Solutions Manual for Molecular Thermodynamics of Fluid-Phase Equilibria 3rd Edition by Prausnitz | Complete Step-by-Step Solutions (All Chapters)

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All 12 Chapters Covered




SOLUTION MANUAL

, C O N T E N T S




Preface i

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

, S O L U T I O N S T O P R O B L E M S


C H A P T E R 2




1. From problem statement, we want to find bP / T g v
.
Using the product-rule,
FG PIJ FG vIJ FG PIJ
H T K H T K H v K
v
P T

By definition,
F I
1 v
P G J
v H T K P

and
F vI
1
T G
v H P K
J T

Then,
F P I P 1.8  10 5
GH T KJ T 5.32  10 6
33.8 bar DC-1
v

Integrating the above equation and assuming P and T constant over the temperature range,
we obtain
P
P T
T

For T = 1C, we get

P 33.8 bar




1

, 2 Solutions Manual




2. Given the equation of state, FV I
PG
H n bJK RT
we find:
FGS IJ FG PIJ nR
H V K H T K V nb
T V

FG S IJ FG VIJ nR
H P K H T K P
T P

FG U IJ TFG PIJ P 0
H V K H T K T V

F U I F UI F V I
H GP K J HGV KJ H GP K J 0
FG H IJ TFG V IJ V nb
T T T



H P K H T K T

For an isothermal change,

S
z V2 FG P IJ
H T K
V 1 V
dV nR ln
V2 nb

V1 nb

P1
nR ln
P2


 U
z
LF U I GF VJI O dP 0
P2

NMGH V JK H P K QP
z
P1 T T


L F V I O
P2

H
NMTGH T JK VPQ dP nbbP P g
P1
2 1

F P IJ
P

G H TS nbbP P g nRT ln G 1


HP K 2 1
2


A U TS nRT ln G
F P IJ 1
HP K 2