Solutions Manual to Accompany
Molecular Thermodynamics of FluidPhase Equilibria Third Edition
John M. Prausnitz Rüdiger N. Lichtenthaler Edmundo Gomes de Azevedo
Prentice Hall PTR Upper Saddle River, New Jersey 07458 http://www.phptr.com
2000 Prentice Hall PTR Prentice Hall, Inc. Upper Saddle River, NJ 07458
Electronic Typesetting: Edmundo Gomes de Azevedo
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Printed in the United States of America 10 9 8 7 6 5 4 3 2
ISBN 0130183881
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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
1.
b
From problem statement, we want to find wP / wT Using the productrule,
FG wP IJ H wT K
v
g
v
.
FG wv IJ FG wP IJ H wT K H w v K P
By definition,
FG IJ H K
1 wv v wT
DP and
NT
T
P
FG IJ H K
1 wv v wP
2
T
Then,
FG wP IJ H wT K
v
DP NT
. u 10 5 18 5.32 u 10 6
33.8 bar D C 1
Integrating the above equation and assuming DPand NT constant over the temperature range, we obtain 'P
DP 'T NT
'P
33.8 bar
For 'T = 1qC, we get
1
2
2.
Solutions Manual
Given the equation of state, P
FG V bIJ Hn K
RT
we find:
FG wS IJ H wV K FG wS IJ H wP K FG wU IJ H wV K FG wU IJ H wP K FG wH IJ H wP K
T
T
T
For an isothermal change,
z FGH V2
'S
V1
V
T
T
FG wP IJ nR H wT K V nb F wV I nR G J H wT K P F wP I TG J P 0 H wT K FG wU IJ FG wV IJ 0 H wV K H wP K F wV IJ V nb T G H wT K P
V
T
wP wT
IJ K
dV V
T
V nb nR ln 2 V1 nb
P nR ln 1 P2
z
LMFG wU IJ FG wV IJ OP dP 0 MNH wV K H wP K PQ LMT FG wV IJ V OP dP nbbP P g MN H wT K PQ FP I 'H T'S nbb P P g nRT lnG J HP K FP I 'A 'U T'S nRT lnG J HP K
'U
'H
z
P1
T
T
P2
P1
'G
P2
2
1
P
2
1
1
2
1 2
3
3.
Solutions Manual
This entropy calculation corresponds to a series of steps as follows: s3 (vapor, T = 298.15 K P = 1 bar)
s1 (saturated liq. T = 298.15 K P = 0.03168 bar)
's1o 2
' vap s
s2 (saturated vapor, T = 298.15 K P = 0.03168 bar)
s3
's1o 2 's2o 3 s1
' vap h
(2436) u (18.015) 298.15
T 's2o 3
Because v
z LMMN FGH P3
P2
IJ K
147.19 J K 1 mol 1
OP PQ
wv dP wT P
RT (ideal gas), P 's2 o 3
§P · R ln ¨ 3 ¸ © P2 ¹ § 1.0 · (8.31451) u ln ¨ ¸ © 0.03168 ¹ 28.70 J K 1 mol 1
s3
s0 (H2 O, vapor) 147.19 28.70 69.96 188.45 J K1 mol 1
4
4.
Solutions Manual
Because D
RT v, P
P
RT Dv
RT 2 3 / v2 v
or P
§ wP · ¨ wv ¸ © ¹T
RTv 2 2v 2 3 v 3
RT v (v 3 6) (2v 2 3 v 3 )2
2.3 L mol 1 , T = 373.15 K, R = 0.0831451 bar L K1 mol1, and molar mass is 100 g
As v mol1,
FG wP IJ H wv K w2
g c kv 2
FG H
1
FG wP IJ H wv K
kg m N
s2
3.3245 bar L1 mol
3.3245 u 10 8 Pa m3 mol
T
T
IJ u (1.4) u FG 1 K H 100 u 10
IJ F K GH
mol m3 u 2.3 u 10 3 3 kg mol
24,621 m2 s 2
w 157 m s 1
5.
Assume a threestep process:
I u FG 3.3245 u 10 JK H 2
8
N mol m2
m3
IJ K
5
Solutions Manual
Isothermal expansion to v = f (ideal gas state)
(1)
(2)
v= V
(2)
Isochoric (v is constant) cooling to T2
(3)
(1)
(T2 ,v2 )
Isothermal compression to v2
(3)
(T1 ,v1 ) T
For an isentropic process, 's
's1 's2 's3
Because s = s(v, T),
ds
FG ws IJ H wv K
dv T
FG ws IJ H wT K
dT v
or
FG wP IJ H wT K
ds
dv v
cv dT T
by using the relations
FG wS IJ FG wP IJ H wv K H wT K T
FG ws IJ H wT K
FG IJ H K
1 wu T wT
v
(Maxwell relation) v
v
cv T
then, 's
³
v f § wP ·
v1
v 2 § wP · T2 c 0 v dT d v ¨ wT ¸ ¨ ¸ dv T1 T v f © wT ¹v © ¹v
³
³
Using van der Waals’ equation of state,
P
RT a v b v2
FG wP IJ H wT K Thus,
v
R vb
6
Solutions Manual
T2 c 0 §v b· v dT R ln ¨ 2 ¸ v b T T © 1 ¹ 1
³
's
To simplify, assume
cv0
c 0p R RT2 P2
v2
Then, RT2 b P2 ln v1 b
§ c0p R · § T · ¨ ¸ ln ¨ 1 ¸ ¨ R ¸ © T2 ¹ © ¹
ª (82.0578) u (T2 ) 45 º ln « » 600 45 ¬ ¼
T2
(3.029) u ln
203 K
6.
P
Because
b2 v2
RT a v b v2
623.15 T2
RT v
LM O a P 1 MM b RTv PP MN1 v PQ
1,
FG1 b IJ H vK
1
1
b b2 " v v2
Thus,
P
LM FG MN H
IJ K
OP PQ
RT a 1 b2 " 1 b RT v v 2 v
or
Pv RT
FG H
1 b
IJ K
a 1 b2 " RT v v 2
7
Solutions Manual
Because
Pv RT
z
1
B C " v v2
the second virial coefficient for van der Waals equation is given by b
B
7.
a RT
Starting with
FG wu IJ H wP K
du Tds Pdv T T
FG ws IJ H wP K
P T
FG wv IJ H wT K
T
FG wv IJ H wP K
P P
T
FG wv IJ H wP K
As v
RT B P
RT a b P T2
FG wv IJ H wT K FG wv IJ H wT K
R 2a P T3
P
Then,
FG wu IJ H wP K 'u
T
T S
RT P2
2a T2
§ 2a ·
³0 ¨© W2 ¸¹ dP 'u
2a S W2
T
8
8.
Solutions Manual
The equation
FG P n IJ bv mg H v2T 1/2 K
RT
can be rewritten as
(PT 1/ 2 )v 3 (PmT 1/ 2 )v 2 nv nm
FG H
v3 m
FG H
IJ K
RT 3/2v 2
IJ K
RT 2 n nm v v P PT 1/ 2 PT 1/ 2
(1)
At the critical point, there are three equal roots for v = vc, or, equivalently,
FG wP IJ H wv K
bv v g
3
c
F w PI GH wv JK 2
2
T Tc
T Tc
v 3 3v c v 2 3v c 2v v c 3
(2)
Comparing Eqs. (1) and (2) at the critical point, m
RTc Pc
3v c
(3)
n
3v 2c
(4)
v c3
(5)
PcTc1/ 2 nm PcTc1/ 2
From Eqs. (3), (4), and (5) we obtain m
vc
n
3RTc 8Pc
3v 2c PcTc1/ 2
The equation of state may be rewritten:
vc 3 or
(6)
m
RTc 8Pc
27 R 2Tc5/ 2 64 Pc
9
Solutions Manual
P
RT v
z
Pv RT
F I GG 1 n JJ GH 1 mv RT 3/2v JK
or 1 m 1 v
n RT 3/2 v
From critical data, 0.0428 L mol 1
m
63.78 bar (L mol 1 )2 K1/2
n
At 100qC and at v = (6.948)u(44)/1000 = 0.3057 L mol1, z = 0.815
This value of z gives P = 82.7 bar. Tables of Din for carbon dioxide at 100qC and v = 6.948 cm3 g1, give P = 81.1 bar or z = 0.799.
9.
We want to find the molar internal energy u(T , v) based on a reference state chosen so that u(T0 , v o f)
Then, u(T , v)
u(T , v) u(T0 , v o f) u(T , v) u(T , v o f) u(T , v o f) u(T0 , v o f )
lim
z
v
vof v f
FG wu IJ H wv K
z FGH T
IJ K
wu dT T v v w vof T0 f
dv lim T
Schematically we have: v =v
1
2
3
Ideal gas Ref. state Ideal gas (T0 , v
)
Intermediate state Ideal gas (T, v )
State of interest Real gas (T, v)
(1)
10
Solutions Manual
In Eq. (1) we are taking 1 mol of gas from the reference state 1 to the state of interest 3 through an intermediate state 2, characterized by temperature T and volume v o f, in a twostep process consisting of an isochoric step and an isothermal step. In the step 1 o 2 the gas is infinitely rarified, and hence exhibits ideal gas behavior. Then, the second integral in Eq. (1) gives: T
§ wu · dT ¨ ¸ v of T0 © wv ¹ v vf
³
lim
because for an ideal gas c 0p cv0
³
T
T0
cv0 dT
³
T
T0
(c0p R) dT
(c0p R)(T T0 )
(2)
R and because, by the problem statement, the heat capacity at
constant pressure of the gas is temperature independent. We have now to calculate the first integral in Eq. (1). To make this calculation, we first transform the derivative involved in the integral to one expressed in terms of volumetric properties. By the fundamental equation for internal energy (see Table 21 of the text),
FG wu IJ T FG wP IJ P H wv K T H wv K T
(3)
Making the derivative using the equation of state give we obtain
FG wu IJ H wv K T
RT RT a v b v b v(v b)
a v(v b)
(4)
Then, v
§ wu · ¨ ¸ dv v f © wv ¹
³
a v ª 1 1º dv b v f ¬« v b v ¼»
³
a ª § vb · v º « ln ¨ » ¸ ln b ¬ © vf b ¹ vf ¼
(5) a § v b vf · ln ¨ ¸ b © v vf b ¹
and
z FGH v
lim
vof v f
wu wv
IJ dv KT
FG H
vb a ln v b
IJ K
(6)
Combining Eqs. (1), (2) and (6) we obtain the desired expression for the molar internal energy, u(T , v)
FG H
a vb (c 0p R)(T T0 ) ln b v
IJ K
11
Solutions Manual
10. ln J w
A(1 xw )2
J w o 1 as xw o 1
such that
Using GibbsDuhem equation, xw d ln J w xs d ln J s
or, because dx w
dx s ( xw xs
1) , xw
d ln J w dx w
d ln J w dx w
xs
d ln J s dx s
2 A(1 x w )(1)
2 A(1 xw )
Then, 2 Axs (1 xs ) dxs xs
d ln J s
ln J s
³0
d ln J s
2 A
³0
(1 xs ) dxs
§ x2 · 2 A ¨ x s s ¸ ¨ 2 ¸¹ ©
ln J s
ln J s
11.
xs
2 A(1 xs ) dxs
A( x w2 1)
Henry’s law for component 1, at constant temperature, is f1
(for 0 x1 a)
k1 x1
where k1 is Henry’s constant. For a liquid phase in equilibrium with its vapor, fi L gas law, fiV
yi P.
Henry’s law can then be written: y1P
k1 x1
Taking logarithms this becomes ln( y1P)
ln k1 ln x1
Differentiation at constant temperature gives
fiV . If the vapor phase obeys ideal
12
Solutions Manual
d ln( y1P) dx1
d ln x1 dx1
1 x1
Using the GibbsDuhem equation x1
d ln P1 d ln P2 x2 dx1 dx1
gives 1 x2
or, because dx 2
d ln( y2 P) dx1
dx1 , x2
d ln( y2 P) dx2
1
or, d ln(y2 P)
d ln x2
Integration gives ln( y2 P)
ln x2 ln C
where lnC is the constant of integration. For x2
1 , y2
1 , and P
P2s . This gives C
P2s and we may write
ln x2 ln P2s
ln( y2 P)
ln( x2 P2s )
or y2 P
P2
x 2 P2s
[for (1 a) x 2 1 ]
which is Raoult’s law for component 2.
12.
Starting from dgi
RTd ln fi ,
'gi
P* o P
f (at P) RT ln i fi (at P* )
(P* is a low pressure where gas i is ideal)
From the Steam Tables we obtain 'h and 's at T and P to calculate 'g from
'g
'h T 's
13
Solutions Manual
Choose P * = 1 bar.
'h 's
196 J g 1
h70 bar h1 bar
s70 bar s1 bar
2.215 J g1 D C 1
Then, at 320qC, 'g
1117.8 J g 1
20137 J mol 1
Thus,
b
ln fi 70 bar, 320qC
g
b
20137 8.31451 u 59315 .
g b
g
4.08
or f = 59.1 bar
13.
The virial equation for a van der Waals gas can be written (as shown in Problem 6) RT a b P RT
v
(1)
At the Boyle temperature, B
b
a RT
b
a RT
or
The Boyle temperature then, is given by
TB
a bR
The JouleThomson coefficient is
P or
FG wT IJ H wP K H
(2)
14
Solutions Manual
P
FG wT IJ FG wT IJ H wH K H wP K FG wP IJ H wH K
P
FG wH IJ H wP K
T
cp
H
T
Because
FG wH IJ H wP K T and because cp is never zero, when P
v T
0,
FG wH IJ H wP K T
FG wv IJ H wT K P 0.
Substitution in Eq. (1) gives
§ wH · ¨ wP ¸ © ¹T
RT a § RT a · b ¨ P RT © P RT ¸¹
b
2a RT
The inversion temperature is TJT
2a Rb
TJT
2TB
f1G
f1L
Comparison with Eq. (2) gives
14.
At equilibrium,
where subscript 1 stands for the solute. At constant pressure, a change in temperature may be represented by
F d ln f I GH dT JK G 1
dT P
F d ln f I GH dT JK L 1
dT
(1)
P
Since the solvent is nonvolatile, f1G (at constant pressure) depends only on T (gas composition does not change.) However, f 1L (at constant pressure) depends on T and x1 (or ln x1 ):
15
Solutions Manual
F d ln f1 I GH dT JK L
F w ln f1 I GH wT JK L
dT P
F w ln f1 I GH w ln x1 JK L
dT P, x
d ln x1
(2)
T ,P
Further,
F d ln f GH dT F d ln f GH wT
I JK I JK
G 1
L 1
h10 h1G RT 2
P
h10 h1L RT 2
P, x
(3)
(4)
where: h10 = idealgas enthalpy of 1;
h1G = realgas enthalpy of 1; h1L = partial molar enthalpy of 1 in the liquid phase.
Assuming Henry’s law, f1 x1
constant
or
F w ln f1 I GH w ln x1 JK L
1
(5)
T ,P
Substituting Eqs. (2), (3), (4), and (5) into Eq. (1), we obtain d ln x1 d (1 / T )
'h1 R
From physical reasoning we expect h1G ! h1L . Therefore x1 falls with rising temperature. This is true for most cases but not always.
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
1.
3
The Gibbs energy of a mixture can be related to the partial molar Gibbs energies by
¦ yi e gi gio j m
g go
(1)
i 1
Since, at constant temperature, dg
RTd ln f , we may integrate to obtain o
g go
RT ln f mixt RT ln f mixt
or g go
RT ln f mixt RT ln P
(2)
where subscript mixt stands for mixture. For a component in a solution, d gi RTd ln fi . Integration gives gi gio
gi gio
RT ln fio
(3)
RT ln( yi P)
Substituting Eqs. (2) and (3) into Eq. (1) gives m
ln f mixt ln P
m
¦ yi ln fi ¦ yi ln( yi P) i 1
ln f mixt ln P
i 1
m
FfI
m
i 1
i
i 1
¦ yi lnGH yi JK ¦ yi ln P
(4)
Because
17
18
Solutions Manual
m
¦ yi ln P
ln P
i 1
Fm yI iJ GH ¦ i 1 K
ln P
Eq. (4) becomes ln f mixt
Assuming the Lewis fule, fi
m
FfI
i 1
i
¦ yi lnGH yi JK
yi f pure i , Eq. (5) becomes m
¦ yi ln fpure i
ln f mixt
i 1
or m
yi fpure i
f mixt
i=1
2.
As shown in Problem 1, ln f mixt
m
FfI
i 1
i
¦ yi lnGH yi JK
This result is rigorous. It does not assume the Lewis fugacity rule. Using fugacity coefficients, fi
M i yi P
and ln f mixt fmixt
y A ln M A yB ln M B ln P y
y
M AA M BB P (0.65)0.25 u (0.90)0.75 u (50)
fmixt
41.5 bar
(5)
19
Solutions Manual
3.
