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

All 12 Chapters Covered
eg eg eg




SOLUTION MANUAL
eg

, Ce g e g Oe g e g Ne g e g Te g e g Ee g e g Ne g e g
Te g e g S




Preface i

Solutionseg toeg Problemseg Chaptereg 2 1

Solutionseg toeg Problemseg Chaptereg 3 17

Solutionseg toeg Problemseg Chaptereg 4 29

Solutionseg toeg Problemseg Chaptereg 5 49

Solutionseg toeg Problemseg Chaptereg 6 81

Solutionseg toeg Problemseg Chaptereg 7 107

Solutionseg toeg Problemseg Chaptereg 8 121

Solutionseg toeg Problemseg Chaptereg 9 133

Solutionseg toeg Problemseg Chaptereg 10 153

Solutionseg toeg Problemseg Chaptereg 11 165

Solutionseg toeg Problemseg Chaptereg 12 177

, Se g e g Oe g e g Le g e g Ue g e g Te g e g Ie g e g Oe g e g Ne g e g S Te g e g O
Pe g e g Re g e g Oe g e g Be g e g Le g e g Ee g e g Me g e g S

Ce g e g He g e g Ae g e g Pe g e g Te g e g Ee g e g R

2




1. Frome g problemegstatement,egweegwantegtoegfinde g Peg/eg (
)
Teg .egUsingegtheegproduct-rule,
veg

jjHP 7jj
eg eg
= −
jjv 7j jjP 7j
eg eg


T
eg eg
e g eg

v H T j H vj
eg
eg eg e g
eg


P
eg
eg
T

Bye g definition,
j 7
(1) egveg
egP = j
eg
j
v H T j eg eg
P

and
1eg eg v j 7
T =eg−eg eg
veg egPeg T Hj jj
Then,

jjP 7j =
e
g eg eg
egPeg 1.8egeg10−5
=eg33.8egbare g ∘C-1
H T j
eg

eg eg
v =
eg 5.32egeg10−
6
T

Integratinge g thee g abovee g equatione g ande g assuminge g Pe g ande g Te g constante g overe g thee g tempe
raturee g range,egweegobtain
eg
Peg =e g PegegT
T

Fore g Te g =e g 1C,e g wee
g get
Peg=eg33.8egbar




1

, 2 Solutionse g Manu
al




2. Givene g thee g equatione g ofe g st
ate,
jV 7 eg



Pj
H n − bjj = RTeg e g
eg
eg
eg




wee g find:
jj S 7j = jjP 7j = nR
eg eg eg e g e g


H V j H T j V − nb
e g eg e
g eg eg

eg eg eg eg eg eg
T V

jjS 7j = −jjV 7j = − nR
eg eg


H P j H T j P
eg e g eg eg eg eg eg

eg eg eg eg
T P

jjU 7j = T jjP 7j − P = 0
eg eg


H V j H T j
eg e g eg eg eg e g eg eg eg

eg eg eg eg
T V

jjU 7j = jjU 7j jjV 7j = 0
eg eg eg

H P j H V j H P j
eg
eg eg
e g eg


T
eg
eg
eg eg
eg eg
T
eg eg
T
e g eg




jj H =eg−T
jjV 7j+ V eg
=egnb
7 H H T
ge g
e
eg eg eg e g
eg
eg e


Pj j
eg eg
T
j
eg
eg
Fore g ane g isothermale g chan g

ge,

Seg=eg
z V2eg


Ve 1g
jjP j7
H T j
eg
eg


eg eg
V
V −egnb
dVe g =egnReglneg 2e g
V1eg−egnb

egP
=e g −nReglneg 1
P2



jHj 7 V
 Ueg
=
z
rj −U jV 7 y P2eg



jH P j Qj dP =
eg
eg


e
eg eg eg




Ljj P1e g
0 T
eg eg
Te g eg
eg eg




z
g


r jV 7 j y
P2eg

jL−T HjT jj +QV jdP = nb(P −
e
g eg
 H eg eg eg eg eg eg eg eg 2eg eg
= P 1e g Peg
j P 7j
eg


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


HP j 2 1
eg 2eg




A = U − TS = −nRT ln j
j eg
eg eg
HP eg eg eg eg eg
eg e g

P 7
j 1eg j
2