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
2000 Prentice Hall PTR Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Electronic Typesetting: Edmundo Gomes de Azevedo
All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.
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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
S O L U T I O N S T O P R O B L E M S
C H A P T E R
2
1. From problem statement, we want to find
b
wP/wTg
_{v}. Using the productrule,w w
F
HG
I
KJ
w wF
HG
I
KJ
wwF
HG
I
KJ
P T T P P T v v v By definition, DP P T w wF
HG
I
KJ
1 v v and NT T P w wF
HG
I
KJ
1 v v Then, w wF
HG
I
KJ
u_{u} P T P T v D N 1 8 10 5 32 10 33 8 5 6 . . . bar C 1 DIntegrating the above equation and assumingD_{P}and N_{T} constant over the temperature range, we obtain 'P P T D N 'T For'T = 1qC, we get 'P 33.8 bar 1
2 Solutions Manual
2. Given the equation of state,
P V n b RT
F
HG
I
KJ
we find: w wF
HG
I
KJ
wwF
HG
I
KJ
S V P T nR V nb T V w wF
HG
I
KJ
w wF
HG
I
KJ
S P V T nR P T P w wF
HG
I
KJ
wwF
HG
I
KJ
U V T P T P T V 0 w wF
HG
I
KJ
wwF
HG
I
KJ
wwF
HG
I
KJ
U P U V V P T T T 0 w wF
HG
I
KJ
w wF
HG
I
KJ
H P T V T V nb TFor an isothermal change,
'S P T dV nR V nb V nb nR P P V V V _{w} w
F
HG
I
KJ
z
_{1}2 _{2} 1 1 2 ln ln 'U U V V P dP T T P P w wF
HG
I
KJ
wwF
HG
I
KJ
L
NMM
O
QPP
z
1 2 0 'H T V T _{P} V dP nb P P P P w wF
HG
I
KJ
L
NMM
O
QPP
z
2 1 1 2b
g
'G 'H T S' nb P P nRT P P _{2} _{1}F
HG
1I
KJ
2b
g
ln 'A 'U T S' nRT P P lnF
HG
1I
KJ
23 Solutions Manual
3. This entropy calculation corresponds to a series of steps as follows:
s T P (saturated liq. = 298.15 K = 0.03168 bar) 1 s T P (saturated vapor, = 298.15 K = 0.03168 bar) 2 s T P (vapor, = 298.15 K = 1 bar) 3 s_{3} 's_{1}_{o}_{2}'s_{2}_{o}_{3}s_{1} vap 1 1 1 2 vap (2436) (18.015) 147.19 J K mol 298.15 h s s T o ' _{u} ' ' 's T _{P} dP P P 2 3 2 3 o w w
F
HG
I
KJ
L
N
MM
O
Q
PP
z
v Because v RT P (ideal gas), 3 2 3 2 1 1 ln 1.0 (8.31451) ln 0.03168 28.70 J K mol P s R P o § · ' _{¨} _{¸} © ¹ § _{u ¨} © ¹ · ¸ 0 3 (H O, vapor)2 147.19 28.70 69.96 s s 1 1 188.45 J K mol4 Solutions Manual 4. Because D RT P v, P RT RT D v 2 3/_{v}2 _{v} or P RT v v v 2 2 3 2 3 3 2 3 ( 6) (2 3 ) T P RT w § · ¨_{w} ¸ © ¹ v v v _{v} _{v} 2 ,
As T = 373.15 K, R = 0.0831451 bar L K1_{ mol}1_{, and molar mass is 100 g }
mol1, v _{2 3}_{. L mol}1 w w
F
HG
PI
KJ
u T v 3 3245 3 3245 10 8. bar L mol1 . Pa m 3mol
w g k_{c} P T 2 2 3 3 2 8 1 1 4 1 100 10 2 3 10 3 3245 10 24 621 w w
F
HG
I
KJ
F
HG
I
KJ
u u uF
HG
I
KJ
uF
_{HG}
uI
_{KJ}
uF
HG
uI
KJ
v v kg m N s mol kg m mol N mol m m m s 2 3 2 3 2 2 ( . ) . . , w _{157 m s}15 Solutions Manual
(1) Isothermal expansion tov = f (ideal gas state)
(2) Isochoric (v is constant) cooling to T_{2} (3) Isothermal compression to v_{2} V T v = (1) (2) (3) (T v, ) (T v, ) 1 1 2 2
For an isentropic process,
's 's_{1}'s_{2}'s_{3} 0 Because s = s(v, T), ds s d s T dT T w w
F
HG
I
KJ
w wF
HG
I
KJ
v v _{v} or ds P T d c T dT w wF
HG
I
KJ
v v vby using the relations
w w
F
HG
I
KJ
wwF
HG
I
KJ
S P T T v _{v} (Maxwell relation) w wF
HG
I
KJ
wwF
HG
I
KJ
s T T u T c T v v v 1 then, 2 2 1 1 0 T T c P P s d dT T T T f f w w § · § · ' _{¨} _{¸} _{¨} _{¸} w w © ¹ © ¹³
vv _{v} v³
v³
vv _{v}dvUsing van der Waals’ equation of state, P RT b a v v2 w w
F
HG
I
KJ
P T R b v v Thus,6 Solutions Manual 2 1 0 2 1 ln T T b c s R d b T § · ' _{¨} _{¸} © ¹
³
v v v T To simplify, assume cv0 c0pR v2 2 2 RT P Then, 2 0 2 1 1 2 ln p ln RT b c R P T b R T _{§} _{·} § · ¨ ¸ _{¨} _{¸} ¨ ¸ _{©} _{¹} © ¹ v 2 2 (82.0578) ( ) 45 623.15 ln (3.029) ln 600 45 T T u ª º u « _{} » ¬ ¼ T2 203 K 6. P RT b a RT b a RT _{}L
N
MM
MM
O
Q
PP
PP
