need to derive an expression for 1 'S/'z 2 T. This can be done with the appropriate Maxwell relationship.

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314 ► Chapter 5. The Thermodynamic Web (c) Develop an expression that relates the change in internal energy to the changes in temperature and length. For U 5 U 1 T, z 2 , find dU. (d) Consider the relative energetic and entropic contributions to the isothermal extension of EPR. The energetic force (the component of the force that, on isothermal extension of the rubber, increases the internal energy) is given by: FU 5 a

'U b 'z T

while the entropic force is given by: FS 5 2Ta

'S b 'z T

Come up with expressions for FU and Fs for EPR. (e) If the change from the unstretched state to the stretched state above occurred adiabatically, would the temperature of the EPR go up, stay the same, or go down? Explain. 5.59 The process in Example 5.2 indicates that we need to put work into the system during an expansion process. Determine whether this result is possible (in a thermodynamic sense); if it is, explain this result physically. 5.60 Gas A expands through an adiabatic turbine. The inlet stream flows in at 100 bar and 600 K while the outlet is at 20 bar and 445 K. Calculate the work produced by the turbine. The following data are available for gas A. The ideal gas heat capacity for this process is: cP 5 30.0 1 0.02T where cP is in [J/(mol K)] and T is in [K]. PvT data has been fit to the following equation: P 1 v 2 b 2 5 RT 1

aP2 T

where, a 5 0.001 3 1 m3K 2 / 1 bar mol 2 4 and, b 5 8 3 1025 3 m3 /mol 4 3 5.61 A vessel with a total volume of 0.1 m has two compartments separated by a membrane. Compartment A has a volume of 0.005 m3 and contains 1 mol of ethane. Compartment B has a volume of 0.095 m3 and is evacuated. The vessel sits in an isothermal bath at 100°C. The membrane ruptures, filling the entire vessel with ethane. Assuming ethane can be described by van der Waals equation of state, answer the following questions: (a) What do you expect the sign of DU to be? Explain based on your understanding of intermolecular interactions. (b) Determine the values for DU, Q, and W. (c) Calculate the change in entropy of the system, surroundings, and universe. (d) If ethane is replaced with n-hexane, do you expect the magnitude of Q to increase, decrease, or stay the same? Explain.

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► CHAPTER

6 Phase Equilibria I: Problem Formulation Learning Objectives To demonstrate mastery of the material in Chapter 6, you should be able to: ► Explain why it is convenient to use the thermodynamic property Gibbs energy to determine pure species phase equilibrium. Discuss the balance between energetic and entropic effects at equilibrium. ► Apply the fundamental property relation for Gibbs energy and other tools of the thermodynamic web to predict how the pressure of a pure species in phase equilibrium changes with temperature and how other properties change in relation to one another. Write the Clapeyron equation and use it to relate T and P for a pure species in phase equilibrium. Derive the Clausius– Clapeyron equation for vapor–liquid mixtures, and state the assumptions used. Relate the Clausius–Clapeyron equation to the Antoine equation. ► Apply thermodynamics to mixtures. Write the differential for any extensive property, dK, in terms of m 1 2 independent variables, where m is the number of species in the mixture. Define and find values for pure species properties, total solution properties, partial molar properties, and property changes of mixing. ► Define a partial molar property and describe its role in determining the properties of mixtures. Calculate the value of a partial molar property for a species in a mixture from analytical and graphical methods. Apply the Gibbs–Duhem equation to relate the partial molar properties of different species. ► Relate the volume, enthalpy, and entropy changes of mixing to the relative intermolecular interactions of like (i-i) and unlike (i-j) interactions. Define the enthalpy of solution. Calculate the enthalpy of mixing from the enthalpy of solution or vice versa. ► Identify the role of the chemical potential—that is, the partial molar Gibbs energy—as the chemical criteria for equilibrium.

►6.1 INTRODUCTION So far, we have used thermodynamics to form relationships between the states of a system that undergoes certain processes. We can apply the first and second law to both reversible and irreversible processes to get information about (1) how much power is needed 315

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316 ► Chapter 6. Phase Equilibria I: Problem Formulation or obtained, (2) how much heat has been absorbed or dissipated, or (3) the value of an unknown property (e.g., T) of the final (or initial) state. In the remaining chapters, we examine another type of problem that we can also use thermodynamics to address—the composition a mixture obtains when it reaches equilibrium between coexisting phases or in the presence of chemical reactions. Chemical and biological engineers routinely deal with processes through which species chemically react to form a desired product(s). This product must then be separated from the other by-products as well as any reactants that remain. Typical separation schemes involve contact or formation of different phases through which one species of a mixture preferentially segregates. Separations technology is also a major concern in cleaning contaminated environments. Therefore, it is desirable to be able to estimate the degree to which species will react and the degree to which a given species will transfer into a different phase as a function of process conditions. These problems lead to the second major branch of thermodynamics, which we will now formulate. It deals only with equilibrium systems. It should be pointed out that this branch still uses the same observations of nature (conservation of energy and directionality) that we have already studied. In these problems, however, we wish to calculate how species distribute among phases when more than one phase is present (phase equilibria) or what types of species are formed and how much of each type is produced as systems approach equilibrium when the molecules in the system chemically react (chemical reaction equilibria). We will consider phase equilibria first. These calculations are restricted to equilibrium systems; therefore, they give information on the direction of the driving force for a given system (i.e., the system will spontaneously move toward its equilibrium state) but no information on the rate at which it will reach equilibrium.

The Phase Equilibria Problem A generic representation of the phase equilibria problem is illustrated in Figure 6.1. In this picture, a and b can represent any phase: solid, liquid, or vapor. We may be interested in any of the following: vapor–liquid, liquid–liquid, liquid–solid, gas–solid, or solid–solid equilibrium. Can you think of an example of each type? We consider a closed system, since strictly speaking only closed systems can be in thermodynamic equilibrium. In an open system, mass flows into and out of the boundary. For mass to flow, some type of driving force, such as a pressure gradient, is necessary. However, we cannot simultaneously have a pressure gradient and mechanical equilibrium (equal pressure). Thus, we will develop our formalism for closed systems. Equilibrium analysis still plays an important part in open systems, since it tells us the driving force for the transfer of species from one phase to another.

T

α

P

n α1,n α2,n α3,...,n αi,...n αm ? n β1,n β2,n β3,...,n βi,...n βm

β

Figure 6.1 Generic phase equilibria problem.

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6.1 Introduction ◄ 317

As we learned in Chapter 1, for a system to be in thermal equilibrium, there are no temperature gradients in the system. Similarly, in mechanical equilibrium there cannot be a pressure gradient. Therefore, we can write. Ta 5 Tb and,

Pa 5 Pb

Thermal equilibrium

(6.1)

Mechanical equilibrium

(6.2)

These two criteria for equilibrium are obvious and therefore straightforward to formulate; they deal with measurable properties. To see that these conditions represent criteria for equilibrium, you may ask, for example, “What would happen if Ta . Tb? ” Energy flows from hot to cold and will, therefore, flow from phase a to b until the temperatures equilibrate. A similar argument can be made for pressure and mechanical equilibrium.1 We will take thermal and mechanical equilibrium for granted in the following discussion; hence, in formulating the phase equilibria problem, we need only measure the temperature and pressure of one phase, and these values must apply to the entire system. This concept is illustrated schematically in Figure 6.1, where temperature and pressure measurements that are made only to phase a apply to the entire system. A piston–cylinder assembly is used to remind us that the system must be able to change in volume to accommodate thermal and mechanical equilibrium. The driving force for species transfer is not so apparent. This chapter will focus on the following questions: • What is the criterion for chemical equilibria of any species i? a

?i 5 ?ib

for chemical equilibrium?

• How do I use these criteria to solve phase equilibria problems (with T and P known)? Before we begin, note that neither mole fraction nor concentration, both measurable properties, represents the driving force for species transfer between phases (as temperature difference represents the driving force for energy transfer between phases). For example, consider an air–water system in phase equilibrium between the vapor and liquid phases. It would be absurd to think that oxygen will transfer from the air into the water until the mole fraction was 0.21 in the water, or conversely that water would transfer into the air until the vapor was almost all water! Unfortunately, the thermodynamic property that drives a system toward chemical equilibrium, unlike thermal or mechanical equilibrium, is not a measurable property. There are two distinct issues imbedded in the problem described above. (1) We must learn why different phases coexist, be it for pure species or for mixtures. (2) We must also develop a formalism to account for mixtures and changing composition; up to this point, we have only dealt with constant-composition systems. Rather than tackle both these tasks simultaneously, our approach is to isolate each problem, solve it, and then integrate them together to solve the entire phase equilibria problem. Figure 6.2 illustrates our solution strategy. In Section 6.2, we will learn why two phases coexist at equilibrium. We will do this for the simplest case possible—a pure species. In Section 6.3, we will learn how to carefully (and formally) describe the thermodynamics of mixtures. Once we have learned these two concepts, they will be integrated together in Section 6.4 to formulate the solution to the problem posed in Figure 6.1. 1

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These criteria can be shown formally through the maximization of entropy.

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318 ► Chapter 6. Phase Equilibria I: Problem Formulation

T

P

α

T

Section 6.2

n α1,n α2,n α3,...,n αi,...n αm

n αi ?

? n β1,n β2,n β3,...,n βi,...n βm

β

β

n βi

Multicomponent phase equilibrium

Pure species phase equilibrium

T T

P

α

P

Section 6.3

α

Section 6.4

P

n α1,n α2,n α3,...,n αi,...n αm

n 1,n 2,n 3,...,n i,...n m

?

Thermodynamics of mixtures

n β1,n β2,n β3,...,n βi,...n βm β Multicomponent phase equilibrium

Figure 6.2 Reduction of the multicomponent phase equilibria problem.

►6.2 PURE SPECIES PHASE EQUILIBRIUM Gibbs Energy as a Criterion for Chemical Equilibrium In this section, we will begin our discussion of equilibrium systems by considering phase equilibrium of a pure species (as shown in the upper right of Figure 6.2). Figure 6.3 shows the the PT projection of the PvT surface that was introduced in Section 1.6. Three coexistence lines are denoted; these lines indicate the pairs of P and T values where more than one phase can be present at equilibrium. At all other values of P and T, a single phase is most stable, and we will not have phase equilibrium. Our objective in this section is to develop a criterion that tells us when two or more phases can coexist at equilibrium or, conversely, when only a single phase is stable. The second law of thermodynamics can give us some insight into answering this question. We

P Critical point Solid

Liquid Coexistence lines Vapor T

Figure 6.3 Identification of the coexistence lines on the PT projection of the PvT surface. Along these lines, two phases can coexist in phase equilibrium.

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6.2 Pure Species Phase Equilibrium ◄ 319

saw in Chapter 3 that entropy relates to the directionality of a process. Equilibrium represents the state at which the system has no tendency to change (i.e., no further directionality). The equilibrium state occurs when the entropy of the universe is a maximum. Thus, we can determine if a pure species at temperature T and pressure P will exist in a single phase or if it can exist with two (or three) phases present at equilibrium by calculating in which case the entropy of the universe is greater. However, this approach is inconvenient because we need to do calculations of the entropy of the entire universe (i.e., we need to calculate changes in entropy of the surroundings as well as the system). We would prefer a way to tell if we have phase equilibrium by considering only properties of the system. In this section, we use a combination of the first and the second laws to develop a new property, Gibbs energy, G. This property is useful because by just looking at the Gibbs energy of each phase in the system, we can determine if either phase is more stable or, alternatively, if the two phases can coexist at equilibrium. We will now explore how Gibbs energy provides this information. Consider a closed system composed of pure species i in mechanical and thermal equilibrium and, therefore, at constant T and P. In most phase equilibria problems, the only work is Pv work. If we assume there is only Pv work, the first law in differential form can be written as: dUi 5 dQ 1 dW 5 dQ 2 PdVi We introduce the subscript “i” to denote that the analysis is performed for pure species i. The rationale for this nomenclature will be discussed further when mixtures are addressed in Section 6.3. For mechanical equilibrium, we have constant pressure, that is, dP 5 0. Therefore, we can subtract VidP from the right-hand side of the previous equation to get: dUi 5 dQ 2 PdVi 2 VidP 5 dQ 2 d 1 PVi 2 Bringing the d 1 PVi 2 term to the left-hand side and applying the definition of enthalpy gives: dHi 5 dQ The second law can be written: dSi $

dQ T

where the inequality holds for irreversible (directional) processes while the equality holds for reversible processes. Solving for dQ, in each of the previous two equations and rearranging gives: 0 $ dHi 2 TdSi Thermal equilibrium requires that we be at constant temperature 1 dT 5 0 2 ; thus, the term SdT may be subtracted from the right-hand side as follows: 0 $ dHi 2 TdSi 2 SidT 5 d 1 Hi 2 TSi 2 We recognize the group of variables on the right-hand side from Chapter 5 as the derived thermodynamic property Gibbs energy, Gi: Gi ; Hi 2 TSi

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(5.3)

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320 ► Chapter 6. Phase Equilibria I: Problem Formulation Hence, combination of the first and second laws attributes the following behavior to our closed system: 0 $ 1 dGi 2 T,P

(6.3)

The subscripts “T” and “P” remind us that this expression is valid only at constant temperature and pressure, which are the criteria for thermal and mechanical equilibrium, respectively. Equation (6.3) says that for a spontaneous process, the Gibbs energy of a system at constant pressure and temperature always gets smaller (or stays the same); it never increases.2 The system wants to minimize its Gibbs energy. Equilibrium is the state at which the system no longer changes properties; therefore, equilibrium occurs at minimum Gibbs energy. If we have two phases a and b, we can write the total Gibbs energy of pure species i as: Gi 5 nai gai 1 nbi gbi

(6.4)

where nai and nbi refer to the number of moles of i in phases a and b, respectively. Differentiating this expression and applying the inequality represented in Equation (6.3), we get: 0

0

dGi 5 d 1 nai gai 1 nbi gbi 2 5 nai dgai 1 gai dnai 1 nbi dgbi 1 gbi dnbi # 0 Two independent properties constrain the state of each phase of the pure substance i; thus, at a given T and P, gai and gbi are constant. Consequently, the first and third terms go to zero. Since we have a closed system, a species leaving one phase must be added to the other phase, so, dnai 5 2dnbi Substitution into the previous equation gives: 1 gbi 2 gai 2 dnbi # 0 From this equation, we can infer how the species in a system respond to approach equilibrium. Consider a system that initially has species in both phases a and b. If gbi is larger than gai , dnbi must be less than zero to satisfy the inequality. Physically, species i will transfer from phase b to phase a, lowering the Gibbs energy, Gi, of the system. Species will transfer until only phase a is present, and we will not have phase equilibrium.

2

Alternatively, the development of Gibbs energy can also include the more general case of non-Pv work, W* . The analog to Equation (6.3) would then read: dW* $ 1 dGi 2 T,P

(6.3*)

Integration of Equation (6.3*) shows that a process that raises the Gibbs energy is possible, but it will not occur spontaneously; rather, it will only occur with a commensurate input of work, W*, of at least the same amount that the Gibbs energy rises. This relationship can tell us, for example, the minimum amount of work required to separate two gases that have spontaneously mixed. Conversely, Equation (6.3*) provides us the maximum useful work we can get from a spontaneous process at constant T and P. Section 9.6 treats a case where non-Pv work is important.

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6.2 Pure Species Phase Equilibrium ◄ 321

Conversely if gai is larger than gbi , only phase b will be present at equilibrium. However, if the Gibbs energies of both phases are equal, this equation becomes an equality and the system has no impetus to change. This condition represents equilibrium. Thus, the criterion for chemical equilibrium is when the Gibbs energy is at the minimum: gai 5 gbi

(6.5)

Note the similarity between Equation (6.5) and Equations (6.1) and (6.2). The derived variable, Gibbs energy, although not directly measurable, provides the same information with regard to chemical equilibrium that temperature does for thermal equilibrium and pressure for mechanical equilibrium!

Roles of Energy and Entropy in Phase Equilibria We will look at the simple system shown in Figure 6.4 to explore the implications of the criterion discussed above. Specifically, we will examine the interplay between energetic effects and entropic effects in determining the Gibbs energy. Figure 6.4 shows a system in which an ideal gas and a perfect crystal coexist at temperature T and pressure P. The solid consists of pure a, while the ideal gas is comprised of both a and b. Species b is noncondensable and is not incorporated into the solid lattice. We will explore how the Gibbs energy of this system depends on the fraction of a that is vaporized. As we have just seen, the system obtains equilibrium in the state where the Gibbs energy is minimized. However, Gibbs energy can be lowered by either lowering h or increasing s. The right side of Figure 6.4 shows plots of various properties of a in this system vs. fraction of species a in the vapor. In these plots, enthalpy is related to the energetics of the system. Recall from Section 2.6 that enthalpy is the appropriate property to quantify the energetics of a closed system at constant pressure, as it combines internal energy with Pv work. The entropy is multiplied by T to give it units of energy, so we can plot it on the same graph. The Gibbs energy is found by subtracting Ts from h. First, consider the enthalpy of the system. Since we have an ideal gas, there is no intermolecular potential energy in the gas phase, just molecular kinetic energy, which is

b a

Ideal gas a and b

a b a

b

a

Joules/mole

P

T

ha Tsa ga = ha − Tsa

Perfect crystal a 0 Equilibrium dga = 0

a in vapor 1 total a

Figure 6.4 A schematic of a simple phase equilibria problem and a plot of some thermodynamic properties at a temperature where both solid and vapor phases coexist at equilibrium.