Purecomponent saturation pressures show that water is relatively nonvolatile at 25qC. Under these conditions the mole fraction of ethane in the vapor phase (yE) is close to unity. Henry’s law applies: fE
H (T ) x E
The equilibrium condition is f EV
f EL
or yE M E P
At 1 bar, M E  1 and H (T )
H (T ) x E
P / xE :
H (T )
1
3.03 u 10 4 bar
0.33 u 10 4
At 35 bar we must calculate ME:
z
P
z 1 dP 0 P
ln M E
Using z
1 7.63 u 10 3 P 7.22 u 10 5 P 2
we obtain ME
0.733
Because Henry’s constant H is not a strong function of pressure, xE
xE
4.
x Ethane
MEP H
fE H
(0.733) u (35) 3.03 u 10 4
8.47 u 10 4
The change in chemical potential can be written, 'P1
P1 P10
§ RT ln ¨ ¨ ©
f1 · ¸ f10 ¸¹
( f10
1 bar )
(1)
20
Solutions Manual
1
The chemical potential may be defined as : § wG0 § wG · ¨ ¨ ¸ ¨ w n1 © w n1 ¹T ,P,n2 ©
P1 P10
· ¸ ¸ ¹T ,n2
(2) § w A0 § wA · ¨ ¨ ¸ ¨ w n1 © w n1 ¹T ,V ,n2 ©
· RT ¸ ¸ ¹T ,n2
Combining Eqs. (1) and (2): 1 § w'A · 1 ¨ ¸ RT © w n1 ¹T ,V ,n 2
ln f1
Using total volume, V 'A RT
nT v , nT nT ln
n1 n2 ,
FG V IJ n1 lnFG V IJ n2 lnFG V IJ H V n b K H n1RT K H n2 RT K T
Taking the partial derivative and substituting gives ln
f1 y1 RT vb
or
b1 vb
FG H
y1RT b1 exp vb vb
f1
IJ K
The same expression for the fugacity can be obtained with an alternative (but equivalent) derivation:
1
A0
U 0 TS 0 ;
P10
§ wG0 ¨ ¨ w n1 ©
G0
U 0 PV 0 TS 0 ;
§ wU0 ¨ ¨ w n1 ©
· ¸ ¸ ¹T ,n j z1
PV 0
nT RT
· § w S0 RT T ¨ ¸ ¸ ¨ w n1 ¹T ,n j z1 ©
· ¸ ¸ ¹T ,n j z1
and § w A0 ¨ ¨ w n1 ©
· ¸ ¸ ¹T ,n j z1
§ wU0 ¨ ¨ w n1 ©
· § w S0 RT T ¨ ¸ ¸ ¨ w n1 ¹T ,n j z1 ©
then P10
§ w A0 ¨ ¨ w n1 ©
· RT ¸ ¸ ¹T ,n j z1
· ¸ ¸ ¹T ,n j z1
21
Solutions Manual
§ f · RT ln ¨ 1 ¸ ¨ f0 ¸ © 1 ¹
P1 P10
By definition, P1
FG w A IJ H wn K
( f10
1 bar)
. The Helmholtz energy change 'a can be written as
1 T ,V ,n j z1
nT 'a
A
¦ ni ai0 i
A
¦ ni (P10 RT ) i
Then, § wnT 'a · ¨ ¸ © wn1 ¹n j z1,T ,V
§ wA · P10 RT ¨ ¸ n w © 1 ¹n j z1,T ,V P1 P10 RT
and P1 P10 RT
ª w(nT 'a / RT ) º 1 « » wn1 ¬ ¼ n j z1,T ,V
Using the equation for 'a , ª w(nT 'a / RT ) º « » wn1 ¬ ¼ n j z1,T ,V
ln
n b V V ln 1 T 1 V nT b V nT b n1RT
or ln f1
§ P P0 ln ¨ 1 1 ¨ © RT
ln
· ¸ ¸ ¹
ª w(nT 'a / RT ) º 1 « » wn1 ¬ ¼ n j z1,T ,V
n1 RT n b T 1 V nT b V nT b § n b · n1RT exp ¨ T 1 ¸ V nT b © V nT b ¹
f1
Hence, f1
y1RT § b · exp ¨ 1 ¸ vb ©vb¹
22
Solutions Manual
5. a)
Starting with Eq. (351): For a pure component (ni
Pi
G ; ni
Pi0
nT ) :
Gi0 ni
Because
G U PV TS Pi0
Pi0 Tsi0 RT
(1)
P RT · V PV Pi0 Tsi0 ¨ ¸ dV RT ln n V n RT ni V © i i ¹
(2)
From Eq. (352), Pi
³
f§
From Eqs. (1) and (2), Pi Pi0
f§
PV ni RT P RT · V ¨ ¸ dV RT ln V ¹ ni RT ni V © ni
³
But, RT ln fi
Pi Pi0
and V ni RT
zi P
Substitution gives § f· RT ln ¨ ¸ © P ¹i
b)
Starting with Eq. (353): For a pure component, yi
f§
P RT · ¨ ¸ dV RT ln zi RT ( zi 1) V ¹ V © ni
³
1. To use Eq. (353), we must calculate
FG wP IJ H wni K T ,V ,pure component i
Pressure P is a function of T, V, and ni and
23
Solutions Manual
§ wP · § wni ¨ ¸ ¨ © wni ¹V © wV
· § wV · ¸ ¨ wP ¸ ¹ni ¹P ©
1
FG wP IJ FG wV IJ FG wP IJ H wni K V H wni K P H wV K ni FG IJ OP H K PQ
LM MN
P P V wP ni ni ni wV n i
But,
FG IJ H K
P V wP ni ni wV n i
LM N
OP Q
1 w (PV ) ni wV n i
Then, f§
³
f
P 1 ni RT dV d ( PV ) ni PV V ni
wP · ¨ ¸ w V © ni ¹T ,V ,n j z1
³
³
f
P PV dV RT ni V ni
³
f
P dV RT ( z 1) n V i
³ Now Eq. (354) follows directly.
6.
The solubility of water in oil is described by f1
H (T ) x1
Henry’s constant can be evaluated at 1 bar where f1 Then, H (T )
f1 x1
1 35 u 10 4
286 bar
1 bar .
(t
140 D C)
To obtain f1 at 410 bar and 140qC, use the Steam Tables (e.g., Keenen and Keyes). Alternatively, get f at saturation (3.615 bar) and use the Poynting factor to correct to 410 bar. At 140qC,
24
Solutions Manual
RT ln f1
'g1o410 bar
'h1o410 bar T 's1o410 bar
282 J g 1
RT ln f1
ln f1
(from Steam Tables)
(282) u (18) (8.31451) u (413)
1.48
Then, f1 = 4.4 bar at 410 bar and 140qC and x1
f1 H (T )
4.4 286
0.0154
7. To = 300 K
Tf = 300 K
Po = ?
Pf = 1 bar
Applying an energy balance to this process, 'h
hf ho
This may be analyzed on a PT plot. Assume 1 mole of gas passing through the valve. P
To ,Po
Tf ,Pf
I
III II T
A threestep process applies: I. Isothermal expansion to the idealgas state. II. Isobaric cooling of the ideal gas. III. Compression to the final pressure. For this process,
25
Solutions Manual
'h
'hI 'hII 'hIII
Since h = h(T,P),
FG wh IJ d P FG wh IJ dT H w P K T H wT K P
dh
'h
T f Pf
³
dh
To Po
ª § wv · º «v T ¨ ¸ » dP Po ¬« © wT ¹ P ¼»T
³
o
³
Tf
To
c0p d T
³
Pf
ª § wv · º «v T ¨ ¸ » dP © wT ¹ P ¼»T ¬« f
But, RT 10 5 50 P T
v Then,
FG wv IJ H wT K P
R 10 5 P T2
and c 0p
'h
F 50 2 u 10 GH T o
5
yA c 0p,A yBc 0p,B
I b P g y e JK
0 A c p,A
o
jb
g FGH
yBc 0p,B Tf To 50
I JK
2 u 10 5 Pf Tf
Substitution gives (616.7) u Po
(33.5 J mol 1 K 1 ) u (10 bar cm 3 J 1 ) u (200 300 K) (950) u (1 bar )
Po = 55.9 bar
8.
From the GibbsHelmholtz equation:
w
FG g IJ HTK wT
h T2
26
Solutions Manual
or, alternately,
FG 'g IJ HTK F 1I wG J HTK
w
'h
h real h ideal
(1)
Because, at constant temperature,
b g
d 'g
RTd ln
FG f IJ H PK
(2)
we may substitute Eq. (2) into Eq. (1) to obtain
FG f IJ H PK F 1I wG J HTK
w ln R
'h
From the empirical relation given,
ln
f P
0.067P
30.7 0.416 P 2 P 0.0012P 2 T T
FG f IJ H PK F 1I wG J HTK
w ln R
30.7P 0.416 P 2
'h R
At P = 30 bar,
'h = (8.31451)u[(30.7)u(30) + (0.416)u(30)2] 'h = 4545 J mol1
9.
Consider mixing as a threestep process: (I) Expand isothermally to idealgas state. (II) Mix ideal gas. (III) Compress mixture isothermally. Starting with
du
LM FG wP IJ POP dv MN H wT K v PQ
cv dT T
27
Solutions Manual
FG wu IJ H wv K T Because, P
FG w P IJ H wT K
T
P v
RT a , v b v2
FG wu IJ H wv K T
a v2
Integration of this equation to the idealgas state (v 'u
z
f
v
a
v
2
dv
f) gives
a v
Therefore, 'uI
x1a1 x 2 a2 v1 v2
5914 J mol 1
[(1 bar)u(1 cm3)  0.1 Joule]
'uII
'uIII
(because is the mixing of ideal gases)
amixt v mixt
[ x12 a11 2 x1x2 a11a22 u (1 0.1) x22 a22 ] x1v1 x2 v 2
5550 J mol 1
' mix h
' mix u
'uI 'uII 'uIII
364 J mol 1
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
1.
4
At 25 Å we can neglect repulsive forces. The attractive forces are London forces and induced dipolar forces; we neglect (small) quadrupolar forces. (There are no dipoledipole forces since N2 is nonpolar.) Let 1 stand for N2 and 2 stand for NH3. London force: *12
F12
d* dr
3 D 1D 2 I1I 2 2 r 6 I1 I 2
9D 1D 2 I1I 2 r 7 I1 I 2
Since, D1
17.6 u 10 25 cm3
D2
22.6 u 10 25 cm3
I1 = 15.5 eV = 2.48 u 10 18 N m . u 10 18 N m I 2 = 11.5 eV = 184
then London F12
62.0 u 10 18 N
Induced dipole force:
29
30
Solutions Manual
*12
F12
D1P 22 r6
6D1P 22 r7
1 D 1 u 10 18 (erg cm 3 )1/ 2 1.47 D 1.47 u 10 18 (erg cm 3 )1/ 2
P2
ind F12
3.8 u 10 18 (erg cm 3 )1/ 2
Neglecting all forces due to quadrupoles (and higher poles),
2.
F tot
F London F ind
F tot
65.8 u 10 18 N
From the LennardJones model: V6
Attractive potential = 4H
Attractive force =
d* dr
r6
24 H
* V6 r7
Assume force of form, Force =
d* dr
FG H IJ V H kK r
(constant )
6
7
Using corresponding states:
FG H JI H kK
(constant ) u Tc D Tc
where D and E are universal constants.
V6
(constant u v c ) 2
E v c2
31
Solutions Manual
(V CH 4 ) 6 (rCH 4 ) 7
(constant ) (H / k ) CH 4 (constant ) (H / k ) B
force CH 4 force substance B
(V B ) 6 (rB ) 7
E (v cCH )2 D(TcCH )
8
2 u 10 force substance B
4
D (TcB )
4
u
(1u 10
7
cm)7
E(v cB ) 2 (2 u 10 7 cm)7
Force substance B = 4 u 10 10 dyne Force = 4 u 10 15 N
3. 8 u 10 16 erg
*AA
By the molecular theory of corresponding states: *ii Hi
*BB *AA Since H / k
FG H HH
B A
IJ ML f b2g PO K MN f b2g PQ
§ r · f¨ ¸ © Vi ¹
HB HA
(r
2V)
( f is a universal function)
0.77Tc (taking the generalized function f as the LennardJones (126) poten
tial), * BB
* AA
TcB TcA
(8 u 10 16 erg) u (180 K/ 120 K) *BB
12.0 u 10 16 erg 12.0 u 10 23 J
32
Solutions Manual
4.
For dipoledipole interaction: *(dd )
PiP j (4 SH or 3 )
f (T i , T j , I)
with
f (Ti , T j , I)
2 cos Ti cos T j sin Ti sin T j cos(Ii I j )
For the relative orientation: j
i
i Ti Tj
j
Ti = 0º
Ii = 0º
Tj = 180º
Ij = 0º
Ij
PiP j
*(dd )
(4 SH o
r 3)
u [2 u 1 u (1)]
For Pi = Pj = 1.08 D = 3.603u1030 C m and r = 0.5u109 m, ( 2) u (3.603 u 10
* ( dd )
(4 S) u (8.8542 u 10
1.87 × 10
21
12
30 2
)
) u (0.5 u 10
9 3
)
J
For the relative orientation: j
i
Ti = 0º
Tj = 90º
Ii = 0º
Ij = 0º
For the dipoleinduced dipole interaction: *(ddi )
P i2D j 2(4 SH o ) 2 r 6
For the relative orientation:
(3 cos 2 T i 1)
P 2j D i 2(4 SH o ) 2 r 6
(3 cos 2 T j 1)
2P i P j (4 SH or 3 )
33
Solutions Manual
j
i
*(ddi )
Ti = 0º
Tj = 90º
ª (3.603 u 1030 )2 u (2.60 u 1030 ) º u (4) » 2u « 10 9 6 ¬« 2 u (1.1124 u 10 ) u (0.5 u 10 ) ¼» 7.77 u 1023 J  8.0 u 1023 J
For the relative orientation: j
i
Ti = 0º
*(ddi )
Tj = 90º
(3.603 u 1030 )2 u (2.60 u 1030 )
u (4 1)
2 u (1.1124 u 1010 ) u (0.5 u109 )6
4.85 u 1023 J  4.8 u 1023 J
5.
The energy required to remove the molecule from the solution is 1
E
Hr
3.5
a3 a
FG Hr 1 IJ P 2 H 2Hr 1K
3.0 u 108 cm
P
2D
(See, for example, C. J. E. Böttcher, 1952, The Theory of Electric Polarization, Elsevier) E = 4.61u1021 J/molecule = 2777 J mol1
6. a) The critical temperatures and critical volumes of N2 and CO are very similar, more similar than those for N2 and argon (see Table J4 of App. J). Therefore, we expect N2/CO mixtures to follow Amagat’s law more closely than N2/Ar mixtures.
34
Solutions Manual
b) Using a harmonic oscillator model for CO, F = Kx, where F is the force, x is the displacement (vibration) of nuclei from equilibrium position and K is the force constant. This constant may be measured by relating it to characteristic frequency X through: X
1 2S
b
K mC mO mC m O
g
where mC and mO are, respectively, the masses of carbon and oxygen atoms. Infrared spectrum will show strong absorption at X. Argon has only translational degrees of freedom while CO has, in addition, rotational and vibrational degrees of freedom. Therefore, the specific heat of CO is larger than that of argon.
7.
Electron affinity is the energy released when an electron is added to a neutral atom (or molecule). Ionization potential is the energy required to remove an electron from a neutral atom (or molecule). Lewis acid = electron acceptor (high electron affinity). Lewis base = electron donor (low ionization potential). Aromatics are better Lewis bases than paraffin. To extract aromatics from paraffins we want a good Lewis acid. SO2 is a better a Lewis acid than ammonia.
8.
From Debye’s equation: ª H 1º v« r » ¬ Hr 2 ¼
Total polarization
4 S N AD 3
§ P2 · 4 SNA ¨ ¸ ¨ ¸ 3 © 3kT ¹
Static polarization independent of T
Measurement of molar volume, v , and relative permitivity, H r , in a dilute solution as a function of T, allows P to be determined (plot total polarization versus 1/T; slope gives P).
35
Solutions Manual
9.
We compare the attractive part of the LJ potential (r >> V) with the London formula. The attractive LJ potential is
FG V11 IJ 6 HrK F V I6 4H 22 G 22 J HrK F V I6 4H12 G 12 J HrK
*11
4 H11
*22
*12
We assume that V12
1/ 2(V11 V22 ) . The London formula is
*11
*22
*12
FG H
3D12 I1 4r 6 3D 22 I 2 4r 6
3 D 1D 2 2 r6
IJ FG I1I2 JI K H I1 I 2 K
Substitution gives
H12
Only when V11
1/ 2
(H11H22 )
V 22 and I1
ª º V11V22 » « «1 V V » 22 ¼» ¬« 2 11
6
ª º I1I2 » « «1 I I » ¬« 2 1 2 ¼»
I 2 do we obtain H12
(H11H22 )1/ 2
Notice that both correction factors (in brackets) are equal to or less than unity. Thus, in general, H12 d (H11H22 )1/ 2
10.
See Pimentel and McClellan, The Hydrogen Bond, Freeman (1960). Phenol has a higher boiling point and a higher energy of vaporization than other substituted benzenes such as toluene or chlorobenzene. Phenol is more soluble in water than other substituted benzenes. Distribution experiments show that phenol is strongly associated when dissolved in
36
Solutions Manual
nonpolar solvents like CCl4. Infrared spectra show absorption at a frequency corresponding to the –OH " H hydrogen bond.
11.
J acetone CCl ! J acetone CHCl because acetone can hydrogenbond with chloroform but not 4 3 with carbon tetrachloride.
12. a) CHCl3
Chloroform is the best solvent due to hydrogen bonding which is not present in pure chloroform or in the polyether (PPD). Chlorobenzene is the next best solvent due to its high polarizability and it is a Lewis acid while PPD is a Lewis base.
Cyclohexane is worst due to its low polarizability.
nbutanol is probably a poor solvent for PPD. Although it can hydrogenbond with PPD, this requires breaking the Hbonding network between nbutanol molecules. tbutanol is probably better. Steric hindrance prevents it from forming Hbonding networks; therefore, it readily exchanges one Hbond for another when mixed with PPD. The lower boiling point of tbutanol supports the view that it exhibits weaker hydrogen bonding with itself than does nbutanol.
b) Cellulose nitrate (nitrocellulose) has two polar groups: ONO2 and OH. For maximum solubility, we want one solvent that can “hook up” with the ONO2 group (e.g., an aromatic hydrocarbon) and another one for the OH group (e.g., an alcohol or a ketone). c)
Using the result of Problem 5,
37
Solutions Manual
1
E
a3
FG Hr 1 IJ P 2 H 2Hr 1K
At 20qC, the dielectric constants are Hr (CCl 4 )
Hr (C8 H18 ) 1.948
2.238
Thus, H r (CCl 4 ) H r (C 8 H18 )
117 .
It takes more energy to evaporate HCN from CCl4 than from octane.
13.
At 170qC and 25 bar:
z H2 is above 1 H2
zamine is well below 1
1
z HCl is slightly below 1
z HCl
a)
1
yamine
A mixture of amine and H2 is expected to exhibit positive deviations from Amagat’s law.
b) Since amine and HCl can complex, mixtures will exhibit negative deviations from Amagat’s law. c) z argon
1 z z HCl
yargon
1
Solutions Manual
38
The strong dipoledipole attractive forces between HCl molecules cause z HCl 1 , while argon is nearly ideal. Addition of argon to HCl greatly reduces the attractive forces experienced by the HCl molecules, and the mixture rapidly approaches ideality with addition of argon. Addition of HCl to Ar causes induced dipole attractive forces to arise in argon, but these forces are much smaller than the dipoledipole forces lost upon addition of Ar to HCl. Thus the curve is convex upwards.