v v v v v 2 1 1 Because b 2 2 1 v , 1 1 1 _{2} 2F
HG
I
KJ
b b b v v v " Thus, P RT b a RT bF
HG
I
KJ
L
NMM
O
QPP
v 1 v v 1 2 2 " or P RT b a RT b v v v 1F
HG
I
KJ
1 2 2 "7 Solutions Manual Because z P RT B C v v v 1 2 "
the second virial coefficient for van der Waals equation is given by
B b a RT 7. Starting with du TdsPdv w w
F
HG
I
KJ
wwF
HG
I
KJ
w wF
HG
I
KJ
w wF
HG
I
KJ
w wF
HG
I
KJ
u P T s P P P T T P P T T P T v v v T As v RT P B RT P b a T2 w wF
HG
vI
KJ
T R P a T P 2 3 w wF
HG
vI
KJ
T RT P T 2 Then, w wF
HG
uI
KJ
P a T T 2 2 2 0 2a u S § ·d ' _{ ¨} _{¸} W © ¹³
P 2 2a u S ' W8 Solutions Manual 8. The equation P n T m RT
F
HG
I
KJ
v2 1 2/b
vg
can be rewritten as (PT1 2/ )v3(PmT1 2/ )v2nvnm RT3/2 2v v3 v2 v 1 2 1 2 0F
HG
mRTI
KJ
F
HG
I
KJ
P n PT nm PT / / (1)At the critical point, there are three equal roots for v = v_{c}, or, equivalently, w w
F
HG
PI
KJ
F
_{HG}
w_{w} PI
_{KJ}
T Tc _{T Tc} v v 2 2 0 (2) v v _{c} v v v_{c} v v v_{c} _{c}b
g
3 3 _{3} 2 _{3} 2 3 _{0}Comparing Eqs. (1) and (2) at the critical point, m RT P c c c 3v (3) n P T_{c c}1 2 c 2 3 / v (4) nm P T_{c c}1 2 c 3 / v (5)
From Eqs. (3), (4), and (5) we obtain m vc 3 (6) vc c c c c RT P m RT P 3 8 or 8 n P T R T P c c c c c 3 27 64 2 1 2 2 5 2 v / /
9 Solutions Manual P RT m n RT
F
H
GG
G
I
K
JJ
J
v v v 1 1 3/2 or z P RT m n RT v v v 1 1 3/2From critical data,
m 0 0428. L mol1 n 63 78. bar (L mol1 2) K1/2 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_{ g}1_{, 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 T( _{0},v o f ) 0 Then, u T u T u T u T u T u T u T u d u T dT T T T ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) lim lim v v v v v v v v v v _{v} v v _{v v} o f o f o f o f w w
F
HG
I
KJ
w wF
HG
I
KJ
ofz
_{f} ofz
_{ f} 0 0 0 (1) Schematically we have: 1 2 Ref. state Ideal gas ( , T v ) 3 0 v = v Ideal gas Intermediate state Ideal gas ( ,T v ) State of interest Real gas ( , )Tv10 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 volumev o f, in a twostep process consisting of an isochoric step and an isothermal step.
In the step 1o2 the gas is infinitely rarified, and hence exhibits ideal gas behavior. Then, the second integral in Eq. (1) gives:
0 0 0 0 0 0 0 lim ( ) ( )( T T T p p T T T u dT c dT c R dT c R T T f of w § · ¨_{w} ¸ © ¹
³
³
³
v v v _{v v} ) (2)because for an ideal gas and because, by the problem statement, the heat capacity at constant pressure of the gas is temperature independent.
c0_{p}c_{v}0 R
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), w w
F
HG
I
KJ
wwF
HG
I
KJ
u T P P T T v v (3)Making the derivative using the equation of state give we obtain w w
F
HG
I
KJ
u RT b RT b a b a b T v v v v v( ) v v( ) (4) Then, 1 1 ln ln ln u a a b d d b b b b a b b b f f f f f f ª § · º w § · ª _{} º _{} « ¨ ¸ » ¨_{w} ¸ « _{} » _{} © ¹ ¬ ¼ ¬ © ¹ ¼ § · ¨ ¸ © ¹³
vv³
vv v v v v v v v v v v v v v (5) and lim ln v _{v} v v v v v of _{f} w wF
HG
I
KJ
F
HG
I
KJ
z
u d a b b T (6)Combining Eqs. (1), (2) and (6) we obtain the desired expression for the molar internal energy, u T c R T T a b b p ( , )v ( )( ) ln v v
F
HG
I
KJ
0 011 Solutions Manual
10.
such that
2
lnJ _{w} A(1x_{w}) J_{w}o 1 as x_{w}o 1 Using GibbsDuhem equation,
ln ln 0 w w s s x d J x d J or, because dx_{w} dx_{s} (x_{w}x_{s} 1), x d dx x d dx w w w s s s lnJ lnJ d dx A x A x w w w w ln ( )( ) ( ) J 2 1 1 2 1 Then, 2 (1 ) ln _{s} s s _{s} 2 (1 _{s}) s Ax x d dx A x J x dxs ln 0 ln 2 0 (1 ) s xs s s s d A x J J
³
³
dx 2 ln 2 2 s s s x A x§ · J ¨ ¸ ¨ ¸ © ¹ lnJs A x( w2 1)11. Henry’s law for component 1, at constant temperature, is f_{1} k x_{1 1} (for 0x_{1}a) where k_{1}is Henry’s constant.