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322 ► Chapter 6. Phase Equilibria I: Problem Formulation solely a function of temperature. On the other hand, the solid is more stable than the gas and has lower enthalpy due to the attractive forces between bonded a atoms. The enthalpy in the entire system can be related to the fraction of atoms in the solid. Consider the vaporization process. As a bond in the solid is broken to increase the fraction of a in the vapor, the system increases in molecular potential energy by an amount proportionate to the bond enthalpy. Each atom that leaves the solid increases the enthalpy in the system by the same amount. This proportionate increase manifests itself in a linear relation between the enthalpy and the fraction of a in the vapor phase, as illustrated in Figure 6.4. Now consider the entropy of the system. The entropy will not increase linearly, as enthalpy did. In the case that all a is in the solid and all the vapor is b, we have as ordered a system as we possible can. The first atom of solid that evaporates will cause a large increase in “disorder,” or more precisely the gas will be able to take on many more different configurations. As more and more atoms vaporize, they enter a gas that has more and more a; therefore, for each additional atom, the additional increase in entropy is less and less. Intuitively, we can see that there is a lot more randomness introduced by the first atom leaving a pure solid and going into the vapor than when there are already many a atoms in the gas phase.3 This relationship was described in Chapter 3 and can be formally developed using statistical mechanics. It is also plotted in Figure 6.4. In this system, there are two tendencies that oppose each other. The perfect crystal with no fraction of a in vapor is the system of minimum energy (or enthalpy), but the gas with all a vaporized maximizes entropy. To see the state that the system will obtain at equilibrium, we need to determine the extent to which each effect dominates. To do so, we need only consider the Gibbs energy 1 ga 5 ha 2 Tsa 2 for the system. The optimal compromise between minimum enthalpy and maximum entropy occurs with part solid and part vapor. At low vapor fractions, the increase in entropy caused by introducing a into the gas phase more than offsets the increase in enthalpy caused by breaking a bond in the solid. Therefore, the Gibbs energy is lowered, so this process will occur spontaneously, and the system will tend to sublimate. Conversely, at high vapor fractions, the increase in stability caused by forming a solid more than adequately compensates for the decrease in entropy associated with losing an atom from the vapor; thus, the gas will tend to crystallize. These two effects are indicated by arrows in Figure 6.4. The system will exist with two phases in equilibrium in between these extremes with gsa 5 gva 4 as illustrated. Now consider a much lower temperature. Since entropy is multiplied by T, the effect of randomness (entropy) cannot compete with the minimization of enthalpy; in this case, the equilibrium state (minimum g) occurs when the system exists as a pure solid with no a in vapor. Hence all the vapor crystallizes. Conversely, at very high temperatures, the effect of entropy dominates and only vapor exists. This behavior is consistent with our experience. Solids exist at low temperature and as the temperature is increased, they sublime (or melt). The second law states that entropy goes to a maximum; yet if our system were at maximum entropy, the solid phase would not exist. How do you resolve this paradox? [Hint: How is the temperature being kept constant?] 3

Alternatively, we can view the increase in entropy when species a goes into the gas without reference to species b. Since the gas is ideal, species a does not know that b is there. However, the partial pressure of a increases as more and more a goes into vapor phase. In Example 6.11, we will quantitatively show that the increase in entropy goes as ya lnya and is, indeed, not linear. 4 Intuitively, we can see that this equilibrium problem can be cast as a “pure” species problem since the gas phase is ideal and a does not know b is there. After we treat mixtures, in general, we will formally see that this relation holds in this special case. It is presented here to help form a conceptual framework for g.

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6.2 Pure Species Phase Equilibrium ◄ 323

EXAMPLE 6.1 Role of Gibbs Energy in Biological Systems–Qualitative

We have seen that the Gibbs energy determines whether a process can occur spontaneously. This concept can be applied to understand aspects of biological systems. Use the Gibbs energy to show why the proteins that control complex living organisms are not stable at high ambient temperatures. SOLUTION The structure of proteins can be considered in different levels of organization. Proteins are long-chain polymers in which a sequence of amino acids are linked by peptide bonds—a bond between the carbon atom in the carboxyl group from one amino acid and the nitrogen atom in the amino group from the neighboring amino acid. This chain forms the primary structure of the protein. The polymer molecule then folds back upon itself, forming intramolecular coulombic, hydrogen, and van der Waals bonds between different, non-neighboring amino acids in the chain. The first folding level forms the secondary structure of the protein. Common secondary structures include a -helices and b-pleated sheets. The third and fourth folding levels form the tertiary and quaternary structure of the protein. The resulting welldefined structure has the precise chemical properties that enable the protein to perform its specific function. We consider a native protein (n) with its structure intact and a denatured protein (d) that no longer has its well-defined structure and cannot perform its specified function. The native protein remains intact only within a limited temperature range above room temperature. We can understand this result in terms of the Gibbs energy of the protein. The intramolecular bonds that define the structure of a protein make it energetically favorable relative to its denatured analog in which some of the bonds of the higher levels of organization are broken; thus, hd . hn

(E6.1A)

On the other hand, the denatured protein is no longer limited by its specific constrained three-dimensional structure and may undertake many more possible configurations. Hence, its entropy is higher, that is: sd . sn

(E6.1B)

To see whether a protein will spontaneously denature, we consider the Gibbs energy difference between its native and denatured forms. Applying the definition of Gibbs energy: Dg 5 gd 2 gn 5 1 hd 2 hn 2 2 T 1 sd 2 sn 2

(E6.1C)

The sign of the Gibbs energy change in Equation (E6.1C) depends on the temperature of the system. At ambient temperature, the first term in Equation (E6.1C) dominates and inspection of Equation (E6.1A) shows Dg . 0. Thus, the protein will not spontaneously denature. As the temperature becomes higher, the second term becomes more important and Dg , 0 since the entropy of the denatured protein is higher. Thus, at higher temperatures, proteins spontaneously denature. The trade-off between the energetically favored hydrogen bonds and electrostatic and van der Waals interactions and the entropically favored randomness of the denatured state determines the temperature at which a protein is no longer stable. In biological systems, this balance commonly occurs between 50°and 70°C. We will calculate this temperature for the protein lysozyme in Example 6.2. The above analysis is simplified since we have not considered the interactions of the protein with the solution in which it sits; yet it is essentially valid. However, in many cases the solvent–protein interactions form an important component in understanding the behavior of these systems; we will have to wait until we learn about the thermodynamics of mixtures to address this more complicated case.

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324 ► Chapter 6. Phase Equilibria I: Problem Formulation

EXAMPLE 6.2 Thermodynamic Prediction for the Temperatures at Which a Protein Unfolds

We can use the principles of phase equilibrium to learn about the stability of proteins in biological systems. In this example, we consider the phase equilibrium of the protein lysozyme (l) between its native phase, n, and its denatured phase, d, where unfolding occurs. The following thermochemical data are available. The heat capacity for each state is independent of temperature and is given by: Denatured: cP,d 5 15.1 B

kJ mol K

R

Native: cP,n 5 6.0 B

kJ mol K

R

At 25°C, the enthalpy and entropy differences between native and denatured states are given by: hd 2 hn 5 242 B

kJ mol

R

sd 2 sn 5 618 B

and

J mol K

R

Answer the following questions: A. At 25°C, which phase is stable? Justify your answer. B. Plot the difference in Gibbs energy from the native state to the denatured state 1 gd 2 gn 2 versus temperature, T. On the plot label the following: i. The heat-denaturation temperature, which is given as the temperature above which the native protein is no longer thermodynamically stable. ii. The cold-denaturation temperature, which is given as the temperature below which the native protein is no longer thermodynamically stable. iii. The temperature that the native state is most stable. SOLUTION A schematic of these two phases at equilibrium is shown in Figure E6.2A. In its native phase, the polymer folds back over itself with the molecules forming secondary bonds, as shown in the inset in the lower left of the figure. When it becomes denatured, it unfolds as shown in the upper right. As discussed in Example 6.1, the additional bonding in the native state gives it a lower enthalpy relative to the denatured state; however, the native state is also more constrained in movement giving it a lower entropy. A. As we have just learned, the thermodynamic property Gibbs energy, g, characterizes the trade-off between enthalpy and entropy and tells us which phase is more stable. Hence, we need to calculate the difference in Gibbs energy between the protein’s

P

T

1

26

d n dl

1

72 84 66

40

95 56 110

40

124 95

110

84

n

26

65

72 58

124

Native molecule

Denatured molecule

n nl Protein lysozyme phase equilibrium

Figure E6.2A Schematic of the equilibrium between the native phase, n, and the denatured phase, d, of the protein lysozyme. The insets show molecular representations of the protein, which unfolds in the denatured phase.

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6.2 Pure Species Phase Equilibrium ◄ 325

denatured state and the native state. Using the values for enthalpy and entropy differences provided earlier, we get: gd 2 gn 5 1 hd 2 hn 2 2 T 1 sd 2 sn 2 5 24,200 2 298 3 618 5 57,800 B

B.

J mol

R

Because the change in Gibbs energy is positive, the Gibbs energy in the native state is lower than the denatured state. Therefore, the native state is more stable, and the protein will not spontaneously unfold. To solve for the difference in Gibbs energy as a function of temperature, it is convenient to first specify a hypothetical path. As is often the case in creating hypothetical paths, We can take one of many approaches. We illustrate one approach next, but two other possibilities are developed in Problem 6.40. To calculate the difference in Gibbs energy at any temperature, we can calculate the differences in enthalpy and entropy separately and then use the definition, g 5 h 2 Ts. Figure E6.2B shows the three-step process for enthalpy (left) and entropy (right). We start each path at the native state at temperature, T. The first step is to change the temperature in that phase to 298 K. This is a “sensible heat” calculation, as shown in Sections 2.6 and 3.6. In this way, each path goes through a temperature of 298K, where we can use the provided data in the next step. In the third step, we change the temperature of the denatured protein back to T. Hence, the limits of integration for step 3 are opposite those of step 1. hd − hn

native

T

sd − sn denatured

native

T ∫ cP,d dT 298

298 ∫ cP,ndT T

298c P,n ∫ dT T T

(hd − hn)298

denatured

T c P,d ∫ dT 298 T

T = 298[K] native

T

T = 298[K] denatured

native

(sd − sn)298

denatured

Figure E6.2B Thermodynamic paths to calculate differences in enthalpy (left) and entropy (right) between the denatured state and the native state at any temperature.

Applying the paths shown in Figure E6.2B, we get: 298

hd 2 hn 5 3 cP,n dT 1 1 hd 2 hn T

T

2 298

1 3 cP,d dT 298

5 1 hd 2 hn 2 298 1 1 cP,d 2 cP,n 2 1 T 2 298 2

(E6.2A) (Continued)

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326 ► Chapter 6. Phase Equilibria I: Problem Formulation

298

T

cP,n

sd 2 sn 5 3 dT 1 1 sd 2 sn T

2 298

cP,d 1 3 dT 5 1 sd 2 sn 2 298 T 298

T

1 1 cP,d 2 cP,n 2 ln ¢

T ≤ 298

(E6.2B)

where we took the heat capacity out of the integral because it is constant. Inserting Equations (E6.2A) and (E6.2B) into the definition for Gibbs energy gives: gd 2 gs 5 1 hd 2 hn 2 2 T 1 sd 2 sn 2 5 1 hd 2 hn 2 298 1 1 cP,d 2 cP,n 2 1 T 2 298 2 2 TB 1 sd 2 sn 2 298 1 1 cP,d 2 cP,n 2 ln ¢

T ≤R 298

We can use the values for the thermochemical data presented earlier to get: gd 2 gs 5 242 1 9.1 1 T 2 298 2 2 TB0.618 1 9.1 ln ¢

T ≤R 298

(E6.2C)

where the numerical values are in units of kJ, mol, and K, as appropriate. Equation (E6.2C) can be used to plot gd 2 gn vs. T, as shown in Figure E6.2C. The Gibbs energy difference increases, goes through a maximum, and then decreases. From this plot, we see that there is a “window” in which the native state of the protein has a lower Gibbs energy than the denatured state and is, therefore, stable. This region is shown by the shaded area. The points where the curve crosses zero demark the cold- and heat-denaturation temperatures. The cold-denaturation temperature and heat-denature temperature are found by setting Equation (E6.2C) to zero. The values are found to be 218 K and 343 K, respectively. In between these temperatures the native protein is stable. Outside this range, the protein unfolds and becomes denatured. The maximum value for gd 2 gn occurs at 278 K. This value can be found either by setting the derivative of Equation (E6.2C) equal to zero and solving for T, or by simply finding where the function reaches a maximum numerically (solver in Excel can be useful for this approach). Maximum Stability, 278 K

75

25

−25 −50 −75 −100 200

225

Heat-denaturation temperature, 343 K

0 Cold-denaturation temperature, 218 K

gd −gn [kJ/mol]

50

250

300 275 Temperature [K]

325

350

375

Figure E6.2C Difference in Gibbs energy between the native phase, n, and the denatured phase, d, of the protein lysozyme. The shaded area represents the range over which the native protein has a lower Gibbs energy and is stable.

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6.2 Pure Species Phase Equilibrium ◄ 327

The Relationship Between Saturation Pressure and Temperature: The Clapeyron Equation In Section 1.5, we learned that P and T are not independent for a pure species that exists in two phases at equilibrium. We now wish to come up with an expression relating the pressure at which two phases can coexist to the temperature of the system. This expression will allow us to calculate, for example, how the saturation pressure changes with temperature. Recall that the saturation pressure, Pisat, is defined as the unique pressure at which pure species boils at a given temperature. Before we begin, we will qualitatively look at the issues of this problem in the context discussed above. We begin with the criterion for equilibrium between two phases: gai 5 gbi

(6.5)

Again, a and b can represent the vapor, liquid, or solid phases. Since two intensive variables completely specify the state of the system, the value of g for each phase is constrained at a given T and P. Thus, we can plot surfaces of Gibbs energy for each phase, as the left side of Figure 6.5 illustrates. The intersection of the two surfaces, the so-called coexistence line, represents the conditions where Equation (6.5) is satisfied and the two phases are in equilibrium. This is the same line you see on the PT projection of the PvT surface in Figure 6.3. To the left of the coexistence line, phase b has lower Gibbs energy and will represent the phase at which the Gibbs energy is at a minimum. Conversely, to the right, only phase a will exist. At the given set of conditions for P and T on the coexistence line, however, both phases exhibit identical Gibbs energies; thus, phases a and b coexist in phase equilibrium. Which phase is more random? Which phase has stronger intermolecular attraction? Let us say we know the equilibrium temperature and pressure of a two phase system, such as at state 1 shown on the PT diagram on the right side of Figure 6.5. We would like to be able to calculate how much the pressure will need to change for any given temperature change so that the system remains in phase equilibrium. The approach we take is shown in Figure 6.5. For a differential change in the equilibrium temperature dT, the change in equilibrium pressure, dP, can be calculated, because Equation (6.5) must be

g

P

Coexistence line Phase β

α

Phase β has lower g and is more stable

Coexistence line (phase equilibrium)

g αi = g βi

dg Stable 1

β

g α + dg α = g β + dg β i

i

i

i

Phase α

P Phase α is more stable

T

T

Figure 6.5 (left) Intersection of two Gibbs surfaces for phases a and b to form a coexistence line; (right) calculation strategy to determine how the equilibrium pressure changes with temperature along a coexistence line.

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328 ► Chapter 6. Phase Equilibria I: Problem Formulation valid along both points on the coexistence line. Hence, both gai 1 dgai 5 gbi 1 dgbi and gai 5 gbi . Subtraction gives: dg ai 5 dgbi

(6.6)

As we saw in Chapter 5, the thermodynamic web provides a useful vehicle for relating derived thermodynamic properties to measured properties. Applying the fundamental property relation for g [Equation (5.9)] to each phase, we get: vai dP 2 sai dT 5 vbi dP 2 sbi dT The phases have been omitted on T and P. Why? Rearrangement yields: sa 2 sbi dP 5 ia dT vi 2 vbi

(6.7)

Now we can also apply Equation (6.5): gai 5 gbi or using the definition of Gibbs energy, Equation (5.3) we get: hai 2 Tsai 5 hbi 2 Tsbi Solving for the difference in entropy: sai 2 sbi 5

hai 2 hbi T

Substitution into Equation (6.7) yields the Clapeyron equation: ha 2 hbi dP 5 ai 1 vi 2 vbi 2 T dT

(6.8)

The Clapeyron equation relates the slope of the coexistence curve to the enthalpy and volume changes of phase transition, both experimentally accessible properties! Can you think of how to measure these values? In other words, it tells us the pressure change, dP, which is necessary to maintain phase equilibria of a substance when the temperature has changed by dT.

Pure Component Vapor–Liquid Equilibrium: The Clausius–Clapeyron Equation Next we consider the specific case of vapor–liquid equilibrium. In this case, the molar volume of the liquid is often negligible compared to the volume of the vapor: vli V vvi or vl < 0

1 Assumption I 2

This assumption implies we are down the liquid–vapor dome shown in Figure 1.6, away from the critical point. If additionally we consider the vapor to obey the ideal gas model, vvi 5

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RT P

1 Assumption II 2

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6.2 Pure Species Phase Equilibrium ◄ 329

the coexistence equation for vapor–liquid equilibrium [Equation (6.8)] becomes: Psat dPsat i Dhvap,i i 5 dT RT2 where Dhvap,i 5 hvi 2 hli and Psat i represent the enthalpy of vaporization and the saturation pressure, respectively, of species i at temperature T. Separating variables, Dhvap,idT dPsat i 5 Psat RT2 i

(6.9a)

Equation (6.9a) is called the Clausius–Clapeyron equation. It can be rewritten in the form: d lnPsat i 5 2

Dhvap,i R

1 da b T

(6.9b)

If we assume the enthalpy of vaporization is independent of temperature, that is, Dhvap,i 2 Dhvap,i 1 T 2

(Assumption III)

we can either definitely integrate Equation (6.9b) between state 1 and state 2 to get: ln

Dhvap,i 1 Psat 1 2 52 B 2 R sat P1 R T2 T1

(6.10)

or write the indefinite integral of Equation (6.9b): ln Psat i 5 const 2

Dhvap,i RT

(6.11)

In fact, Assumption III is not valid over large temperature ranges. The enthalpy of vaporization decreases as temperature increases, so Equations (6.10) and (6.11) can be used only over a limited temperature range. However, in surprisingly many cases, the error introduced by Assumption III is approximately offset by the errors of Assumptions I and II, leading to linear behavior of ln Psat vs. T21 over a larger range than would be originally surmised. Saturation pressure correlations are commonly reported in terms of the Antoine equation: lnPsat i 5 A 2

A C1T

(6.12)

Here A, B, and C are empirical parameters that are available for many fluids. Values for Antoine constants can be found in Appendix A.1. The Antoine equation, an empirical equation, is strikingly similar to Equation (6.11). The Antoine equation brings back a similar theme to that discussed with equations of state (Chapter 4). When an empirical equation’s form reflects the basic physics that it is trying to describe, it tends to work better. Why do you think the Antoine equation works better than the Clausius–Clapeyron equation in correlating saturation pressures? There are several more complex correlations reported in the literature for Psat as a function of T. These forms will not be covered in this text.

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330 ► Chapter 6. Phase Equilibria I: Problem Formulation

EXAMPLE 6.3 Estimation of the Enthalpy of Vaporization from Measured Data

Trimethyl gallium, Ga 1 CH3 2 3, can be used as a feed gas to grow films of GaAs. Estimate the enthalpy of vaporization of Ga 1 CH3 2 3 from the data of saturation pressure vs. temperature given in Table E6.3.5 21, the slope SOLUTION Examination of Equation (6.11) suggests that if we plot ln Psat i vs. T will give 2 1 Dhvap, Ga1CH323 /R 2 . The data in Table E6.3 are plotted in such a manner in Figure E6.3. A least-squares linear regression is also shown in Figure E6.3. The high correlation coefficient implies Dhvap, Ga1CH323 is constant in this temperature range. Taking the slope of the line, we get:

2

Dhvap,Ga 1CH323 R

5 24222.1 3 K 4

Solving for the enthalpy of vaporization gives: Dhvap, Ga1CH323 5 35.1 3 kJ/mol 4 For comparison, a value measured by static bomb combustion calorimetry has been reported as 33.1 kJ/mol, a difference of 6.0%. TABLE E6.3

Saturation Pressure Data for Ga 1 CH3 2 3 Pisat 3 kPa 4

T [K] 250

2.04

260

3.3

270

7.15

280

12.37

290

20.45

300

32.48

310

49.75 4.5 4

In(Pisat [kPa])

3.5 3

InPisat = − 4222.1T −1 + 17.556 R 2 = 0.9973

2.5 2 1.5 1 0.5 0 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 0.0038 0.0039 1 [K−1] T

Figure E6.3 5

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0.004

Plot of data in Table E6.2 and a least-squares linear ﬁt of the data.