14. a) Acetylene has acidic hydrogen atoms while ethane does not. Acetylene can therefore complex with DMF, explaining its higher solubility. No complexing occurs with octane. b) At the lower pressure (3 bar), the gasphase is nearly ideal. There are few interactions between benzene and methane (or hydrogen). Therefore, benzene feels equally “comfortable” in both gases. However, at 40 bar there are many more interactions between benzene and methane (or hydrogen) in the gas phase. Now benzene does care about the nature of its surroundings. Because methane has a larger polarizability than hydrogen, benzene feels more “comfortable” with methane than with hydrogen. Therefore, KB (in methane) > KB (in hydrogen). c) Under the same conditions, CO2 experiences stronger attractive forces with methane than with hydrogen due to differences in polarizability. This means that CO2 is more “comfortable” in methane than in H2 and therefore has a lower fugacity that explains the condensation in H2 but not in CH4. d) It is appropriate to look at this from a correspondingstates viewpoint. At 100°C, for ethane TR  12 . , for helium, TR  80 . At lower values of TR (near unity) the molecules have an average thermal (kinetic) energy on the order of H (because TR is on the order of kT / H ). The colliding molecules (and molecules near one another, of which there will be many at 50 bar) can therefore be significantly affected by the attractive portion of the potential, leading to z < 1. At higher TR , the molecules have such high thermal energies that they are not significantly affected by the attractive part. The molecules look like hard spheres to one another, and only the repulsive part of the potential is important. This leads to z > 1. e) Chlorobenzene would probably be best although cyclohexane might be good too because both are polar and thus can interact favorably with the polar segment of poly(vinyl chloride). Ethanol is not good because it hydrogen bonds with itself and nheptane will be poor because it is nonpolar. f)
i) Dipole. ii) Octopole. iii) Quadrupole. iv) Octopole.
39
Solutions Manual
g) Lowering the temperature lowers the vapor pressure of heptane and that tends to lower solvent losses due to evaporation. However, at 0qC and at 600 psia, the gas phase is strongly nonideal, becoming more nonideal as temperature falls. As the temperature falls, the solubility of heptane in highpressure ethane and propane rises due to increased attraction between heptane and ethane on propane. In this case, the effect of increased gasphase nonideality is more important than the effect of decreased vapor pressure.
15. a)
They are listed in Page 106 of the textbook: 1. 2. 3. 4.
Q can be factored so that Qint is independent of density. Classical (rather than quantum) statistical mechanics is applicable. *total ¦ (*Pairs ) (pairwise additivity). * / H F (r / V ) (universal functionality).
b) In general, assumption 4 is violated. But if we fix the core size to be a fixed fraction of the collision diameter, then Kihara potential is a 2parameter (V, H) potential that satisfies corresponding states. c) Hydrogen (at least at low temperatures) has a de Broglie wavelength large enough so that quantum effects must be considered and therefore assumption 2 is violated. Assumption 1 is probably pretty good for H2; assumption 4 is violated slightly. All substances violate assumption 3, but H2 isn’t very polarizable so it might be closer than the average substance to pairwise additivity. d)
Corresponding states (and thermodynamics in general) can only give us functions such as
c p c 0p . Values of c 0p (for isolated molecules) cannot be computed by these methods, because the contributions to c 0p (rotation, vibration, translational kinetic energy) appear in Qint and the kinetic energy factor, not in the configuration integral.
16.
Let D represent the phase inside the droplet and E the surrounding phase. Schematically we have for the initial state and for the final (equilibrium) state:
40
Solutions Manual
D
NO3
NO3
+
Na
K
NO3
NO3
D
+
Na
K+
+
+
K+
Na
2
Lys
Lys2
E
E
Equilibrium state
Initial state
Because the molar mass of lysozyme is above the membrane’s cutoff point, lysozyme cannot diffuse across the membrane. Let: G represent the change in K+ concentration in D; M represent the change in Na+ concentration in E. The final concentrations (f) of all the species in D and E are: In D:
In E:
c fD
Lys 2
c fE
Lys2
c 0E
c fD
Lys2
K
c 0D G K
c fE K
c fD c fE
G
Na
c fD NO
M
Na
c 0D G M NO 3
3
c 0E M Na
c fE
NO 3
c 0E G M NO3
The equilibrium equations for the two nitrates are PE PE K NO
P D P D K NO 3
3
(1) PE PE Na NO3
PD PD Na NO3
Similar to the derivation in the text (pages 102103), Eq. (1) yields cE cE K NO 3
c D c D K NO 3
(2) E
E
c c Na NO
cD
3
cD Na NO 3
where, for clarity, superscript f has been removed from all the concentrations. Substituting the definitions of G and M gives:
FH c0D GKI FG c0D G MIJ K H NO3 K FH c0E MIK FG c0E G MIJ Na H NO3 K where
FG H
(G ) c 0E G M NO 3
FG H
IJ K
(M) c 0D G M NO 3
IJ K
(3)
41
Solutions Manual
c 0D K
0.01 mol 1 kg water
c0D 
NO3
9.970 u 10 3 mol L1
(using the mass density of water at 25D C: 0.997 g cm3 )
and
c 0E Na
c 0D 
NO3
0.01 mol L1
Solving for G and M gives
4.985 u 10 3 mol L1
G
(4) 3
M
1
5.000 u 10 mol L
Because both solutions are dilute, we can replace the activities of the solvent by the corresponding mole fractions. The osmotic pressure is thus given by [cf. Eq. (450) of the text] E RT x s ln vs x sD
S
(5)
where xs is the mole fraction of the solvent (water) given by xsD
xwD
D D D 1 ( xNO x x ) 3
K
Na
xE + K
xE Na
(6) xsE
E xw
E 1 ( xNO 3
E xLys )
Because solutions are dilute, we expand the logarithmic terms in Eq. (5), making the approximation ln(1 A)  A : RT vw
S
E E E D D D ª ( x E º «¬ NO3 xK xNa xLys ) ( xNO3 xK xNa ) »¼
Again, because solutions are very dilute, ci c  i c c ¦i w
xi
i
with v w cw  1
Therefore, with these simplifying assumptions, the osmotic pressure is given by S
D D D RT [cNO c c 3
K
Na
E E E cNO c c 3
K
Na
E cLys ]
Using the relationships with the original concentrations, we have
42
Solutions Manual
0E RT [(c 0D G M ) (c 0D G) (M) (c 0E G M ) (c 0E M) (cLys )]
S
K
NO3
Na
NO3
Because c 0D
c 0D
c 0E
c 0E
K
NO3
Na
NO3
we obtain S
0E RT (2c 0D 2c 0E cLys 4G 4 M ) K
(7)
Na
The lysozyme concentration is 0E cLys
2g 1L
(2 / 14,000) mol 1L
1.429 u 10 4 mol L1
Substitution of values in Eq. (7) gives the osmotic pressure S
(8314.51 Pa L mol 1 K 1 ) u (298 K) u [(2 u 9.97 u 10 3 ) (2 u 0.01) (1.429 u 10 4 ) (4 u 4.985 u 10 3 ) (4 u 5.000 u 10 3 ) (mol L1 )] 354 Pa
17.
Because only water can diffuse across the membrane, we apply directly Eq. (441) derived in the text:
ln aw
ln( xw J w )
Sv pure w
(1)
RT
where subscript w indicates water. Since the aqueous solution in part D is dilute in the sense of Raoult’s law, J w  1 . This reduces Eq. (1) to: ln xw or equivalently,
ln(1 xA xA2 )  ( xA xA2 )
Sv pure w
RT
(2)
43
Solutions Manual
RT ( x A xA 2 )
S
At T = 300 K, v pure w
v pure w
0.018069 mol L1.
Mole fractions xA and xA2 can be calculated from the dimerization constant and the mass balance on protein A: K
vw
Mw dw
aA2
10 5
2 aA
18.015 0.997
(5 / 5,000) mol A (1000 / 18.069) mol water

xA2
(4)
2 xA
18.069 cm3 mol 1
181 . u 10 5
x A 2 x A2
(5)
Solving Eqs. (4) and (5) simultaneously gives 7.34 u 10 6
xA xA2
5.38 u 10 6
Substituting these mole fractions in Eq. (3), we obtain S
(0.0831451) u (300) u (5.38 u 10 6 7.34 u 10 6 ) (0.018069) 0.01756 bar
1756 Pa
18. a)
From Eq. (445): S c2
RT RTB*c2 " M2
Ploting S / c2 (with S in pascal and c2 in g L1) as a function of c2, we obtain the protein’s molecular weight from the intercept and the second virial osmotic coefficient from the slope.
44
Solutions Manual
50
S /c2, Pa L g 1
47.5 45 42.5 40 37.5 35 0
20
40 1 BSA Concentration, g L
60
From a leastsquare fitting we obtain: Intercept = RT/M2 = 35.25 Pa L g1 = 35.25 Pa m3 kg1 from which we obtain
M2
(29815 . ) u (8.314) 35.25
RT 35.25
70.321 kg mol 1
70,321 g mol 1
Slope = 0.196 = RTB*. Therefore, B* = 7.92u108 L mol g2. The protein’s specific volume is given by the ratio molecular volume/molecular mass. The mass per particle is m
M2 NA
70,321 6.022 u 1023
1.17 u 10 19 g molecule 1
Because protein molecule is considered spherical, the actual volume of the particle is 1/4 of the excluded volume. Therefore the actual volume of the spherical particle is 118 . u 10 24 4
2.95 u 10 25 m3 molecule 1
which corresponds to a molecular radius of 4.13u109 m or 4.13 nm. For the specific volume: 2.95 u 10 25 117 . u 10 19
2.52 u 10 6 m3 g 1
2.52 cm3 g 1
b) Comparison of this value with the nonsolvated value of 0.75 cm3 g1, indicates that the particle is hydrated.
45
Solutions Manual
c) Plotting S / c2 (with S in pascal and c2 in g L1) as a function of c2 for the data at pH = 7.00, we obtain: 60 57.5
S /c2, Pa L g 1
55 52.5
50 47.5
45 42.5
40 15
25
35 45 BSA Concentration, g L1
55
Slope = 0.3317, which is steeper that at pH = 5.34, originating a larger second virial osmotic coefficient: B* = 1.338u107 L mol g2 = 13.38u108 L mol g2. At pH = 7.00, the protein is charged. The charged protein particles require counterions so electroneutrality is obtained. The counterions form an ion atmosphere around a central protein particle and therefore this particles and its surrounding ion atmosphere have a larger excluded volume than the uncharged particle. It is the difference between the value of B* at pH = 7.00 and that at pH = 5.37 for the uncharged molecule gives the contribution of the charge to B*: (13.38 7.92) u108 L mol g2 =
1000 z 2 4 M22U1mMX
In this equation, we take the solution mass density U1  Uwater  1 g cm 3 . Moreover, M2 = 70,321 g mol1, and mMX  0.15 mol kg1: z2
4 u (5.46 u 10 8 ) u (70,321) 2 u (10 . ) u (0.15 u 10 3 )  162 1000
or z r13. Because pH is higher than the protein’s isoelectric point, the BSA must be negatively charged. Hence, z = 13. 19. a)
From Eq. (445) and according to the data: S c c0
RT RTB* (c c0 ) " M2
46
Solutions Manual
Ploting S / (c c0 ) (with S in pascal and c2 in g L1) as a function of c c0, we obtain the solute’s molecular weight from the intercept and the second virial osmotic coefficient from the slope. 18.0
S/(cc0), Pa L g
1
17.5
17.0
16.5
16.0
30
40
50
(cc 0), g L
60
1
From a leastsquare fitting we obtain: Intercept = RT/M2 = 14.658 Pa L g1 = 14.658 Pa m3 kg1 from which we obtain M2
RT 14.659
(298.15) u (8.314) 14.659
169.109 kg mol1
169,109 g mol 1
Slope = 0.053495 = RTB*. Therefore, B* = 2.16u108 L mol g2. b) The number of molecules in the aggregate is obtained by comparison of the molecular weight of the original ether (M = 390 g mol1) with that obtained in a):
Number of molecules in the aggregate =
169,109  434 390
Assuming that the colloidal particles are spherical, we obtain the molar volume of the aggregates from the value of the second virial osmotic coefficient B*. It can be shown (see, e.g., Principles of Colloid and Surface Chemistry, 1997, P.C. Hiemenz, R. Rajagopalan, 3rd. Ed., Marcel Dekker) that B* is related to the excluded volume Vex through B*
NA V ex 2 M22
Solutions Manual
47
From a), B* = 2.16u108 L mol g2 = 2.16u105 m3 mol kg2 and M2 = 169,109 g mol1 = 169.109 kg mol1. The above equation gives Vex that is 4 times the actual volume of the particles (we assume that the aggregates are spherical). Calculation gives for the aggregate’s volume, 5.13u1025 m3 (or 30.90 dm3 mol1) with a radius of 4.97u109 m or 5 nm.
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
1.
5
Initial pressure Pi:  for n1 moles of gas 1 at constant T and V: PiV n1RT
B 1 n1 11 V
or
Pi
n1RT n12 RTB11 V V2
Final pressure Pf:  after addition of n2 moles of gas 2 at same T and V:
Pf where (n1 n2 ) 2 B
(n1 n2 ) RT (n1 n2 )2 RTB V V2
n12 B11 2n1n2 B12 n22 B22
Pressure change 'P :
'P
Pf Pi
n2 RT RT (2n1n2 B12 n22 B22 ) V V2
Solving for B12:
B12
· 1 § V 2 'P n2V n22 B22 ¸ ¨ ¸ 2n1n2 ¨© RT ¹
49
50
Solutions Manual
2.
For precipitation to occur,
s
V fCO ! fCO 2
s
To obtain fCO , 2
s
ln
fCO
60 1 (27.6) dP (83.1451) u (173) 0.1392
³
2
(0.1392)
s
fCO
2
0.156 bar (at 60 bar, 173 K)
2
Next, find the vapor mole fraction of CO2 that is in equilibrium with the solid at the specified P and T: V fCO
yCO2
2
M CO2 P
Using the virial equation for the vapor, ln M VCO
2
2 ( yCO2 BCO2 yH2 BH2 CO2 ) ln z v
Because yCO2 1 , we may make the approximations
z
z H2
v
and
v H2
z H2 RT P
From data for H2 (see App. C) at –100qC,
8.8 cm3 mol 1
BH2 which indicates that z H2 From correlations:
1.
460 cm3 mol 1
BCO2 BCO2 H2
321 . cm3 mol 1
At equilibrium V fCO
and then
2
s
fCO
2
51
Solutions Manual
s
yCO2
fCO
2
2P ª ½ yCO2 BCO2 (1 yCO2 ) BH2 CO2 º¼ ¾ P exp ® ¬ ¯ RT ¿
yCO2
0.00344
Because yCO2 0.01 at equilibrium, CO2 precipitates. To find out how much, assume solid is pure CO2. Let n be the number of moles of CO2 left in the gas phase. From the mass balance and, as basis, 1 mole of mixture, 0.00344
n
n n 0.99
0.003417
The number of moles precipitating is
0.01 0.003417 0.0066 moles CO2
3. To = 313 K
Tf = ?
Po = 70 bar
Pf = 1 bar
V L . Condensation will occur in the outlet if fCO ! fCO 2
2
First it is necessary to find the outlet temperature, assuming no condensation. JouleThomson throttling is an isenthalpic process that may be analyzed for 1 mole of gas through a 3step process: I, III: Isothermal pressure changes.
P
To ,Po
II: Isobaric temperature change.
I Tf ,Pf III P=0
II T
52
Solutions Manual
Then 'htotal
z LMMN 0
'hI
Po
(1)
FG wv IJ OP dP H wT K P PQ
(2)
v To Tf
³T
'hII
z LMMN Pf
'hIII
'hI 'hII 'hIII
c p,mixt dT
(3)
o
v Tf
FG wv IJ OP dP H wT K P PQ
(4)
Assuming that the volumetric properties of the gaseous mixture are given by the virial equation of state truncated after the second term, v
RT Bmixt P
then,
FG H
FG wv IJ H wT K P
IJ K
dBmixt R P dT P
where Bmixt is the second virial coefficient of the mixture. Because y12 B11 2 y1 y2 B12 y22 B22
Bmixt
2 = CO2)
(1 = CH4; dBmixt dT
y12
(5)
dB11 dB dB 2 y1 y2 12 y22 22 dT dT dT
(6)
If c (0 )
B
c (1) c (2) T T2
then dB dT
c (1) T
2
2c (2) T3
(7)
Assume c 0p, mixt
then,
yCH 4 c 0p,CH yCO2 c 0p,CO 4 2
(8)
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Solutions Manual
ª
§ dBmixt dT
³P «¬ Bmixt To ¨© o
§ 2c(1) 3c(2) (0) ¨ cmixt mixt mixt ¨ To To2 ©
Tf Pf ª · º § dBmixt c p,mixt dT « Bmixt Tf ¨ ¸ » dP To 0 ¬ ¹P ¼ © dT
³
³
· º ¸ » dP ¹P ¼
· § 2c(1) 3c(2) (0) ¸ ( Po ) c p,mixt (Tf To ) ¨ cmixt mixt mixt ¸ ¨ Tf Tf2 ¹ ©
· ¸ ( Pf ) ¸ ¹
From data: As y1
0.3 , c p,mixt
0.7 , y2
36.22 J K 1 mol1 according to Eq. (8).
From Eqs. (5), (6) and (7),
Substitution in Eq. (9) gives Tf
(0 ) cmixt
41849 .
(1) cmixt
18683
( 2) cmixt
34.12 u 10 5
278.4 K .
Second, the fugacities of liquid and vapor phases may be calculated.
