Thermodynamics in materials science, second edition

Thermodynamics in Materials Science Other Materials Science Textbooks Physical Metallurgy, by William F Hosford Selection of Engineering Materials and Adhesives, by Lawrence W Fisher, PE Advanced Mechanics of Materials and Applied Elasticity, by Anthony Armena`kas, Ph.D Engineering Designs with Polymers and Composites, James C Gerdeen, Harold W Lord, and Ronald A.L Rorrer Modern Ceramic Engineering, Third Edition, by David Richerson COMING SOON Introduction to Chemical Polymer Science: A Problem Solving Guide, by Manas Chanda Second Edition Thermodynamics in Materials Science Robert DeHoff Boca Raton London New York A CRC title, part of the Taylor & Francis imprint, a member of the Taylor & Francis Group, the academic division of T&F Informa plc CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2006 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S Government works Version Date: 20110713 International Standard Book Number-13: 978-1-4200-0585-1 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under U.S Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers For permission to photocopy or use material electronically from this work, please access www.copyright com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400 CCC is a not-for-profit organization that provides licenses and registration for a variety of users For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com Preface to the Second Edition The presentation of the principles and strategies at the heart of thermodynamics has been retained from the first edition The principles and laws, the definitions, the criterion for equilibrium, and the strategies for deriving relationships among variables and for finding the conditions for equilibrium are all intact There is a new emphasis on the structure of thermodynamics, introduced in a new Chapter 1, which provides a visualization of how all of these components integrate to solve problems There is a new emphasis on the main goal of thermodynamics in materials science, which is the use of thermochemical databases to generate maps of equilibrium states, such as phase diagrams, predominance diagrams, and Pourbaix corrosion diagrams There are many other useful applications of thermodynamic information, but the equilibrium maps are clearly the most widely used tools in the field Although computer software to convert database information into equilibrium maps was available at the writing of the first edition, such software now comes with more comprehensive databases and breadth of application and, perhaps most importantly, user-friendliness It is also more widely available for student use as materials science programs have acquired it for use in their research or teaching The CALPHAD origins of these programs is dealt with in Chapters to 10 There is a danger that applications in this field may achieve a “black box” status in which the results of all this software, in the form of equilibrium maps and other information, come to be used without an understanding of their origins In industry, this kind of information may be a component in a decision-making process that has millions of dollars on the line It is crucial that the connections between the results and the fundamentals provided by this kind of text be maintained Preface to the First Edition In his classic paper in 1883, J Willard Gibbs completed the apparatus called phenomenological thermodynamics, which is used in engineering and science to describe and understand what determines how matter behaves This work is all the more remarkable in the light of the enormous expansion of our knowledge in science and technology in the 20th century