Instant download and all chapters Solutions Manual Molecular Thermodynamics of Fluid-Phase Equilibria 3rdEdition John M. Prausnitz, Rudiger N. Lichtenthaler, Edmundo Gomes de Azevedo https://testbankdata.com/download/solutions-manual-molecular-thermodynamics-fluid-phase-equilibria-3rd-edition -john-m-prausnitz-rudiger-n-lichtenthaler-edmundo-gomes-de-azevedo/ SOLUTIONS MANUAL Molecular Thermodynamics of Fluid-Phase Equilibria, Third Edition John M. Prausnitz Rudiger N. Lichtenthaler Edmundo Gomes de Azvedo C O N T E N T S Preface 1 Solutions to Problems Chapter 2 1 Solutions to Problems Chapter 3 17 Solutions to Problems Chapter 4 29 Solutions to Problems Chapter 5 49 Solutions to Problems Chapter 6 81 Solutions to Problems Chapter 7 107 Solutions to Problems Chapter 8 121 Solutions to Problems Chapter 9 133 Solutions to Problems Chapter 10 153 Solutions to Problems Chapter 11 165 Solutions to Problems Chapter 12 177 PREFACE This manual provides detailed solutions to all problems presented in the third edition of the textbook Molecular Thermodynamics of Fluid-Phase Equi- libria (Prentice-Hall, 1999). This manual was prepared according to the corrections presented on the third ( and subse- quent) printing of the textbook. For those who have copies of the first and second printings, an errata is available for download at http://www.phptr.com/ptrbooks/ptr_OJ39777458.html or at http://alfa. ist. utl.pt/-ejsga. Although we have carefully checked all solutions presented here, we would appreciate be- ing informed of any misprints or errors that the text may contain. We gratefully acknowledge the contribution of Van Nguyen that checked all the solutions presented here. Berkeley, August 1999 J. M Prausnitz Berkeley, California R. N. Lichtenthaler Heidelberg, Germany
[email protected] E. Gomes de Azevedo Lisbon, Portugal
[email protected] httpt/ralfa.ist.utl.pu-ejsga SOLUTIONS T O PROBLEMS CHAPTER 2 1. From problem statement, we want to find (i>P I i>T)v . Using the product-rule, By definition, exp =.!.(av) v "iJT p and Kr =-.!.(av)r V i)p Then, ("idT)p)v = ap = 1.8x10-5 = 33.8 bar oc-1 Kr 532 x 10-6 Integrating the above equation and assuming ap and Kr constant over the temperature range, we obtain For l:!..T = 1 °C, we get llP = 33.8 bar 1 2 Chapter 2: Classical Thennodynamics of Phase Equilibria 2. Given the equation of state, we find: (as) dV (aP) v T = i)T nR = V - nb (au) ,,.(aP) -P= = av r -' lar v 0 (au) ap -(au) T av (av)apT = T 0 (an) aP t =-.,rav)+v -'lar =nb For an isothermal change, AS=JV2(i)p) dV=nRln V2-nb vt v sr Vj +nb R =-nRln- 1 P2 t.A =AU-TM= -nRTin( �) Solutions Manual 3 3. This entropy calculation corresponds to a series of steps as follows: S3(wpcr, T=298.15K P= 1 ter) S1 (saturated liq. S2 (saturated wpor, T=298.15K T=298.15K P = 0.03168 ter) P= 0.03168 ter) .dvaph _ (2436)x(18.015) _ 1 - 47 . 19 JK-1 mo1-1 A- _ A _ 1�2 - Ll.vaps- Ll..)" - T 298.15 Because v = RT (ideal gas), p &2--,3 =-Rln( � J -{8.31451)xln( ) = 1.0 0.03168 =-28.70 JK-1 moi-1 = 147.19-28.70+69.96 = 188.45 JIC1 mol-l Documents Similar To Solutions Manual Molecular Thermodynamics of Fluid-Phase Equilibria 3rd Edition John M. Prausnitz, Rudiger N. 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