L fCO
ª P vL º CO2 s Ms exp « » xCO2 J CO2 PCO d P s 2 CO2 « PCO » RT 2 »¼ ¬«
³
2
This equation may be simplified assuming that xCO2 , J CO2 , M sCO equal to unity. 2 s 3 1 L 39.8 bar and v At 278 K, P 49.0 cm mol . CO2
CO 2
L fCO
2
(39.8) u exp
LM (49.0) u (1 39.8) OP . ) u (278) Q N (831451
36.6 bar
Fugacity of vapor is calculated from V fCO
2
M CO2 yCO2 P
with ln MCO2
P ª 2( yCO BCO yCH BCO CH ) Bmixt º ¬ ¼ RT 2 2 4 2 4
ln MCO2
>2 u (0.3) u (139) 2 u (0.7) u (77) (69)@
P RT
(9)
54
Solutions Manual
M CO2
0.995  1
V fCO
0.3 bar
Then 2
Because V L fCO fCO 2
2
no condensation occurs.
4.
The Stockmayer potential is *
4H
LMF V I 12 F V I 6 OP P 2 MNGH r JK GH ( r JK PQ r 3 g(T1, T2 , I2 I1 )
where P is the dipole moment.
T2
T1
I
We can write the potential in dimensionless form: * H
f
F r , P2 I GH V HV 3 JK
where f is a universal function.
Therefore, we can write the compressibility factor z in terms of the reduced quantities: z
f (T , P , P )
with ~ T
kT H
55
Solutions Manual
P
~ P
PV3 H
P HV 3
5. a) For acetylene: Tc 308.3 K , Z 0.184 . At 0qC, TR (see, e.g., AIChE J., 21: 510 [1975]):
0.886  0.90 . Using LeeKesler charts
's (0 ) / R
3.993
zV(0 )
0.78
z L(0 )
0.10
's (1) / R
3.856
zV(1)
0.11
z L(1)
0.04
' vap s
(8.31451) u [3.993 (0.184) u (3.856)] 391 . J K 1 mol 1 ' vap h T ' vap s (273) u (39.1) 10.67 kJ mol 1
' vapu
' vap h RT (zV z L ) 10.67 (8.31451) u (273) u (0.76 0.092) u 10 3 ' vapu
915 . kJ mol 1
b) C 4 H10 : Tc
At 461 K:
425.2 K
N 2 : Tc
126.2 K
Pc
38.0 bar
Pc
33.7 bar
Z
0.193
Z
0.04
vc
255 cm3 mol1
vc
89.5 cm3 mol1
TR
TR
1084 .
3.65
Using the PitzerTsonopoulos equation (see Sec. 5.7):
BC 4 H10
267 cm3 mol1
BN2
For B12 : Z 12
1 (Z 1 Z 2 ) 2
0.1165
15.5 cm3 mol1
56
Solutions Manual
(Tc1Tc2 )1 2
Tc12 zc12
Pc12
Then, B12
( TR12
2316 . K
0.291 0.08Z12
0.2817
zc12 RTc12
34.3 bar
1 1/ 3 (v v1c/ 3 ) 3 2 8 c1
216 . cm3 mol1
Bmixt
177.2 cm3 mol 1
y12 B11 2 y1y2 B12 y22 B22
c) CH 4 (1): Tc
190.6 K
N 2 (2): Tc
126.2 K
H 2 (3): Tc
Pc
46.0 bar
Pc
33.7 bar
Pc
13.0 bar
Z
0.008
Z
Z
0.22
vc
99.0 cm3 mol 1
vc
0.040
89.5 cm3 mol 1
vc
33.2 K
65.0 cm3 mol 1
At 200 K:
TR
158 .
TR
6.02
TR
105 .
At 100 bar:
PR
2.97
PR
2.17
PR
7.69
Using the mixing rules suggested by Lee and Kesler: 1 ¦ ¦ x j x k (v1c/j3 v1c/k3 )3 8 j k
v c,mixt
Tc,mixt
1 ¦ ¦ x j x k (v1c/j3 v1c/k3 )3 (Tc j Tck )1/ 2 8v c j k Zmixt
¦ x jZ j # 0 j
Pc,mixt
(0.291 0.08Z mixt ) RTc,mixt / v c,mixt
v c,mixt
84.1 cm 3 mol 1
Tc,mixt
111.38 K ( TR
180 . )
Pc,mixt
31.47 bar ( PR
318 . )
1 hmixture (h1 h2 h3 ). 3 Using the LeeKesler charts,
Enthalpy of mixing = h E
1990 . )
57
Solutions Manual
1 922 (5385 1383 0) 3
hE
1334 J mol 1
6.
L/G P t
L
5
G
40 bar 25 qC
40% N2 50% H2 10% C3 H8
From the mass balance for C 3 H 8 : yinG
yout G x out (L [ yin yout ]G ) yout G
0.10 x C3
0.01898
0.05 yinG
0.005 5.005 xC3
mole fraction of C3H8 in effluent oil.
To find the driving force, note that
Pi
PC3
yC3 P total
(0.10) u (40)
Pi*
fC*3 / M *C3
PC*3
4 bar
where
fC*3
x C3 H
(0.01898) u (53.3) 1012 . bar
and ln M *C3
2P
z * RT
[ yC3 BC3 yN2 BC3 N2 yH2 BC3 H2 ] ln z *
Obtain virial coefficients from one of the generalized correlations:
58
Solutions Manual
400.8 cm3 mol1
BC3
We may estimate z * * yC 3
To find * yC 3
,
* yN , 2
BC3 N2
73.5 cm3 mol1
BC3 H2
3.5 cm3 mol1
0.95 (feed value). * * and yH , we know that ( yN / y* ) = 4/5. As first guess, assume 2 2 H2
0.10. Then we calculate,
M *C3
1012 . (40) u (0.855)
* 0.855 and yC 3
0.0292
This gives * y* yN H2 2
0.9708
or
* yN 2
0.4308 , y*H2
0.540
With these y * ’s, calculate M *C3 again:
M *C3
0.924
and
* yC 3
0.027
That is close enough. Thus, PC*3
1012 . 0.924
Now we must check the assumption z * sumption is close enough.
bar 1095 . 0.95 was correct. Using the virial equation, the as
Driving force = (PC3 PC*3 )
7.
Since fiV
(4.00 1095 . )
fiL , s v f ( P Psolv ) xi Hi,solv exp i RT
yi Mi P s As Psolv
0,
yi xi Using the virial equation,
Hi exp(vif P / RT ) Mi P
2.91 bar
59
Solutions Manual
ln M1
2 ( y1B11 y2 B12 ) ln z mixt v
ln M 2
2 ( y2 B22 y1B12 ) ln z mixt v
Because Pv RT v2
B RT RT v mixt P P
3.23 cm3 mol1
Bmixt v
B 1 mixt v
524 cm3 mol1 , z mixt
10067 .
Thus, M1
y1 x1
M2
0.8316 (100) u exp
0.9845
LM (60) u (50) OP . )Q N (313) u (8314
(50) u (0.8316)
2.70
and y2 x2 D 2,1
8.
2131 .
FG y2 IJ FG x1 IJ H x2 K H y1 K
7.89
For methane (1) and methanol (2), we may write f 1V
f 1L
f 2V
f 2L
Neglecting the Poynting corrections [Note: the Poynting Correction is 1.035. Including this, we get y2 = 0.00268], y1M1P
x1H1,2
60
Solutions Manual
x 2 J 2 P2s M s2
y2 M 2 P
Because x2 = 1, assume J 2 1 . Use virial equation to get fugacity coefficients:
M s2 Assuming y1
1 , y2
exp
B22 P2s RT
(4068) u (0.0401) (8314 . ) u (273)
exp
0.993
0 as first estimate, M1
M2
0.954
0.783
Thus, x1
0.0187
(1 x1 )P2s M s2
y2
Using now y2
M 1P H1,2
0.00250
M2P
0.00250 , get M1
0.954
M2
0.770
0.01867
y2
0.00255
and x1
This calculation is important to determine solvent losses in natural gas absorbers using methanol as solvent.
9.
2(monomer)
dimer
The equilibrium constant is
Ka
ad
fd / fd0
(am )2
( fm / fm0 )2
2 fd §¨ fm0 fm2 ¨© fd0
· ¸ ¸ ¹
where ad is the activity of the dimer, and am is the activity of the monomer. 2
The quantity f m0 / fd0 is a constant that depends on T, but not on P or y. Then,
61
Solutions Manual
k f m2
fd
where k is a constant.
10.
CH3
CH3 CH
CH3
O
CH CH3
CH3CH2
Diisopropyl ether
O
CH2CH2CH2CH3
Ethyl butyl ether
HCl can associate with the ether’s nonbonded electron pairs. However, the diisopropyl ether will offer some steric hindrance. The crosscoefficient, B12 , is a measure of association. Both virial coefficients will be negative; B12 for ethyl butyl ether/HCl will be more negative.
11.
Let D be the fraction of molecules that are dimerized at equilibrium. 2A 1 D
nT
A2 D/2
1 D
yA
2
yA
D 2
1
D 2
D/2 1 D / 2 1 D 1 D / 2
By assuming the vapor to be an ideal gas, we may write K
PA2
yA2 P
( PA )
2
( yA P)
(D / 2)(1 D / 2)
2
(1 D)2 P
At the saturation pressure, P = 2.026 bar and yA
0.493
D
0.6726
62
Solutions Manual
Then fAV
fAL
(0.493) u (2.026)
yA P
0.999 bar
The pressure effect on fugacity is given by
FG w ln f IJ H wP K
vi RT
i
T
Assuming the liquid to be incompressible in the range Ps to 50 bar, ln
fA (50 bar) fA (2.026 bar)
v A 'P RT
fA (50 bar) 1.1 bar
12. a)
The RedlichKwong equation is z
FG H
Pv RT
v 1 a 1 . 5 v b RT vb
If z is expanded in powers of 1/ v : 1
z
B C " v v2
This gives
b
B
b2
C
a RT 1.5 ab RT 1.5
But, B'
C'
B / RT
C B2 (RT )2
Substitution gives B'
FG H
1 a b RT RT 1.5
IJ K
IJ K
63
Solutions Manual
a
C'
3 3.5
R T
b)
FG 3b a IJ H RT K 1.5
Using an equation that gives fugacities from volumetric data, we obtain ln M1
P RT
ª y2 a 2 y1a1 y22 a2 2 y22 (a1a2 )1/ 2 º « b1 1 1 » RT 1.5 «¬ »¼
Evaluate a and b using critical data: Ethylene (1):
Tc = 282.4 K
Pc = 50.4 bar
Nitrogen (2):
Tc = 126.2 K
Pc = 33.7 bar
a1 = 7.86u107 bar cm6 K1/2 mol2 a2 = 1.57u107 b1 = 40.4 cm3 mol1 Substitution gives M1
0.845 ,
y1M1P
f1
13.
8.44 bar
Using the virial equation,
P
RT RTBmixt v v2
with y12 B11 2 y1 y2 B12 y22 B22
Bmixt
(1)
For maximum pressure,
FG wP IJ H wy K T 1
,v
Substituting Eq. (1) into Eq. (2) gives ( y2 § wBmixt · ¨ ¸ © wy1 ¹T
At maximum,
FG H
IJ K
RT wBmixt wy1 T v2
(2)
1 y1 ) :
2 y1 B11 2 y2 B12 2 y1 B22 2 B22
64
Solutions Manual
B22 B11 B22 B12
y1
(3)
Using the correlations, B11 B22
126.7 cm3 mol1 (ethylene) 12.5 cm3 mol1 (argon) 45.9 cm3 mol1
B12
Substitution in Eq. (3) gives y1 = 0.134
14.
Consider a 3step process: Mixed ideal gases
Pure ideal gases Tf
Tf
P = 20.7 bar I
Mixed gases Tf
Pure gases Ti
The overall enthalpy change is zero: 'H I 'H II 'H III 'H I
where
n1'H1 n2 'H 2
'H1
c 0p1 (Tf Ti )
H0 H (RTc1 ) RTc1
'H 2
c 0p2 (Tf Ti )
H0 H (RTc2 ) RTc2
H0 H is evaluated using LeeKesler Tables. RTc
(1 = hydrogen; 2 = ethylene)
65
Solutions Manual
For heat capacities, we can estimate
'H II
c 0p1
28.6 J K1 mol1
c 0p2
43.7 J K1 mol1
(because we are mixing ideal gas)
F H H I RT GH RTc JK c 0
'HIII
, mixt
,mixt
evaluated at TR = Tf / Tc,mixt, PR = 20.7/Pc,mixt
Find Tf by trial and error: Tf
15.
247 K
At equilibrium,
fAV yA M A P
s
fA
(1 xCO2 )PAs M sA exp
z
s
P
vA
s PA
RT
dP
Assuming xCO2  0 , ( y A P)MA
PAs MsA
s
ª v ( P Ps ) º A » exp « A RT « » ¬ ¼
But MsA
§ B Ps exp ¨ AA A ¨ RT ©
· ¸ 1.00 ¸ ¹
Because yA yCO2 , yCO2 # 1
MA yA P PAs
or
P º ª exp «(2 BA CO2 BCO2 ) RT »¼ ¬
s
ª v ( P P s ) (2 BA CO BCO )P º 2 2 A » exp « A RT RT « » ¬ ¼
66
Solutions Manual
s
s
ª s y Pº « PA v A RT ln As » /(v A 2 BA CO2 BCO2 ) PA ¼» ¬«
P
Substitution gives P = 68.7 bar 19 . u 10 4 1 and assumption yA  0 is correct.
For this pressure, yA
16.
Let 1 = ethylene and 2 = naphthalene. As, at equilibrium,
s
f2V
f2 and as
s
f2
P2s M s2 exp f2V
we may write
P22
RT
dP
y2 M 2 P
s
Using idealgas law: M s2
s
As v 2
128.174 1.145
M V2
1
111.94 cm 3mol 1 ,
y2
b)
v2
(1 x1 )P2s M s2 exp[v 2 (P P2s ) / RT ] / M V2 P
y2 a)
z
s
P
s
ª v (P Ps ) º 2 » / P 1.1u 10 5 P2s exp « 2 RT « » ¬ ¼
Using VDW constants: a a1 a2
27 R 2 Tc2 / 64 Pc
4.62 u106 bar cm6 mol2 4.03 u107 bar cm6 mol2
b
RTc / 8 Pc
b1
58.23 cm3 mol1
b2
192.05 cm3 mol1
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Solutions Manual
b1 a1 / RT
B11
122.2 cm3 mol1
1382 cm3 mol1
B22
Because y2 2( y1B12 y2 B22 ) Bmixt @
b12 a12 / RT
B12
b12
with b12
12514 . cm3 mol1
and
P P # (2 B12 B11 ) RT RT
1 / 2(b1 b2 ) and a12
(a1a2 )
a12 = 1.36u107 bar cm6 mol2
Then B12 = 406 cm3 mol1. MV2 y2
P º ª exp «(2 B12 B11 ) RT »¼ ¬
0.446
(2.80u104) u exp [(111.9u30)/(83.1451u308)]/(0.446u30) y2 = 2.4u105
17.
s
Water will condense if f HV O ! f H O . 2 2 Thus, the maximum moisture content, yH2O , is given by
f HV
2O
yH2O M VH O P 2
s
fH
x H2O PHs O 2
2O
exp
z
P
Ps H 2O
s
vH O 2 dP RT
Assuming that the condensate is pure (solid) water,
x H2O Then
1
and
M sH
2O
1
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Solutions Manual
( yH2O MVH
2O
) u (30 bar)
ª 1.95 · º § (18 / 0.92) u ¨ 30 « (1.95 torr) 750.06 ¸¹ » © » exp « (750.06 torr/bar) « (83.1451) u (263.15) » « » ¬ ¼ y H 2O
8.90 u 10 5 M VH
2O
Let 1 = N2, 2 = O2, and 3 = H2O. To get M VH
2O
use the virial equation of state: ln MVH O 2
ln M3
[2( y1 B13 y2 B23 y3 B33 ) Bmixt ]
P RT
with Bmixt
¦ ¦ yi y j Bij i
j
Assume y3 3):
OP PQ
f
(2) V
76
Solutions Manual
B
LM MN
2 V3 SN AV 3 2SAN A 3 kT (n 3)V n
OP PQ
LM N
OP Q
2 2 3A SN AV 3 SN AV 3 3 3 kT (n 3)V n
(3)
b) From Eq. (3) we see that it is the attractive part of the potential that causes negative B and is responsible for the temperature dependence of B [the first term on the right hand side of Eq. (3) is independent of temperature].
23.
Substitution of the squarewell potential [Eq. (539)] into Eq. (517) gives B
2SN A
³
ª 2SN A « ¬
f
(1 e*(r ) / kT ) r 2 dr
³
V
(1 ef / kT )r 2 dr
ª V3 2SN A « ¬« 3
³
R'
V
³
R'
V
(1 eH / kT )r 2 dr
f
º (1 e0 / kT )r 2 dr » R' ¼
³
º (1 eH / kT )r 2 dr 0 » ¼»
ª V 3 § R '3 V 3 · º 2SN A « ¨ ¸ (1 eH / kT ) » 3 ¸¹ «¬ 3 ¨© 3 »¼ 2 2 SN AV3 SN A (1 eH / kT )( R '3 V3 ) 3 3
In the equation above, R' = RV = 1.55V. For argon, V = 0.2989 nm = 0.2989u109 m, = 141.06 K, and R' = 1.55V = 4.633u1010 m. The above equation gives for T = 273.15 K,
B
H/k
3.368 u 10 5 (6.202 u 10 5 ) 2.834 u 10 5 m 3 mol 1  28 cm 3 mol 1
The calculated value compares relatively well with the experimental B for argon at the same temperature: Bexp = 22.08 cm3 mol1.
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Solutions Manual
24.
Substitution of Sutherland potential [Eq. (537)] in Eq. (519) with N = NA gives
B(T )
ª 2SN A « ¬
³
V
(1 ef / kT ) r 2 dr
³
f
V
6 º (1 eK / kTr ) r 2 dr » ¼
(1) 2 SN AV3 3 § K exp ¨ © kTr 6
³
f
V
6
(1 eK / kTr ) r 2 dr
· K K2 K3 K4 " ¸  1 kTr 6 2(kT )2 r12 6(kT )3 r18 24(kT )4 r 24 ¹
[ ex  1 x
(2)
x2 x3 x 4 "] 2! 3! 4!