(Via NIST) J. F. Sackman, and L. H. Long, Trans. Faraday Soc., 54, 1797 (1958).

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6.2 Pure Species Phase Equilibrium ◄ 331

EXAMPLE 6.4 Verification of the “Equal Area” Rule Presented in Section 4.3

In the discussion of cubic equations of state (Section 4.3), it was stated that the vapor–liquid dome could be constructed from a subcritical isotherm generated by the equation of state. The saturation pressure was identified as the line that divided the isotherm into equal areas, as shown in Figure E6.4. Verify that this statement is consistent with the criteria for equilibrium developed above. SOLUTION The vapor–liquid equilibrium criterion developed in this section states that the saturated vapor and the saturated liquid have equal Gibbs energies, that is, gvi 5 gli or,

gvi 2 gli 5 0

(E6.4)

Another way to write Equation (E6.4) is that the integral, from saturated liquid to saturated vapor, of the differential Gibbs energy must be zero: giv

3 dg 5 0 gil

We can apply the fundamental property relation for dg, that is, Equation (5.9): 0

sat. vap.

3 3 vidP 2 sidT 4 5 0

Pressure

sat. liq.

P sat Equal areas v sat(liq)

v sat(vap)

Volume

Figure E6.4 Subcritical isotherm given by a cubic equation of state divided into equal areas.

(Continued)

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332 ► Chapter 6. Phase Equilibria I: Problem Formulation

The second term goes to zero, since we are evaluating the integral along an isotherm; hence, dT 5 0. Using the product rule, we get: viv

sat. vap.

3 vidP 5

viv Pisatvi 0 vil

2 3 Pdvi 5 0 vil

sat. liq.

or, viv

Pisat 1 vvi 2 vli 2 2 3 Pdvi 5 0 vil

The resulting expression can be interpreted in regard to the plot in Figure E6.4. The integrated v l area from vli to vvi on the plot of P vs. v is equal to the product Psat i 1 vi 2 vi 2 ; therefore, the saturation pressure is the line that divides the isotherm into equal areas, resulting in a net area under the saturation line matching the net area above it.

EXAMPLE 6.5 Change in the Melting Point of a Solid Nanoparticle

The field of nanotechnology is an emerging area for chemical and biological engineers. Consider nickel nanoparticles for use as a catalyst to grow carbon nanotubes. The normal melting temperature, Tm, of nickel is 1728 K. In a nanoparticle, the curved surface leads to a force acting tangentially to the surface of the particles, which changes the Gibbs energy of the solid. The magnitude of this effect is determined by the surface tension, s. The differential change in Gibbs energy can be written to account for surface tension in terms of curvature, or inverse radius (1/R). In this case, the fundamental property relation given by Equation (5.9) can be modified as follows: 1 dgi 5 2sidT 1 vidP 1 2visid ¢ ≤ R The following properties are available for Ni: surface tension of solid Ni:

ssNi 5 1.75 B

density of solid Ni:

rsNi 5 8,900 B

Enthalpy of fusion:

Dhfus,Ni 5 217.48 B

J m2 kg m3 kJ

mol

R

R

R

Consider Ni nanoparticles with radius of approximately 2 nm 1 2 3 1029 m 2 . You are interested in whether they might melt when they are processed at 750°C and 1 atm. For simplicity, you may assume that the system contians only pure nickel nanoparticles. At 750°C and 1 atm, what is the phase of pure 2 nm Ni nanoparticles at equilibrium? State any assumptions that you make. SOLUTION Figure E6.5A shows a schematic of phase equilibrium between solid nickel and liquid nickel. On the left of the diagram, bulk (macroscopic) Ni is shown, and the Ni nanoparticles

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6.3 Thermodynamics of Mixtures ◄ 333

in which we are interested are shown on the right. We wish to calculate the temperature at which the nanoparticles melt. To do this calculation, we can develop a relationship between (1/R) and T along a coexistence line in a similar way to how we developed a relationship between P and T for the Clapeyron equation. A schematic path for such a calculation is shown in Figure E6.5B. Along the coexistence line the differential change in Gibbs energy of the solid must equal that of the liquid: dgs 5 dgl Applying the modified form of the Equation (5.9) yields: 0 0 0 1 1 s s l l l s s 2sNidT 1 vNidP 1 2vNi sNi d ¢ ≤ 5 2sNidT 1 vNidP 1 2vNil sNi d ¢ ≤ R R where the differential term in pressure is set equal to zero because the system is at constant pressure, and the differential term for curvature of the liquid is also set to zero. Gathering like terms, we get: 1 s 1 ssNi 2 slNi 2 dT 5 2vsNisNi d¢ ≤ R Solid–Liquid Equilibrium

(E6.5A)

Solid–Liquid Equilibrium of nanopalticles of radius, R Tm

Liquid

s s Liquid s s s s s s s s s s s s s s s s

n lNi

R

Solid

Solid

n sNi

T=?

Figure E6.5A Schematic of the equilibrium between solid and liquid nickel. The schematic on the left shows the system for bulk nickel, which melts at Tm. The case on the right shows nickel nanoparticles of radius, R. P = const

State 1 (T1, 1/R1)

Tm

s

Temperature

dg

liquid

= l

dg

Solid

State 2 (T2, 1/R2)

T

1 =0 R1

Figure E6.5B

Curvature

1 R2

Coexistence line from the bulk state (1) to the nanocrystalline state (2). (Continued)

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334 ► Chapter 6. Phase Equilibria I: Problem Formulation

We do not have data for the entropy change from the liquid to the solid, but we do have data for the enthalpy change for bulk Ni. These quantities can be related by equating the Gibbs energies (as was done in the development of the Clapeyron equation): s

l 1 ssNi 2 sNi 25

l

1 hNi 2 hNi 2 T

5

Dhfus,Ni Tm

Substituting this result back into Equation (E6.5A) and solving for dT gives: dT 5

2vsNissNiTm 1 d¢ ≤ Dhfus,Ni R

(E6.5B)

Finally, we can integrate Equation (E6.5B) from the bulk state to the state associated with the solid nanoparticles of 2 nm: T

1/R

s

s

2vNisNiTm 1 3 dT 5 Dh 3 d¢ R ≤ fus,Ni

(E6.5C)

1/R50

Tm

We can calculate the molar density of the solid by using the density of solid nickel provided earlier and the molecular weight of Ni, 58.69 [g/mol]:

vsNi 5

MWNi 5 rsNi

58.69 B 8,900 B

kg m3

g mol

R

R 3 1,000 B

g kg

5 6.6 3 1026 B R

m3 R mol

Integrating Equation (E6.5C) and solving for the temperature at which the nickel nanoparticles melt, we get: T 5 Tm 1

2vsNissNiTm 1 2vsNissNi 1 ¢ ≤ 5 Tm B1 1 ¢ ≤ R 5 587 K Dhfus,Ni R Dhfus,Ni R

Thermodynamically, the nickel nanoparticles will melt at the processing temperature of 750°C! Analyses such as the one illustrated in this example have helped researchers understand seemingly anomalous data in processing carbon nanotubes. This example also illustrates how property behavior can significantly change at the nanoscale.

►6.3 THERMODYNAMICS OF MIXTURES Introduction In this section, we explore how to formally treat thermodynamic properties of species in mixtures. In terms of possible combinations of intermolecular interactions, mixtures are inherently more complex than pure species. For a pure species i, all the intermolecular interactions are identical. The resulting thermodynamic properties—such as Vi, Ui, Si, Hi, and Gi —are a manifestation of those interactions. Since a mixture contains more than one species, its properties are determined only in part by an average of each of the pure species (i-i) interactions. We must now also take into account how each of the species interacts with the other species in the mixture, that is, the unlike (i-j) interactions. Thus

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6.3 Thermodynamics of Mixtures ◄ 335

67 ml 50 ml

+

= 20 ml

Pure ethanol

Pure water

Mixture of ethanol and water

Figure 6.6 Mixing of ethanol and water.

the properties of a mixture depend on the nature and amount of each of the species in the mixture. The values of the mixture’s properties will be affected not only by how those species behave by themselves but also by how they interact with each other. Consider, for example, the experiment depicted in Figure 6.6. If we mix 50 ml of ethanol with 20 ml of water at 25°C and measure the resulting volume of solution, as careful as we might be, we get 67 ml! (Try this yourself.) Where has the other 3 ml gone? The solution has “shrunk” because the ethanol and water can pack together more tightly than can each species by itself. This is due to the nature of the hydrogen bonding involved in the structure of the liquid. We still have the same total mass, since mass is a conserved quantity; however, the mixture volume is different from the sum of the pure species volume. The difference in the mixture is based on the nature of the unlike ethanol–water interactions and the fact that they are different from the water–water or ethanol–ethanol pure species interactions. This example shows that in the treatment of multicomponent mixtures, it is important to realize that species in solution can behave quite differently than they do by themselves, depending on the chemical nature of their neighbors in solution. This behavior will affect all the thermodynamic properties of the solution. When a species becomes part of a mixture, it loses its identity; yet it still contributes to the properties of the mixture, since the total solution properties of the mixture depend on the amount present of each species and its resultant interactions. We wish to develop a way to account for how much of a solution property (V, H, U, S, G . . .) can be assigned to each species. We do this through a new formalism: the partial molar property.

Partial Molar Properties The state postulate tells us that if we specify two intensive properties for any pure species, we constrain the state of a single-phase system. For extensive properties, we must additionally specify the total number of moles.6 In Chapter 5, we learned how to mathematically describe any intensive thermodynamic property in terms of partial derivatives of two independent, intensive properties. Since we are now concerned with thermal and mechanical equilibrium, it makes sense to choose T and P as the independent, intensive properties. We wish to extend the formulation to mixtures with changing composition. In addition to specifying two independent properties, we must also consider the number of moles of each species in the mixture. We now wish to specify the extensive thermodynamic property of the entire mixture, K, where we use the symbol K to represent any possible extensive thermodynamic property, that is, K 5 V, H, U, S, G, and so on. In essence, by using K, we avoid repetitive derivations by treating the problem in general. If we were to divide K by the total 6

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Or an analogous quantity that specifies the size of the system.

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336 ► Chapter 6. Phase Equilibria I: Problem Formulation number of moles in the system, we would get the intensive property k 5 v, h, u, s, g, and so on. We call K (or k) the total solution property. Mathematically, we can write the extensive total solution property K in terms of T, P, and the number of moles of m different species: K 5 K 1 T, P, n1, n2, c, ni, c, nm 2

(6.13)

for example V 5 V 1 T, P, n1, n2, c, ni, c, nm 2 C H 5 H 1 T, P, n1, n2, c, ni, c, nm 2 S ( Note that Equation (6.13) is an extension and generalization of the equation on page 18 to systems of m species. We need to know m 1 2 independent quantities to completely specify K. In Equation (6.13) we specifically choose the system temperature, pressure, and number of moles of each of the m species. Once these measured properties are specified, the state of the system is constrained and all the extensive properties, K, take specific values. The differential of K can then be written as the sum of partial derivatives of each of these independent variables, as follows: dK 5 a

m 'K 'K 'K ≤ dni b dT 1 a b dP 1 a ¢ 'T P,ni 'P T,ni i51 'ni T,P,nj 2 1

(6.14)

for example, 'V b dT 1 'T P,ni E 'H dH 5 a b dT 1 'T P,ni dV 5 a

m 'V 'V dni b dP 1 a a b 'P T,ni i51 'ni T,P,nj 2 1 m U 'H 'H ≤ dni a b dP 1 a ¢ 'P T,ni i51 'ni T,P,nj 2 1 (

a

We use the notation nj21 to specify that we are holding the number of moles of all 1 m 2 1 2 species except species i constant when we take the partial derivative with respect to ni. It is convenient to define a new thermodynamic function, the partial molar property, Ki : Ki ; ¢ always in terms of ni never xi

'K ≤ 'ni T,P,nj 2 1

always hold the intensive properties P and T constant

(6.15) the number of moles of all other species except i are held constant

for example, Vi 5 ¢

'V ≤ 'ni T,P,nj 2 1

F V 'H Hi 5 ¢ ≤ 'ni T,P,nj 2 1 (

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6.3 Thermodynamics of Mixtures ◄ 337

A partial molar property is always defined at constant temperature and pressure, two of the criteria for phase equilibrium. Partial molar properties are also defined with respect to number of moles. The number of moles of all other j species in the mixture are held constant; it is only the number of moles of species i that is changed. A common mistake in working with partial molar properties is to erroneously replace number of moles with mole fractions. We must realize, however, that Equation (6.15) does not simply convert to mole fraction, that is, Ki 2

1 'K ¢ ≤ nT 'xi T,P,xj 2 i

In changing the number of moles of species i, we change the mole fractions of all the other species in the mixture as well, since the sum of the mole fractions must equal 1. Placing Equation (6.15) into Equation (6.14), the total differential of the variable K becomes:

dK 5 a

m 'K 'K b dT 1 a b dP 1 a Kidni 'T P,ni 'P T,ni i51

for example, m 'V 'V b dT 1 a b dP 1 a Vidni 'T P,ni 'P T,ni i51 m E U 'H 'H dH 5 a b dT 1 a b dP 1 a Hidni 'T P,ni 'P T,ni i51 (

dV 5 a

At constant temperature and pressure, this equation reduces to dK 5 a Kidni

(6.16)

If, in addition to keeping T and P constant, we also keep the composition of the mixture constant (i.e., the mole fraction of all m species), then the partial molar properties are constant. In this case, we can integrate Equation (6.16) to get: K 5 a Kini 1 C where C is a constant of integration. To determine C, we can use either a physical argument or a more mathematical argument involving Euler’s theorem on homogeneous functions. The physical argument follows; the mathematical argument is presented in Example 6.6. The intensive property, k, depends only on temperature, pressure, and composition of the m species present. Thus, k depends only on the relative amounts of each species in the system. On the other hand, the extensive property K is linearly dependent on the total amount of species present. For example, let’s consider volume, a property that can exist in intensive form as k 5 v or in extensive form as K 5 V. If the number of all species in the system doubles at constant T and P (i.e., n1 S 2n1, n2 S 2n2, c, nm S 2nm), the molar volume, v, remains the same. The extensive property, V, however, will double.

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338 ► Chapter 6. Phase Equilibria I: Problem Formulation Similarly, if the number of all the m chemical species are cut in half, v remains the same but V will be cut in half. The number of species can be divided in half an arbitrary number of times and v will remain unchanged. However, in the limit of an infinitesimally small number of species, the extensive property, V, will go to 0. In general, the extensive property K must go to zero as a ni goes to zero; thus, the constant of integration becomes 0, and we have: K 5 a niKi

(6.17)

for example, V 5 a niVi FH 5 a niHiV ( or, dividing by the total number of moles, k5

K 5 a xiKi ntotal

(6.18)

for example, v 5 a xiVi Fh 5 a xiHiV ( where k is the corresponding intensive property to K and xi is the mole fraction of species i. Both Equations (6.17) and (6.18) are consistent with the physical argument above. Physical Interpretation of Ki Equation (6.17) indicates that the extensive total solution property K is equal to the sum of the partial molar properties of its constituent species, each adjusted in proportion to the quantity of that species present. Similarly, Equation (6.18) shows that the intensive solution property k is simply the weighted average of the partial molar properties of each of the species present. The partial molar property, Ki, can then be thought of as species i’s contribution to the total solution property, K. We can logically extend this thought to interpret a partial molar property as though it represents the intensive property value of an individual species as it exists in solution. In contrast, the pure species property, ki, indicates how an individual species acts when it is by itself. The difference Ki 2 kl compares how the species behaves in the mixture to how it behaves by itself. If this number is zero, the species behaves identically in the mixture to how it behaves as a pure species. In contrast, if this number is large, the species interactions in the mixture are quite different from when it is by itself.

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6.3 Thermodynamics of Mixtures ◄ 339

T,P

T,P Mixture of water (w ) and ethanol (e)

Pure water (w ) Δnw vw = 18

ml mol

Process

V2 = V1 + ΔV nw, 2 = nw, 1 + Δnw ne, 2 = ne, 1

V1 nw, 1 ne, 1 State 1

ΔV Partial molar volume: ∂V ΔV ≅ Vw = ∂n Δn w w T,P,ne ml = 16.9 mol

State 2

Figure 6.7 Schematic of experimental determination of the partial molar volume of water in a specified mixture of ethanol (e) and water (w).

Figure 6.7 presents a schematic representation of a process to measure the partial molar volume of water (w) for the mixture that is shown on the right side of Figure 6.6. To determine the partial molar volume of water for the well-defined composition of nw (moles of water) and ne (moles of ethanol) at a temperature T and pressure P, we measure the volume change, DV, for an incremental addition of Dnw moles of water while holding the number of moles ethanol, the temperature, and the pressure constant. The partial molar volume is then given by the change in volume divided by the change in number of moles of water. For the composition in Figure 6.6, we get Vw 5 16.9 3 ml/mol 4 , which is less than the pure species molar volume of water, vw 5 18 3 ml/mol 4 . Thus, the contribution of water to the volume of this mixture is less than its contribution to the volume when it exists as a pure species. This result indicates that the inclusion of water–ethanol interactions leads to a closer packing than exclusively water–water interactions. As we change the composition of the mixture, we change the relative amounts of water– ethanol and water–water interactions, so the value of the partial molar volume will change. While Figure 6.7 specifically illustrates the partial molar volume, similar depictions can be drawn for any partial molar property.