During the last century, hundreds of books have been written on thermodynamics In most cases, these texts were directed at students in a particular field Thermodynamics plays a key role in biology, chemistry, physics, chemical and mechanical engineering, and materials science Each presentation offered its own slant to its intended audience Several of these texts are classics that have endured for decades, experiencing many revisions and many printings An author who undertakes an introductory text in thermodynamics in the face of this history had better be sure of his subject, and have something unique to say After teaching introductory thermodynamics to materials scientists for nearly three decades at both the graduate and undergraduate levels, I am convinced that the approach used in this text is unique and in many ways better than that available elsewhere Thermodynamics in Materials Science is an introductory text intended primarily for use in a first course in thermodynamics in materials science curricula However, the treatment is sufficiently general so that the text has potential applications in chemistry, chemical engineering, and physics, as well as materials science The treatment is sufficiently rigorous and the content sufficiently broad to provide a basis for a second course either for the advanced undergraduate or graduate level Thermodynamics is a discipline that supplies science with a broad array of relationships between the properties that matter exhibits as it changes its condition All of these relationships derive from a very few, very general and pervasive principles (the laws of thermodynamics) and the repetitive application of a very few, very general strategies It is not a collection of independent equations conjured out of misty vapors by an all-knowing mystic for each new application There is a structure to thermodynamics that is elegant and, once contemplated, reasonably simple The approach that undergirds the presentation in this text emphasizes the connections between the foundations and the working relations that permit the solution of practical problems In this emphasis, and in its execution, it is unique among its competitors The difference is crucial to the student seeing the subject for the first time Most texts spend a significant amount of print and the student’s time in presenting the laws of thermodynamics and in laying out arguments that justify the laws and lend intuitive interpretation to them This presentation is based on the recognition that such diversions are a significant waste of time and effort for the student and, what is worse, are usually confusing to the uninitiated Worse still, students may be left with an inadequate intuition that merely serves to mislead them when they attempt to apply it to complex systems Thermodynamics is fundamentally a rational subject, rich with deductions and derivations Intuition in thermodynamics is not for the uninitiated Thus, the laws are presented as fait accompli: “great accomplishments of the 19th century” that distilled a broad range of scientific observation and experience into succinct statements that reflect how the world works It is best at this beginning stage that these laws be presented with clear statements of their content, without the perpetual motion arguments, Carnot cycle, and other intuitive trappings The most significant departure of this text from other works lies in the treatment of the concept of equilibrium in complex physical systems, and in the presentation of a general strategy for finding the conditions for equilibrium in such systems A general criterion for equilibrium is developed directly from the second law of thermodynamics The mathematical