We now have to replace the approximate result [Eq. (2)] in Eq. (1) and perform the necessary integrations. The result is:
B(T )
2SN A K 2SN A K 2 2SN A K 3 2SN A K 4 2 SN AV 3 " 3 3kTV 3 18(kT ) 2 V 9 90(kT )3 V15 504(kT ) 4 V 21
This equation is best solved using an appropriate computer software such as Mathematica, TKSolver, MathCad, etc. Making the necessary programming we obtain at 373 K, B(methane) = 20 cm3 mol1 B(npentane) = 634 cm3 mol1
In both cases, the agreement with experiment is very good.
25.
The equation of equilibrium for helium is f1L
f1R
where superscripts L and R stand for left and right compartments, respectively. Equivalently,
(1)
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Solutions Manual
y1L M1L P L
y1R M1R P R
(2)
Equation (533) of the text gives the fugacity coefficients in both compartments from the volumeexplicit virial equation of state: ln M1
2( y1B11 y2 B12 ) Bmixt
P RT
(3)
with Bmixt
y12 B11 2 y1 y2 B12 y22 B22
(4)
Further, we also have material balances y1L y2L
1
(5) y1R y3R
1
Applying Eqs. (3), (4), and (5) to both compartments yields ln M1R
{2 ª¬ y L B 1
L L L º ª L 2 11 (1 y1 ) B12 ¼ ¬( y1 ) B11 2 y1 (1 y1 ) B12
PL (1 y1L )2 B22 º ¼ RT
}
(6) ln M1L
{2 ª¬ y R B 1
R R R º ª R 2 11 (1 y1 ) B13 ¼ ¬( y1 ) B11 2 y1 (1 y1 ) B13
PR (1 y1R )2 B33 º ¼ RT
}
Total mole balance on helium gives n1L n1R
0.02 mol
(7)
Combining with mass balances on ethane and nitrogen gives
y1L
n1L 0.99 n1L (8)
y1R
n1R
0.02 n1L
0.99 n1R
. n1L 101
Substituting Eqs. (6) and (8) into Eq. (2) yields n1L
0.013 mol
Combining this result and Eq. (8) gives y1L
0.013
y1R
0.007
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Solutions Manual
26.
Because the equilibrium constant is independent of pressure, the more probable reaction is the one that satisfies this condition. (1)
Assuming reaction (a) is more probable:
With this scheme, concentration of (HF) 6 is negligible compared to those of ( HF) and (HF) 4 . The equilibrium constant K (a ) is K (a )
y( HF ) 4 y(4HF ) P 3
y( HF ) 4 1 y( HF ) 4
4
(1) P3
Total mass balance for ( HF) gives y( HF) 4 nT (4 u M HF ) 1 y( HF) 4 nT M HF
V u U HF
(2)
where V is the total volume; U HF and M HF are the mass density and the molar mass of hydrogen fluoride, respectively; nT is the total number of moles that can be calculated by assuming that the gas phase is ideal: nT
PV RT
(3)
Substituting Eqs. (2) and (3) into Eq. (1) yields
FG U RT 1IJ HPuM K LM1 (1 / 3)F U RT 1I OP GH P u M JK PQ MN (1 / 3)
K (a )
HF
HF
(4)
4
HF
P
3
HF
Applying Eq. (4) at the two pressures, 1.42 and 2.84 bar, we obtain for K (a ) : P (bar)
UHF (g/L)
K(a)
1.42 2.84
1.40 5.45
0.0595 0.453
Because K (a ) depends on pressure, reaction (a) cannot be the more probable one. Next we need to check for the pressure independence of K ( b ) . (2)
Assuming reaction (b) is more probable:
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Solutions Manual
In this case, concentration of (HF) 4 is negligible compared to those of ( HF) and (HF) 6 . The equilibrium constant K ( b ) is K( b)
y( HF )6 y(6HF ) P 5
y( HF )6 1 y( HF )6
6
(5) P5
Mass balance for HF in this case is: y( HF )6 nT (6 u M HF ) 1 y( HF )6 nT M HF
V u U HF
(6)
where all terms are defined in Eq. (2). Substitution of Eqs. (3) and (6) into Eq. (5) gives
FG U RT 1IJ HPuM K MMNL1 (1 / 5)FGH PUu MRT 1IJK OPQP (1 / 5)
K ( b)
HF
HF
(7)
6
HF
P5
HF
Corresponding values of K ( b ) at 1.42 and 2.84 bar are: P (bar)
UHF (g/L)
K(b)
1.42 2.84
1.40 5.45
0.017 0.017
Because K ( b ) is independent of pressure, reaction (b) is the more probable.
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
1.
The three equations of equilibrium (in addition to T L y1M1V P
x1J 1 f1L
y2M V2 P
x2 J 2 f2L
y3M V3 P
x3 J 3 f3L
T V and P L
6
P V ) are
with (assuming the liquid incompressible) f1L
v L (P P1s ) P1s M1s exp 1 RT L
L
We write similar expressions for f 2 and f 3 . For M V we may write RT ln M1V
P(2 y1B11 2 y2 B12 2 y3 B13 Bmixt )
and similar expressions for M V2 and M V3 . In these equations, Bmixt
y12 B11 y22 B22 y32 B33 2 y1 y2 B12 2 y1 y3 B13 2 y2 y3 B23
81
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Solutions Manual
2.
Given g E
Ax1 x 2 with P1s / P2s
1.649 , and assuming ideal vapor,
y1P
J1
x1P1s y2 P
J2
At azeotrope, x1
x 2 P2s
y1: ln J 1
ln
ln J 2
ln
P P1s P P2s
or
ln
J1 J2
ln
P2s P1s
0.5
From the g E expression, ln J 1
A 2 x RT 2
ln J 2
A 2 x RT 1
and
Then J ln 1 J2
A RT
A 2 ( x x12 ) RT 2
0.5 x22
x12
1 4 x2 2
or x2
1 RT 2 4A
Because 0 d x 2 d 1 , 1 2
1 RT 1 d d 2 4A 2
Thus, if  A t RT , an azeotrope exists.
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Solutions Manual
3.
From the plot Pxy we can see the unusual behavior of this system: 1. There is a double azeotrope 2. Liquid and vapor curves are very close to each other.
75
Pressure, kPa
74
73
72
Liquid (
)
Vapor (
)
0.2
0.4
0.6
0.8
1.0
x1, y1
4.
Neglecting vapor phase nonidealities, P
x1J 1P1s x 2 J 2 P2s
(1)
At the maximum, § wP · ¨ ¸ © wx1 ¹T
§ wJ · § wJ · J1P1s x1P1s ¨ 1 ¸ x2 P2s ¨ 2 ¸ J 2 P2s w x © wx2 ¹T © 1 ¹T
From the GibbsDuhem equation (at constant T and low pressure),
(2)
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Solutions Manual
§ w ln J1 · § w ln J 2 · x1 ¨ ¸ x2 ¨ ¸ © wx1 ¹T © wx1 ¹T
or x1 § wJ1 · ¨ ¸ J1 © wx1 ¹T
x2 § wJ 2 ¨ J 2 © wx2
· ¸ ¹T
(3)
Substituting Eq. (3) into Eq. (2) and simplifying, § x wJ · ( J1P1s J 2 P2s ) ¨ 1 1 1 ¸ © J1 wx1 ¹
There are two possibilities: (1)
J 1P1s J 2 P2s
J 2 P2s
1
Then
J 1P1s D
( y2 / x2 ) ( y1 / x1 )
D
1 corresponds to an azeotrope
(2)
x wJ 1 1 1 J 1 wx1
The solution to this differential equation is x1J 1 constant . To find the constant, use the boundary condition J 1 1 when x1 Hence J 1 x1 1. †
x1J 1P1s if J 1 x1 1, then y1 must be 1. Hence, the curve Px goes through a maximum at x1 trivial one).
As y1P
5.
1.
1. This is also an azeotrope (but a
Given gE
A12 x1 x 2 A13 x1 x 3 A14 x1 x 4 A23 x2 x3 A24 x 2 x 4 A34 x3 x 4
where
†
This may not be immediately obvious. But J1 x1 is the activity, and the activity of component 1 cannot reach unity for any x1 less
than one because the solution will split into two phases of lower activity. See Fig. 625 in the text.
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Solutions Manual
x1
n1 / nT
x2
x2 / nT
# with nT n1 n2 n3 n4 the total number of moles. Because
F wn g I GH wn JK E
T
RT ln J 1
1
P,T ,n2 ,n3 ,n4
we find RT ln J1
A12 x22 A13 x32 A14 x42 x2 x3 ( A12 A13 A23 ) x2 x4 ( A12 A14 A24 ) x3 x4 ( A13 A14 A34 )
6.
Calculate Ty giving pressure and for x = 0.1, 0.2, …, 0.9 bubblepoint calculation. We have to solve the equilibrium equations: M i yi P
J i xi Pis
(1)
Because total pressure is low (below atmospheric) we assume vapor phase as ideal: M i  1 . The activity coefficients are obtained from the equation for GE given in the data. Using Eq. (647) of the text we obtain ln J 1
21 . x 22
ln J 2
21 . x12
(2)
As the pressure is fixed, temperature varies along with x1 (and y1) and is bounded by the saturation temperatures of the two components. These can be easily obtained from the vaporpressure equations. They are given in the form, ln P s
A
B T C
(3)
from which we obtain the saturation temperature
Ts
B A ln P s
C
(4)
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Solutions Manual
For Ps = 30 kPa, we obtain T1s
387.26 K for cyclohexanone (1) and T2s
415.59 K for
phenol (2). To obtain the Tx1y1 diagram we assign values for the liquid mole fraction x1. Total pressure is P
J 1 x1P1s J 2 x2 P2s
P1s
P
or (5)
P2s
J 1 x1 J 1 x1 s P1
To start the calculation we make an initial estimate of the temperature: x1T1s x 2T2s
T
(6)
0.5 u 387.26 0.5 u 415.59
For example, let us fix x1 = 0.5: T
401.42 K
With this temperature we obtain P1s and P2s from Eq. (3), the purecomponent vapor pressure equations: P1s J1
47.243 kPa and P2s
exp(21 . u 0.52 )
0.592 and J 2 P1s
Next we recalculate
17.918 kPa Because we fixed x1, Eqs. (2) give
0.592 .
73.482 kPa from Eq. (5), which in turn gives a new temperature,
T = 416.34 K, from the pure cyclohexanone vapor pressure equation. The sequence of calculations is now repeated for this new temperature (we assume here that activity coefficients are independent of temperature), yielding: P2s
30.786 kPa; P1s
T1s
415.34 K
71.426 kPa
[from Eq.(5)]
[from Eq.(6)]
x x x
P1s
71.552 kPa;
T1s 415.40 K;
P2s
29.798 kPa
After these values, the change in temperature is small and therefore additional iterations leads to no significant further change in the remaining values. We can now calculate the vapor phase mole fraction from y1
x1J 1P1s P
(0.5) u (0.592) u (71552 . ) 30
0.706
The whole process is repeated for a new liquid mole fraction. The following figure shows the computed Tx1y1 diagram for this system at 30 kPa.
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Solutions Manual
430
V
Temperature/K
420
410
L+V 400
L 390
380 0.0
0.2
0.4
0.6
0.8
1.0
Cyclohexanone Mole Fraction
Similarly, with the data calculated we can easily draw the corresponding y1x1 diagram, shown in the figure below.
Vapor Mole Fraction Cyclohexanone
1
0 0.0
0.2
0.4
0.6
0.8
1.0
Liquid Mole Fracton Cyclohexanone
As both figures show, this system has an azeotrope at T az  421 K and for the composition x1az  0.3 .
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Solutions Manual
7.
Assume g E
Ax1 x 2 , where A is a function of temperature. Then, RT ln J 1f
RT ln J 2f
A
10 T 273
A RT
But ln J 1f
0.15
Because
At x1
x2
w( g E / T ) wT
h E
w( g E / T ) wT
10 Rx1 x 2
h E
(T 273) 2
T2
T2
0.5 and T = 333K, hE
641 J mol 1
' mix h
8. a) From the equation for H w we can obtain the infinite dilution partial molar enthalpy of water in sulfuric acid solutions at 293 K and 1 bar as:
Hwf
b)
lim Hw
xw o 0
lim Hw
x A o1
41.44
kJ mol1
The mixing process is schematically shown below.
A
HW
HA
W
W
Cooling coils
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Solutions Manual
Taking the liquid in the vessel as the system, a first law balance gives for this flow process: dQ dW H Adn A H w dnw
dU
where work W is dW pressure. Then,
PdV , done on environment by the rising liquid level, under constant
d (U PV ) HA
dH
dQ H Adn A H w dnw
Integrating between initial state (empty vessel) and final state (full vessel), because H A (T , P) of the pure acid and H w H w (T , P) of the pure water are constant, we obtain H
But H
Q n A H A nw H w
or
Q
H n A H A nw H w
n A H A nw Hw , and the equation above becomes Q
n A (H A H A ) nw (H w H w )
n A 'H A nw 'H w
'H
(1)
where nA = 1 mol and nw = 2 mol in the final state. In Eq. (1), the quantity (H w H w ) is given by the equation given in the data, because the reference state in that equation has been chosen to be pure water at system T and P: Hw Hw
134 x 2A (1 0.7983x A ) 2
kJ mol 1
(2)
We need now to calculate the quantity (H A H A ) , knowing (Hw Hw ) . This can be done by using the GibbsDuhem equation. At constant T and P: x Ad H A x w d H w
dHA
xw d Hw xA
1 xA d Hw xA
(3)
Differentiating Eq. (2), at constant T and P, we obtain: d Hw
(134 kJ mol 1 ) u
2 x A (1 0.7983x A ) 2 2(0.7983) x 2A (1 0.7983x A ) (1 0.7983x A ) 4
dx A
(4) 268 x A (1 0.7983x A )
3
dx A
(kJ mol 1 )
Therefore, from Eqs. (3) and (4), dH A
268(1 x A ) (1 0.7983x A ) 3
dx A
(kJ mol 1 )
Integrating Eq. (5) between composition xA and composition xA = 1 (pure acid) gives
(5)
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Solutions Manual
H A H A (x A
1)
z
HA HA
xA
(1 0.7983x A ) 3
1
LM 335.71(x 0.1263) OP MN (1 0.7983x ) PQ A
268(1 x A )
dx A (6)
xA
2
1
74.51(1 x A ) 2 (1 0.7983x A ) 2
(kJ mol) 1
We can now calculate the heat load Q in Eq. (1). Setting n = nA + nw = 3 mol, and using Eqs. (2) and (6) in (1), Q
ª 74.51x A (1 x A )2 134 x 2A (1 x A ) º n« » (1 0.7983x A )2 »¼ «¬ (1 0.7983x A )2 ª 74.51x A (1 x A )(1 x A 1.7983x A ) º n« » (1 0.7983x A )2 ¬« ¼» ª 74.51x A (1 x A ) º n« » ¬ 1 0.7983x A ¼
(7)
(kJ mol 1 )
Substitution of n = 3 mol and xA = 1/3 gives the desired heat load:
Q
39.23 kJ
Q is negative because heat is removed from the system.
9. a) Yes, it’s possible. Slight positive deviations merely mean that the physical interaction between SO2 and C4H8 makes a larger contribution to the excess Gibbs energy than does the chemical interaction. b) E gSO
2 isobutene
E ! gSO
2 n
butene 2
because the tendency to complex (which tends to make gE negative) is stronger with nbutene2. Steric hindrance in isobutene is larger than in nbutene2.
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Solutions Manual
10.
The suggested procedure is to integrate numerically a suitable form of the GibbsDuhem equation. At low pressures, we may write the GibbsDuhem equation:
FG IJ x2 FG wJ 2 IJ H K T J 2 H wx 2 K T
x1 wJ 1 J 1 wx1
By assuming idealgas behavior, J1
y1P
P1
x1P1s
x1P1s
wJ 1 wx1
1 x1P1s
FG wP1 IJ P1 H wx1 K x12 P1s
Similarly, wJ 2 wx1
1
x2 P2s
FG wP2 IJ P2 H wx2 K x22 P2s
Substituting we find x1 wP1 P1 wx1
Because P
P1 P2 , dP
x2 wP2 P2 wx 2
dP1 dP2 , then wP2 wx2
wP [ 1x P ] wx2 1 2 1 x1P2
'P2 'x 2
'P 1 x 2 P1 'x 2 1 x1P2
In different form:
For Px data, we choose a small 'x2 (say 0.05) and integrate to find 'P2 and thus P2 . We obtain P1 by difference: P1 P P2 . This method is described by Boissanas, quoted in Prigogine and Defay, Chemical Thermodynamics, page 346. It gives good agreement with experimental partialpressure data for this particular system.
11.
For a binary system, the Wilson equation gives
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Solutions Manual
ln J1
ª /12 º / 21 ln( x1 x2 /12 ) x2 « » ¬ x1 /12 x2 x2 x1/ 21 ¼
(1)
ln J 2
ª /12 º / 21 ln( x2 x1/ 21 ) x1 « » ¬ x1 x2 /12 x2 x1/ 21 ¼
(2)
At infinite dilution these equations become
For J 1f
12.0, J 2f
ln J 1f
ln /12 (1 / 21 )
(3)
ln J 2f
ln / 21 (/12 1)
(4)
3.89, solve Eqs. (3) and (4) to find, / 21
0.6185
/ 12
0.1220
Assuming idealgas behavior and neglecting Poynting correction, we may write:
P
y1P
x1J 1P1s
(5)
y2 P
x2 J 2 P2s
(6)
x1J 1P1s x 2 J 2 P2s
From Perry’s, the saturation pressures at 45qC are: P1s P2s
0.188 bar 0.0958 bar
To construct the Pxy diagram: 1. Choose x1 (or x2) 2. Calculate y1 (or y2) from Eq. (5) and using Eqs. (1) and (2) 3. Calculate P from Eq. (7).
12.
The solution procedure would be: 1. Find P1s and P2s at each T. 2. At this low pressure, assume idealgas behavior and neglect Poynting correction:
(7)
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Solutions Manual
y1P
x1J1P1s
y2 P
x2 J 2 P2s
For J’s use the Wilson equation with two parameters: /12 and / 21 . Assume that (O 11 O 12 ) and (O 22 O 12 ) are independent of temperature. /12 and / 21 are, however, temperaturedependent as given by Eqs. (6107) and (6108).