EXAMPLE 6.6 Integration of Equation 6.16

Mathematically verify that integration of Equation (6.16) leads to Equations (6.18) or (6.17) (a) starting with the expression for dK; (b) based on applying Euler’s theorem to Equation (6.13). SOLUTION (a) We begin with the expression for dK: dK 5 a

m 'K 'K b dT 1 a b dP 1 a Kidni 'T P,ni 'P T,ni i51

(E6.6A)

Recognizing that K 5 nTk and ni 5 nTxi, we can rewrite Equation (E6.6A) as: nTdk 1 kdnT 5 nT a

'k 'k b dT 1 nT a b dP 'T P,ni 'P T,ni

m

m

i51

i51

1 a KixidnT 1 a KinTdxi

(E6.6B)

(Continued)

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340 ► Chapter 6. Phase Equilibria I: Problem Formulation

Collecting terms in Equation (E6.6B) of nT and dnT, we get: Bdk 2 a

m 'k 'k b dT 2 a b dP 2 a Kidxi RnT 'T P,ni 'P T,ni i51 m

1 Bk 2 a Kixi RdnT 5 0

(E6.6C)

i51

We can now make an argument similar to the physical argument given above. The total size of the system should not affect how the system is affected by changes in composition; thus nT and dnT are independent of each other. For Equation (E6.6C) to hold, in general, each of the terms in the bracket must be zero. Hence, m

k 5 a Kixi

(E6.6D)

i51

and, dk 5 a

m 'k 'k b dT 1 a b dP 1 a Kdxi 'T P,ni 'P T,ni i51

(E6.6E)

Equation (E6.6D) is identical to Equation (6.18). (b) We start by multiplying the number of moles in a system by an arbitrary amount a. At a given T and P, the extensive property K should also be increased by that amount: aK 5 K 1 T, P, an1, an2, c, ani, c, anm 2

(E6.6F)

Euler’s theorem would describe Equation (E6.6F) by saying that the extensive thermodynamic quantity K is a first-order, homogeneous function of ni. Differentiating Equation (E6.6F) by a, we get: B

' 1 aK 2 'K 'K R 5 K 5 n1 B R 1 n2 B R 1c 'a ' 1 an1 2 T,P,ni ' 1 an2 2 T,P,ni T,P 1 ni B

'K 'K R 1 c1 nm B R 2 1 1 ' ani T,P,nj 2 i ' anm 2 T,P,nj 2 i

(E6.6G)

where we have applied the chain rule to get the expression on the right-hand side of Equation (E6.6G). Equation (E6.6G) should be valid for any a, so at a 5 1, we get:

'K K 5 a ni B R 5 a niKi 'n T,P,n i

j2i

which is identical to Equation (6.17).

The Gibbs–Duhem Equation The Gibbs–Duhem equation provides a very useful relationship between the partial molar properties of different species in a mixture. It results from mathematical manipulation of property relations. The approach is similar to that used in Chapter 5 to develop relationships between properties. The reason the Gibbs–Duhem equation is so useful is that it provides constraints between the partial molar properties of different species in a mixture. For example, in a binary mixture, if we know the values for a partial molar

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6.3 Thermodynamics of Mixtures ◄ 341

property of one of the species, we can apply the Gibbs–Duhem equation to simply calculate the partial molar property values for the other species. The formulation of the Gibbs–Duhem equation follows. We begin with Equation (6.17), the definition of a partial molar property: K 5 a niKi

(6.17)

We differentiate Equation (6.17) at constant T and P: dKPT 5 a 3 nidKi 1 Kidni 4 where the subscript “PT” indicates that these properties are held constant. But from Equation (6.16), we know: dKPT 5 a Kidni

(6.16)

For the previous two equations to both be true, in general: 0 5 a nidKi

Const T and P

(6.19)

for example, 0 5 a nidVi ≥0 5

a nidHi (

¥

Equation (6.19) is the Gibbs–Duhem equation. Its straightforward derivation should not overshadow its tremendous utility. To see the usefulness of the Gibbs–Duhem equation, let’s examine the scenario where we wish to find the partial molar volume of species b in a binary solution when we know the partial molar volume of species a, Va, as a function of composition. If we apply Equation (6.19) to the property volume, we get: 0 5 a nidVi

Const T and P

For a binary system, we can differentiate with respect to xa to get: 0 5 na

dVa dVb 1 nb dxa dxa

If we divide by nT, rearrange, and integrate: Vb 5 23

xa dVa ¢ ≤ dxa 1 2 xa dxa

Thus, if we have an expression for (or plot of) the partial molar volume of species a vs. mole fraction, we can apply this equation to get the corresponding expression for species b. The expressions for partial molar properties are not independent but rather constrained by the Gibbs–Duhem equation. Such an interelation makes sense from a molecular perspective. The partial molar properties are governed by how a species behaves in the mixture. We expect the partial molar properties of a and b to be related since it is the same

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342 ► Chapter 6. Phase Equilibria I: Problem Formulation interaction between a and b that determines the difference between how each of these species behaves in the mixture as compared to by themselves, as pure species.

Summary of the Different Types of Thermodynamic Properties There are many different types of properties of which to keep track in mixtures. In this section, we review our nomenclature and see how we keep track of the different types of properties. We consider total solution properties, pure species properties, and partial molar properties. Total Solution Properties The total solution properties are the properties of the entire mixture. They are written as: extensive

K: V, G, U, H, S, c

intensive

k 5 K/ntotal 5 v, g, u, h, s, c

What is V for the experiment in Figure 6.6? Pure Species Properties The pure species properties are the properties of any one of the species in the mixture as it exists as a pure species at the temperature, pressure, and phase of the mixture. We denote a pure species property with a subscript “i”. In general, for species i, we have: extensive:

Ki: Vi, Gi, Ui, Hi, Si, c

intensive:

ki 5

Ki 5 vi, gi, ui, hi, si, c ni

What is Vwater for the experiment in Figure 6.6? Partial Molar Properties Partial molar properties can be viewed as the specific contribution of species i to the total solution property, as discussed in the previous section. They are written: Ki: Vi, Gi, Ui, Hi, Si, c Examination of Equations (6.17) and (6.18) shows that these properties must be intensive and, in general, Ki 2 ki We can envision two limiting cases. First, consider when the mole fraction of species i goes to 1. In this case, any given molecule of i will interact only with other molecules of i. Hence, the solution properties will match those of the pure species, and, Ki 5 ki

lim xi h 1

Second, we can imagine the case where species i becomes more and more dilute. In that case, a molecule of species i will not have any like species with which it interacts; rather, it will interact only with unlike species. We call this case that of infinite dilution and write the partial molar property as: Ki 5 Ki`

c06.indd 342

lim xi h 0

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6.3 Thermodynamics of Mixtures ◄ 343

EXAMPLE 6.7 Types of Thermodynamic Properties in Figure 6.6

Label the volumes depicted in Figure 6.6 according to the different types of thermodynamic properties depicted in this section. SOLUTION The experiment depicted in Figure 6.6 is now labeled in Figure E6.7 according to the different types of properties defined above. Properties identified include the molar volume 1 ve, vw 2 and the extensive volume of each pure species 1 Ve, Vw 2 , as well as the partial molar volumes 1 Ve, Vw 2 and total solution volume of the mixture (V).

67 ml 50 ml

=

+

neve or Ve

20 ml

nwvw or Vw Pure ethanol

neVe

or V

nwVw

Pure water

Mixture of ethanol and water

Figure E6.7 Nomenclature of different types of volume in mixing of ethanol (e) and water (w).

Property Changes of Mixing A property change of mixing, DKmix, describes how much a given property changes as a result of the mixing process. It is defined as the difference between the total solution property in the mixture and the sum of the pure species properties of its constituents, each in proportion to how much is present in the mixture. Mathematically, the property change of mixing is given by: DKmix 5 K 2 a niki

(6.20)

for example, DVmix 5 V 2 a nivi U E DHmix 5 H 2 a nihi (

where the pure species properties, ki, are defined at the temperature and pressure of the mixture. What is DVmix for the experiment in Figure 6.6? Substituting Equation (6.17) into Equation (6.20) gives: DKmix 5 a niKi 2 a niki 5 a ni 1 Ki 2 ki 2

(6.21)

for example, DVmix 5 a ni 1 Vi 2 vi 2 U E DHmix 5 a ni 1 Hi 2 hi 2 (

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344 ► Chapter 6. Phase Equilibria I: Problem Formulation Equation (6.21) shows that the property change of mixing is given by the proportionate sum of the difference between the partial molar property and the pure species property for each of the species in the mixture. This result is not surprising, since we interpret a partial molar property of a given species as the contribution of that species to the mixture, while the pure species property is indicative of how that species behaves by itself. In the special case where all the partial molar properties equal the pure species properties—that is Ki 5 ki —the property change of mixing is zero. Analogously for intensive properties, we get: Dkmix 5 k 2 a xiki

(6.22)

for example, Dvmix 5 v 2 a xivi U EDh 5 h 2 mix a xihi ( and, Dkmix 5 a xi 1 Ki 2 ki 2

(6.23)

for example, Dvmix 5 a xi 1 Vi 2 vi 2 EDh 5 U mix a x i 1 H i 2 hi 2 ( We can interpret the property change of mixing by considering a hypothetical process in which pure species undergo an isothermal, isobaric mixing process. On a molecular level, the property change of mixing reflects how the interactions between unlike species in the mixture compare to the like interactions of the pure species they replaced. For example, a volume change of mixing results from differences in how closely species in a mixture can pack together in comparison to how they pack as pure species. A negative Dvmix will result when the unlike interactions “pull” the species in the mixture closer together, while a positive Dvmix will result when the unlike interactions “push” the species apart when they are mixed. Negative volume changes of mixing tend to imply that the unlike interactions are more attractive than the like interactions they replace, while positive volume changes result when the unlike interactions are not as attractive. If the species behave identically in the mixture as they did as pure species, Dvmix is zero. We see a negative Dvmix in the ethanol–water mixture depicted in Figures 6.7 and E6.7 since the two species can pack together more closely in the mixture as compared to when they are by themselves. Enthalpy of Mixing Similarly, we can understand the enthalpy change of mixing in terms of this hypothetical mixing process at constant T and P. Again, we use the idea that at constant pressure, enthalpy is the appropriate thermodynamic property to characterize energetic

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6.3 Thermodynamics of Mixtures ◄ 345

interactions.7 Thus, Dhmix quantifies the difference between the energetic interactions of the species in the mixture to those of the pure species. In general, the enthalpy of mixing is negative when the species in the mixture are more stable than their pure species counterparts. Conversely, positive enthalpies of mixing result when the species are less stable in the solution than as pure species. When the energetic interactions of the mixture are identical to those of the pure species, Dhmix is zero. For example, consider a liquid mixture of water 1 H2O 2 and sulfuric acid 1 H2SO4 2 . Sulfuric acid will dissociate in water, forming positively charged hydrogen ions and single and doubly negative charged sulfate ions. Figure 6.8 schematically compares the energetic interactions of water as a pure species compared to water in the mixture. The top left of the figure shows pure water hydrogen-bonding to other molecules of water at the mixture T and P. The characteristic energy of this interaction is given by its pure species liquid enthalpy, hH2O, which is depicted on the left of the energy-level diagram at the bottom of the figure. At the top right of the figure, water molecules are shown orienting around a negatively charged bisulfate ion in the mixture. The characteristic enthalpy of water in the mixture is given by its partial molar enthalpy, HH2O. The electrostatic interactions that form are more stable, energetically, than the hydrogen bonds depicted for the pure species at the left of the figure. Thus, the energy-level diagram at the bottom of the figure shows HH2O at a lower (more stable) value than as a pure species. Similar arguments can be made about the decrease in enthalpy of sulfuric acid as it dissociates in the mixture.8 Inspection of Equation (6.23) shows that since HH2O , hH2O and HH2SO4 , hH2SO4, then Dhmix , 0. In fact, the energetics of this binary mixture have been well studied, and the mixing process is highly exothermic. For example, at 21°C, the enthalpy of mixing of sulfuric acid and water can be fit to the following equation:9 Dhmix 5 274.40xH2SO4 xH2O 1 1 2 0.561xH2SO4 2 3 kJ/mol 4

(6.24)

Figure 6.9 shows the enthalpy and volume changes of mixing for water and methanol at 0°, 19.7°, and 42.4°C. The enthalpy of mixing is negative, indicating that the species Pure water

Water mixed with sulfuric acid

δ O

δ2− O H H δ+ δ+ HSO4

H δ+

−

H

H

δ2− O

Energy

H δ+

δ+

H δ+

δ+

H δ+

H δ+

H δ+

δ2− O

δ2− O

δ2− O

H δ+

H δ+

δ2− O

H δ+

H δ+

2−

hH

2O

HH

2O

7

Section 2.6 relates enthalpy to energy changes in a closed system at constant P.

8

The enthalpy of mixing also includes the ionization energy. Equation fit to data reported by W. D. Ross, Chem. Eng. Progr., 43, 314 (1952).

9

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Figure 6.8 Energetic interactions of molecules of water as a pure species and interacting with HSO2 4 in a sulfuric acid–water mixture.

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346 ► Chapter 6. Phase Equilibria I: Problem Formulation 0

0.00

T = 42.4°C

cm3 mol

−400

T = 19.7°C

Δvmix

Δhmix

J mol

−200

−600 −800

−1000 −1200 0

T = 0°C 0.2

0.4

0.6

0.8

1.0

−0.40

−0.80 −1.20 0

0.2

xH

2O

0.4

0.6

0.8

1.0

xH

2O

Figure 6.9 Enthalpy and volume changes of mixing of a binary mixture of water and methanol at 0°, 19.7°, and 42°C.

in the mixture are energetically more stable than they are by themselves, as pure species. Likewise, the volume change of mixing associated with these species is negative, indicating that the species in the mixture pack together more tightly.10 Again, we can understand the mixing behavior in this system in terms of intermolecular interactions. The strongest energetic interaction is hydrogen bonding; however, these species also manifest van der Waals forces. Both water and methanol can form hydrogen bonds as pure species as well as in the mixture. Apparently, this hydrogen-bonded network can pack more efficiently with both species present in a mixture as compared to when they are separate, as pure species. Since the species are closer, the van der Waals interactions are stronger and the energy is lower. The order of magnitude of this effect is 102 3 J/mol 4 , far less than the mixing effects of a sulfuric acid–water system, which arises from the presence of point charges of the ions in solution. The enthalpy of mixing decreases as the temperature increases. However, the functionality of Dhmix with mole fraction is similar at all three temperatures. By contrast, the volume change of mixing is practically identical for all three temperatures. Figure 6.10 shows the enthalpy of mixing for cyclohexane 1 C6H12 2 and toluene at 18°C. In this case, the enthalpy of mixing is positive. Its order of magnitude is 102 3 J/mol 4 , which is comparable to a methanol–water mixture. The dominant energetic interaction in this nonpolar system is dispersion. The magnitude of unlike dispersion interactions is often well approximated as the geometric mean of the corresponding like interactions. With such a relation, the mixture will always be less stable than the weighted average of the pure species,11 thus, nonpolar mixtures usually exhibit positive enthalpies of mixing. Figure 6.11 shows the enthalpy of mixing for chloroform and methanol at 25°C. This figure shows unusual behavior in that the enthalpy of mixing changes sign with changing chloroform mole fraction. At low chloroform mole fractions, the enthalpy of mixing is negative, while at high mole fractions it is positive. This behavior can be understood in terms of the superposition of two competing effects. Over all composition ranges, there is behavior similar to the one described for the system of cyclohexane–toluene, where the average of the like interactions is more favorable than that of the unlike interactions. This effect contributes a positive value to Dhmix. However, the specific and directional nature of the hydrogen bonds involved also play a role in the energetics of this system.

c06.indd 346

10

In the case of volume changes of mixing, all three curves overlap.

11

This point will be discussed more in Chapter 7. See Example 7.9.

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6.3 Thermodynamics of Mixtures ◄ 347 750 625

Δhmix

J mol

500 375 250 125 0 0

0.1

0.2

0.3

0.4

0.5 xC H

0.6

0.7

0.8

0.9

1

6 12

Figure 6.10 Enthalpy of mixing of a binary mixture of cyclohexane 1 C6H12 2 and toluene at 18°C. 500

Δhmix

J mol

250 0 −250 −500 −750

0

0.1

0.2

0.3

0.4

0.5 xCHCl

0.6

0.7

0.8

0.9

1

3

Figure 6.11 Enthalpy of mixing of a binary mixture of chloroform 1 CHCl3 2 and methanol at 25°C.

Methanol has both an electronegative O with two sets of lone pairs of electrons as well as one H atom to contribute to H-bonding, while CCl3H contributes only an H atom. Thus, at a CCl3H mole fraction of roughly one-third, the number of hydrogen and number of oxygen lone pairs match, and the system can form the most possible H-bonds and, therefore, is most stable. This composition represents the minimum in the enthalpy of mixing in Figure 6.11.

EXAMPLE 6.8 Calculation of Dhmix from Experimental Data

An experiment is performed to measure the enthalpy of mixing of chloroform, CHCl3, and acetone, C3H6O. In this experiment, pure species inlet streams of different compositions are mixed together in an insulated mixer at steady-state. This mixing process is exothermic, and the heat that is removed in order to keep the system at a constant temperature of 14°C is measured. The measured data are presented in Table E6.8A. Based on these data, calculate the enthalpy of mixing vs. mole fraction and plot the result. SOLUTION A schematic of the system is shown in Figure E6.8A. Chloroform is labeled species 1 and acetone species 2. The weight fractions, wi, of the inlet streams and enthalpies of all the (Continued)

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348 ► Chapter 6. Phase Equilibria I: Problem Formulation

TABLE E6.8A Chloroform

A Heat Evolved vs. Weight Percentage

Weight % CHCl3

Heat evolved 3 J/g 4 , 1 2q^ 2

10 20 30 40 50 60 70 80 90

4.77 9.83 14.31 19.38 23.27 25.53 25.07 21.55 13.56

Source: E. W. Washburn (ed.), International Critical Tables (Vol. V) (New York: McGraw-Hill, 1929).

streams are labeled. The mole fraction can be calculated from the weight fraction as follows:

x1 5

w1 MW1 w1 w2 1 MW1 MW2

where MWi is the molecular weight of species i. A first-law balance gives: # 0 5 1 n# 1h1 1 n# 2h2 2 2 1 n# 1 1 n# 2 2 h 1 Q

(E6.8A)

(E6.8B)

We need to write Equation (E6.8B) in terms of what we are looking for, Dhmix, and what we have measured, 1 2q^ 2 : Dhmix 5 h 2 1 x1h1 1 x2h2 2 5

# Q

5 q 5 2 1 2q^ 2 MW 1 n# 1 1 n# 2 2

(E6.8C)

where the average molecular weight is given by MW 5 x1MW1 1 x2MW2. Table E6.8B gives enthalpy of mixing vs. mole fraction, as calculated from Equations (E6.8A) and (E6.8C), respectively, using data from Table E6.8A. These data are plotted in Figure E6.8B. A large negative enthalpy of mixing results from mixing chloroform and acetone. This result is consistent with the energetic interactions in the system. The dominant unlike interaction

CHCl3

w1 h1

T = 14°C CHCl3 and C3H6O

C3H6O

w2 h2

h

−q^

Figure E6.8A Schematic representation of experiment to measure enthalpy of mixing.

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6.3 Thermodynamics of Mixtures ◄ 349

TABLE E6.8B Enthalpy of Mixing vs. Mole Fraction Chloroform, Using Data from Table E6.8A x1

Dhmix 3 J/mol 4

0 0.050 0.107 0.170 0.242 0.323 0.418 0.527 0.657 0.811 1.000

0 2291.9 2636.6 2984.1 21,420.7 21,826.4 22,156.1 22,291.4 22,146.2 21,483.5 0

0

J mol

−1000

Δhmix

−500

−1500 −2000 −2500 0

0.1

0.2

0.3

0.4

0.5 xCHCl

0.6

0.7

0.8

0.9

1

3

Figure E6.8B Enthalpy of mixing of a binary mixture of chloroform 1 CHCl3 2 and acetone at 14°C. results from hydrogen bonding between the H of chloroform and O of acetone (see Section 4.2). Neither of the pure species forms hydrogen bonds; thus, the unlike interactions are more stable than the like interactions they replace, and a large negative enthalpy of mixing results.

EXAMPLE 6.9 Calculation of Final Temperature for a Mixing Process

An insulated piston–cylinder assembly contains two compartments divided by a partition. As shown in Figure E6.9A, the top compartment initially contains 2 moles of pure liquid 1, whereas the bottom contains a mixture of 4 moles of liquid 2 and 4 moles of liquid 3. The temperature is initially 4°C, and the pressure is 1 bar. The piston–cylinder assembly is well insulated. If the partition is removed and the species are allowed to mix and come to equilibrium, determine the final temperature. The following data are available. The heat capacity of each of the three species is: cp,1 5 27.5 B

J

mol # K

R, cp,2 5 25 B

J

mol # K

R, cp,3 5 20.0 B

J

mol # K

R

(Continued)

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350 ► Chapter 6. Phase Equilibria I: Problem Formulation

1 bar

4°C n1 = 2 mol Partition

n2 = 4 mol n3 = 4 mol

Initial State

Figure E6.9A Scematic of the mixing process.

The enthalpy of mixing for a ternary mixture of 1, 2, and 3 is given by:

Dhmix 5 400x1x2 1 800x1x3 1 80x2x3 1 37.5x1x2x3 B

J mol

R

SOLUTION Applying the first law to this adiabatic process, we get: DH 5 0 Because enthalpy is a property that depends only on initial and final states, we are free to choose a convenient hypothetical path. One such path is illustrated in Figure E6.9B, where we first isothermally mix pure 1 to the mixture of 2 and 3 and then change the temperature of the mixture. In drawing this path, we consider that for the endothermic enthalpy of mixing relation provided earlier, further mixing will lead to a temperature decrease. Along this path, we can expand the first law as follows: DH 5 0 5 DHmix 1 DHsens

ΔHmix

T1 = 4°C

ΔH = 0

ΔHsens

T2 = ???

Pure 1 Mixed 2, 3

Mixed 1,2,3

Figure E6.9B Two-step thermodynamic path to solve the problem. The enthalpy change for this process is decomposed into the enthalpy of mixing, DHmix, and the enthalpy due to a temperature change within the liquid phase (sensible heat), DHsens.

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6.3 Thermodynamics of Mixtures ◄ 351

where DHmix represents the enthalpy change due to mixing and DHsens represents the enthalpy change due to a decrease in temperature (sensible heat). To calculate the enthalpy of mixing, it is useful to consider another hypothetical path. In this case, we need to recognize the thermochemical data are in the form of the difference between the pure species (unmixed state) and the mixture in the form of the given equation for Dhmix. However, both the initial state and the final state contain mixtures of more than one species. Therefore, we pick the hypothetical path shown in Figure E6.9C where we first “unmix” the initial state into its pure species components and then mix all three components to arrive at the final state. As a procedural note, in this case it is useful to formulate the problem in terms of extensive properties (i.e., DHmix) rather than intensive properties (i.e., Dhmix) because the “unmixing” process and the mixing process contain different numbers of moles (8 vs. 10 mol). We can now calculate:

DHmix 5 DHIunmix 1 DHII mix

(E6.9A)

First, we find DHIunmix, which describes the enthalpy change in unmixing species 2 and 3 and is equal in magnitude but opposite in sign to the enthalpy change in mixing species 2 and 3. DHIunmix 5 2DHImix 5 2 1 n2 1 n3 2 Dhmix We use the given form for Dhmix: Dhmix 5 400x1x2 1 800x1x3 1 80x2x3 1 37.5x1x2x3 B

J mol

R

In this case, only species 2 and 3 are involved in the process so: n1 5 0 3 mol 4 , n2 5 4 3 mol 4 , n3 5 4 3 mol 4 , x1 5 0, x2 5 0.5, x3 5 0.5 and, DHIunmix 5 2160 3 J 4 Next, we determine DHII mix as follows: DHII mix 5 1 n1 1 n2 1 n3 2 Dhmix with: n1 5 2 3 mol 4 , n2 5 4 3 mol 4 , n3 5 4 3 mol 4 , x1 5 0.2, x2 5 0.4, x3 5 0.4 so, DHII mix 5 1100 3 J 4 To find the overall enthalpy of mixing, we add the two contributions: DHmix 5 DHIunmix 1 DHII mix 5 2160 1 1100 5 940 3 J 4

(E6.9B)

To determine the final temperature, we can relate the heat capacity values in the problem to DHsens. T2

DHsens 5 3 1 n1cp,1 1 n2cp,2 1 n3cp,3 2 dT

(E6.9C)

T1

(Continued)

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352 ► Chapter 6. Phase Equilibria I: Problem Formulation

1 bar

4°C

1 bar

4°C

n1 = 2 mol n1 = 2 mol

Partition

n2 = 4 mol

n2 = 4 mol

n3 = 4 mol

ΔH mix

Initial State

ΔH Iunmix = −ΔH Imix

n3 = 4 mol

Final State

4°C

1 bar

(n2 + n3)

II ΔH mix (n1 + n2 + n3)

n1 = 2 mol Partition

n2 = 4 mol

Partition

n3 = 4 mol

Unmixed State

Figure E6.9C Thermodynamic path to determine the enthalpy of mixing. The enthalpy change is first determined by “unmixing” species 2 and 3 and then mixing all the species. Inserting numerical values gives: DHsens 5 235 1 T2 2 277 2 3 J 4 Finally, substituting Equations (E6.9B) and (E6.9C) into Equation (E6.9A) gives: DH 5 0 5 DHmix 1 DHsens 5 940 1 235 1 T2 2 277 2

Solving for temperature, we get, T2 5 273K.

Enthalpy of Solution The energetic interactions characteristic of the mixing process are often reported as the | , instead of the enthalpy of mixing. The enthalpy of solution enthalpy of solution, Dh s corresponds to the enthalpy change when 1 mole of pure solute is mixed in n moles of pure solvent. It is defined per mole of solute as opposed to enthalpy of mixing, which is | can be determined calorimetrically, written per total moles of solution. The value of Dh s | and since Dh s is convenient to measure in the laboratory, reported values are often in this form. | for several solutes in water at 25°C. Example 6.10 illustrates Table 6.1 reports Dh s how to calculate values for the enthalpy of mixing when enthalpy of solution data are available. The enthalpy of solution is also commonly used when the solute exists in the

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6.3 Thermodynamics of Mixtures ◄ 353

solid or gas phase as a pure species. In this case, the enthalpy difference associated to | . For example, solids held together with ionic the phase change is also manifest in Dh s bonds—for example, salts such as NaCl—demonstrate positive enthalpies of solution. In this case, the ions in the pure species solid are close together. When they dissolve into the mixture, the energy increases as the ions get father away from one another. Can you think of a practical application for a system with a large positive enthalpy of solution? On the other hand, gases tend to have negative enthalpies of solution, as the species are closer to their neighbors in the mixture and exhibit stronger attraction. The first six species listed in the Table 6.1 are acids. Their enthalpies of solution differ appreciably, corresponding to different degrees of dissociation. The stronger the acid, the greater the extent to which it disassociates into ionic species and the larger the electrostatic interactions that result. For example, acetic acid, C2H4O2, is a weak acid and has a positive enthalpy of solution at higher solute concentrations. On the other | for sulfuric acid is always large and negative due to the energetic interactions hand, Dh s discussed earlier. The other species reported in Table 6.1 give a flavor of the energetic mixing effects of other types of solutes. Gaseous NH3 and solid NaOH form basic solutions; an alcohol 1 C2H5OH 2 and salts (ZnCl2 and NaCl) are also reported.

EXAMPLE 6.10 Calculation of Dhmix | from Dh s

| , for nitric acid in water at 18°C. Find Table 6.1 presents data for the enthalpy of solution, Dh s the corresponding values for the enthalpy of mixing vs. mole fraction of solute. SOLUTION Let the solute, HNO3, be species 1 and the solvent, H2O, be species 2. The enthalpy of solution can be written by dividing the enthalpy of mixing by the mole fraction solute: | 5 Dhmix Dh s x1

(E6.10A)

The mole fraction of solute is defined as the number of moles of solute, 1, divided by the total number of moles in the system, 1 1 n: x1 5

1 11n

(E6.10B)

Substituting Equation (E6.10B) in (E6.10A) and rearranging gives: Dhmix 5

| Dh s 11 1 n2

(E6.10C)

Table E6.10 presents values for Dhmix as calculated from Equation (E6.10C) vs. mole fraction as calculated from Equation (E6.10B) for the data presented in Table 6.1. The order of table entries has been reversed so that the values for mole fraction appear in ascending order. TABLE E6.10 x1

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Enthalpy of Mixing vs. Mole Fraction HNO3, Using Data from Table 6.1 Dhmix 3 J/mol 4 21,556

0.048 0.091

22,895

0.167

24,788

0.200

25,396

0.250

26,075

0.333

26,694

0.500

26,556

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354

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213,113 220,083 224,301 226,978 228,727 231,840 232,669 232,744 232,748 233,338

231,087 244,936 252,007 257,070 261,045 270,040 274,517 276,358 276,986 299,203

H2SO4 226,225 248,819 256,852 261,204 264,049 269,488 271,777 273,729 273,848 275,189

HCl

H3PO4 C2H4O2 NH3

R NaOH

C2H5OH ZnCl2

2812 2,510 628 229,539 245,886 2837 21,799 669 232,049 246,798 24,184 586 232,761 228,886 22,787 247,187 26,276 233,263 234,430 23,757 228,870 247,384 27,531 377 233,598 237,761 24,694 232,677 247,719 29,874 2209 234,267 242,501 27,364 240,083 247,840 210,962 2628 234,434 242,865 28,866 246,191 247,949 211,966 21,130 234,518 242,526 29,975 255,020 248,057 212,468 21,276 234,560 242,334 210,268 261,086 260,501 21,435 234,644 242,869 271,463

HF

mol solute

J

1,941 2,671 3,732 4,109 3,891

NaCl

Source: F. D. Rossini et al., Selected Values of Physical and Thermodynamic Properties of Hydrocarbons and Related Compounds (Pittsburgh: Carnegu Press, 1953).

1 2 3 4 5 10 20 50 100 `

n 3 mol H2O 4 HNO3

| Dh s B

TABLE 6.1 Enthalpy of Solution of Different Species in Water at 25°C

6.3 Thermodynamics of Mixtures ◄ 355

Entropy of Mixing While the enthalpy of mixing is characteristic of the how the energetic interactions change upon mixing, the entropy of mixing, Dsmix, characterizes the increase in the number of configuration induced by the mixing process. We can interpret Dsmix by again considering a hypothetical process in which pure species undergo an isothermal, isobaric mixing process. Entropy is proportional to the number of possible molecular configurations that a state can exhibit. Since there are always many more ways to configure the species in a mixture as compared to a pure species, Dsmix is always positive. In Example 6.11, we calculate the entropy of mixing for a binary mixture using the ideal gas model. The result from that example can be generalized for an ideal gas mixture of m species to give: m

Dsmix 5 2R a yi, ln yi

(6.25)

i51

In contrast, the enthalpy and volume changes of mixing of an ideal gas are identically zero, since an ideal gas does not exhibit any intermolecular interactions and the molecules size is negligible. Often it is assumed that liquid solutions mix completely randomly, and so the entropy of mixing follows the ideal gas relation given by Equation (6.25). Such a liquid is said to form a regular solution. Species that interact exclusively through van der Waals forces can form regular solutions. However, mixtures whose species significantly differ in size deviate from regular solution behavior. Chemical effects such as association and solvation can also lead to deviation from regular solution behavior. For example, the association reaction between chloroform and acetone leads to mixtures that are structured, and the entropy change of mixing is significantly less than what would be predicted for a regular solution.

EXAMPLE 6.11 Dsmix, for an Ideal Gas Mixture

Develop an expression for the entropy change of mixing for a binary ideal gas mixture. SOLUTION This example is a generalization of Example 3.9. A schematic of the process represented by DSmix is shown in Figure E6.11A. By definition the temperature and pressure of the pure species must be the same as the mixture. Therefore, the (extensive) volume of the mixture will be larger than each of the pure species. We are free to choose any path to calculate the entropies in Equation (E6.11A). One possible solution path is shown in Figure 6.11B. In the first step (Step I), each of the pure species, a and b, are isothermally expanded to the size of the container of the mixture. During this process, the pressure drops to pa and pb, the partial pressure of species a and b in the mixture. The next step (Step II) is superimposing both these expanded systems. For this process, DSmix 5 DSIa 1 DSIb 1 DSII

(E6.11A)

We begin by applying Equation (3.22) at constant temperature to species a: DsIa 5 2R ln

pa P

5 2R ln

yaP P

5 2R ln ya

For the extensive entropy we multiply by the number of moles of a to get: DSIa 5 naDsaI 5 2Rna ln ya

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(E6.11B)

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356 ► Chapter 6. Phase Equilibria I: Problem Formulation

Similarly for b: Dsb 5 2R ln yb and, DSIb 5 nbDsIb 5 2Rnb ln yb

(E6.11C)

For step II, we recognize that ideal gas a does not know b is there and vice versa. Thus, the properties of each individual species do not change and: II DSII 5 naDsII a 1 nbDsb 5 0

(E6.11D)

The path shown in Figure E6.11B has interesting implications in terms of the molecular viewpoint of the entropy increase for species a. Increased entropy is not directly from b mixing in

P,T

P,T

b

a

a a

a

a b

a

b

a b

+

a

a

P,T

b a

S

Sb,pure

Sa,pure

a

a

a

b

Figure E6.11A Process representative of DSmix for a binary system of species a and b.

P,T

a

pa ,T

a a

a

ΔS Ia = na ΔS Ia

a

a a a

a

a

a

a

P,T

a

Sa,pure

a

P,T

ΔS lIb =

pb ,T

a

b a

0 b

a a

a

b a

S b

b

ΔS Ib = nb ΔS Ib

b b

b

b

Sb,pure

Figure E6.11B

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Solution path to calculate DSmix for a binary system of species a and b.

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6.3 Thermodynamics of Mixtures ◄ 357

a, but rather that species a has more room in which to move so that there is more uncertainty where it is; therefore, information is lost, and the entropy is higher. Substituting Equations (E6.11B), (E6.11C), and (E6.11D) in (E6.11A) gives: DSmix 5 2R 1 na ln ya 1 nb ln yb 2 To find the intensive entropy change of mixing, we divide by the total number of moles: Dsmix 5

DSmix 5 2R 1 ya ln ya 1 yb ln yb 2 na 1 nb

In general, when there are m species, we can sum together similar contributions from each species to get: Dsmix 5 2R a yi ln yi Equation (E6.11E) is presented in the text as Equation (6.25).

Determination of Partial Molar Properties We have introduced a new type of property, the partial molar property. This property tells us about the contribution of a given species to the properties of a mixture. Our next question is: How do we obtain values for these partial molar properties? There are several ways in which to accomplish this task. In this section, we consider two examples of how we might calculate a partial molar property: by analytical means when we have an equation that describes the total solution property or by graphical means from plots of total solution data. Energy-related properties such as enthalpy or internal energy must be defined with respect to a reference value. In this case, it is often convenient to consider the partial molar property change of mixing, which can be written as:

DKmix,i 5 ¢

' a 1 niki 2 'DKmix 'K ≤ 5¢ ≤ 2C S 'ni 'ni T,P,nj 2 i T,P,nj 2 i 'ni

T,P,nj 2 i

where Equation (6.20) was used. Since the pure species properties are constant at a given temperature and pressure, every term for the derivative in the sum becomes zero except the one associated with ni. Therefore, the partial molar property change of mixing becomes: DKmix,i 5 Ki 2 ki

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(6.26)

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358 ► Chapter 6. Phase Equilibria I: Problem Formulation for example, DHmix,i 5 Hi 2 hi ≥ DGmix,i 5 Gi 2 gi ¥ ( Inspection of Equation (6.26) shows that DKmix,i gives the relative value of how species i behaves in a mixture to how it behaves by itself as a pure species. In this form, the pure species property forms a reference state to which the particular partial molar property is referred. Analytical Determination of Partial Molar Properties Often an analytical expression for the total solution property, k, is known as a function of composition. In that case, the partial molar property, Ki, can be found by differentiation of the extensive expression for K with respect to ni, holding T, P, and the number of moles of the other j species constant, as prescribed by Equation (6.15). We illustrate this method by showing how to calculate partial molar volumes for a binary mixture of species 1 and 2 with the virial equation of state. The virial equation can be written in terms of the total solution molar volume using Equations (4.27) and (4.43): v5

BmixP RT RT B1 1 R5 1 y21B11 1 2y1y2B12 1 y22B22 P RT P

If we know the virial coefficients, B11 and B22, and the cross-virial coefficient, B12, we can solve for the partial molar volumes of each species in the mixture. We first write the virial equation in terms of extensive volume and number of moles: V 5 1 n1 1 n2 2 v 5 1 n1 1 n2 2

n2B11 1 2n1n2B12 1 n22B22 RT 1 1 1 n1 1 n2 2 P

where we have used y1 5 n1 / 1 n1 1 n2 2 and y2 5 n2 / 1 n1 1 n2 2 . Differentiation gives: V1 5 ¢

2n1B11 1 2n2B12 n2B11 1 2n1n2B12 1 n22B22 'V RT ≤ 5 1 2 1 1 n1 1 n2 2 1 n1 1 n2 2 2 'n1 T,P,n2 P

We can simplify this expression to get: V1 5

RT 1 1 y21 1 2y1y2 2 B11 1 2y22B12 2 y22B22 P

Similarly, the partial molar volume of species 2 can be written: V2 5

RT 2 y21B11 1 2y21B12 1 1 y22 1 2y1y2 2 B22 P

To obtain a value for the pure species molar volume, we set y2 5 0 in the virial equation to give: RT v1 5 1 B11 P This expression can also be obtained through the expression for V1 above, since V1 5 v1 in lim x1 h 1.

