solutions manual for Molecular Thermodynamics of Fluid Phase Equilibria, Third Edition by Prausnitz

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 (P / T ) . v
Using the product-rule,

jjHP7j j = −jjHv7jj jj P7j
T v T H vj
P T

By definition,
(1) v j 7
P = j j
v H T j P

and
j 7
1 v
T = − j j
v H P j T

Then,


jjHPTj7j = 
 −5
P
= 1.8  10 = 33.8 bar ∘C-1
−6
v T 5.32  10

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, jV 7
Pj
H n − bjj = RT
we find:
jj S 7j = jj P7j = nR
H V j H T j V − nb
T V

jj S 7j = −jj V 7j = − nR
H P j H T j
T
P P

jj U 7j = Tjj P7j − P = 0
H V j H T j T V

j U7 j U7 j V 7
Hj P j j = Hj V jj Hj P j j = 0
T T T

jj H7j = −T
jj V 7j +V = nb
H P j T
H T j
For an isothermal change,

S =
z V2 jj P j7
V1 H T j V
dV = nR ln
V2 − nb

V1 − nb

P1
= −nR ln
P2


U=
z
rj −U 7 jj V 7j y dP = 0
P1
P2

LjHj V jj H P j Qj T T



H =
zr j V 7 y
P2


Lj−T Hj T jj +QV j dP = nb(P − P )
P1 P
j P 7j
2 1


G = H − TS = nb(P − P )− nRT ln j 1


HP j 2 1
2


A = U − TS = −nRT ln j
j P 7j 1
HP j 2