3. Assume value of (O 11 O 12 ) and (O 22 O 12 ) and calculate the total pressure: Pcalc
x1J 1P1s x2 J 2 P2s
4. Repeat; assuming new values. Keep repeating until Pcalc is very close to 0.5 bar for every point; that is until n
¦ (Pcalc P)2
is a minimum
i 1
where n is the number of data points.
13. a)
2Butanone: H H
H
H
C
C
C
C
H
O
H
H
H
Cyclohexane:
(CH2)6 6 groups CH2: R = 0.6744; Q = 0.540 Molecule
Group
Number
R
Q
2Butanone
CH3CO CH3 CH2 CH2
1 1 1 6
1.6724 0.9011 0.6744 0.6744
1.488 0.848 0.540 0.540
Cyclohexane
b) We use UNIFAC activity coefficient equations to calculate J1 and J2 for the equimolar mixture at 75ºC (for a detailed example of a similar UNIFAC calculation see Chapter 8 of The Prop
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Solutions Manual
erties of Gases and Liquids by R.C. Reid, J.M. Prausnitz, B. E Poling (4th. Ed., McGrawHill, 1988). We obtain: ln J 1
ln J 1comb ln J 1res
0.01228 0.2595
0.27238
J1
1.31
ln J 2
ln J 2comb ln J 2res
0.01415 0.3420
0.35615
J2
1.43
c) Using UNIFAC we can calculate the activity coefficients as a function of composition at 75ºC. Total pressure is calculated from x1J 1P1s x 2 J 2 P2s
P
and the vaporphase composition from x1J 1P1s
y1
Using P1s
Antoine
vapor
pressure
0.8695 bar and for cyclohexane
P2s
P
equations
at
75ºC,
we
obtain
for
2butanone
0.8651 bar .
The following figures show the calculated results in the form of Px1y1 and y1x1 diagrams. 1.3
L 1.2
Pressure/bar
1.1
L+V
L+V
1.0
V
0.9 0.8 0.7 0.6 0.0
0.2
0.4
0.6
2Butanone Mole Fraction
0.8
1.0
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Solutions Manual
Vapor Mole Fraction 2Butanone
1
0 0.0
0.2
0.4
0.6
0.8
1.0
Liquid Mole Fracton 2Butanone
The table below shows the calculated activity coefficients from UNIFAC, vapor composition and total pressure. x1
J1
J2
y1
P/bar
0 0.2 0.4 0.5 0.6 0.8 1.0
4.38 2.27 1.51 1.32 1.18 1.04 1.00
1.00 1.07 1.27 1.42 1.62 2.19 3.10
0 0.351 0.447 0.485 0.528 0.660 1.0
0.8511 1.124 1.175 1.178 1.169 1.096 0.8695
Comparison of the calculated J’s in this table with those given in the data, indicate that the latter are actually UNIFAC predictions and not experimental data. In the tables, at x1 = 0 and x1 =1 the activity coefficients listed are, respectively, J 1f and J 2f . UNIFAC predicts J 1f which compares well with the experimental ebulliometry data at 77.6ºC,
14.
The UNIQUAC equation is gE
E E gcombinatorial gresidual
J 1f
3.70.
4.38 ,
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Solutions Manual
E gcomb RT
E gres RT
) ) T T z x1 ln 1 x2 ln 2 ( x1q1 ln 1 x2 q2 ln 2 ) )1 )2 x1 x2 2
x1q1 ln(T1 T2 W21 ) x2 q2 ln(T2 T1W12 )
)1
x1r1 x1r1 x2 r2
T1
x1q1 x1q1 x2 q2
W12
a exp( 12 ) T
W21
a exp( 21 ) T
The condition for instability of a binary liquid mixture is
F w 2 ' mix g I 0 GH wx 2 JK P,T
(1)
where ' mix g is the molar change in Gibbs energy upon mixing, or § w 2 gE ¨ ¨ wx 2 © 1
· § 1 1 · RT ¨ ¸ 0 ¸ ¸ © x1 x2 ¹ ¹ P,T
Incipient instability occurs at § w 2 ' mix g · ¨ ¸ ¨ wx 2 ¸ © ¹ P,T
§ w 3' mix g · ¨ ¸ ¨ wx 3 ¸ © ¹ P,T
and
Given x1 , x 2 and all parameters, we could determine if Eq. (1) is satisfied. However, the procedure is long and tedious. It is easier to graph ' mix g over the composition range and to look for inflection points. For the data given, phase separation occurs at 40qC.
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Solutions Manual
xnHexane 0
0.4
0.2
0.6
0.8
100
' mix g (J mol1)
200
300
400
500 223.15 K 233.15 K 243.15 K 253.15 K 263.15 K
15.
If the two curves cross ' mix g / RT is zero because ' mix g
' mix h T ' mix s
This is not possible, because 'mixg must always be negative for two liquids to be miscible.
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Solutions Manual
16.
To relate J if to Hi, j :
fi
Ji
xi fi0
then, J if
§ f lim ¨ i xi o0 ¨ x f 0 © i i
· ¸¸ ¹
§ f · lim ¨ i ¸ 0 fi © xi ¹ 1
Hi, j fi0
Assuming fi0 # Pis , J if
Hi, j
J 1f
H1,2
2 107 .
1869 .
J 2f
H 2,1
16 . 133 .
1203 .
Pis
then, P1s
P2s
Using the van Laar equations, ln J1
Ac § Ac x1 · ¨1 c ¸ B x2 ¹ ©
ln J 2
2
Bc § B c x2 · ¨1 c ¸ A x1 ¹ ©
2
we get ln J 1f
Ac
0.625
ln J 2f
Bc
0.185
To solve for vapor composition (assuming ideal vapor and neglecting Poynting corrections), y1P
x1J1P1s
y2 P
x2 J 2 P2s
P
y1P y2 P
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Solutions Manual
At x1
x2
0.5, ln J1
ln J 2
0.625
§ 0.625 · ¨ 1 0.185 ¸ © ¹
2
0.185
§ 0.185 · ¨ 1 0.625 ¸ © ¹
2
0.033
J1
1.033
0.110
J2
1.116
Therefore,
17.
y1P
(0.5) u (1.033) u (1.07)
0.55 bar
y2 P
(0.5) u (1.116) u (1.33)
0.74 bar
P
0.55 0.74 1.29 bar
y1
0.55 / 1.29
y2
0.574
0.426
To estimate the vaporphase composition, assume ideal vapor: yi P
xi J i Pis
(i=1, 2, 3)
Then, P
y1P y2 P y3 P
To find the activity coefficients, assume that g E / RT is given by a sum of Margules terms [Eq. (6149)]. Then, ln J 1
A12 c x 22 A13 c x32 ( A12 c A13 c A23 c ) x2 x3
(1)
ln J 2
A12 c x12 A23 c x32 ( A12 c A23 c A13 c ) x1 x 3
(2)
ln J 3
A13 c x12 A23 c x 22 ( A13 c A23 c A12 c ) x1 x2
(3)
We can find A12 c , A13 c , A23 c from binary data.
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Solutions Manual
From (12) binary: A RT
ln J 1f A12 c
c At 300K, A12
§ 320 · (0.262) u ¨ ¸ © 300 ¹
ln(13 . )
0.262
at 320K
0.262
0.280 (assuming regular solution).
From (13) binary: At azeotrope x1 y1P
x1J1P1s
y3 P
x3 J 3 P3s
J1
ln J1
At x3
0.5,
A RT
c A13
y1 , x3
P / P1s
J3
y3
1.126
A 3 x2 RT
0.475.
From (23) binary: A 2 x RT 3
ln J 2
At incipient instability,
A RT c
2
or A R
With x1
x2
2T c
A RT
§ A ¨ © RT c
A23
1.80
·§ Tc ¸ ¨¨ ¹© T
· ¸ ¸ ¹
§ 270 · (2) u ¨ ¸ 1.80 © 300 ¹
x3 , from Eqs. (1), (2) and (3):
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Solutions Manual
ln J 1
0.0322
J1
0.968
ln J 2
0.409
J1
1506 .
ln J 3
0.475
J1
1607 .
y1P
(1 / 3) u (0.968) u (0.533)
0.172 bar
y2 P
(1 / 3) u (1506 . ) u (0.400)
0.201 bar
y3 P
(1 / 3) u (1607 . ) u (0.533)
0.286 bar
P = 0.172 + 0.201 + 0.286 = 0.659 bar Then,
18.
y1
0.261
y2
0.305
y3
0.434
Using the 3suffix Margules equation, g E / RT
x1 x2 [ A B( x1 x2 )]
we obtain ln J 1
( A 3B) x 22 4 Bx23
ln J 2
( A 3B) x12 4 Bx13
At infinite dilution, ln J 1f
A B
ln J 2f
A B
which gives A = 1.89
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Solutions Manual
B = 0.34 For instability to occur,
F w2g I GH wx 2 JK 1 E
RT P,T
FG 1 1 IJ 0 H x1 x2 K
Rewriting g E as gE wg E wx1
RT [( A B) x1 ( A 3B) x12 2Bx13 ]
RT [( A B) 2( A 3B) x1 6 Bx12 ]
w2g E wx12
RT [2( A 3B) 12Bx1 ]
Thus, the condition for instability (at constant T) is: ª 1 1 º RT « 2 A 6 B 12 Bx1 »0 x1 1 x1 ¼ ¬
Finding the zeros of the function in brackets, x1
0.421 and
x1
0.352 in the range
Thus, instability at T is in the range
0.352 < x1 < 0.421
19. a)
At the azeotrope
FG wP IJ H wx A K T With g E / RT of the form g E / RT
Ax A x B , ln J A
Ax B2
ln J B
Ax 2A
0 < x1 < 1.
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Solutions Manual
Assuming an ideal vapor phase,
§ wP · ¨ ¸ © wx A ¹
x A J A PAs
yB P
x B J B PBs
P
x A J A PAs x B J B PBs
P
x A exp( Ax B2 )PAs x B exp( Ax 2A )PBs
PAs exp( Ax B2 )(1 2 x A x B A) PBs exp( Ax 2A )(1 2 x B x A A)
PAs exp( Ax B2 ) Ax B2
yA P
PBs exp( Ax 2A )
ln( PBs / PAs ) Ax 2A
At 30qC,
Then, x A
PAs
0.235 bar;
PBs
0.658 bar;
A = 0.415
PAs
0.539 bar;
PBs
0.658 bar;
A = 0.415
PAs
1.119 bar;
PBs
1.367 bar;
A = 0.330
0.30.
At 50qC,
Then x A
0.26.
At 70qC,
Then x A
b)
0.20.
Assuming ideal vapor,
At azeotrope, x A
y A , xB
JA
y A P / x A PAs
JB
y A P / x B PBs
yB . Then
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Solutions Manual
JA
P / PAs
JB
P / PBs
Taking the ratio JA / JB ln( J A / J B )
PBs / PAs ln PBs ln PAs 4050 4050 11.92 T T
12.12
0.20 Because ln J A
Ax B2
ln( J A / J B )
A
ln J B
and
1 5 10 x A
A( x B2 x 2A )
Ax 2A , 0.2
because 0 < x A < 1
If A > 0.2 there is an azeotrope. The pure component boiling points are: tbA
67qC
tbB
61qC
In the range 61qC < t < 67qC, A is always larger than 0.2. Therefore, the azeotrope exists.
c) The enthalpy of mixing equation cannot be totally consistent since the expression for g E is quadratic in mole fraction and the expression for ' mix h is cubic. However, they may be close. To check this, we use the GibbsHelmholtz equation: wg E / RT wT
wg E / RT wT
h E RT 2
gE RT
A(T ) x A x B
x A xB
wA # x A x B (0.00425) wT
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Solutions Manual
Because h E
' mix h,
' mix h (1) RT
(323) u (0.00425) x A x B
1373 . x A xB
The other data indicate ' mix h (2) RT
(1020 . 0.112 x A ) x A x B
Looking at selected values: xA
'mixh(1)/RT
'mixh(2)/RT
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.123 0.220 0.288 0.330 0.343 0.330 0.288 0.220 0.123
0.093 0.167 0.221 0.256 0.269 0.261 0.231 0.178 0.101
The above shows the degree of inconsistency of the two sets of data.
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
1.
7
Using regularsolution theory and data for A in CS2 we find the solubility parameter for A. Then, we predict vaporliquid equilibria for the A/toluene system. Let B refer to toluene and C refer to CS2. From regularsolution theory,
RT ln J A
b
v A ) 2C G A G C
g
2
(for A in CS2)
Further, assuming ideal vapor phase, we have yA P
JA
PA
xA J A PAs
8 (0.5) u (13.3)
1203 .
or ln J A
0.185
Then, 1/ 2
G A GC
ª RT ln J A º r« » ¬ vA ¼
1 )C
with vA
200 cm3 mol1 GA GC
and
)C
0.234
r 6.30 (J cm 3 )1/ 2
107
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Solutions Manual
For liquid hydrocarbons, G is approximately 1218 (J cm3)1/2. Therefore, we take the smaller value. For A in toluene,
RT ln J A
b
2 vA)B GA GB
g
2
or JA
RT ln J B
118 .
b
v B) 2A G A G B
g
2
or JB
137 .
For ideal vapor, P
PA PB
x A J A PAs xB J B PBs
25.3 kPa
Hence,
2.
yA
0.31
yB
0.69
Excess properties ( h E , s E ) are defined in reference to an ideal (in the sense of Raoult’s law) mixture of pure components. The partial molar quantities h E and s E are the contributions to these excess properties per differential amount added to the solution. The “pure” acetic acid is highly dimerized, so as the first bits go into solution thse dimers must be broken up. This will require energy ( h1E ! 0 ) and will increase the entropy more than is accounted for by the ideal mixing term ( s1E ! 0 ). As x1 gets larger, some dimer will begin to exist in the solution, so these effects will diminish. Therefore, at small x1, the curves should look something like this:
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Solutions Manual
s1
h1
3.
0.5
x1
0.5
x1
The K factors for hexane (1) and benzene (2) (neglecting Poynting corrections and assuming idealvapor phase) are: J 1P1s
K1
P J 2 P2s
K2
P
where x1J 1P1s x 2 J 2 P2s
P
Using regularsolution theory,
RT ln J 1 RT ln J 2
b v ) bG
g G g
v1) 22 G1 G 2 2
2 1
1
2
2
2
The volume fractions are )1
0.389
)2
132 .
J2
0.611
From the above equations, J1
1.08
P = (0.3)u(1.32)u(0.533) + (0.7)u(1.08)u(0.380) = 0.498 bar K1
KC6H14
(1.32) u (0.533) 0.498
K2
KC6H6
(1.08) u (0.38) 0.498
1.41
0.82
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Solutions Manual
4.
Ether and pentachloroethane hydrogen bond with each other (but not with themselves). Then, g E 0 . x
1
gE
5.
The relative volatility of A and B is D A,B
( yA / x A ) ( yB / x B )
Assuming ideal vapor phase and neglecting Poynting corrections,
At the azeotrope, x A
yA P
x A J A PAs
yB P
x B J B PBs
yA and
D A,B
1
J A PAs J B PBs
From regularsolution theory,
b bG
RT ln J A
2 vA)B GA GB
RT ln J B
v B) 2A
A
GB
g g
2
2
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Solutions Manual
Because v A
v B , J A J B , then PAs For the ternary mixture,
PBs .
v i ( G i G )2
RT ln J i
G
¦ )i Gi i
As ) A Then
)B
0.2 and ) C
0.6 , G
ln J A
(100) u (14.3 17.2)2 (8.31451) u (300)
JA
1.40
ln J B
(100) u (16.4 17.2)2 (8.31451) u (300)
JB
1.03
J A PAs
D A,B
6.
17.2 (J cm 3 )1/ 2 .
J B PBs
JA JB
1.40 1.03
1.36
Assuming ideal vapor phase and neglecting Poynting corrections,
P
y1P
x1J 1P1s
y2 P
x2 J 2 P2s
x1J 1P1s x 2 J 2 P2s
Using regularsolution theory,
Because v1
As v1
v2
RT ln J1
v1) 22 (G1 G2 )2
RT ln J 2
v 2 )12 (G1 G2 )2
v 2 , we can rewrite [Eqs. (725) and (726)]:
160 cm3 mol1,
RT ln J1
v1x22 (G1 G2 )2
RT ln J 2
v 2 x12 (G1 G2 )2
(1)
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Solutions Manual
J1
exp(0.616x22 )
J2
exp(0.616 x12 )
Substitution in Eq. (1) gives P
0.533exp(0.616 x22 ) 0.800 exp(0.616 x12 )
At the azeotrope,
FG wP IJ H wx1 K T
Thus, after differentiation, 0.616 x 22
0.40547 0.16 x12
Solving for x1, x1
7.
0.171
Neglecting vaporphase nonidealities and Poynting corrections, the total pressure, P, is P
x A J A PAs x B J B PBs
Because the two fluids are similar in size, simple and nonpolar, we can assume that J’s are given by twosuffix Margules equations:
As xA
xB
ln J A
A 2 x RT B
ln J B
A 2 x RT A
0.5, P
0.667
LM FG A IJ OP u b0.427 0.493g N H 4RT K Q
0.5 u exp
A = 4696 J mol1 From Eq. (6144) of the text, Tc
A 2R
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Solutions Manual
Tc
282 K
If one considers the effect of nonrandomness (based on the quasichemical approximation), Eq. (7110) gives A 2.23R
Tc
(assuming that
A 2R
w 2k
253 K
constant and that the temperature dependence is given by ln J propor
tional to 1/T). Thus, random mixing predicts a value higher than that given by quasichemical theory. The observed consolute temperature is likely to be lower than both.
8. V
yi = ?
Let: 1 = benzene z1= z2= 0.5
2 = nbutane
L
There are three unknowns: x1 , y1, and V / F . To solve for them, we use two equilibrium equations and one mass balance. Assuming ideal vapor and neglecting Poynting corrections: y1P
(1 y1 )P
z1
x1J 1P1s
(1 x1 )J 2 P2s
FG V IJ y1 FG1 V IJ x1 H FK H FK
Using regularsolution theory,
RT ln Ji
vi )2j (Gi G j )2
xi = ?