For a liquid phase in equilibrium with its vapor, _{i}L f_{i}V. If the vapor phase obeys
idealV
f gas law,
1
f_{i} y P_{i} .
Henry’s law can then be written:
y P_{1} k x_{1} Taking logarithms this becomes
1
1 1
ln(y P) lnk lnx Differentiation at constant temperature gives
12 Solutions Manual 1 1 1 1 1 ) ln 1 y P d x dx dx x ln( d
Using the GibbsDuhem equation x d P dx x d P dx 1 1 1 2 2 1 0 ln ln gives 2 2 1 ln( ) 1 x d y P 0 dx dx_{2} dx_{1}, or, because 2 2 2 ln( ) 1 d y P x dx or, ln(y P_{2} ) ln _{2} d d x Integration gives 2 2 ln(y P) lnx lnC lnC
where is the constant of integration.
For , and . This gives and we may write
2 x_{2} , y1 _{2} 1 P P_{2}s C P_{2}s 2 2 2 2 ln(y P) lnx lnPs ln(x Ps) or [for y P_{2} P_{2} x P_{2 2}s (1a)x_{2}1] which is Raoult’s law for component 2.
12. Starting from dg_{i} RTdlnf_{i},
*
* *
(at )
ln ( is a low pressure where
(at ) gas is ideal) P P i i i f P g RT P f P i o '
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 = 1 bar. 1 5 bar J g1DC Then, at 320qC, P* 'h h_{70}_{bar}h_{1}_{bar} 196J g1 1bar 2 21. 's s_{70} s 'g 1117 8. J g1 20137J mol1 Thus, ln . . . f_{i 70} _{320} 20137 8 31451 59315 4 08 bar, qC _{u}
b
g
_{b}
_{g b}
_{g}
or f = 59.1 bar13. The virial equation for a van der Waals gas can be written (as shown in Problem 6) v RT
P b
a
RT (1)
At the Boyle temperature,
B b a RT 0 or b a RT The Boyle temperature then, is given by
T a
bR
B (2)
is The JouleThomson coefficient
P w w
F
HG
TPI
KJ
_{H}14 Solutions Manual P w w
F
HG
I
KJ
w wF
HG
I
KJ
w wF
HG
I
KJ
w wF
HG
I
KJ
T P T H P H H P c H P T T p Because w wF
HG
I
KJ
w wF
HG
I
KJ
H P _{T} v T T _{P} v 0 P , w wF
HG
I
KJ
and because c_{p} is never zero, when H P _{T} 0 . Substitution in Eq. (1) gives
2 T H RT a RT a b P P RT P R a b RT w § · § ¨_{w} ¸ ¨ © ¹ © T · ¸ ¹
The inversion temperature is
T 2a Rb
JT
Comparison with Eq. (2) gives
JT 2 B
T T
14. At equilibrium,
G
1 1
where subscript 1 stands for the solute.
At constant pressure, a change in temperature may be represented by fL f d f dT 1 _{dT} d f1 dT dT G P L P ln ln
F
HG
I
KJ
F
HG
I
KJ
(1)depends only on T (gas composition does not change.) However, (at constant pressure) depends on T and x_{1} (or ):
Since the solvent is nonvolatile, f_{1}G (at constant pressure)
15 Solutions Manual d f dT dT f T dT f x d x L P L P x L T P ln ln ln ln ln , , 1 1 1 1 1
F
HG
I
KJ
wwF
HG
I
KJ
w wF
HG
I
KJ
(2) Further, dlnf_{1}G h_{1}0 h_{1}G dT RT P 2F
HG
I
K
J
(3) d f T RT 1 1 2 w h h L P x L ln , 0 1F
HG
KJ
I
(4) where: = idealgas enthalpy of 1; = realgas enthalpy of 1; h_{1}0 h_{1}Gh_{1}L = partial molar enthalpy of 1 in the liquid phase. Assuming Henry’s law,
f x 1 1 constant or w w
F
HG
lnlnI
KJ
, f x L T P 1 1 1 (5)Substituting Eqs. (2), (3), (4), and (5) into Eq. (1), we obtain d x d T h R ln ( / ) 1 1 1 '
From physical reasoning we expect h_{1}G !h_{1}L. Therefore 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
3
1. The Gibbs energy of a mixture can be related to the partial molar Gibbs energies by g g y g_{i} _{i} g_{i} i m o
_{¦}
e
j
1 (1)Since, at constant temperature,
o
ln ,
dg RTd f we may integrate to obtain
or
ggo RT fmixt RT fmixt o
ln ln
ggo RTlnfmixtRTlnP (2)
where subscript mixt stands for mixture.