procedure for deriving the equations that describe the internal condition when it is at equilibrium is then presented with rigor It is the central viewpoint of this text that, since all of the “working equations” of thermodynamics are mathematical statements of these internal conditions for equilibrium, establishment of the connection between these conditions and first principles is crucial to a working understanding of thermodynamics Indeed, the remainder of the text is a series of applications of this general strategy to the derivation of the condition for equilibrium in systems of increasing complexity, together with strategies for applying these equations to solve problems of practical interest to the student With each increment in the level of sophistication being treated, new parts of the apparatus of thermodynamics are introduced and developed as they are needed The general strategy for getting to the working equations is the same for all of these applications Thus, the connection to the fundamental principles is visible for each new development Furthermore, this connection can be maintained without introducing any mathematical or conceptual shortcuts Repetition builds confidence; rigor builds competence One early chapter introduces the concepts of statistical thermodynamics This subject is treated as an algorithm for converting an atomic model for the behavior of the system, formulated as a list of the possible states that each atom may exhibit, into values of all of the thermodynamic properties of the system The strategy for deriving the conditions for equilibrium in this case applies to the derivation of the Boltzmann distribution function, which reports how the atoms are distributed over the energy levels when the system attains equilibrium The algorithm is then illustrated for the ideal gas model and the Einstein model for a crystal Statistical thermodynamics is used very little in subsequent chapters because the classes of systems that are the domain of materials science tend to be too complex for tractable treatment, much less for presentation to first-time students of the subject Most chapters contain several illustrative examples, designed to emphasize the strategies that connect principles to hard numerical answers Each chapter ends with a summary that reviews the important concepts, strategies, and relationships that it contains Each chapter also ends with a collection of homework problems, many of which are designed so that they are best solved using a personal computer: the astute student may find it useful to write some more general programs that can be used repeatedly as the level of sophistication increases Examples of homework problems will be drawn more or less uniformly from the major classes of materials: ceramics, metals, polymers, electronic materials, and composites This approach serves to illustrate the power of the concepts, laws, and strategies of phenomenological thermodynamics by demonstrating that they can be applied to all states of matter The experience gained in 25 years of teaching an undergraduate course in thermodynamics in materials science, together with more than 15 years of teaching a graduate course in the same area, has resulted in an approach to the topic that is unique The approach accents rigor, generality, and structure in developing the concepts and strategies that make up thermodynamics because the connections between first principles and practical problem solutions are sharply illuminated; the first-time student can hope not only to apply thermodynamics to the sophisticated end of systems that are the bread and butter of materials science, but to understand their application It is a pleasure to