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Solutions Manual
with v1 = 92 cm3 mol1 and v 2 = 106 cm3 mol1. Then, J1
exp(0.828) 22 )
J2
exp(0.950)12 )
Substitution gives y1 (1 y1 )
0.368x1 exp(0.828) 22 )
(1)
4.76(1 x1 ) exp(0.950)12 )
(2)
FG V IJ y1 FG1 V IJ x1 H FK H FK
0.5
(3)
) 1 and ) 2 are related to x1 and x2 by Eqs. (725) and (726). To solve Eqs. (1), (2), and (3) for x1, y1, and V / F, assume first that J i x1
y1
0.856
0.315
V F
1. This gives,
0.658
A second approximation ( J i z 1 ) gives
x1
9.
xC6H6
0.94
y1
0.35
V F
0.741
As derived in Sec. 7.2 of the text, the regularsolution equations can be written in the van Laar form gE
Ax1 x 2 A x1 x 2 B
(1)
where parameters A and B are related to purecomponent liquid molar volume and solubility parameters as follows [Eqs. (738) and (739)]: A v A (G A G B ) 2 (2) B
v B (G A G B ) 2
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Solutions Manual
Substituting the given liquid molar volumes and solubility parameters, we obtain A
(120 cm3 mol 1 ) u [(18 12) 2 J cm 3 ]
4320 J mol 1 (3)
B
(180
cm3 mol 1 ) u [(18 12) 2 J
cm 3 ]
6480 J
mol 1
As discussed in Section 612, the temperature and composition at the consolute point are found from solving:
FG w ln aA IJ FG w2 ln aA IJ H w xA K T ,P H w xA2 K T ,P
(4)
Upon substitution of Eq. (1) into Eq. (4), the results are given in Eq. (6146) in the text: c xA
[( A / B) 2 1 ( A / B)]1/ 2 ( A / B) 1 ( A / B)
(5) Tc
c c 2 xA (1 x A )( A2 / B) c c 3 R[( A / B) xA (1 x A )]
where superscript c denotes consolute. Substituting Eq. (3) into Eq. (5), we finally obtain
10.
c xA
0.646
Tc
328 K
For each phase we choose the standardstate fugacity for cyclohexane as its pure subcooled liquid at 25ºC. The equation of equilibrium is x 3(1) J (31)
x 3(2) J (32)
(1)
where subscript 3 denotes cyclohexane and superscripts (1) and (2) denote, respectively, carbon disulfide phase and perfluoronheptane phase. Rearrangement of Eq. (1) gives x (1) K{ 3 x 3(2)
J (32) J (31)
(2)
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Solutions Manual
Because phase (1) contains only carbon disulfide and a trace amount of cyclohexane, whereas phase (2) contains only perfluoronheptane and a trace amount of cylohexane, J (31) and J (32) are essentially activity coefficients at the infinitedilution limit of cyclohexane.
Hence, we can write
K{
x3(1) x 3(2)
LM J (32) OP f MN J (31) PQ
(3)
where superscript f denotes the infinitedilution limit of cyclohexane. From the regularsolution theory [Eq. (737)], J (31) º ln ª J (1) ¬ 3 ¼
f
and J (32)
f
v3 (G1 G3 )2 RT
f
v3 ( G 2 G 3 )2 RT
f
are given by:
(4) º ln ª J (2) ¬ 3 ¼
Substituting the purecomponent liquid molar volumes and solubility parameters, we obtain º ln ª J (1) ¬ 3 ¼
f
(109 cm3 mol1 )
u ª(20.5 16.8)2 J cm3 º ¼ (8.314 J mol1 K1 ) u (298 K) ¬
0.602 (10)
º ln ª J (2) ¬ 3 ¼
f
3
1
(109 cm mol )
u ª(12.3 16.8)2 J cm3 º ¼ (8.314 J mol1 K1 ) u (298 K) ¬
0.891
Substituting Eq. (5) into Eq. (3), we have
K
11.
1.34
From the definition of the solubility parameter, G, G2
'ucomplete vaporization vL
[Complete vaporization means going from saturated liquid to ideal gas at constant T.] Then,
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Solutions Manual
F h0 h L I RT RT (1 z L ) GH RTc JK c
G2
z L RT Ps
LM h0 h L OP MM RTL c 1 1L PPPRs z MM z TR PP N Q
2
G Pc
Because
h0 h L RTc
f (TR , Z )
and zV
zL
f (TR , Z)
PRs
f (TR , Z)
f (TR , Z)
for TR d 1,
G2 Pc
b g
b g b
f (0) TR Z f (1) TR higher terms
g
[Reference: Lyckman et al., 1965, Chem. Eng. Sci., 20: 703].
12. a) Pure methanol is hydrogenbonded to dimers, trimers, etc. In dilute solution (in isooctante), methanol is a monomer. For an orderofmagnitude estimate, we can assume that, to make a monomer, approximately one hydrogen bond must be broken. Thus h E = 12 kJ mol1. b) From solubility parameters we get (roughly) an endothermic heat of 263 J mol1. The molar specific heat is (roughly) 125 J K1 mol1 . Thus, 't  2D C. c)
We want a Lewis acid that can hook on to the double bond in hexene. Good Solvents are: Dimethyl sulfoxide Sulfur Dioxide Acetonitrile
U V W
Strong Lewis acids
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Solutions Manual
Poor Solvents are: Ammonia
UV W
Aniline
Weak Lewis acids
It is also important that the solvent should not be something that prefers selfinteraction than that with hexene molecules. Strongly hydrogenbonded liquids (e.g. water and most alcohols) would therefore be poor solvents.
13.
Let (HA) be the acid. In ionized form, H+ + A
HA
Hexane (HA)
H
(HA)
A + H+
W
Water
Equilibrium constants are defined as
K1
K2
bHAg bHAg
W
(H )(H ) (HA)W
H
(A )2 (HA)W
In hexane: CH = (HA)H In water:
CW = (HA)W + (A) = (HA)W +
b g
K 2 HA
b g
W
b g
K1 HA H K1K 2 HA H CW
Thus,
K1C H K1K 2C H
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Solutions Manual
CH
FG1 K H
1
CW
1
K1K2
CH CW
IJ K
with
1
a
K1K2 and K1
b
14. Benzene AB
1/3 AT
AW Water
Assume constant distribution coefficient and “reaction” equilibrium. AB AW
K1
AT
b AB g3
K2
In water: CW
AW
In benzene: CB CB
CB CW
AB 3 AT
3 K1C W 3K 2 K13C W
2 K1 3K 2 K13C W
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Solutions Manual
Thus, plot
CB 2 . as a function of CW CW
Slope is 3K 2 K13 and intercept is K1.
15. a) Pure CH2Cl2 and acetone do not hydrogen bond themselves but some hydrogen bonding is likely to occur between dissimilar pairs, which explains the negative deviations from Raoult’s law observed for this system. Pure methanol is highly hydrogen bonded. However, in dilute solutions of CH2Cl2, methanol exists primarily as monomer. Hydrogen bonding between methanol and CH2Cl2 is likely to be weak. (Note that at infinite dilution, activity coefficient J 1f indicates the effect on a molecule 1 when surrounded by molecules of the other component).
b) Nitroethane has a large dipole moment. Both nhexane and benzene are nonpolar but, due to S electrons, benzene is more polarizable. Therefore, we expect nitroethane/benzene interactions to be stronger than those for nitroethane/nhexane. Thus Jnitroethane in nhexane is larger than that in benzene. c) Both CHCl3 and methanol are polar and slightly acidic. Although methanol has a slightly higher dipole moment, CHCl3 is likely to solvate the coal tar more easily because methanol tends to form strong hydrogen bonds with itself.
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
1.
8
Given
FG1 1IJ F H rK
ln *1f
For r >>1, ln *1f
1 F
If F = 0.44, ln *1f
*1f
1.44
4.22
As defined, *1f
a1 )1
a1 10 4
4.22
a1
4.22 u 10 4
Because
a1 P1
P1 P1s
(4.22 u 10 4 ) u (4.49)
For a nonvolatile polymer, P2
0.0019 bar
0. Therefore, P # P1
0.0019 bar
121
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Solutions Manual
2.
We can use the data for the Henry’slaw region to evaluate the Flory interaction parameter, F, and then predict results at higher concentration. Let: 1 = solvent 2 = polymer wi = weight fraction ) i = volume fraction Then, w1 U1
v1 v1 v 2
)1
w1 w2 U1 U2
In the Henry’s law region, w1 o 0, a1 )1
w2 o 1
f1
w1H1,2
)1 f10
)1 f10
But, as w1 o 0, )1
w1U 2 / U1
)2
1
Then, a1 )1
If f10
U1H1,2 U2 f10
P1s ,
a1 )1
f1
U1H1,2
)1 f10
U2 P1s
(0.783) u (18.3) (111 . ) u (3340 / 760)
From FloryHuggins theory (r is large), a ln 1 )1
) 2 F) 22
If )1 o 1 ,
ln(2.94) 1 F
F
0.078
2.94
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Solutions Manual
At higher concentrations (w1 = 0.5), assume F z F(w1 ) :
)2
(0.5) u (1/1.11) (0.5) u (1/1.11) (0.5) u (1/ 0.783)
()1
0.414
0.586)
Then, a ln 1 )1 a1 )1 P  f1
0.414 0.078 u (0.414) 2
153 .
a1
or
a1 f10  a1P1s
0.898
(0.898) u (3340)
3000 torr
3000 torr  3.9 bar
P
3. a)
The generalized van der Waals partition is given by [Eq. (839)] Q T , V , N
1 § Vf · ¨ ¸ N ! ©¨ / 3 ¹¸
N
ª
Eo · º ¸» © 2kT ¹ ¼
>qext (V )@N >qint (T )@N «exp §¨ ¬
N
(1)
Following the discussion on pages 442 and 443 of the textbook, we further have Vf /3 Eo 2
qext
§ Vf · ¨¨ 3 ¸¸ ©/ ¹
rc
rsK 2v
(2) Vf
v
ª§ v ·1/ 3 º Wr v* «¨ ¸ 1» «© v* ¹ » ¬ ¼ V Nr
Substituting Eq. (2) into Eq. (1) yields
3
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Solutions Manual
ln N ! Nrc ln / 3 Nrc ln(W r v* )
ln Q
(3) ª§ v ·1/ 3 º NrsK 3Nrc ln «¨ ¸ 1» N ln qint * 2vkT «© v ¹ » ¬ ¼
Because the first, second, third and fifth terms on the righthand side of Eq. (3) are only functions of temperature, the equation of state is given by 1 § w ln Q · ¨ ¸ Nr © w v ¹ N ,T
§ w ln Q · ¨ wV ¸ © ¹ N ,T
P kT
(4) 1 ª 3Nrc § 1 v 2 / 3 · NrsK § 1 « ¨ ¸ ¨ Nr « (v 1/ 3 1) ¨ 3 v*1/ 3 ¸ 2kT © v 2 © ¹ ¬
·º ¸» ¹ »¼
We can rewrite Eq. (4) as v 1/ 3
P v T
1/ 3
v
1 T v 1
(5)
where the reduced properties are defined by T
T
T* P
P
*
P
2v*ckT sK 2
2v * P sK
(6)
v
v
v*
Equation (5) is the Flory equation of state [Eq. (845) of the textbook].
b)
The configurational partition function [Eqs. (882) and (883] Q
§ E · QC exp ¨ ¸ © kT ¹
(7) QC
(constant) N
/ v* ) N
(V 1 N 0 ! N ! (V / v* ) N (r 1)
where
V
v*
N0 r N
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Solutions Manual
z E = H Nr 2
ª rN º « » «¬ (V / v* ) »¼
(8)
Combining Eqs. (7) and (8) gives ln Q
ln(constant)
§V · V ln ¨ ¸ v* © v * ¹ v* V
§V · ln N 0 ! ln N ! N (r 1) ln ¨ ¸ © v* ¹
(9)
§V · H* (v*r N )2 ¨ ¸ V 2 © v* ¹
where H*
z H 2 kT
(10) ª§ V · º ln N 0 ! ln «¨ ¸ r N » ! ¬© v* ¹ ¼ ª§ V · º ª§ V · º ª§ V · º «¨ * ¸ r N » ln «¨ * ¸ r N » «¨ * ¸ r N » ¬© v ¹ ¼ ¬© v ¹ ¼ ¬© v ¹ ¼
The equation of state is thus given by P kT
§ w ln Q · ¨ wV ¸ © ¹ N ,T § V · § V · v* (1/ v* ) 1 ln ¨ ¸ ¨ ¸ V v* v* © v * ¹ © v* ¹ 1
(11)
1 v*
1 §V · §V · ln ¨ rN ¸¨ rN ¸ V * * * ¹ ©v ¹ ©v rN v * v
N (r 1)
We can rewrite
1/ v*
§ V 2 · H* (v*r N )2 ¨ ¸ ¨ v* ¸ V / v* © ¹ 1/ v*
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Solutions Manual
§ PV · § 1 · ¨ kT ¸ ¨ r N ¸ © ¹© ¹
P v T V § · ¨ ¸ * §V · §V · v ¸ v ln ¨ ¸ v ln ¨ r N ¸ (v 1) ¨ ¨ V rN ¸ © v* ¹ © v* ¹ ¨ * ¸ ©v ¹ 1 * rN v 1 H r V / v*
(12)
1 ª § 1 ·º 1 1 v ln ¨ 1 ¸ » r «¬ © v ¹ ¼ T v
where T
P
v =
T T*
T z H / 2k
P
P
P*
z H / 2v*
(13)
v v*
Equation (12) is the SanchezLacombe latticefluid equation of state [Eq. (884) of the textbook].
4.
The FloryHuggins equation for the activity of the solvent [Eq. (811)] is ln a1
2 § 1· ln(1 )*2 ) ¨ 1 ¸ )*2 F)*2 r © ¹
Conditions for incipient instability give [analogous to Eqs. (6141) and (6142)]: § w ln a1 · ¨ ¸ ¨ w)* ¸ © 1 ¹ P,T § 2 · ¨ w ln a1 ¸ ¨¨ *2 ¸¸ © w)1 ¹ P,T
Equivalently, we have
(1)
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Solutions Manual
§ w ln a1 · ¨ ¸ ¨ w )* ¸ 2 ¹ P,T ©
(2) § 2 · ¨ w ln a1 ¸ ¨¨ * 2 ¸¸ © w) 2 ¹ P,T
because )1* )*2
1
Substituting Eq. (1) into Eq. (2), we obtain 1 c 1 )*2
c § 1· ¨ 1 ¸ 2Fc )*2 © r¹
§ ¨ 1 ¨¨ *c © 1 )2
2
· ¸ 2F c ¸¸ ¹
where superscript c stands for critical. Hence, we obtain )*2
Fc
5.
c
1 1 r1/ 2 1§ 1 · ¨ 1 1/ 2 ¸ 2© r ¹
2
The FloryHuggins equation for the activity coefficient of HMDS (1) [Eq. (812)] with F is ln J1
ª § 1· º § 1· ln «1 ¨ 1 ¸ )*2 » ¨ 1 ¸ )*2 ¬ © r¹ ¼ © r¹
Using data in Table 85 of the text, calculated molecular characteristic volumes V * , ratios of molecular segments r and activity coefficients of HMDS (at )*2 lowing table:
0.8 ) are given in the fol
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Solutions Manual
Substance
V*
v*sp Mn
r
(cm3 mol1)
HMDS PDMS PDMS PDMS PDMS PDMS PDMS PDMS
3 10 20 100 350 1000 f
V2* / V1*
3.21 5.13 8.09 21.61 34.07 40.70 f
162.30 521.53 832.89 1313.8 3507.8 5530.1 6604.8 f
ln J1 ()*2
0.8)
0.249 0.389 0.528 0.677 0.722 0.735 0.809
As we increase the molecular weight of PDMS, ln J1 becomes more negative. J1 is smaller than unity and increasingly deviates from unity as the molecular weight of PDMS is increased. This example illustrates the effect of differences in molecular sizes of HMDS and various PDMS with F 0 (Fig. 83).
6.
The flux of gas i through the membrane is given by, Eq. (8118) Ji
Di G G G SiF PiF SiP PiP GM
(1)
Because solubility coefficients for both O2 and N2 in the feed and permeate are assumed to be equal and the permeate pressure is vacuum, Eq. (1) reduces to
Ji
Di SiG PiF GM (2)
Di SiG GM
yi PF
where yi and PF ( PF 2 u 105 Pa) denote, respectively, the mole fraction of component i and the total pressure of the feed. For the feed mixture (air) we have
yO2 # 0.21 (3)
yN2 # 0.79 Substituting Eq. (3) and the given data for membrane thickness, solubility and diffusion coefficients into Eq. (2), the corresponding fluxes of O2 and N2 are
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Solutions Manual
J O2
0.148 u 10 5
m3 m2 s1
J N2
0.197 u 10 5
m3 m2 s1
The separation factor is defined by [Eq. (8121)] G DO2 SO
2
D O2 / N 2
G DN2 SN 2
2.79
Although DO2 / N2 ! 1 , the net flux of N2 is larger than that of O2 due to the difference in partial pressures in the feed.
7.
If the feed pressure were low, we could use Eq. (8112) to calculate J1 , the flux of carbon dioxide, and J2 , the flux of methane. Equation (8113) then gives the composition (y) of the permeate. However, because the pressure of the feed is high, we must allow for the effect of pressure on nonideality of the gas phase. Equation (8111) is J1
D1 M c1F c1MP GM
(1)
where c1M is the concentration of carbon dioxide in the membrane; subscripts F and P refer to feed and permeate. To find c1MF , we use the equilibrium relation
( y1PM1 )F
ª § v1P · º M « H1c1 exp ¨ ¸» © RT ¹ ¼ F ¬
(2)
where H1 and v1 are Henry’s constant and partial molar volume for carbon dioxide in the membrane, both at 300 K and 100 bar. Fugacity coefficient M1 is given by the virial equation of state, truncated after the second virial coefficient [Eq. (533)]:
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Solutions Manual
§ 2 · P ¨2 yi B1i Bmixt ¸ ¨ ¸ RT © i 1 ¹
¦
ln M1
(3) 2
2
¦¦ yi y j Bij
Bmixt
i 1j 1
Substituting the given temperature, pressure and second virial coefficients into Eq. (3), we obtain M1
0.729
Substituting Eq. (4) and all other given data into Eq. (2) yields 0.314 mol L1
c1MF
(4)
To find c1MP we use the equilibrium relation ( y1PM1 )P
( H1c1M )P
(5)
y1P 19
mol L1
(6)
where PP 1 bar and M1P 1. Hence, Eq. (5) reduces to c1MP
The quantity y1P is an unknown in this problem. Substituting Eqs. (4) and (6) into Eq. (1) gives
§ mol · J1 ¨ ¸ ¨ cm2 s ¸ © ¹
y 5 u 106 § · 0.314 u 103 1P u 10 3 ¸ ¨ 0.1 © 19 ¹
(7)
Applying the same procedure for methane (2), we obtain § mol · J2 ¨ ¸ ¨ cm2 s ¸ © ¹
y 50 u 10 6 § · 0.155 u 10 3 2 P u 10 3 ¸ ¨ 0.1 © 50 ¹
(8)
The (steadystate) mole fraction of carbon dioxide in the permeate is given by [Eq. (8113)] y1P
J1 J1 J2
(9)
Further, the mass conservation gives y1P y2 P
1
(10)
Substituting Eqs. (7), (8) and (10) into Eq. (9) gives y1 P = 0.168 for carbon dioxide in the permeate. Therefore, for methane in the permeate, y2 P = 0.832.