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6.3 Thermodynamics of Mixtures ◄ 359

The volume change of mixing is given by: Dvmix 5 v 2 1 y1v1 1 y2v2 2 Substituting in for v, v1, and v2: Dvmix 5 B

RT RT RT 1 y21B11 1 2y1y2B12 1 y22B22 R 2 By1 ¢ 1 B11 ≤ 1 y2 ¢ 1 B22 ≤ R P P P

or simplifying: Dvmix 5 2y1y2 BB12 2

B11 1 B22 R 2

(6.27)

Inspection of the term in brackets on the right-hand side of Equation (6.27) shows that the magnitude and sign of Dvmix are determined by comparing the strength of the unlike interactions, given by B12, to that of the like interactions, given by the average of B11 and B22. If the unlike term is stronger (i.e., has a larger-magnitude negative number), the volume change of mixing is negative, whereas a positive change results when the like interactions are stronger.

EXAMPLE 6.12 Calculation of Mixture Properties Using the Virial Equation

Consider a binary mixture of 10 mole% chloroform (1) in acetone (2) at 333 K and 10 bar. The second virial coefficients for this system are reported to be B11 5 2910, B22 5 21330, and B12 5 22005 cm3 /mol. Determine v1, V1, and Dvmix. SOLUTION Applying the equations for the pure species and partial molar volumes above, we get: RT 1 B11 5 1860 3 cm3 /mol 4 P

(E6.12A)

RT 1 1 x21 1 2x1x2 2 B11 1 x22B12 2 x22B22 5 991 3 cm3 /mol 4 P

(E6.12B)

v1 5 and, V1 5

where we represent the liquid mole fractions by xi instead of yi. Using Equation (6.27) gives Dvmix 5 x1x2 1 2B12 2 B11 2 B22 2 5 2372 3 cm3 /mol 4

(E6.12C)

Comparing Equations (E6.12A) and (E6.12B), we see that the molar volume of pure chloroform, vi, is about double its contribution to the molar volume of the mixture with 10% chloroform, as indicated by V1. This result can be explained in terms of hydrogen bonding between H in chloroform and carbonyl in acetone. This association reaction “pulls” the species in the mixture closer to each other (see Section 4.2 and Example 6.8). The result of this interaction between chloroform and acetone is a reduced solution volume; therefore, Dvmix is negative. It should be noted that the interaction between chloroform and acetone is unusually strong in comparison to the pure species interactions and is not typical of most mixtures.

12 In fact, this system is chosen for illustration because its pronounced interactions depict the graphical approach well.

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360 ► Chapter 6. Phase Equilibria I: Problem Formulation

EXAMPLE 6.13 Calculation of HH2SO4 and HH2O

Develop expressions for the partial molar enthalpies of sulfuric acid and water in a binary mixture at 21°C. The pure species enthalpies are 1.596 [kJ/mol] and 1.591 [kJ/mol], respectively, and the enthalpy of mixing is given by Equation (6.24). Calculate their values for an equimolar mixture of sulfuric acid and water. Plot HH2SO4 and HH2O vs. xH2SO4. SOLUTION We can write the total solution enthalpy as: h 5 hH2SO4xH2SO4 1 hH2OxH2O 1 Dhmix or, numerically, h 5 1.596xH2SO4 1 1.591xH2O 2 74.40xH2SO4xH2O 1 1 2 0.561xH2SO4 2 3 kJ/mol 4

(E6.13A)

To apply the definition of a partial molar property, Equation (6.15), we need to write Equation (E6.13A) in terms of the extensive enthalpy and the number of moles of sulfuric acid and water: H 5 nTh 5 1.596nH2SO4 1 1.591nH2O 2 74.40

1 41.74

nH2SO4nH2O nH2SO4 1 nH2O

n2H2SO4nH2O

(E6.13B)

1 nH2SO4 1 nH2O 2 2

Differentiating Equation (E6.13B) with respect to nH2SO4 gives the partial molar enthalpy of sulfuric acid: HH2SO4 5 ¢

2 2 nH2SO4nH nH 'H 2O 2O ≤ 5 1.596 2 74.40 1 83.48 1 nH2SO4 1 nH2O 2 2 'nH2SO4 T,P,nH O 1 1 nH2SO4 1 nH2O 2 3 2

10,000

Partial molar enthalpy [J/mol]

0

HH2O

−10,000 −20,000 −30,000 −40,000 −50,000 −60,000

HH2SO4

−70,000 −80,000 0

0.1

0.2

0.3

0.4

0.5 xH SO 2

0.6

0.7

0.8

0.9

1

4

Figure E6.13 Partial molar enthalpies of water and sulfuric acid at 21°C.

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6.3 Thermodynamics of Mixtures ◄ 361

and, simplifying, 2 2 3 kJ/mol 4 HH2SO4 5 1.596 2 74.40xH 1 83.48xH2SO4xH 2O 2O

(E6.13C)

Similarly, for water we get: HH2O 5 1.591 2 74.40x2H2SO4 1 41.74x2H2SO4 1 1 2 2xH2O 2 3 kJ/mol 4

(E6.13D)

A plot of the partial molar enthalpies calculated from these expressions is presented in Figure E6.13. For an equimolar mixture 1 xH2SO4 5 xH2O 5 0.5 2 , Equations (E6.13C) and (E6.13D) give: HH2SO4 5 26.6 3 kJ/mol 4

EXAMPLE 6.14 Use of the Gibbs– Duhem Equation to Relate Partial Molar Properties

and HH2O 5 217.0 3 kJ/mol 4

Verify that the expressions developed in Example 6.12 for the partial molar enthalpies of sulfuric acid and water in a binary mixture at 21°C satisfy the Gibbs–Duhem equation. SOLUTION The Gibbs–Duhem equation can be written for the partial molar enthalpies in this system as: 0 5 nH2SO4dHH2SO4 1 nH2OdHH2O

(E6.14A)

Differentiating Equation (E6.14A) with respect to mole fraction of sulfuric acid and dividing by the total number of moles gives: 0 5 xH2SO4

dHH2SO4 dxH2SO4

1 xH 2 O

dHH2O

(E6.14B)

dxH2SO4

We now take the derivatives of the two expressions in Equation (E6.14B) by using Equations (E6.13C) and (E6.13D). First, the derivative of the partial molar enthalpy of sulfuric acid gives: dHH2SO4 dxH2SO4

5 2148.80xH2O

dxH2O dxH2SO4

1 166.96xH2OxH2SO4

2 1 83.48xH 2O

dxH2O dxH2SO4 (E6.14C)

However, the change in number of moles of water with respect to the number of moles of sulfuric acid is given by: dxH2O dxH2SO4

5 21

(Continued)

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362 ► Chapter 6. Phase Equilibria I: Problem Formulation

Thus, Equation (E6.14C) becomes: dHH2SO4 dxH2SO4

2 5 148.80xH2O 2 166.96xH2OxH2SO4 1 83.48xH 2O

2 5 218.16xH2O 1 250.44xH 2O

(E6.14D)

where we have used xH2SO4 5 1 1 2 xH2O 2 . Finally, multiplying Equation (E6.14D) by mole fraction of sulfuric acid gives: xH2SO4

dHH2SO4 dxH2SO4

2 5 218.16xH2OxH2SO4 1 250.44xH x 2O H2SO4

(E6.14E)

Likewise, for the derivative of the partial molar enthalpy of water, we get: dHH2O dxH2SO4 and,

5 2148.80xH2SO4 1 83.48 3 xH2SO4 1 1 2 2xH2O 2 1 x2H2SO4 4 5 xH2SO4 1 18.16 2 250.44xH2O 2 xH 2 O

dHH2O dxH2SO4

2 5 18.16xH2OxH2SO4 2 250.44xH x 2O H2SO4

(E6.14F)

Inspection of Equations (E6.14E) and (E6.14F) shows that Equation (E6.14B) is satisfied.

Graphical Determination of Partial Molar Properties Say we want to calculate the partial molar volume (or any other partial molar property) for a binary mixture when we have a graph of the molar volume (or whatever molar property) vs. mole fraction of one component, as shown in Figure 6.12. The values in this figure are taken from the chloroform (1)–acetone (2) binary mixture discussed in Example 6.12. Recall that the unlike interaction, in this case, is unusually large.12 Applying Equation (6.18) to this case: v 5 x1V1 1 x2V2

(6.28)

Substituting x1 5 1 2 x2 yields: v 5 1 1 2 x2 2 V1 1 x2V2 Differentiating with respect to x2, multiplying by x2, and applying the Gibbs–Duhem equation gives: dv x2 5 2x2V1 1 x2V2 dx2 or,

x2

dv 5 2V1 1 1 x1V1 1 x2V2 2 dx2

Using the definition for v in Equation (6.28) and rearranging, we get: v 5 V1 1 x2 intercept

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dv dx2

(6.29)

slope

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6.3 Thermodynamics of Mixtures ◄ 363 2000 1750

v

cm3 mol

1500 1250 1000 750 0

0.1

0.2

0.3

0.4

0.5 x2

0.6

0.7

0.8

0.9

1

Figure 6.12 Experimental data of the molar volume for a binary system of components 1 and 2

If v is plotted vs. x2, Equation (6.29) represents a straight line with slope dv/dx2 and intercept V1. Therefore, the partial molar volume for any composition, x2, can be found by drawing a line tangent to the curve and taking the value of the intercept, as shown for x2 5 0.7 in Figure 6.13. Analogously, V2 can be found by taking the value of the line at x2 5 1. This method is descriptively referred to as the tangent-intercept method.13 We can generalize Equation (6.29) to get: K1 5 k 2 x2

dk dx2

(6.30)

for example, dv dx2 FH1 5 h 2 x2 dh V dx2 ( V1 5 v 2 x2

Equation (6.30) holds only for binary mixtures. For the generalization to mixtures with more than two components, see Problem 6.71. It is interesting to consider the limiting cases of the partial molar volume of species 1 in terms of composition of 1. In the limit as x1 goes to 1, we have all 1 and no 2 in the mixture. In this case, the partial molar volume just equals the pure species molar volume: V1 5 v1 lim x1 h 1 This result is the logical consequence of our interpretation of a partial molar property. If only species 1 is present, it must contribute entirely to the solution properties. Therefore, it must be equal to the pure solution property. The tangent-intercept method can also be applied to the partial molar property change of mixing; if Dvmix is plotted vs. x2, the intercept gives 1 DVmix 2 1 5 V1 2 v1.

13

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364 ► Chapter 6. Phase Equilibria I: Problem Formulation 2000 v1 1750 1500

v

cm3 mol

Δv mix

v2 V2

1250 1000 V1 750 0

(x2) mix = 0.7 0.1

0.2

0.3

0.4

0.5 x2

0.6

0.7

0.8

0.9

1

Figure 6.13 Determination of partial molar volumes from graphical data for a binary mixture.

In the other extreme, consider one molecule of species 1 placed in a solution of 2. In this case, 1 interacts only with molecules of 2 and may have a partial molar volume very different from the first case if the nature of the 1-2 interaction differs from the 1-1 ` interaction. We call this the limit of infinite dilution, V1 . `

V1 5 V1 lim x1

h

0

`

Both limits are shown in Figure 6.14. Note: V1 2 v1. 2000 v1

lim x1→ 1

v [cm3 mol]

lim x1→ 0

v2

1000

V∞ 1 0 0

0.1

0.2

0.3

0.4

0.5 x2

0.6

0.7

0.8

0.9

1

Figure 6.14 Limiting cases of partial molar volumes. (Note that the y-axis scale is different from Figures 6.12 and 6.13.)

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6.3 Thermodynamics of Mixtures ◄ 365

EXAMPLE 6.15 Alternative Calculation of HH2SO4 and HH2O

Develop an expression for the partial molar enthalpy of sulfuric acid in water at 21°C using Equation (6.30). SOLUTION If we apply Equation (6.30) to this system, we get: HH2SO4 5 h 2 xH2O

dh dxH2O

(E6.15A)

Thus, we begin with the expression for the molar enthalpy given by Equation (E6.13A). h 5 1.596xH2SO4 1 1.591xH2O 2 74.40xH2SO4xH2O 1 1 2 0.561xH2SO4 2 3 kJ/mol 4

(E6.15B)

Differentiating Equation (E6.15B) gives: dh 5 21.596 1 1.591 2 74.40 3 1 xH2SO4 2 xH2O 2 1 1 2 0.561xH2SO4 2 1 0.561xH2SO4xH2O 4 dxH2O (E6.15C) Substituting Equations (E6.15B) and (E6.15C) into (E6.15A) gives: 2 2 1 1 2 0.561xH2SO4 2 1 41.74xH2SO4xH HH2SO4 5 1.596 1 xH2SO4 1 xH2O 2 2 74.40xH 2O 2O

2 2 3 kJ/mol 4 HH2SO4 5 1.596 2 74.40xH 1 83.48xH2SO4xH 2O 2O

(E6.15D)

Expressions (E6.15D) and (E6.13C) are identical.

EXAMPLE 6.16 Graphical Determination of HH2SO4 and HH2O

Graphically determine values for the partial molar enthalpies of sulfuric acid and water in an equimolar mixture at 21°C by plotting Equation (E6.13A). SOLUTION A plot of Equation (E6.13A) is shown in Figure E6.16. Also illustrated in the figure is the tangent line to the plot at a mole fraction of 0.5. The values of the intercepts of the tangent lines give: HH2SO4 5 26700 3 J/mol 4 and HH2O 5 217,100 3 J/mol 4 These values agree with the numbers obtained analytically in Example 6.13. (Continued)

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366 ► Chapter 6. Phase Equilibria I: Problem Formulation

0 Enthalpy of solution −5,000 HH SO 4

−10,000

h

J mol

2

Tangent at x = 0.5

−15,000

HH

2O

−20,000 0

0.2

0.4

0.6

xH

0.8

1

2O

Figure E6.16 Graphical determination of values for the partial molar enthalpies of sulfuric acid and water.

Relations Among Partial Molar Quantities Partial molar properties can be related to each other, further extending the thermodynamic web. For example, consider the total solution enthalpy: H 5 U 1 PV Differentiating with respect to ni at constant T, P, and nj, we get: ¢

'U 'PV 'H ≤ 5¢ ≤ 1¢ ≤ 'ni T,P,nj 2 i 'ni T,P,nj 2 i 'ni T,P,nj 2 i

Since the pressure is constant, ¢

'H 'U 'V ≤ 5¢ ≤ 1 P¢ ≤ 'ni T,P,nj 2 i 'ni T,P,nj 2 i 'ni T,P,nj 2 i

Applying the definition of a partial molar quantity, we get: Hi 5 Ui 2 PVi

(6.31)

Note the similarity between Equation (6.31), which applies to the partial molar enthalpy, and the total solution enthalpy. It is straightforward to show that the similar derived property relations described in Chapter 5 likewise hold: Gi 5 Hi 2 TSi Ai 5 Ui 2 TSi We can also find relations for the partial molar Gibbs energy analogous to the Maxwell relations discussed in Chapter 5. We begin by applying the expression for the differential change of Gibbs energy in a mixture: dG 5 a

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m 'G 'G b dT 1 a b dP 1 a Gidni 'T P,ni 'P T,ni i51

(6.32)

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6.4 Multicomponent Phase Equilibria ◄ 367

When there is no change in composition, dni 5 0, the above relationship must reduce to Equation (5.9). Therefore, we can write Equation (6.32) as: m

dG 5 2SdT 1 VdP 1 a Gidni

(6.33)

i51

By equating the second derivatives of the first and third terms (as we did for the Maxwell relations), we get: ¢

'Gi ≤ 5 2Si 'T P,ni

(6.34)

A convenient form of the temperature dependence of the partial molar Gibbs energy is given by taking the partial derivative of 1 Gi/T 2 with respect to T at constant P: '¢ §

Gi ≤ T

'T

¥

5

Gi 2TSi 2 Gi Hi 1 'Gi ¢ ≤ 2 25 52 2 2 T 'T P,ni T T T

(6.35)

P,ni

where first the chain rule was applied, then Equation (6.34) was used. Cross-differentiation of the second and third terms of Equation (6.33) yields: ¢

'Gi ≤ 5 Vi 'P T,ni

(6.36)

The role of the partial molar Gibbs energy in phase equilibria will be discussed in the next section. For now, we can see that the two equations above are useful in determining the pressure and temperature dependence of this quantity.

►6.4 MULTICOMPONENT PHASE EQUILIBRIA The Chemical Potential—The Criteria for Chemical Equilibrium We are now ready to combine the phase equilibria criteria developed in Section 6.2 with our description for mixtures in Section 6.3 to synthesize the complete phase equilibria problem (see Figure 6.2). We begin with our criterion for chemical equilibria, Equation (6.3). Writing the differential change in the total Gibbs energy as the sum of the differential change in each phase gives: dG 5 0 5 dGa 1 dGb Substituting the fundamental property relation given by Equation (6.33) to each phase, we get: m

a

m

0 5 B2SdT 1 VdP 1 a Gidni R 1 B2SdT 1 VdP 1 a Gidni R i51

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b

(6.37)

i51

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368 ► Chapter 6. Phase Equilibria I: Problem Formulation We can now apply our criteria for thermal equilibrium and mechanical equilibrium. For thermal equilibrium, we have: Ta 5 Tb Thus, if we have a differential change in the temperature of phase a, it must be matched by a corresponding differential change in the temperature of phase b: dTa 5 dTb Similarly, the criteria for mechanical equilibrium can be applied to differential changes in pressure to get: dPa 5 dPb Applying the criteria for thermal and mechanical equilibrium to Equation (6.32) gives: a

m

m

b

0 5 B a Gidni R 1 B a Gidni R i51

i51

The partial molar Gibbs energy is such an important quantity in chemical equilibria that it is given a special name: the chemical potential, mi: mi ; ¢

'G ≤ 'ni T,P,nj 2 i

(6.38)

We will see shortly why. It is important to remember that chemical potential and partial molar Gibbs energy are synonymous. In terms of chemical potential, we have: 0 5 a mai dnai 1 a mbi dnbi m

m

However, since we have a closed system, any species leaving phase a will enter phase b. Hence, dnai 5 2dnbi Thus, we get: 0 5 a 1 mai 2 mbi 2 dnai m

Now, for this equation to be true in general, mai 5 mbi

(6.39)

Equation (6.39) applies to all m species in the system; that is, there are m different equations here. By comparing Equation (6.39) to Equation (6.5), it is evident that the criterion for chemical equilibrium in mixtures is obtained by replacing the pure species quantity, the molar Gibbs energy, gi, with the contribution of species i to the Gibbs energy of the mixture, that is, its partial molar Gibbs energy, Gi 5 mi. This realization could be intuited by our interpretation of a partial molar quantity in Section 6.3.

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6.4 Multicomponent Phase Equilibria ◄ 369

Thigh

Energy transfer Tlow (a)

μi,high

Mass transfer of species i

Figure 6.15 Conceptual illustration of the analogy between (a) temperature as the driving force for energy transfer and (b) chemical potential as the driving force for mass transfer.