For a component in a solution, d g_{i} RTdlnf_{i}. Integration gives g_{i}g_{i}o RTlnf_{i}o
o _{ln(} _{)}
i i i
g g RT y P (3)
Substituting Eqs. (2) and (3) into Eq. (1) gives
f P y_{i} i m i m mixt
¦
1 1 lnf_{i}_{¦}
y_{i}ln(y P_{i} ) ln ln lnf_{mixt} lnP y ln f _{i}ln y y P i i i i m i m_{¦}
F
HG
I
KJ
_{¦}
1 1 (4) Because 17Solutions Manual 18 ln
¦
y_{i} P P¦
y P i m i i ln ln mF I
HG KJ
1 1 Eq. (4) becomes lnf y ln f y i i i i m mixtF
HG
I
KJ
¦
1 (5)Assuming the Lewis fule, fi y fi pure_{i}, Eq. (5) becomes
or lnf y_{i}lnf _{i} i m mixt
¦
pure 1 f fyi _{i} m mixt pure i=1
2. As shown in Problem 1, lnf y ln f y i i i i m mixtF
HG
I
KJ
¦
1This result is rigorous. It does not assume the Lewis fugacity rule. Using fugacity coefficients,
P f_{i} M_{i i}y and P lnf_{mixt} y_{A}lnM_{A}y_{B}lnM_{B}ln A B y y B A f_{mixt} M M P 0.25 0.75 (0.65) u(0.90) u(50) fmixt 41.5 bar
Solutions Manual 19
3. Purecomponent saturation pressures show that water is relatively nonvolatile at 25qC. Under these conditions the mole fraction of ethane in the vapor phase (y_{E}) is close to unity.
Henry’s law applies:
f_{E} H T x( ) _{E} The equilibrium condition is
or f_{E}V f_{E}L y_{E}M_{E}P H T x( ) _{E} At 1 bar, ME 1 and H T( ) P x/ E: H T( ) . . u u 1 0 33 10 3 03 10 4 4_{ bar}
At 35 bar we must calculate M_{E}:
lnM_{E}
z
z P dP P _{1} 0 Using P2 we obtain z 1 7 63 10. u 3P7 22 10. u 5 M_{E} 0 733.Because Henry’s constant H is not a strong function of pressure, P H x f H E E E M x_{E} x_{Ethane} u ) . 3 03 1 u u ( .0 733 (35) 04 8.47 10 4
4. The change in chemical potential can be written,
0 1 1 1 1 _{0} 1 ln f RT f § · ¨ ¸ 'P P P ¨ ¸ © ¹ (f_{1}0 1 bar) (1)
Solutions Manual 20
The chemical potential may be defined as1:
2 n_{2} 2 _{2} 0 1 1 1 _{, ,} 1 _{,} 0 1 _{, ,} 1 _{,} T P n _{T} T V n _{T n} G G n n A A RT n n § · §w · w P P ¨ ¸ ¨_{¨} ¸_{¸} w w © ¹ _{©} _{¹} § · §w · w ¨ ¸ ¨ ¸ _{¨} _{¸} w w © ¹ _{©} _{¹} (2) 0
Combining Eqs. (1) and (2):
2 1 1 _{, ,} 1 ln 1 T V n A f RT n §w' · ¨ ¸ w © ¹ V n_{T}v , n_{T} n1n2,
Using total volume, 'A RT n Vn b_{T} n RT n RT V n V n V T
F
H
K
G
IJ
F
H
G
K
IJ
F
H
K
1 2Taking the partial derivative and substituting gives
G
IJ
ln _{1}ln _{2}ln 1 1 1 ln f b y RT b b v v or f y RT b b b 1 1 1F
HG
I
KJ
v exp vThe same expression for the fugacity can be obtained with an alternative (but equivalent) derivation: 1 0 0 0_{;} 0 0 0 0_{;} 0 T A U TS G U PV TS PV n RT 0 0 0 0 1 G 1 _{,} _{1} 1 1 _{,} _{1} j j j T n T n U S RT T n n n , 1 T n z z z · §_{w} · §_{w} · ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨_{w} ¸ © ¹ © ¹ © ¹ § w P ¨ ¨ _{w} ¸ _{w} and 0 0 0 1 _{,} _{1} 1 _{,} _{1} 1 _{,} _{1} j j T n T n T n A U S RT T n n n j z z z §_{w} · §_{w} · §_{w} · ¨ ¸ ¨ ¸ ¨ ¸ ¨_{w} ¸ ¨ _{w} ¸ ¨_{w} ¸ © ¹ © ¹ © ¹ then 0 0 1 1 _{,} _{1} j T n A RT n z §_{w} · P ¨ ¸ ¨w ¸ © ¹
Solutions Manual 21 0 1 0 1 1 _{0} 1 1 ln f ( 1 bar) RT f f § · ¨ ¸ P P ¨ ¸ © ¹ P1 1 _{1} w w
F
HG
I
KJ
z A n T V n j, ,. The Helmholtz energy change 'a
By definition, can be written as
0 0 1 ( ) T i i i i i n a A n a A n RT ' P
¦
¦
Then, 1, , 1, , 0 1 1 1 0 1 1 j T V j T V T n n n a A RT n n RT z z §w ' · §w · P ¨ _{w} ¸ ¨_{w} ¸ © ¹ © ¹ P P and 1 0 1 1 ªw(nT'a RT/ )º 1 _{, ,} 1 j n T V RT n z P P ¬ ¼Using the equation for
« _{w} » 'a , 1 1 1 , , 1 ( / ) ln ln 1 j T T T T n T V n a RT V V n b n V n b n RT z ªw ' º « _{w} » _{} ¬ ¼ Vn b or 1 1 _{, ,} j nz T V 0 1 1 1 1 1 ( / ) ln ln 1 ln T T T T n a RT f RT n n RT n b V n b V n b §_{P P} · _{ª}_{w} _{'} _{º} ¨ ¸ _{«} _{»} ¨ ¸ _{¬} _{w} _{¼} © ¹ 1 1 1 exp T T T n RT n b f V n b V n b § · ¨ ¸ _{©} _{¹} Hence, 1 1 1 exp y RT b f b b § · ¨ ¸ © ¹ v v
Solutions Manual 22
5.