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Solutions Manual
8. a)
Flux of water through the membrane is given by Eq. (8128): Jw
ª v w ( PF PP ) º ½ § permeability · L L ¨ thickness ¸ u ® xwF xwP exp « »¾ RT © ¹ ¯ ¬ ¼¿
(1)
L where xwP 1 (pure water in the permeate); PP Pws 0.0312 atm. To calculate concentration of water in the feed, we use
( Pw )F
( xwL Pws )F
(2)
with Pws
0.0312 atm (1 0.0184) u (0.0312) atm
PwF
Therefore, we obtain L xwF
0.9816
Because the permeate is pure water, we obtain vw  vw
18.015 g mol 1 0.997 g cm 3
18.069 cm 3 mol 1
The feed pressure is thus given by 7.2 u 10 4 g cm 2 s1
§ 2.6 u 10 5 g cm cm 2 s1 · ¨ ¸ ¨ ¸ 10 u 10 4 cm © ¹ ª (18.069 cm3 mol 1 ) u ( PF 0.0312 atm) º °½ ° u ®0.9816 (1) u exp « »¾ 1 1 °¯ ¬« (82.06 atm L K mol ) u (298.15 K) ¼» ¿°
Therefore, the feed pressure is PF = 63.93 atm
b) 1u 106 gallons/day
3785.4 m3 /day
0.0438 m3 /s
43.67 kg/s
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Solutions Manual
Flux
7.2 u 10 4 g cm 2 s1
43.67 u 103 g s1 ( A cm 2 )
where A is the membrane area needed. Solving Eq. (3) for the area, we get A
6.07 u 10 7 cm 2
6.07 u 10 3 m 2
6.5 u 10 4 ft 2
(3)
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
1.
9
The solubility product is the equilibrium constant for the reaction Ag+ + ClAqueous solution
AgCl Solid defined as K SP
(a )(a ) Cl Ag
being the standard states the pure solid AgCl and the ideal dilute 1molal aqueous solution for each ion.
a)
Let S be the solubility of AgCl in pure water, in a Ag K SP
J
Ag
S
(a )(a ) Cl Ag
mol AgCl (a molality). kg water
a Cl (J
Ag
J S Cl
)(J ) S 2 Cl
(J r S ) 2
(1)
Because the solution is very dilute, J r  1 and S  K SP and is of the order of 105 molal and therefore the ionic strength is also very low: I  131 . u 10 5 molal. Therefore we may apply the DebyeHückel limiting law. I
1 [m u (1) 2 m u (1) 2 ] 2
m
S
Using Eq. (950a) of the text,
133
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Solutions Manual
log J (rm)
0.510 u (1) u (1) I 1/ 2
0.510 S 1/ 2
(2)
We now replace J r given by Eq. (2) into Eq. (1) and solve for the solubility S: 1/ 2
[(10 0.510 S
172 . u 10 10
)S ]2
S = 1.31 ×105 mol kg1
( J r  10 . )
b) With the addition of NaCl the ionic strength increases and we need to evaluate J r because now the solution is not very dilute and therefore we don't have J r  10 . . Let S be the new solubility of AgCl in this aqueous solution that contains NaCl. The molalities are m Ag
S
m Cl
S 0.01
The total ionic strength (due almost exclusively to NaCl because S is small) is I
1 [S u (1) 2 S u (1) 2 0.01 u (1) 2 0.01 u (1) 2 ]  0.01 mol kg 1 2
We use in this case the extended limiting law [Eq. (952)] with AJ
ln J r
1174 . u (0.01)1/ 2 1 (0.01)1/ 2
Jr
1174 . kg1/2 mol 1/2 :
0.90
As in a), K SP
(a
Ag
)(aCl )
(J
Ag
)( J Cl )(m
Ag
)(mCl )
( J r )2 (S )(S 0.01) 1.72 u 10 10
or substituting J r
0.90 ,
S (S 0.01)
212 . u 10 10
Because S is small and S PUCEP V
T
V
T
L+V
L+V
L’+ V
L L+L’
L
L’ L+L’
1
3.
x
2
x
1
Let A stand for alcohol. For alcohol distributed between phases ' and " ' ' xA JA
" " xA JA
Then, K
F xA' I G "J A o0 H xA K lim
x
F J "A I G'J A o0 H J A K lim
x
At 0qC and 1 bar, ln J 'A
2400 ' )2 (1 x A RT
ln J "A
320 " )2 (1 xA RT
The pressure correction to JA is
J A (P2 ) The temperature correction is:
J A (T2 ) Thus, we can write,
z z
J A (P1 ) exp
J A (T1 ) exp
P2
P1
vE dP RT
T2
h E
T1
RT 2
dT
2
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Solutions Manual
ln J 'A
[
100 2400 16 dP (8.31451) u (273) 1 (8.31451) u (273)
³
303
4800
³273 (8.31451) u (T 2 ) dT ]u (1 xA )
'
2
' 2 0.9178u (1 xA )
[
ln J"A
100 320 10 dP (8.31451) u (273) 1 (8.31451) u (273)
³
303
600
³273 (8.31451) u (T 2 ) dT ]u (1 xA )
" 2
0.0712(1 x"A )2
This gives K
4.
lim
xA o 0
J "A
0.429
J 'A
For pure benzene, neglecting fugacity coefficients and assuming constant density of each phase with respect to pressure,
fBL PBs,L exp with v L
v L (P PBs,L )
87.7 cm3 mol 1 and v
RT
s
s
fB PBs,
s exp v s (P P s ) s, B
RT
77.4 cm 3 mol 1 , as obtained from density data.
ª (87.7) u (200 P s,L ) º B » PBs,L exp « (83.1451) T «¬ »¼
PBs,
s exp ª« (77.4) u (200 PBs,s ) º» «¬
(83.1451)T
»¼
Temperature T can be found from the intercept of the curves obtained by representing each side of the last equation as a function of temperature.
s
In an alternate way, we can express PBs,L and PBs, from vaporpressure equations:
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Solutions Manual
10 (7.96221785/T ) 750.06
PBs,L
PBs,
s
10 (9.846 2310 /T ) 750.06
which, together with the last equation, can be solved for T: Tm(200 bar) = 284.4 K
5.
The RedlichKwong equation is: RT a v b T 1/ 2v(v b)
P
with a
2a 2 zA AA 2zA zBaAB zBaBB
¦ ¦ zi z j aij i
j
¦ zi bi
b
zA bA zBbB
i
Assuming that aAB is given by the geometric rule, aAB
we get for zA
zB
(aAA aBB )1/ 2
0.5, a
4.35 u 10 8 bar (cm3 mol1) K1/2
b = 91.5 cm3 mol1 Substitution in the RK equation gives for total pressures:
P
413 . bar
Because this result is absurd, we use Henry’s constant data to find aAB . For infinitely dilute solutions of A in B,
H A,B
(PM A ) xA
f PBs M B
s At infinite dilution, Ptotal # Ppure B which can be obtained from the RK equation with the
appropriate constants [ a
4.53 u 10 8 bar (cm3 mol1)2 K1/2 and b
82.8 cm3 mol1].
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Solutions Manual
This gives, PBs
113 . bar
Therefore, H A,B
f MA
PBs
7.01 113 .
6.195
For the RK equation, fugacity coefficients are given by: f ln M A
ln
LM N
OP Q
b abA 2aAB v vb vb b Pv A ln ln ln 3/2 3/2 2 v b v b RT b v v vb RT RT b
Using b # bB , a # aB and v # v B (infinite dilution of A). Solving for aAB , aAB
3.963 u 10 8 bar (cm3 mol1)2 K1/2 P
Then, for the mixture, a
4.159 u 10 8 bar (cm3 mol1) K1/2 b
91.5 cm3 mol1
Calculating again the pressure we obtain,
P
6.
4.14 bar
Let 1 = C2H6 and 2 = C6H6. The K factor of component i is defined as Ki
yi xi
M iL (P, x ) M Vi (P, y)
with M
c (0) c (1) P c (2) zi
Thus, we need to solve for P and y1 (or y2). At equilibrium, y1M1V
x1M1L
(1)
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Solutions Manual
(1 y1 )M V2
(1 x1 )M 2L
(2)
From the given equations, we rewrite Eqs.(1) and (2) for x1 = 0.263: (12545 . 2.458 u 10 4 P 0.4091y1 ) y1
11699 . 0.008345P
[0.74265 7.0069 u 10 3 P 0.50456(1 y1 )](1 y1 )
0.24596 0.001874 P
The above equations can be solved (either graphically or numerically) for P and y1:
y1
P
59 atm
0.715
( y2
0.285)
Then,
[From Kay’s data: K1
7.
K1
y1 x1
0.715 0.263
2.72
K2
y2 x2
0.285 0.737
0.387
2.73 and K 2
0.41 ].
The stability criterion is [see Eq. (6131) of text]:
F w2 g E I RT F 1 1 I 0 GH wx 2 JK GH x1 x2 JK 1 T ,P We need an expression for gE valid at high pressures. Because
F wg E I GH wP JK T ,x
we write g E (T , P, x )
g E (T , P
1 atm, x )
(RT ) u (1877 . ) x1 x 2
vE
z z
P E
1 P
1
v dP
x1 x 2 (4.026 0.233 ln P )dP
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Solutions Manual
Thus, g E (T , P, x )
(RT ) u (1877 . ) x1 x 2 (P 1) u (4.026) x1 x 2
(P 1) u (0.233) x1 x2 0.233P (ln P ) x1 x 2 (42043 4.259P 0.233P ln P) x1 x 2
For gE of this form ( g E 6.12):
Ax1 x 2
Ax1 x 2 , where A is a constant), the stability criterion is (see Sec. A !2 RT
or
42043 4.259P 0.233P ln P ! (2) u (82.0578) u (273) Solving for P,
P 1046 atm (or 1060 bar) At pressures higher than 1060 bar, the system splits into two phases. To solve for the composition at a higher pressure, we use: x1' J 1'
(1 x1' )J '2
x1" J 1"
(1 x1" )J "2
where (42043 4.259P 0.233P ln P ) x 2j
RT ln J i
At 1500 atm (or 1520 bar) and 273 K, ln J i
2.0477 x 2j
Thus, x1' exp[2.0477(1 x1' ) 2 ]
(1 x1' ) exp[2.0477( x1' ) 2 ]
x1" exp[2.0477(1 x1" ) 2 ]
(1 x1" ) exp[2.0477( x1" ) 2 ]
Solving (either graphically or numerically), we obtain x1'
0.37
( x2'
0.63)
x1"
0.63
( x2"
0.37)
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Solutions Manual
8.
We want to relate hE to volumetric data. Relations given in Chapter 3 of the text may be used. We write hE at any pressure P relative to hE at 1 bar as:
z LMMN P
h E (P ) h E (1 bar)
vE T
1
F wv E I OP dP GH wT JK P PQ
Thus, we need the above integrand as a function of pressure at 333 K. From volumetric data, using linear regression at each pressure between 323 K and 348 K,
F wv E I GH wT JK 1 bar
F wv E I GH wT JK 100 bar
0.0186
F wv E I GH wT JK 250 bar
0.0154
F wv E I GH wT JK 500 bar
0.01239
0.00963
Using linear interpolation, at 333 K, v E (1 bar) 1.091
v E (100 bar)
0.9638
v E (250 bar)
v E (500 bar)
0.6846
0.8284
If F (P)
vE T
F wv E I GH wT JK P
then: P (bar)
1
100
250
500
F(P) (J bar mol1)
0.5102
0.4164
0.3297
0.2522
Using a trapezoidrule approximation,
³
360
F ( P)dP
128 J mol 1
1
Therefore, at 333K h E (360 bar)
h E (1 bar)
360
³1
F ( P)dP
1445 128
h E (360 bar, 333K)
1317 J mol 1
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Solutions Manual
9.
For condensation to occur, L V fW ! fW
To find the temperature for condensation (at constant pressure and vapor composition), we solve the equilibrium relation L fW
V fW
The liquid phase is assumed to be pure water. Its fugacity is given by L fW
L f pure exp
z
P
s PW
vW dP RT
As a good approximation, let s s L # P s exp v W ( P Pw ) fW W RT
(obtain data from Steam Tables)
Thus, we are neglecting M sW and we assume that (liquid) water is incompressible over the s and P (150 atm). pressure range between PW
The vapor phase is described by an equation of state. Therefore, V fW
yW M VW P
To obtain M VW , we use the RedlichKwong equation of state: P
RT a v b T 1/ 2v(v b)
(1)
from which we obtain ln MW
ln
b v §vb· W (2¦ y j aWj )( RT 1.5b)1 ln ¨ ¸ vb vb © v ¹ j
(2)
b · Pv § vb ¨ ln v v b ¸ ln RT ¹ b ©
abW RT
1.5 2
where v is the molar volume of the mixture and
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Solutions Manual
a
¦¦ yi y j aij i
b
j
¦ yi bi i
In these equations, aW and aCO2 are given as functions of temperature; the crosscoefficient is
aij
(ai (0) a j (0) )1/ 2 0.5R 2T 2.5 K
V Use the following procedure: With a trialanderror procedure, we can calculate f W .
1. 2. 3. 4. 5.
Guess temperature. Calculate v from equation of state [Eq. (1)]. Use T and v (along with P and y) to calculate M W from Eq. (2). Calculate the fugacity of vapor. V with saturation pressure of water at that temperature. Compare f W
Typical results are: T (K)
MW
V fW (atm)
475 500 525 550
0.588 0.633 0.711 0.746
17.7 19.9 21.3 22.4
V and L as a function of temperature (see figure), we see that Plotting f W fW L fW
V fW
at T  482 K. That is the temperature where condensation first occurs (dewpoint temperature of the mixture).
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Solutions Manual
Fugacity of water, atm
100
L fW
75
50
V fW
25
0 400
450
500
550
600
Temperature, K
Fugacity of water in vapor phase and in liquid phase at P = 150 atm.
10. a)
For equilibrium between solid solute and solute dissolved in the supercritical fluid,
s
f2 (P, T ) or
s
d ln f2
f2f (P, T , y2 ) d ln f2f
(1)
where subscript 2 refers to solute and superscript f to fluid phase. Expanding Eq. (1) with respect to T, P and composition (see Sec. 12.4), we obtain (temperature is constant):
F w ln f2s I GG wT JJ H K P, y F w ln f2f I GG wT JJ H K P, y
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Solutions Manual
F w ln f2s I GH wx2 JK P,T
(pure solid solute)
F w ln f2f I F fI GG wP JJ dP GG w lnwy2f2 JJ dy2 H K T ,y H K T ,P
F w ln f2s I dP GH wP JK T
(2)
But because
F w ln f2s I GH wP JK F w ln f I GG wP JJ H K
s
v2
RT
T
f 2
v2f RT T
Equation (2) becomes:
s
(v 2 v2f ) dP RT
F w ln f2f I GG wy2 JJ dy2 H K T ,Pdy
F w ln f2f I GG w ln y2 JJ d(ln y2 ) H K T ,P
(3)
Finally, because
f2f
y2M 2 P
Equation (3) becomes
FG w ln y IJ H wP K
s
2
b)
T
v 2 v2f RT w ln M 2 1 w ln y2 T ,P
FG H
(4)
Maxima (or minima) occur when
FG w ln y IJ H wP K 2
v 2s
IJ K
0 T
Because w ln M 2 / w ln y2 is always greater than –0.4, the above derivative is zero when
v2f . f It is necessary, then, to calculate v2 as a function of pressure.
Using Eq. (1241),
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Solutions Manual
FG wP IJ H wn2 K T ,V ,n1 FG wP IJ H wV K T , all n
v2
and the RedlichKwong equation of state with the mixing rules,
¦¦ xi x j aij
a
i
j
¦ xi bi
b
i
we obtain
v2
b · RT § 1 2 v b ¨© v b ¸¹
2(¦ xi a2i ) ab2 /(v b) i
v(v b)T 1/ 2 RT a ª 2v b º 2 1/ 2 « 2 2» ( v b) T ¬« v (v b) ¼»
Assuming that the fluid phase is almost pure solvent, v, a and b are those for pure solvent 1. Cross parameter a12 is given by: a12
(a11a22 )1/ 2 (1 k12 )
Constants are: bar (cm3 mol 1 )2 K1/2
a11
0.7932 u 108
a22
0.11760 u 1010 bar (cm3 mol 1 )2 K1/2
a12
0.3264 u 109
b1
bar (cm3 mol 1 )2 K1/2
40.683 cm3 mol 1
b2
140.576 cm3 mol 1
Using volumetric data for ethylene at 318 K (IUPAC Tables), and because
s
v2
128174 . 1144 .
the maximum (and minimum) occurs ( v2
112 cm3 mol 1
s
v 2 ) at (see figure below)
minimum = 19 bar
maximum = 478 bar These values are in good agreement with results shown in Fig. 539 of the text.
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Solutions Manual
500
Solubility minimum
Solubility maximum
s
v 2 =112 0
10
v2 (cm3 mol 1)
19 bar 500
100
1000
10,000
P (bar)
478 bar
1000
1500
2000
Partial molar volumes of naphthalene infinitely dilute in ethylene at 318 K calculated from RedlichKwong equation of state with k12 = 0.0182.