acknowledge the help of Heather Klugerman, who provided advice in the more sophisticated aspects of word processing involved in putting together this text Pamela Howell proofread the manuscript with remarkable skill before it was submitted to the publisher David C Martin, University of Michigan, and Monte Poole, University of Cincinnati, offered many helpful comments and suggestions while reviewing the manuscript My thanks to the many students, both graduate and undergraduate, who for many years encouraged me to undertake this text Finally, I am grateful to my wife, Marjorie, who sacrificed many evenings, weekends, and vacations as I disappeared into the den to work on the project The Carnot Cycle 577 Process I: ð dQ ð Q P rev ¼ dQrev ¼ I ¼ RTI ln P1 OVERP0 ¼ R ln T TI TI TI P0 Process II: ð dQ rev ¼0 T Process III: ð dQ ð Q P P rev ¼ dQrev ¼ III ¼ 1OVERTIII RTIII ln ¼ R ln T TIII TIII P2 P2 Process IV: ð dQ rev ¼0 T The result for the system is the sum of these four contributions: ð dQ P P rev ¼ R ln ỵ ỵ R ln ỵ T P0 P2 which can be written ð dQ P P rev ¼ R ln T P0 P2 ðH:8Þ The pressure ratios in this equation, (P2/P1) and ðP0 =P3 Þ; can be expressed simply in terms of the temperatures of the two isothermal steps by applying the adiabatic path Equation H.5 and Equation H.6: ð dQ P rev ¼ R ln T P0 P3 P2 ¼ R ln TI TIII CV ỵRị=R TIII TI CV ỵRị=R H:9ị or dQ rev ẳ R ln1ị ẳ T ðH:10Þ Thus it is demonstrated that, for any Carnot cycle using an ideal gas operating between any pair of temperatures TI and TIII and any choice of independent pressures 578 Thermodynamics in Materials Science P0 and P1 (the other two pressures are determined by the definition of the cycle) ð dQ rev ¼0 T Accordingly, within these restrictions, the integral is a state function and its integrand is an exact differential of a state function Carnot and others went on to demonstrate that every reversible cyclic path that could be devised can be decomposed into an infinite sequence of connected infinitesimal Carnot cycles Since the integral in question is zero for every Carnot cycle it totals to zero for any sum of Carnot cycles Thus, the integral has the character of a state function for every cyclic process that can be imagined for an ideal gas The result was further extended to systems containing arbitrary working substance (i.e., other than an ideal gas) This advancement was achieved by imagining the system with the real working substance coupled to one with an ideal gas working substance so that the energy, heat and work exchanges accompanying the Carnot cycle for each of these two systems are equal and opposite at each step It is concluded that ð dQ rev ¼0 T for any real system taken along an arbitrary path Thus, the result is completely general and this integral is demonstrated to be a state function of general application The symbol widely adopted for this state function is S: the origins of the name, entropy, are obscure For any arbitrary system taken along any reversible path DS ¼ ð dS ¼ ð dQ rev T ðH:11Þ It follows that, for each infinitesimal step along the reversible path dS ¼ dQrev T ðH:12Þ and the general relation between the entropy and the reversible heat absorbed is established The development of the concept of entropy did not end with the Carnot cycle argument The statistical interpretation of the entropy, introduced in Chapter 6, evolved half a century later Notions of configurational entropy represented, for example, by the entropy of mixing of solutions, expanded the concept The quantification of the role of entropy production in irreversible processes took another half a century, and continues to evolve The Carnot cycle lies at the foundation of the development of