a) :
For a pure component Starting with Eq. (351)
(n_{i} n_{T}): 0 0 ; i i i i i G G n n P P Because G UPVTS (1) From Eq. (352), 0 0 0 i i Tsi RT P P 0 0 ln i i V i i P RT V PV dV RT Ts n V n RT n f_{§} _{·} P ¨ ¸ P © ¹
³
i i (2) qs. (1) and (2), From E 0 _{ln} i i i V i i PV n RT P RT V dV RT n V n RT n f_{§} _{·} _{} P P ¨ ¸ © ¹³
i But, 0 ln _{i} _{i} _{i} RT f P P and V n RT z P i i Substitution gives ln ln _{i} ( _{i} 1) V i i f f§ § · P RT RT dV RT z RT z P n V · ¨ ¸ ¨ ¸ © ¹³
© ¹ b) with Eq. (353):For a pure component To use Eq. (353), we must calculate Starting , y_{i} 1. w w
F
HG KJ
nPI
i T V, ,pure component iSolutions Manual 23 1 i i V P n n V P i n P V §w · §w · §w · ¨_{w} ¸ ¨_{w} ¸ ¨_{w} ¸ © ¹ © ¹ © ¹ w w
F
HG
I
KJ
w wF
HG
I
KJ
wwF
HG
I
KJ
w wF
HG
I
KJ
L
N
MM
O
_{Q}
PP
P n V n P V P n P n V n P V i V i _{P} _{ni} i i i _{ni} But, P n V n P V n PV V i i _{ni} i _{ni} w wF
HG
I
KJ
wwL
NM
O
QP
1 ( ) Then, 1 , , 1 ( ) ( 1) i j n RT V i _{T V n} V i i PV V i i V i P P dV d PV n n n P PV dV RT n n P dV RT z n z f f f f §w · ¨ ¸ w © ¹³
³
³
³
³
Now Eq. (354) follows directly.
6. The solubility of water in oil is described by f_{1} H T x( ) _{1}
f_{1} bar1 Henry’s constant can be evaluated at 1 bar where . Then, H T f x ( ) 1 (t ) u 1 4 1 35 10 286bar 140DC
To o _{1} bles (e.g., Keenen and Keyes).
Alternatively, ge e Poynting factor to correct to 410 bar.
At 1
btain f at 410 bar and 140qC, use the Steam Ta t f at saturation (3.615 bar) and use th
Solutions Manual 24
T s
o ' o
(from Steam Tables)
1 1 410 bar 1 ln RT f 'g_{o} 'h 410 bar 1 410 bar RTlnf_{1} 282 J g1 ln ( ) ( ) ( . ) ( ) . f_{1} 282 18 8 31451 413 1 48 u u Then, f_{1} = 4.4 bar at 410 bar and 140qC and
x f H T 1 1 4 4 286 0 0154 ( ) . . 7. T = 300 K_{o} P = ?_{o} T = 300 K_{f} P = 1 bar_{f}
Applying an energy balance to this process, 'h hfho 0
This may be analyzed on a PT plot. Assume 1 mole of gas passing through the valve.
P T I II III o o T P, f f T P, p process applies:
I. Isothermal expansion to the idealgas state. II. Isobaric cooling of the ideal gas.
A threeste
III. Compression to the final pressure. For this process,
Solutions Manual 25 'h 'h_{I}'h_{II}'h_{III} 0 Since h = h(T,P), dh h P d P h T dT T P w w
F
HG
I
KJ
w wF
HG
I
KJ
o o T P o o o f 0 0 0 f f f f T P T P p P _{P} _{T} T _{P} _{T} h d h T d P c d T T dP T T ' ª _{§}_{w} _{·} º ª _{§}_{w} _{·} º « ¨_{w} ¸ » « ¨_{w} ¸ » © ¹ © ¹ « » « » ¬ ¼ ¬ ¼³
³
v v³
³
v v But, v RT P 50 T 105 Then, w wF
HG
TI
K
_{P} P T2 andJ
v R 105 c_{p}0 y c_{A} _{p}0_{,A}y c_{B} 0_{p}_{,B} 'h P y T cp y cp T T T P uF
H
K
uG
I
J
0 0F
_{HG}
50 2 10I
_{KJ}
5 o A B f o f fjb
g
,A ,B Subs o ) u u u33 5J mol 1 K 1 10bar cm J3 1 200 300K 950 1bar Po = 55.9 bar 0 50 2 10 5
b g e
o tituti n gives (616 7. )uP_{o} ( . ) ( ) ( ) ( ) (8. From the GibbsHelmholtz equation:
w
F
HG
I
KJ
w g T T h T2Solutions Manual 26 or, alternately, w
F
HG
I
KJ
wF I
' T h h h 1 eal idealHG KJ
'g T r _{(1)}Because, at constant temperature,
d g RTd f
P '
b g
lnF
_{HG}
I
_{KJ}
(2)we may substitute Eq. (2) into Eq. (1) to obtain
R f P T h w
F
HG
I
KJ
wF
HG
I
KJ
ln 1 'From the empirical relation given,
ln f . . . . P P T P P P T 0 067 30 7 0 0012 2 0 416 2 R f P T P P h R w
F
HG
I
KJ
wF
HG
I
KJ
ln . . 1 30 7 0 416 2 ' AtP = 30 bar, 'h = (8.31451)u[(30.7)u(30) + (0.416)u(30)2_{]} ''h =4545 J mol
19. Consider mixing as a threestep process:
(I) Expand isothermally to idealgas state. (II) Mix ideal gas.
(III) Compress mixture isothermally
.
Starting with du c dT T P T P d w w
F
HG
I
KJ
L
NMM
O
QPP
v v vSolutions Manual 27 w w
F
HG
I
KJ
wwF
HG
I
KJ
u T P T P T v _{v} P RT a , Because, b v v2 w wF
HG
uvI
KJ
_{T} v2 aIntegration of this equation to the idealgas state (v f gives )
'u a d a f
z
_{v} v v v 2 Therefore, 'u_{I} x a1 1x a J mol 1 2 2 2 1 5914 v v [(1 bar)u(1 cm3_{)} _{ 0.1 Joule]}'uII 0 (because is the mixing of ideal gases)
2 2 1 11 1 2 11 22 2 22 mixt III mixt 1 1 2 2 1 [ 2 (1 0.1) 5550 J mol ] x a x x a a x a a u x x u ' v v v
mixh mixu uI uII uIII
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
4
1.
r At 25 Å we can neglect repulsive forces.
The attractive forces are London forces and induced dipola forces; we neglect (small) quadrupolar forces. (There are no dipoledipole forces since N_{2} is nonpolar.)
Let 1 stand for N_{2} and 2 stand for NH_{3}. London force: *12 1 2_{6} 1 2 1 2 3 2 D D r I I I I F d dr r I I I I 12 1 2_{7} 1 2 1 2 9 * D D Since, m m = 15.5 eV = D1 17 6 10. u 25 c 3 D2 22 6 10. u 25c 3 2 4. 18
I
1 8 10u N m 1 84 10. u 18 = 11.5 eV =I
2 N m then 0 10 London _{ } _{u}Induced dipole force:
F_{12} 62. 18 N
Solutions Manual 30 *_{12} 1 2 2 6 D P r F r 12 1 2 2 6D P 7 1 2 1 4 1D u1 1018(erg cm )3 1 2/ P2 . 7 1 47 10u 18 m )3 / /
eglecting all forces due to quadrupoles (and higher poles), .
D (erg c
F_{12}ind 3 8 10. u 18(erg cm3)1 2 N
Ftot FLondonFind Ftot_{ }_{65.8 10}_{u} 18 _{N}
2. From
Attractive potential = the LennardJones model:
4 6 V 6 H r * Attractiv force =e d dr r * 24 6 7 HV
Assume force of form,
Force = d
F
HG
I
KJ
dr k r * (constant) H V 6 7Using corresponding states: H D k T T c c
F
HG KJ
(constant)u whereD and E are universal constants.I
V E 6 2 2 u (constant v ) v c cSolutions Manual 31 force CH force substance B constant constant CH4 CH CH B B 4 4 6 4 7 6 7 ( ) ( ) ( / ) ( / ) ( ) ( ) ( ) ( ) H H V V k k r r B CH4 2 E CH4 B _{B} 8 7 7 2 7 7 2 10 (1 10 cm) force substance B (2 10 cm)
(
)
(
)
_{(}
_{)}
c c _{c} T T D u u u D _{E} u v(
v_{c})
Force substance B = u4 1010 dyne
Force = u4 1015 N
3.