μi,low (b)

The chemical potential is an abstract concept; it cannot directly be measured. However, the relation between chemical potential and mass transport is identical to the relation between temperature and energy transport or pressure and momentum transport. This concept is illustrated in Figure 6.15. In Figure 6.15a, we see two systems of different temperature. If they are placed in contact, there will be energy transfer via heat from high to low temperature until the temperatures become equal and we have reached thermal equilibrium. We could very well call temperature “thermal potential,” since it provides the driving force toward thermal equilibrium. However, we already know this property from physical experience as temperature, so we stick with that name. Figure 6.15b shows the analogous relation between chemical potential and diffusion. Here we see two systems of different chemical potential for species i. In this case, species i will transport from high to low chemical potential until the chemical potentials become equal and we have reached chemical equilibrium. If we know mi for each phase, we know which way species i will tend to transfer. The biggest difficulty in understanding chemical potential is that it is an abstract concept, whereas T and P are measured properties with which we have direct experience. However, we can apply the concept of chemical potential to learn about driving forces for species transfer in much the same manner as we apply temperature to energy transport.

EXAMPLE 6.17 Determination of the Gibbs Phase Rule for Nonreacting Systems

Consider a system at temperature T and pressure P with m species present in p phases. How many measurable properties need to be determined (e.g., T, P, and xi) to constrain the state of the entire system? SOLUTION From Equations (6.1), (6.2), and (6.39), we can construct the following set of equations for m species and p phases: Ta 5 Tb 5 c 5 Tp Pa 5 Pb 5 c 5 Pp ma1 5 mb1 5 c 5 mp1 ma2 5 mb2 5 c 5 mp2

(E6.17A)

( mam 5 mbm 5 c 5 mpm (Continued)

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370 ► Chapter 6. Phase Equilibria I: Problem Formulation Each row in the set of equations above has 1 p 2 1 2 equal signs. Thus, there are a total of 1 p 2 1 2 1 m 1 2 2 equalities in the set of equations above. The chemical potential in a given phase depends on the temperature, pressure, and mole fraction of each of the species present. Since the sum of mole fractions equals 1, we have to know 1 m 2 1 2 mole fractions for each phase along with the temperature and pressure to constrain the state of that phase. Thus we must specify 1 m 1 1 2 p variables to determine the state of the system. The number of variables we can independently pick (the so-called degrees of freedom, I) is obtained by subtracting the total 1 m 1 1 2 p variables we need to specify by the 1 p 2 1 2 1 m 1 2 2 equalities in Equations (E6.17A). Thus, we can independently specify: I 5 1m 1 12p 2 1p 2 12 1m 1 22 5 m 2 p 1 2

(E6.17B)

quantities. Equation (E6.17B) is identical to Equation (1.12). This example does not consider chemical reaction. If we have reactions, we place additional constraints due to the reaction stoichiometry. This case is addressed in Chapter 9.

Temperature and Pressure Dependence of mi In analogy to the case for pure species we saw in Section 6.2, we can determine how a mixture in equilibrium responds to changes in measured variables. We can use the thermodynamic web to relate the change in chemical potential and the criteria for chemical equilibrium between phases with changes in pressure, temperature, and mole fraction. It is convenient to begin by dividing Equation (6.39) by T: mai mbi 5 T T

(6.40)

If we choose temperature, pressure, and mole fraction of species i as independent variables, the change in chemical potential divided by temperature can be related by the following partial differentials: B

' 1 mai /T 2 ' 1 mai /T 2 ' 1 mai /T 2 R dT 1 B R dP 1 B R dxai a 'T 'P 'xi P,xma T, xm T,P

5B

' 1 mbi /T 2 ' 1 mbi /T 2 ' 1 mbi /T 2 R dT 1 B R dP 1 B R dxbi b b 'T 'P 'xi P,xm T, xm T,P

Factoring out (l/T) in the second and third terms on each side gives: B

' 1 mai /T 2 1 'mai 1 'mai R dT 1 B R dP 1 B R dxai a 'T T 'P T,xma T 'xi T,P P,xm

5B

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b b ' 1 mbi /T 2 1 'mi 1 'mi R dxbi R b dT 1 B R b dP 1 B 'T T 'P T,xm T 'xi T,P P,xm

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6.4 Multicomponent Phase Equilibria ◄ 371

Applying Equations (6.35) and (6.36) gives:

2

a a a b Vi Hbi Vbi Hi 1 'mi 1 'mi a dT 2 dP 1 B R dx 5 2 dT 1 dP 1 B R dxbi (6.41) i b a T2 T T 'xi T,P T2 T T 'xi T,P

Equation (6.41) is valid for any two phases a and b, in general. We now consider the specific case of vapor–liquid equilibrium. Denoting the vapor-phase mole fraction, yi, and the liquid-phase mole fraction, xi, Equation (6.41) becomes: 2

Hvi Viv Hli Vli 1 'mvi 1 'mli dT 1 dP 1 B R dy 5 2 dT 1 dP 1 B R dxi (6.42) i T2 T T 'yi T,P T2 T T 'xi T,P

For the case of an ideal gas, we can simplify further. The partial molar enthalpy and the partial molar volume equal the pure species molar enthalpy and volume, respectively, since the interactions of species i in the mixture are the same as its interactions as a pure species: Hvi 5 hvi Vvi 5 vvi 5

(6.43) RT P

(6.44)

Similarly, the change in chemical potential with respect to mole fraction is given by the change in pure species Gibbs energy with respect to mole fraction: ¢

'gi 'yi

≤

T,P

5¢

'hi 'si ≤ 2 T¢ ≤ 'yi T,P 'yi T,P

For an ideal gas, the enthalpy is independent of mole fraction. To find the dependence of entropy, we apply the analogy of Equation (E6.11B). So at constant pressure: R dsi 0 T,P 5 2Rd ln 1 yi 2 5 2 dyi yi Thus, B

'gvi 'mvi RT R 5B R 5 yi 'yi T,P 'yi T,P

(6.45)

Substituting Equations (6.43), (6.44), and (6.45) into Equation (6.42) gives: 2

dyi Hli Vli hvi dP 1 'mli dT 1 R 5 2 2 dT 1 dP 1 B R dxi 1R 2 yi T P T T T 'xi T,P

(6.46)

Equation (6.46) shows how the temperature or pressure of a phase transition for a liquid–ideal gas mixture is related to changes in composition. We will discuss phenomena of boiling point elevation in this context in Chapter 8.

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372 ► Chapter 6. Phase Equilibria I: Problem Formulation

►6.5 SUMMARY

In this chapter, we formulated the criteria for equilibrium between two phases. We labeled the phases generally as a and b, which can represent the vapor, liquid, or solid phases. We reduced this problem into two parts. First, we addressed pure species phase equilibrium. We determined that the derived property Gibbs energy is a minimum at equilibrium, so only when two phases have equal values of Gibbs energy can they coexist. Second, we addressed the thermodynamics of mixtures. We discovered that the partial molar property, Ki, is representative of the contribution of species i to the mixture. In analogy to the case for pure species, the criterion for chemical equilibrium between two phases for species i in a mixture is that the partial molar Gibbs energy, Gi, is equal in the two phases. Since it sets the criteria for chemical equilibrium, we often call the partial molar Gibbs energy the chemical potential, mi. In summary, the criteria for equilibrium between phases a and b can be written:

and,

Ta 5 Tb

Thermal equilibrium

(6.1)

Pa 5 Pb

Mechanical equilibrium

(6.2)

mai 5 mbi

Chemical equilibrium

(6.39)

Gibbs energy establishes the criteria for pure species phase equilibria by accounting for the balance between the system’s tendency to minimize energy and its tendency to maximize entropy. At low temperature, energetic effects become more important, while at higher temperature entropic effects dominate. We can apply the thermodynamic web to relate how the pressure of a system in phase equilibrium varies with temperature. This analysis leads to the Clapeyron equation. Application of the Clapeyron equation to vapor–liquid equilibrium, together with the assumptions of a negligibly small liquid volume and an ideal gas, leads to the Clausius–Clapeyron equation. The integrated form of the Clausius–Clapeyron equation is functionally similar to the Antoine equation, a common empirical equation used to correlate pure species saturation pressures with temperature. Thermodynamic properties of a mixture are affected both by the like (i-i) interactions and by the unlike (i-j) interactions, that is, how each of the species in the mixture interacts with all of the other species it encounters. The total solution property K, 1 K 5 V, H, U, S, G, c 2 , represents a given property of the entire mixture. It can be written as the sum of the partial molar properties of its constituent species, each adjusted in proportion to how much is present: K 5 a niKi

(6.17)

ni

where the partial molar property is defined by: Ki 5 ¢

'K ≤ 'ni T,P,nj 2 i

(6.15)

Additionally, the pure species property, ki, is defined as the value of that property of species i as it exists as a pure species at the same T and P of the mixture. Values of a partial molar property for a species in a mixture can be calculated from an analytical expression by applying Equation (6.15) and by graphical methods, as illustrated in Figure 6.13. In the case of infinite dilution, species i becomes so dilute that a molecule of species i will not have any like species with which it interacts; rather, it will interact only with unlike species. Additionally, partial molar properties of different species in a mixture can be related to one another by the Gibbs–Duhem equation: 0 5 a nidKi Const T and P

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(6.19)

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6.6 Problems ◄ 373 A property change of mixing, DKmix, describes how much a given property changes as a result of a process in which pure species are mixed together. It is written as: DKmix 5 K 2 a niki 5 a n i 1 Ki 2 k i 2

(6.21)

Volume and enthalpy changes of mixing have values of zero when the like and unlike interactions are identical. When the unlike interactions are more attractive, these quantities are negative, while they are positive when unlike interactions are less favorable. Conversely, the entropy of mixing is positive in all cases, since there are many more ways to configure a mixture as compared to the pure species. The energetic interactions characteristic of the mixing process are often reported as | . The enthalpy of solution corresponds to the enthalpy change when the enthalpy of solution Dh s 1 mole of solute is mixed in n moles of solvent.

►6.6 PROBLEMS Conceptual Problems 6.1 A pot of boiling water is open to the atmosphere when, at time 0, the pot is sealed by a tight lid so no gas can escape. With continued heating, the water continues to boil. As best you can, estimate how the temperature changes with time and plot it on the following graph. You may assume that the amount of water that evaporates (the evaporation rate) stays constant with time. Explain your answer. 250

T [°C]

200

150

100

50

0 (lid on)

1

2

3 Time

6.2 The boiling point for species A at 1 bar is reported to be 250 K, and you desire to know the boiling point at 10 bar. Knowing the enthalpy of vaporization, you apply the Clausius-Clapeyron equation and calculate the temperature to be 300 K. However, at that pressure, you also know that species A is not an ideal gas, but rather attractive intermolecular interactions are significant. If you accounted for the attractive interactions, you would find that Species A boils (at less than 300 K, at 300 K, at greater than 300 K, or there is no way to determine). Explain your answer. 6.3 The boiling point for species A at 1 bar is reported to be 250 K, and you desire to know the saturation pressure at 300 K. Knowing the enthalpy of vaporization, you apply the Clausius-Clapeyron equation and calculate the pressure to be 10 bar. However, at that pressure, you also know that species A is not an ideal and that attractive intermolecular interactions are significant. If you accounted for the attractive interactions, you would find that the saturation pressure for Species A is (less than 10 bar, equal to 10 bar, greater than 10 bar, or there is no way to determine) Explain your answer.

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374 ► Chapter 6. Phase Equilibria I: Problem Formulation 6.4 A well-insulated tank with a valve at the top contains saturated water at 5 MPa. The quality of the water is 0.1. (a) What is the ratio of the liquid volume to the vapor volume? (b) The valve is opened, and the vapor is allowed to escape to atmosphere while the water continues to boil. Plot how you think the temperature will change with pressure. Explain your reasoning. 6.5 Your coworker has scribbled down the saturation pressures for a pure species from the solid (sublimation) and liquid (evaporation) as follows: ln Psat 5 2

15800 2 0.76 ln T 1 19.25 T

ln Psat 5 2

15300 2 1.26 ln T 1 21.79 T

and,

However, in his haste, he forgot to note which equation was for sublimation and which was for evaporation. Please help your coworker by determining the correct matches. Explain your reasoning. 6.6 (a) A pure fluid shows the following s vs. T behavior. Draw schematically how the chemical potential would change with temperature. g

s

T

T

(b) A pure substance shows the following v vs. P behavior at constant temperature. Draw schematically how the molar Gibbs energy would change with pressure. Explain your reasoning and describe the important features on your plot. g

v

P

P

6.7 At room temperature, iron exists in the ferrite phase 1 a-Fe 2 . At 912°C, it goes through a phase transformation to the austenite phase 1 g-Fe 2 . Which phase of iron has stronger bonds? Explain.

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6.6 Problems ◄ 375 6.8 Consider the crystallization of species a. The molar Gibbs energy of pure species a vs. temperature at a pressure of 1 bar is shown below. Take the molar volume of species a in the liquid phase to be 20% larger than its molar volume as a solid.

Pure species a

Gibbs energy, g [J/mol]

6,000

4,000

2,000

100

200 Temperature,T (K)

300

Answer the following questions: (a) Identify the location of the freezing point on the diagram above. Identify which section of the plot corresponds to liquid and which part corresponds to solid. What is the temperature at which the liquid crystallizes? What is the Gibbs energy of the liquid at this point? (b) Come up with a value for the entropy of the solid phase and the liquid phase. (c) Consider this process occurring at a much higher pressure. Sketch how the plot above will change. Will the freezing point be higher or lower? Try to be as accurate as possible with the features of your sketch. Write down all your assumptions. 6.9 The slope of the coexistence line between the solid and liquid phases shown in Figure 6.3 is positive. Most pure species behave this way. However, species that expand on freezing (like liquid water to ice) exhibit a negative slope. Using the thermodynamic principles presented in this chapter, show that the slope must be negative. 6.10 For two species, A and B, with a positive enthalpy change of mixing: (a) Are the like interactions or the unlike interactions stronger? Explain. (b) If pure species A and B are mixed adiabatically, will the temperature (increase, stay the same, decrease, or cannot tell)? Explain. (c) If pure species A and B are mixed isothermally, what will be the sign of Q? Explain. (d) If an equamolar mixture of species A and B is mixed adiabatically with pure A, will the temperature (increase, stay the same, or decrease)? Explain. 6.11 The following diagram shows the normal melting point of pure solid 1 to be Tm. Consider now that the same pure solid 1 is in a liquid mixture with four species 1, 2, 3, and 4 as shown on the right. How does the temperature at which 1 will be in equilibrium with liquid, T, compare to the case on the left (T , Tm, T 5 Tm, T . Tm, or you cannot tell without more information)? Explain your answer.

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376 ► Chapter 6. Phase Equilibria I: Problem Formulation

1 atm

Tm

1 atm

T= ?

Liquid

Liquid

n l1

n l1,

Solid

n l2,

n l3,

n l4,

Solid

n s1

n s1

6.12 The normal boiling point of pure liquid 1 is Tb. Consider now that the same pure liquid 1 is in a vapor mixture with itself and another species. How does the temperature at which liquid 1 will be in equilibrium with vapor, T, compare to the case for the pure species (T , Tb, T 5 Tb, T . Tb, or you cannot tell without more information)? Explain your answer. 6.13 The normal melting point of pure solid 1 is Tm. Consider now that the same pure solid 1 is melting into a liquid that initially contains pure 2. How does the temperature at which 1 will melt into the liquid, T, compare to Tm? You may assume that the enthalpy of mixing of liquid 1 and 2 is zero. 6.14 In Example 6.5, we found Ni nanopartibles of 2 nm radius would melt at 587 K, assuming the system contained pure Ni liquid. Consider now 2 nm Ni nanoparticles in equilibrium with a liquid mixture. How would the actual melting temperatature compare to the value using the assumptions in Example 6.5? Explain. 6.15 In this problem you will compare properties of species 1 as a pure liquid at 1 atm and 300 K (System I) to species 1 in a mixture with species 2, 3, and 4 at the same pressure and temperature (System II), as shown in the following figure. Select the correct answer in each case, and explain your reasoning.

1 atm

300 K Liquid

1 atm

300 K Liquid

n l1

n l1, n l2, n l3, n l4,

System I

System II

(a) How does the partial molar enthalpy of species 1 in system I compare to the partial molar I II I II enthalpy in system II (HI1 , HII 1 , H1 5 H1 , H1 . H1 , or need more information)? Assume that all the intermolecular interactions in the mixture are the same. (b) How does the partial molar entropy of species 1 in system I compare to the partial molar I II I II entropy in system II (SI1 , SII 1 , S1 5 S1 , S1 . S1 , or need more information)? Assume that all the intermolecular interactions in the mixture are the same. (c) How does the molar Gibbs energy of species 1 in system I compare to the partial molar Gibbs I II I II energy in system II (GI1 , GII 1 , G1 5 G1 , G1 . G1 , or need more information)? Assume that all the intermolecular interactions in the mixture are the same. (d) How does the partial pure species Gibbs energy of species 1 in system I compare to the pure I II I II species Gibbs energy in system II (gI1 , gII 1 , g1 5 g1 , g1 . g1 , or need more information)? Assume that all the intermolecular interactions in the mixture are the same.

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6.6 Problems ◄ 377 (e) How does the partial molar enthalpy of species 1 in system I compare to the partial molar I II I II enthalpy in system II (HI1 , HII 1 , H1 5 H1 , H1 . H1 , or need more information)? Assume that the intermolecular interactions of species 1 with itself are stronger than with any of the other species in the mixture. 6.16 For a given binary system, the partial molar volume of species 1 is constant. What can you say about species 2? Explain.