thermodynamics; however, its role is largely hidden in the treatment of complex multicomponent, multiphase systems with capillarity and electrochemical effects, which are the bread and butter of materials science I Answers to Homework Problems CONTENTS Numerical Answers to Homework Problems 579 Chapter 579 Chapter 580 Chapter 580 Chapter 581 Chapter 581 Chapter 582 Chapter 583 Chapter 584 Chapter 10 584 Chapter 11 587 Chapter 12 588 Chapter 13 589 Chapter 14 589 Chapter 15 590 NUMERICAL ANSWERS TO HOMEWORK PROBLEMS Homework problems in the text that have mathematical answers (as opposed to text answers) use as a database either: (1) data that are given the statement of the problem, or (2) data obtained from the Appendices of this text These data are intended for use as exercises for homework assignments Thermochemical data evolve with time as new experimental results appear, and new optimizations are applied to existing data For applications in the real world the reader is directed to a variety of databases (some free, some expensive) available through the Internet, which will give results that are not necessarily identical to those in the Appendices, because they are up to date CHAPTER Problem 2.5 dz ẳ 36u2 v cosxịdu ỵ 12u3 cosxịdv 12u3 v sinðxÞdz 579 580 Thermodynamics in Materials Science CHAPTER Problem 3.16 Reaction DS298 (J/mol K) C ỵ O2 ẳ CO2 2Al ỵ 32 O2 ẳ Al2O3 Si ỵ C ẳ SiC C ỵ 12 O2 ẳ CO Si ỵ O2 ẳ SiO2 Si ỵ CO2 ẳ SiC ỵ O2 Al2O3 ỵ 32 Si ẳ 32 SiO2 ỵ 2Al 2Al þ 3CO2 ¼ Al2O3 þ 3CO SiO2 þ 2C ¼ CO2 ỵ SiC CO ỵ 12 O2 ẳ CO2 2.92 2313.04 27.98 89.70 2182.36 210.93 42.56 254.3 191.6 286.35 CHAPTER Problem 4.3 After the isobaric step, V ¼ 25:97 (cc/mol) After the isothermal step, V ¼ 25:76 (cc/mol) Problem 4.4 (a) 55.83 (J/mol K) (b) 2.68 (J/mol K) (c) 96.42 (J/mol K) (d) 2.23 (J/mol K) (e) 42.67 (J/mol K) (f) 95.72 (J/mol K) Problem 4.5 32.14 (J/mol K); 0.74 (J/mol K) Problem 4.6 (a) n ¼ 0:536 (moles); DU ¼ 81 ¼ 72 (l-atm) (b) Tf ¼ 1365 K; DU ¼ 72 (l-atm) Problem 4.7 Function to be plotted: DSðT; PÞ ¼ CP ln T P R ln 300 Answers to Homework Problems 581 Problem 4.8 DUP; Vị ẳ CV PV P0 V0 ị DHP; Vị ẳ CP ðPV P0 V0 Þ Problem 4.9 ›H ›G S ¼ CP CP TSa Problem 4.10 dF ¼ S ỵ PV a Pb Pb C dT dS a Ta P Problem 4.12 (a) Ti ¼ 219:4 K, Tf ¼ 292:5 K (b) Q ¼ 233:77 (J/mol) (c) W ¼ 945:62 (J/mol) (d) DU ¼ 911:84 (J/mol); DH ¼ 1520 (J/mol); DS ¼ 0:217 (J/mol K); DF ¼ 27902 (J/mol); DG ¼ 27294 (J/mol) Problem 4.13 (a) For silver, P ¼ 9:6 £ 106 (atm) (b) For alumina, P ¼ 978 (atm) Problem 4.14 Benchmark values: DG (298 K, 10210 atm) ¼ 57,050 (J/mol) DG (1000 K, 100 atm) ¼ 95,050 (J/mol) Problem 4.15 For a combination of the maximum values for a; b and V; CP CV ¼ 0:435: For a combination of the minimum values, CP CV ¼ 0:011 CHAPTER Problem 5.7 Unconstrained, zmax ¼ at ðx ¼ 2; y ẳ 2ị; Constrained by y ẳ x; z0max ẳ 2:4 at x ẳ 6=5; y ẳ 12=5ị CHAPTER Problem 6.2 List the macrostates states systematically starting with (10, 0, 0), (9, 0, 1), (9, 1, 0), (8, 0, 2), (8, 1, 1), (8, 2, 0), etc 582 Thermodynamics in Materials Science Problem 6.3 (a) 44 ¼ 16 microstates (b) Enumerate systematically There are 10 macrostates Problem 6.4 (a) 43 ¼ 64 (b) 415 ¼ 1.074 £ 109 (c) 154 ¼ 50,625 (d) 3050 ¼ 7.18 £ 1073 (e) 1001000 ¼ (102)1000 ¼ 102000 Problem 6.5 (a) DU ¼ (J/mol) (b) VII =VI ¼ 12 (c) State II is more likely Problem 6.6 (a) Calculator: 10! ¼ 3.629 £ 106, 30! ¼ 2.653 £ 1032, 60! ¼ 8.321 £ 1081 (b) Stirling’s Formula: 4.54 £ 105, 1.927 £ 1031, 4.28 £ 1080 (c) Errors in ln x!: 0.138, 0.035, 0.016 Problem 6.7 DS ¼ 213:05 (J/mol K) ¼ 2.17 £ 10223 (J/atom-K) (Applying Equation 6.9) Problem 6.9 DU ¼ 2311 (J/mol) Problem 6.10 (a) DS ¼ 2:162 (J/mol K) (b) DS ¼ 2:162 (J/mol K) CHAPTER Problem 7.3 mðT; PÞ mð298; 1Þ ẳ 220:79T ln T ỵ 13:16T ỵ 31; 370 ỵ RT ln P Problem 7.6 The error in log P is about 1.5% at the melting point (over a range of about 1500 K) Problem 7.7 DS ¼ 18:6 (J/mol K) Answers to Homework Problems 583 Problem 7.8 T 1L ¼ 1710 K Problem 7.9 DHv ¼ 269; 000 (J/mol) Problem 7.10 Pag G ¼ 2:6 £ 10210 (atm); Pgd G ¼ 9:0 £ 1026 (atm); Pd LG ¼ 7:0 Ê 1025 (atm) Problem 7.11 T;PịbLG ẳ 576 K; 6:38 Ê 10210 atmịị; T;Pị1bG ẳ 500 K; 4:47 Ê 10212 atmị Problem 7.12 Metastable triple point: T; Pị1LG ẳ 569 K; 4:34 £ 10210 atmÞ CHAPTER Problem 8.1 ðWt:%OÞ ¼ 0:044 rO ¼ 0:18 gm of O mol of O ; cO ¼ 0:011 ; gm of soln cc of soln gm of O cc of soln Problem 8.3 Benchmark: DV2 ¼ 54X 21 X2 ; DV1 ¼ 27X1 X 22 Problem 8.5 Plot the results of a Gibbs –Duhem integration: DHPn ¼ 12; 500 XCu XCu ; DHmix ¼ 6250X 2Cu XPn Problem 8.9 Plot the activity of zinc obtained from RT ln aZn ¼ T X ½13; 200ðXAl XZn ị ỵ 19; 200XZn 4000 Al ỵ RT ln XZn 584 Thermodynamics in Materials Science Problem 8.10 Total Properties PMP of A PMP of B 2513 0.565 21.749 2824 2842 2531 1353 0.304 20.942 1521 1511 1363 4667 1.049 23.247 5244 5211 4700 DGxs mix (J/mol) DSxs mix (J/mol K) DVmix (cc/mol) DHmix (J/mol) DUmix (J/mol) DF xs mix (J/mol) Problem 8.13 Plot as a function of temperature g0A ¼ g0B ¼ e213;500=RT Problem 8.14 1AB ¼ 26:111 £ 10220 J=bondị Problem 8.15 Plot fAB ẳ ẵ2XA XB 1:3ị; then fAA ¼ XA fAB =2; fBB ¼ XB fAB =2 Note that fAA and fBB go negative at compositions not too far from XB ¼ 1=2; the model loses physical significance for compositions far from XB ¼ 1=2: Problem 8.16 Plot DHmix 1 N z 1AA ỵ 1BB ị AB 1 ¼ XA fAB ; fBB ¼ XB fAB 2 fAB ¼ fAA CHAPTER No numerical problems CHAPTER 10 Problem 10.1 Given: DGamix {a; a} ¼ 8400X aA X aB ỵ RTX aA ln X aA ỵ X aB ln X aB ị DGLmix {L; L} ẳ 10; 500X LA X LB ỵ RTX LA ln X LA ỵ X LB lnX LB ị Answers to Homework Problems 585 (a) Compute and plot for the {L;L} reference state: a!L DGamix {L; L} ¼ DGamix {a; a} X aA DGA X aB DGB0 a!L DGLmix {L; L} ¼ DGLmix {L; L} (b) Compute and plot for the{L;a} reference state: a!L DGamix {L; a} ¼ DGamix {a; a} X aA DGA DGLmix {L; a} ẳ DGLmix {L; L} ỵ X LB DGB0 a!L (c) Compute and plot for the {a;a} reference state: DGLmix {a; a} DGamix {a; a} ¼ DGamix {a; a} a!L ẳ DGLmix {L; L} ỵ X LA DGA ỵ X LB DGB0 a!L Problem 10.2 (a) Compute and plot: DGbmix {b; b} ẳ 28200X bA X bB ỵ RTX bA ln X bA ỵ X bB ln X bB ị DGLmix {L; L} ẳ 210; 500X LA X LB ỵ RTX LA ln X LA ỵ X LB ln X LB Þ (b) Compute and plot: 6800 ð660 TÞ 660 8200 DGLmix {b; L} ẳ DGLmix {L; L} ỵ X LA 1050 Tị 1050 DGbmix {b; l} ẳ DGbmix {b; b} X bB Problem 10.3 (a) Compute and plot: aB {L; L} ẳ XB e8400=RTịXA (b) Compute and plot: aB {L; a} ¼ aB {L; L}e21200=RT Problem 10.9 (a) X2 ¼ 0:307 to X2 ¼ 0:693 586 Thermodynamics in Materials Science (b) Compute and plot: Dm2a ẳ 12;7001 X2 ị2 ỵ RT ln X2 (c) Compute and plot: dDm2a dX2 ¼ RT 2 £ 12;700ð1 X2 Þ X2 Problem 10.16 Solve the following equations simultaneously: " a2 317 5600X L2 ỵ 11;400X ỵ R1300ịln X L2 X a2 # ¼0 " 1246 5600ð1 X L2 ị2 ỵ 11;4001 X a2 ị2 ỵ R1300ị ln X L2 X a2 # ẳ0 At 1300 K X a2 ¼ 0:763; X L2 ¼ 0:824 Problem 10.17 Tc ¼ 637:5 K Spinodal boundary is given by s# " T X2;spin Tị ẳ 1^ 12 Tc Problem 10.18 X2max ¼ 0:649; Tc ¼ 652:3 K Problem 10.19 TX p ị ẳ 29380X p X p ị ỵ 12838:8ị1 X p ị ỵ 9426:3ịX p 8:81 X p ị ỵ 6:3X p TX p ị ẳ 10;000X p X p ị ỵ 12838:8ị1 X p ị ỵ 9426:3ịX p 8:81 X p ị ỵ 6:3X p Problem 10.20 g00 ¼ 370 Problem 10.21 18,930 K Answers to Homework Problems 587 CHAPTER 11 Problem 11.2 A ¼ 2161;200 (J); products form Problem 11.3 H2 ỵ O2 ẳ H2 O A ẳ 297;700 Jị CO ỵ O2 ẳ CO2 A ẳ 2145;000 Jị CH4 ỵ O2 ẳ CO2 ỵ 2H2 A ẳ 2328;000 Jị Oxygen is consumed, oxides form Problem 11.4 XCO ¼ 0:295; XCO2 ¼ 0:705; XO2 ¼ 1:8 £ 10220 Problem 11.6 (a) DGf Tị ẳ 2239;700 ỵ 94:5T (J/mol) (b) KTị ẳ expẵ2DG0f Tị=RT (c) PO2 ẳ 6:9 Ê 10217 (atm) (d) A ¼ 2136;400 (J/mol) Problem 11.7 (a) DGf ðSiO2 Þ ¼ 2714;600 (J); DGf H2 Oị ẳ 2194;100 (J) (b) DGr ¼ 2326;400 (J) (c) Kr ¼ 7:78 £ 1015 ; (H2/H2O) ¼ 8.76 £ 107 (d) PO2 ¼ 1:6 £ 10235 (atm) (e) PO2 ¼ 1:6 £ 10235 (atm) Problem 11.9 (a) PO2 ¼ £ 10212 (atm) (b) K ¼ £ 1033 (c) (CO/CO2) ¼ £ 105 (d) Yes (e) (H2/H2O) ¼ £ 102 (f) PO2 10227 Problem 11.10 (a) Yes (b) No (c) No (d) Yes 588 Thermodynamics in Materials Science Problem 11.11 Cu2O: A ¼ 2254; 800 (J); CaO: A ¼ 2516; 000 (J) NiS: A ¼ 120; 900 (J); CaCl2: A ¼ 2564; 000 (J) Problem 11.13 Na2 O ỵ O2 ỵ S2 ẳ Na2 SO3 2=5Nb2 O5 ỵ O2 þ 4=5S2 ¼ 4=5NbS2 O5 2=7Cu2 O þ O2 þ 2=7S2 ẳ 4=7CuSO4 1=4Cu2 S ỵ O2 ỵ 1=8S2 ẳ CuSO4 Problem 11.16 If total pressure is atm, for PCO 0:447 oxidation will be prevented Carburization requires PCO 0:96: Thus, if PCO 0:96; both conditions are satised Problem 11.17 Reactions: 2CO ỵ O2 ẳ 2CO2 and 2H2 ỵ O2 ẳ 2H2O (a) equilibrium mole fractions: CO ¼ 0.40; CO2 ¼ 0.09; H2 ¼ 0.377; H2O ¼ 0.133; O2 ¼ 1.6 £ 10220 (b) Affinities: Aẵ2H2 O ẳ 2378; 000 (J); Aẵ2CO2 ẳ 2374; 400 (J) CHAPTER 12 Problem 12.6 Equation of the surface: PðH; Tị ẳ PH ẳ 0; Tịe2gV L =RTịH where PH ¼ 0; TÞ is given by Equation 7.39 Problem 12.7 PH ẳ 0ị ẳ 1:728 Ê 1026 (atm); Pr ẳ 20:5mị ẳ 1:722 Ê 1026 (atm) Note that the vapor pressure is lowered because the curvature is negative Problem 12.9 (a) la ¼ 1:98 £ 1027 (cm); lb ¼ 1:68 £ 1027 (cm) Problem 12.10 (a) XB ¼ 0:109; X bB ¼ 0:127 (b) lb ¼ 1:16 £ 1027 (cm); l1 ¼ 1:18 £ 1027 (cm) Answers to Homework Problems 589 Problem 12.12 0.866 Problem 12.14 550 (ergs/cm2) Problem 12.15 Yes, it will wet the grain boundaries completely Problem 12.16 (a) f ẳ 1638 (b) sc ẳ 4:1mị CHAPTER 13 Problem 13.1 (a) DSv ¼ 8:85 (J/mol K); DHv ¼ 95; 800 (J/mol) Problem 13.2 (a) Xi ¼ exp½7:6=R expẵ2188; 000=RT Problem 13.3 Xvv ẳ Xv expẵ0:89=R expẵ29580=RT Problem 13.4 High range: Xf ẳ expẵ10=R expẵ2550; 000=RT Low range: Xf ẳ expẵ1=R expẵ2350; 000=RT Problem 13.7 (a) M2ỵ0.00501O3 (b) M2O320.00749 (c) M220.0025O3 (d) M2O3ỵ0.00375 CHAPTER 14 Problem 14.1 T ¼ 398:38 K at all x; (b) DS ¼ 0:266 (J/mol K) Problem 14.6 Doubling the isotope ratio requires a centrifuge with 50 cm radius rotating at 20,000 rpm Problem 14.9 (a) DG ¼ 2219:59 (J) (b) DGgrad ¼ 0:235 (J); DGTot ¼ 2219:35 (J) 590 Thermodynamics in Materials Science CHAPTER 15 Problem 15.1 Compute and plot: pffiffiffiffiffiffiffiffiffiffiffiffiffi 2K ^ K 2 4Kc acị ẳ 2c Use ỵ sign Problem 15.2 pH ¼ 5.72 Problem 15.3 5.2 £ 1026 (gm) Problem 15.4 CNaOH ¼ 3:162 £ 1025 (mol/liter) Problem 15.5 1.26 £ 1027 (liters) ¼ 0.126 microliters Problem 15.10 For the compositions given, aMg ¼ 0:006; 0:024; 0:061; 0:160; 0:357; 0:494 Problem 15.12 Formation energies needed as input (J/mol): NiO ẳ 215,900; NiO2 ẳ 215,100; Niỵ þ ¼ 48,200; HNiO2 ¼ 349,200; H2O ¼ 273,200 Triple points (E, pH) occur as follows: (Ni, Niỵ ỵ , NiO) at (2 0.250, 6.09); (Niỵ ỵ , NiO, NiO2) at (0.863, 6.09); (NiO, NiO2, HNiO2 ) at (0.146, 18.21); 22 (Ni, NiO, HNiO2 ) at (2 0.967, 18.21) ... unique and in many ways better than that available elsewhere Thermodynamics in Materials Science is an introductory text intended primarily for use in a first course in thermodynamics in materials. .. used without an understanding of their origins In industry, this kind of information may be a component in a decision-making process that has millions of dollars on the line It is crucial that the... this kind of text be maintained Preface to the First Edition In his classic paper in 1883, J Willard Gibbs completed the apparatus called phenomenological thermodynamics, which is used in engineering