*_{AA} u8 1016 erg By the molecular theory of corresponding states:
( 2 ) ii _{f}§ r · _{r} * i i V ¨ ¸ H _{©}V _{¹} * * BB
F
B B _{( is a universal function)} AAHG
A AI
KJ MM PP
H H H H f f f 2 2b g
L
N
b g
O
Q
the LennardJones (126) potential),
Since H /k 0 77. T_{c} (taking the generalized function f as
B A BB AA 16 ( 8 10 erg) (180 K/ 120 K) c c T T * * u u 12 0 10. 16 *_{BB} u erg 12.0 10u 23 J
Solutions Manual 32
4. For dipoledipole interaction:
*( ) ( ) ( , , ) dd r f i j i j P P SH T T I 4 _{o} 3 with
f( ,T T I_{i} _{j}, ) 2cosT_{i}cosT_{j}sinT_{i}sinT_{j}cos(I_{i}I_{j})
For the relative orientation:
i j i= 0º Ii= 0º T_{j}= 180º I_{j}= 0º T *( ) ( ) [ ( )] ( ) dd r r i j i j u u u P P SH P P SH 4 2 1 1 2 4 3 3 o o
ForP_{i} = P_{j} = 1.08 D = 3.603 10u 30 C m and _{r = 0.5}_{u10}

9_{ m,} 30 2 12 9 3 ( 2) (3.603 10 ) ( ) (4 ) (8.8542 10 ) (0.5 10 ) dd u u * S u u u u 21 1.87 × 10 JFor the relative orientation:
i j Ti= 0º Ii = 0º
T_{j}= 90º I_{j}= 0º
For the dipoleinduced dipole interaction:
*( ) ( ) ( cos ) ( cos ) dd_{i} P Di j _{i} j i _{j} SH T P D SH T 2 2 6 2 2 2 6 2 2 4 _{o} r 3 1 2 4( _{o}) r 3 1
For the relative orientation: i j i Tj T i j i Tj T Ij
Solutions Manual 33 i j _{T} i = 0º Tj = 90º 30 2 30 10 9 6 ( ) 2 2 (1.1124 10 ) (0.5 10 ) i dd 23 (3.603 10 ) (2.60 10 ) (4) 7.77 10 J ª_{} _{u} _{u} _{u} º u * u « » u u u u « » ¬ ¼ u  8.0 10u 23J
or the relative orientation: F i j Ti = 0º Tj = 90º 30 2 30 10 9 6 (3.603 10 ) (2.60 10 ) ( ) (4 1) 2 (1.1124 10 ) (0.5 10 ) i dd u u u u u u u u 23 4.85 10 J u  4.8 10u 23J *
5. The energy required to remove the molecule from the solution is E a r r
F
HG
I
KJ
1 1 2 1 3 2 H H P 3.5 r a 3.0 10 cm 2 D H u Pee, for example, C. J. E. Böttcher, 1952, The Theory of Electric Polarization, Elsevier)
8
(S
E = 4.61u1021_{ J/molecule = 2777 J mol}1
6.
a) The critical temperatures and critical volumes of N_{2} and CO are very similar, more similar than those for N_{2} and argon (see Table J4 of App. J). Therefore, we expect N_{2}/CO mixtures to follow Amagat’s law more closely than N_{2}/Ar mixtures.
Solutions Manual 34
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 S 1 2 K m m m m C O C O
b
g
where m_{C} and m_{O} 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).
romatics are better Lewis bases than paraffin. To extract aromatics from paraffins we want Lewis base = electron donor (low ionization potential).
A
a good Lewis acid. SO_{2} is a better a Lewis acid than ammonia.
8. From Debye’s equation:
2 Static polarization Total polarization independent of 1 4 4 2 3 3 3 r A A r T N N kT § · ªH º P S D S ¨ ¸ «_{H } » _{¨} _{¸} ¬ ¼ _{©} _{¹} v
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).
Solutions Manual 35
9. We compare the attractive part of the LJ potential (r >> V) with the London formula. The attractive LJ potential is
*11 11 11 6 4 H
F
HG
VI
KJ
r *22 22 22 6 4F I
HG K
rJ
H V * 12 6 4 12 12F
HG
rI
K
J
H VWe assume that V_{12} 1/ 2(V V_{11} _{22}). The London formula is
*_{11} 1 2 1 6 3 4 D I r *22 2 2 2 6 3 4 D I r *12 _{6} 1 2 2 3
F
HG
1 2KJ
1 2IF
HG
KJ
r I I Substitution givesI
D D I I 6 11 22 1 2 1/ 2 12 11 22 _{1} _{1} 11 22 1 2 2 2 ( ) I I I I ª _{V V} º ª º « » « » H H H « _{V V} » « _{} » « » « » ¬ ¼ ¬ ¼Only when V11 V22 and I1 I2 do we obtain
22) 1/ 2 12 ( 11 22)
H H H
Notice that both correction factors (in brackets) are equal to or less than unity. Thus, in general,
12 ( 11
H d H H 1/ 2
10. See Pimentel and McClellan, The Hydrogen Bon Phenol has a higher boiling point and a higher en
d, Freeman (1960).
ergy of vaporization than other substituted
benzenes suc ater than other substituted
benzenes sociated when dissolved in
h as toluene or chlorobenzene. Phenol is more soluble in w . Distribution experiments show that phenol is strongly as
Solutions Manual 36
nonpolar v d spectra show absorption at a frequency corresponding to the –OH H hydrogen b
sol ents like CCl_{4}. Infrare
"
ond.11.
rbon tetrachloride.
CCl4 CHCl3
acetone acetone
J ! J because acetone can hydrogenbond with chloroform but not with ca
12. a)
CHCl_{3} 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 p it is a Lewis acid while PPD is a Lewis base.
olarizability and
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 Hhydrogenbonding ne
nbutanol molecules.
twork between
tbutanol is
ports the view that it exhibits weaker h
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 supydrogen bonding with itself than does nbutanol.
b) _{2} and OH. For maximum
solubility, we want one solvent that can “hook up” with the ONO_{2} group (e.g., an aromatic hydrob n o one for the O ro (e.g., an alcohol or a ketone).
c) Using the result of Problem 5,
Cellulose nitrate (nitrocellulose) has two polar groups: ONO car o ) and an ther H g up
Solutions Manual 37 E a r r
F
HG
I
KJ
1 1 2 1 3 2 H H PAt 20qC, the dielectric constants are
4 8 (CCl ) 2.238 (C H ) 1.948 r r H H _{18} Thus, H H r r ( ) ( ) . CCl C H_{8} 4 18 117
It takes more energy to evaporate HCN from CCl_{4} than from octane.