Numerical Problems 6.17 (a) Use the Clausius–Clapeyron equation and data for water at 100°C to develop an expression for the vapor pressure of water as a function of temperature. (b) Plot the expression you came up with on a PT diagram for temperatures from 0.01°C to 100°C. (c) Include data from the steam tables on your plot in part (b) and comment on the adequacy of the Clausius–Clapeyron equation. (d) Repeat parts (b) and (c) for 100°C to 200°C. (e) Repeat parts (a)–(c), but correct for the temperature dependence of Dhvap, using heat capacity data from Appendix A.2. 6.18 What pressure is needed to isothermally compress ice initially at 25°C and 1 bar so that it changes phase? 6.19 One mole of a pure species exists in liquid–vapor equilibrium in a rigid container of volume V 5 1 L, a temperature of 300 K, and a pressure of 1 bar. The enthalpy of vaporization and the second virial coefficient in the pressure expansion are: Dhvap 5 16,628 3 J/mol 4 and Br 5 21 3 1027 3 m3 /J 4 Assume the enthalpy of vaporization does not change with temperature. You may neglect the molar volume of the liquid relative to that of the gas. (a) How many moles of vapor are there? (b) This container is heated until the pressure reaches 21 bar and is allowed to reach equilibrium. Both vapor and liquid phases are still present. Find the final temperature of this system. (c) How many moles of vapor are there now? 6.20 Tired of studying thermo, you come up with the idea of becoming rich by manufacturing diamond from graphite. To do this process at 25°C requires increasing the pressure until graphite and diamond are in equilibrium. The following data are available at 25°C: Dg 1 25°C, 1 atm 2 5 gdiamond 2 ggraphite 5 2866 3 J/mol 4 rdiamond 5 3.51 3 g/cm3 4 rgraphite 5 2.26 3 g/cm3 4 Estimate the pressure at which these two forms of carbon are in equilibrium at 25°C. 6.21 You wish to know the melting temperature of aluminum at 100 bar. You find that at atmospheric pressure, Al melts at 933.45 K and the enthalpy of fusion is: Dhfus 5 210,711 3 J/mol 4 Heat capacity data are given by: clP 5 31.748 3 J/ 1 mol K 2 4 , csP 5 20.068 1 0.0138T 3 J/ 1 mol K 2 4 Take the density of solid aluminum to be 2700 3 kg/m3 4 and liquid to be 2300 3 kg/m3 4 . At what temperature does Al melt at 100 bar?

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378 ► Chapter 6. Phase Equilibria I: Problem Formulation 6.22 The vapor pressure of silver (between 1234 K and 2485 K) is given by the following expression: ln P 5 2

14,260 2 0.458 ln T 1 12.23 T

with P in torr and T in K. Estimate the enthalpy of vaporization at 1500 K. State the assumptions that you make. 6.23 At a temperature of 60.6°C, benzene exerts a saturation pressure of 400 torr. At 80.1°C, its saturation pressure is 760 torr. Using only these data, estimate the enthalpy of vaporization of benzene. Compare it to the reported value of Dhvap 5 35 3 kJ/mol 4 . 6.24 Pure ethanol boils at a temperature of 63.5°C at a pressure of 400 torr. It also boils at 78.4°C and 760 torr. Using only these data, estimate the saturation pressure for ethanol at 100°C. 6.25 An alternative criteria for chemical equilibrium between two phases of pure species i can be written: gi a gi b ¢ ≤ 5¢ ≤ T T Apply the thermodynamic web to show that the partial derivative of this function with respect to temperature at constant pressure is given by:

C

gi hi '¢ ≤ T S 52 2 T 'T P

6.26 At 922 K, the enthalpy of liquid Mg is 26.780 [kJ/mol] and the entropy is 73.888 [J/(mol K)]. Determine the Gibbs energy of liquid Mg at 1300 K. The heat capacity of the liquid is constant over this temperature range and has a value of 32.635 [J /(mol K)]. 6.27 Solid sulfur undergoes a phase transition between the monoclinic (m) and orthorhombic (o) phases at a temperature of 368.3 K and pressure of 1 bar. Calculate the difference in Gibbs energy between monoclinic sulfur and orthorhombic sulfur at 298 K and a pressure of 1 bar. Which phase is more stable at 298 K? Take the entropy in each phase to be given by the following expressions: Monoclinic phase: sm 5 13.8 1 0.066T 3 J/ 1 mol K 2 4

Orthorhombic phase: so 5 11.0 1 0.071T 3 J/ 1 mol K 2 4 6.28 At 900 K, solid Sr has values of enthalpy and entropy of 20.285 [kJ/(mol)] and 91.222 [J/(mol K)], respectively. At 1500 K, liquid Sr has values of enthalpy and entropy of 49.179 [kJ/mol] and 116.64 [J/(mol K)], respectively. The heat capacity for the solid and liquid phases is given by: 1 cP 2 sSr 5 37.656 3 J/ 1 mol K 2 4 and 1 cP 2 lSr 5 35.146 3 J/ 1 mol K 2 4 respectively. Using only these data, determine the temperature of the phase transition between solid and liquid. What is the enthalpy of fusion? The result of Problem 6.25 could be useful. 6.29 At 1100 K, solid SiO2 has values of enthalpy and entropy of 2856.84 3 kJ/mol 4 and 124.51 [J/(mol K)], respectively. At 2500 K, liquid SiO2 has values of enthalpy and entropy of 2738.44 3 kJ/mol 4 and 191.94 [J/(mol K)], respectively. The heat capacities for the solid and liquid phases are given by: 1 cP 2 sSiO2 5 53.466 1 0.02706T 2 1.27 3 1025 T 2 1 2.19 3 1029 T 3 3 J/ 1 mol K 2 4

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6.6 Problems ◄ 379 1 cP 2 lSiO2 5 85.772 3 J/ 1 mol K 2 4 Using only these data, determine the temperature of the phase transition between solid and liquid. What is the enthalpy of fusion? 6.30 Determine the second virial coefficient, B, for CS2 at 100°C from the following data. The saturation pressure of carbon disulfide 1 CS2 2 has been fit to the following equation:

ln Psat CS2 5 62.7839 2

4.7063 3 103 2 6.7794 ln T 1 8.0194 3 1023 T T

sat

where T is in [K] and ln PCS2 is in [Pa]. The enthalpy of vaporization for CS2 at 100°C has been reported as Dhvap, CS2 5 24.050 3 KJ/mol 4 Compare to the reported value of BCS2 5 2492 3 cm3 /mol 4 6.31 Calculate the Gibbs energy of the liquid and vapor phases of water for the following conditions: (a) Saturated water at 100°C (b) 100°C and 50 kPa Are these results consistent with Equation (6.3) and your intepretation of Gibbs energy? 6.32 The saturation pressure of pure solid species A is given by:

sat

ln PA 5 12.0 2

3000. T

sat

where PA is in [bar] and T is in [K]. You wish express the vapor pressure of the liquid in the form: sat

ln PA 5 A 2

B T

You know the enthalpy of fusion is 210.94 kJ/mol. As best you can, determine the constants A and B. For this problem, you can assume that the enthalpy differences for phase changes do not change with T. 6.33 At 1 bar, silver melts at 1233.95 K. The density of the liquid and solid are: rl 5 9,300 B

kg m3

R

and

rs 5 10,500 B

kg m3

R

You may assume these values are constant in this problem. The entropy at the normal melting point is: sl 5 90.885 B

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J mol K

R

and

ss 5 81.730 B

J mol K

R

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380 ► Chapter 6. Phase Equilibria I: Problem Formulation The molecular weight of silver is 107.9 [g/mol]. (a) Calculate the Gibbs energy of fusion for silver at 5,000 bar and 1400 K, assuming the entropy of each phase is constant. (b) Which phase is stable at 5,000 bar and 1400 K? Explain. (c) Account for the temperature dependence of entropy in the calculation for part A. The following heat capacity data are available: clP 5 33.472 B

J mol K

R

and

cPs 5 22.963 1 6.904 3 1023T B

J mol K

R

with T in [K]. (d) What is the melting temperature of silver at 1 bar? At 3,000 bar? 6.34 You need to find the triple point of pure species A and are unable to find its value in any reference books. You can find the following data for A. It sublimes at 200 K and 0.1 bar, and it boils at 250 K and 1 bar. (a) Calculate the temperature and pressure at the triple point. You may assume the enthalpies of sublimation and vaporization are constant. The values are: Dhsub,i 5 13.921 B

kJ mol

R

and

Dhvap,i 5 28.937 B

kJ mol

R

(b) It turns out that the enthalpy changes are better represented in the following forms: Dhsub,i 5 A 1 BT

and

Dhvap,i 5 C 1 DT

where A, B, C, and D are constants. Come up with an expression to find T and P at the triple point with these forms. 6.35 You need to find the enthalpy of sublimation of solid A at 300 K. The following equilibrium vapor pressure measurements have been made on pure A: (1) at 250 K, the pressure is 0.258 bar and (2) at 350 K, the pressure is 2.00 bar. The following heat capacity data are known: csP 5 40 B

J mol K

R

and

cvP 5 40 1 0.1T B

J mol K

R

(a) Calculate the enthalpy of sublimation, assuming Dhsub is constant. (b) Calculate the enthalpy of sublimation, accounting for the temperature variation of Dhsub. (c) Estimate the error in the constant T assumption. 6.36 A TP diagram of carbon is presented in the following figure. The following data are available at 25°C. rdiamond 5 3.51 B

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g cm3

R

and rgraphite 5 2.26 B

g cm3

R

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6.6 Problems ◄ 381 700

Solid III

600

500

P [kbar]

liquid 400

300 diamond 200

100 graphite 2000

1000

3000

4000

5000

T [K ]

Answer the following questions: (a) Identify the region where diamond is the thermodynamically stable phase of carbon. In this region, what can you say about the Gibbs energy of diamond relative to the other phases of carbon? (b) What is the lowest temperature that liquid carbon can exist? What is the pressure at that temperature? Identify the location on the TP diagram. (c) Using the preceding phase diagram, estimate the difference in enthalpy between diamond (d) and graphite (g), Dh 5 hd 2 hg at 300K. You may assume Dh is constant. (d) From the value of Dh in Part (c), which phase has stronger bonds? Explain. Does this make physical sense? (e) Obtain a more accurate estimate of Dh at 300K using the values of specific heat for diamond and graphite: cP,d 5 6.1 B

J mol K

R

and

cP,g 5 8.5 B

J mol K

R

6.37 Consider the use of CF2Cl2 as a dispersing agent for aerosol spray cans. Estimate the pressure a can has to withhold at 40°C. Its enthalpy of vaporization at its normal boiling point (244 K) kJ is Dhvap 5 20.25B R. State your assumptions. mol 6.38 At 1 atm titanium melts at 1941 K and boils at 3560 K. Its triple point pressure is 5.3 Pa. Using only these data, estimate the enthalpy of vaporization of titanium. You will need to think about a reasonable assumption to solve this problem.

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382 ► Chapter 6. Phase Equilibria I: Problem Formulation 6.39 You wish to determine the enthalpy of vaporizaton of species A. You are able to find the following data: Species A has a normal boiling point of 207.3 K, and at 20.0 atm, it boils at 301.5 K. The following equation of state is reported: P5

RT aP 2 v T

with a 5 25 K. As best as you can, estimate Dhvap. State any assumptions that you make. 6.40 In Example 6.2, we developed an expression for gd 2 gn vs. T for the protein lysozyme (l) between its native phase, n, and its denatured phase, d, where unfolding occurs. Develop the same expression from the following different approaches: (a) Use the property relation given by Equation (5.14). (b) Use the results of Problem 6.25. 6.41 Consider an ideal gas mixture at 83.14 kPa and 500 K. It contains 2 moles of species A and 3 moles of species B. Calculate the following:VA, VB, vA, vB, VA, VB, V, v, DVmix, Dvmix. 6.42 For a given binary system at constant T and P, the molar volume (in cm3 /mol) is given by v 5 100ya 1 80yb 1 2.5yayb (a) What is the pure species molar volume for species a, va? (b) Come up with an expression for the partial molar volume, Va, in terms of yb. What is the partial molar volume at infinite dilution, Va`? (c) Is the volume change of mixing, Dvmix, greater than, equal to, or less than 0? Explain. 6.43 Consider a mixture of species 1, 2, and 3. The following equation of state is available for the vapor phase: Pv 5 RT 1 P2 3 A 1 y1 2 y2 2 1 B 4 where, A 1 B 1 5 29.0 3 1025 B R, 5 3.0 3 1025 B R RT atm2 RT atm2 and y1, y2, and y3 are the mole fractions of species 1, 2 and 3, respectively. Consider a vapor mixture with 1 mole of species 1, 2 moles of species 2, and 2 moles of species 3 at a pressure of 50 atm and a temperature of 500 K. Calculate the following quantities: v, V, v1, v2, v3, V1. 6.44 The molar enthalpy of a ternary mixture of species a, b, and c can be described by the following expression: h 5 25000xa 2 3000xb 2 2200xc 2 500xaxbxc 3 J/mol 4 (a) Come up with an expression for Ha (b) Calculate Ha for a solution with 1 mole a, 1 mole b, and 1 mole c. (c) Calculate Ha for a solution with 1 mole a but with no b or c present. (d) Calculate Hb for a solution with 1 mole b but with no a or c present. 6.45 Plot the partial molar volumes of CO2 and C3H8 in a binary mixture at 100°C and 20 bar as a function of mole fraction CO2 using the van der Waals equation of state. 6.46 The Gibbs energy of a binary mixture of species a and species b at 300 K and 10 bar is given by the following expression:

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6.6 Problems ◄ 383 g 5 240xa 2 60xb 1 RT 1 xa ln xa 1 xb ln xb 2 1 5xaxb 3 kJ/mol 4 (a) For a system containing 1 mole of species a and 4 moles of species b, find the following: `

ga, Ga, Ga , DGmix. (b) If the pure species are mixed together adiabatically, do you think the temperature of the system will increase, stay the same, or decrease. Explain, stating any assumptions that you make. 6.47 Consider a binary mixture of species 1 and species 2. A plot of the partial molar volumes in 3 cm3 /mol 4 of species 1 and 2, V1 and V2, vs. mole fraction of species 1 is shown below. For a mixture of 1 mole of species 1 and 4 moles of species 2, determine the following quantities for this mixture: V1, V2, v1, v2, V1, V2, V, v, DVmix.

70 67.5 V2

65 62.5

Vi

60 57.5 55 52.5 50 V1

47.5 45 0

0.1

0.2 0.3 0.4

0.5 0.6 0.7 0.8 0.9 x1

1

6.48 Enthalpies of mixing for binary mixtures of cadmium (Cd) and tin (Sn) have been fit to the following equation at 500°C: Dhmix 5 13,000XCdXSn 3 J/mol 4 where, XCd and XSn are the cadmium and tin mole fractions, respectively. Consider a mixture of 3 moles Cd and 2 moles Sn. (a) Show that: 1 DHmix 2 Cd 5 HCd 2 hCd (b) Based on the equations above, calculate values for HCd 2 hCd and HSn 2 hSn at 500°C. (c) Show that the results are consistent with the Gibbs–Duhem equation. (d) Data from which the above equation was derived are presented below, along with the model fit. Graphically determine values for HCd 2 hCd and HSn 2 hSn at 500°C. Compare your answer to part (b). Show your work.

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384 ► Chapter 6. Phase Equilibria I: Problem Formulation Heat of mixing in cadmium(Cd)-tin (Sn) system

6000 5400 4800

Δhmix

J mol

4200 3600

data

3000 2400 1800

fit to: Δhmix = 13,000 XCdXSn

1200 600 0

0

0.2

0.4

J mol

0.6

0.8

1

xCd

6.49 3 moles of pure water are adiabatically mixed with 1 mole of pure ethanol at a constant pressure of 1 bar. The initial temperatures are the pure components are equal. If the final temperature is measured to be 311.5 K, determine the initial temperature. The enthalpy of mixing between water (1) and ethanol (2) has been reported to be fit by:

Dhmix 5 x1x2 3 2190.0 1 214.7 1 x2 2 x1 2 2 419.4 1 x2 2 x1 2 2 1 383.3 1 x2 2 x1 2 3 R 2 235.4 1 x2 2 x1 2 4 4 3 K 4 6.50 A stream of pure water flowing at 1 mol/s is adiabatically mixed with a stream containing equamolar amounts of water and ethanol, also flowing at 1 mol/s. This steady-state process occurs at a constant pressure of 1 bar. The temperatures of the inlet streams are 298 K. Determine the temperature of the outlet stream. The enthalpy of mixing between water (1) and ethanol (2) is given in Problem 6.49. 6.51 The Gibbs energy of a binary mixture is given by the following expression: Dgmix 5 RT 3 xa ln xa 1 xb ln xb 4 1 1000xaxb B

J mol

R

where R is the ideal gas constant and T is in [K]. For a mixture at 298 K with the following mole fractions of a, calculate Ga 2 ga. (a) xa 5 1 (b) xa 5 0.4 (c) xa 5 0 1 a in infinite dilution 2 6.52 The molar enthalpy of a binary liquid mixture of species 1 and 2 is given by: h 5 x1 1 275 1 75T 2 1 x2 1 125 1 50T 2 1 750x1x2 B

J mol

R

where T is the temperature in [K] (a) What is the enthalpy of mixing, DHmix in [J], of a mixture with 2 mol of 1 and 3 mol of 2 at 20°C? (b) Consider the adiabatic mixing of a stream of pure 1 flowing at 2 mol/s with a stream of pure 2 flowing at 3 mol/s. Both inlet streams are at 20°C. What is the exit temperature of the mixture? 6.53 The molar volume, in 3 cm3 /mol 4 , of a binary mixture of ethanol (1) and ethylene glycol (2) at 25°C is given in the following table.

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6.6 Problems ◄ 385 v [cm3/mol]

x1 0

55.828

0.1092

55.902

0.2244

56.06

0.3321

56.245

0.4393

56.513

0.5499

56.866

0.6529

57.178

0.7818

57.621

0.8686

58.004

1

58.591

Using the graphical method, determine the partial molar volume of ethanol at 25°C for the following mole fractions of species 1: (a) x1 5 0.8 (b) x1 5 0.4 (c) x1 5 0 1 1 in infinite dilution 2 6.54 Fit the data in Problem 6.53 to the form: v 5 ax1 1 bx2 1 cx1x2 (a) Use the analytical method to come up with an expression of the partial molar volume of ethanol. (b) Determine the values at x1 5 0.8, x1 5 0.4, and x1 5 0 (1 in infinite dilution) 6.55 The volume change of mixing, in 3 cm3 /mol 4 , for binary liquid mixture of sulfuric acid (1) and water (2) at 25°C is given by the following expression: Dvmix 5 213.1x1x2 2 2.25x21x2 The density of pure sulfuric acid is 1.8255 3 g/cm3 4 . (a) Plot the partial molar volumes of H2SO4 and H2O versus mole fraction H2SO4. `

(b) What is the value for the partial molar volume of water at infinite dilution, V2 ? (c) Comment on the numerical results in Part (b) and relate it to what you think is physically happening in the system. 6.56 The molar enthalpy of a binary liquid mixture of species 1 and 2 is given by: h 5 1500x1 1 800x2 1 1 25.0x1 1 35.0x2 2 11.86x1x2 2 T B

J mol

R

where T is the temperature in [K]. This expression is valid in the temperature range from 280 K to 360 K. Please answer the following questions: (a) You wish to dilute and cool a process stream containing 50 mol% species 1 and 50 mol% species 2. Consider the adiabatic mixing of a stream a (containing 50 mol% species 1 and 50 mol% species 2) with a molar flow rate of 2 mol/s at 75°C and 1 bar with a stream b of pure species 1 flowing at a rate of 3 mol/s at 20°C and 1 bar. What is the exit temperature of the outlet stream? You may assume steady-state. (b) What is the partial molar enthalpy of species 1, H1,