Solutions Manual Molecular Thermodynamics of Fluid-Phase Equilibria 3rd Edition John M. Prausnitz, Rudiger N. Lichtenthaler, Edmundo Gomes de Azevedo

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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 Equilibria (Prentice-Hall, 1999). This manual was prepared according to the corrections presented on the third ( and subsequent) 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 being 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,

= 1.8x10-5 = 33.8 bar oc-1 ("idT)p)v = ap 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

2.

Chapter 2: Classical Thennodynamics of Phase Equilibria

Given the equation of state,

we find:

(aP) v d (as) V

T

= i)T

nR = V - nb

,,.(aP) -P= av r -' lar v (au) = ap -(au) av (av) ap (au) =-.,rav)+v aP -'lar (an) =

T

0

T

0

T

=nb

t

For an isothermal change, AS=JV2(i)p) dV=nRln V2-nb v Vj +nb vt

sr

R

=-nRln- 1

P2

t.A =AU-TM=

-nRTin( �)

3

Solutions Manual

3.

This entropy calculation corresponds to a series of steps as follows: S3(wpcr, T=298.15K P= 1 ter)

S1

S2 (saturated wpor,

(saturated liq. T=298.15K P = 0.03168 ter)

A-

_ A

T=298.15K P= 0.03168 ter)

_

1�2 - Ll.vaps-

Ll..)"

.dvaph _ (2436)x(18.015) _ 1 - 47 . 19 JK-1 mo1-1 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

)