13. At 170qC and 25 bar: is above 1 is well below 1 z_{H2} z_{amine} z 1 H2 HCl 0 1 yamine z_{HCl} is slightly below 1
a) A mixture of amine and H is expected to exhibit positive deviations from Amagat’s la_{2} w.
b) Since amine and HCl can complex, mixtures will exhibit negative deviations from Amagat’s
c) law. z 1 0 1 y_{argon} HCl z argon z
Solutions Manual 38
The strong dipoledipole attractive forces between HCl molecules cause , while argon is nearly ideal. Addition of arargon to HCl greatly reduces the attractive forces experienced by the HCl molecules, and the mixture rapidly approaches ideality with addition of argon. Addition l to Ar causes induced dipole attractive forces to arise in argon, but these forces are much smaller than the dipoledipole forces lost upon add
upwards.
z_{HCl} 1
of HC
ition of Ar to HCl. Thus the curve is convex
14.
a) has acidic hydrogen atoms while ethane does not. Acetylene can therefore complex with DMF, explaining its higher solubility. No comcomplexing occurs with octane.
lower pressure (3 bar), the gasphase is nearly ideal. There are few interactions beene and methane (or hydrogen). Therefore, benzbeene feels equally “comfortable” in both gases.
However, at 40 bar there are many more interactions betwee
drogen) in the gas phase. Now benzene does care about the nature of its surroundings. Because meth han hydrogen, benzene feels more “comfortable” with methane than with hydrogen. Therefore, K_{B} (in methane) > K_{B} (in hydrogen).
c) ctive forces with methane than
with hydrogen due to differences in _{2} is more “comfortable” in methane than in H_{2} and therefore has a lower fugacity that explains the condensation in H_{2} but not in CH_{4}.
d) It is appropriate to look at this from a correspondingstates viewpoint. At 100°C, for ethane , for helium,
At lower values of (near unity) the molecules have an average thermal (kinetic) energy on the order of H (because is on the order of
Acetylene
b) At the tween benz
n benzene and methane (or hyane has a larger polarizability t
Under the same conditions, CO_{2} experiences stronger attra polarizability. This means that CO
T_{R} 12. T_{R} 80 . T_{R}
T_{R} kT /H ). The colliding molecules (and molecules near one another, of which there will be many therefore be significantly affected by the attractive portion of the potential, leading to < 1. At higher , 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 de). ot good because it hydrogen bonds with itself and nheptane will be poor because it is nonpolar.
le. iv) Octopole.
at 50 bar) can
z T_{R}
both are polar and thus can interact favorably with the polar segment of poly(vinyl chlori Ethanol is n
f) i) Dipole. ii) Octopole. iii) Quadrupo
Solutions Manual 39
g) Lowering the temperature lowers the vapor pressure vent losses due to evaporation. However, at 0qC and at
ideal, 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 e.
of heptane and that tends to lower sol600 psia, the gas phase is strongly
nonethane on propan 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:
int dent of density.
2. Classical (rather than quantum) sta
4. ) (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 H_{2}; assumption 4 is violated slightly. All substances violate assumption 3, but H_{2}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 0. Values of (for isolated molecules) cannot be computed by these methods, because
(rotation, vibration, translational kinetic energy) appear in Q_{int} and the 1. Q can be factored so that Q is indepen
tistical mechanics is applicable. 3. *_{total}
_{¦}
(*_{Pairs}) (pairwise additivity).* /H F r( /V
c0p
cp cp
the contributions to c0_{p}
kinetic energy factor, not in the configuration integral.
16. Le
Schematically we have for the initial state and for the final (equilibrium) state: tD represent the phase inside the droplet and E the surrounding phase.
Solutions Manual 40 NO_{3} Na+ Lys2 K+ 3 NO
Initial state Equilibrium state
K+ NO_{3} Na+ 3 NO Na+ Lys2 K+ E D D E
Because the molar mass of lysozyme is above the membrane’s cutoff point, lysozyme
canG represent the change in K+_{ concentration in}_{D;}
centrations (f) of all the species inD and E are:
2 D
K
not diffuse across the membrane. Let:
M represent the change in Na+_{ concentration in}_{E.}
he final con T InD: c 0 cf c Lys f K _{NO3} D 0D _{G} _{c}fD _{M} _{c}f _{c} _{} Na _{NO3} D D _{G} _{M} 0 E E E 0 In E: c c Lys2 Lys2 cK G cNa cNa f f fE 0E _{} E 0E _{ }_{G} _{M} q. (1) yields Na _{NO3} Na _{NO3} E E D D E E D D (2)
where, for clarity, superscript f has been removed from all the concentrations.
NO3 Na _{NO3} _{NO3} M c c NO3 NO3 f
The equilibrium equations for the two nitrates are
PD PD PE PE
K _{NO3} K _{NO3}
PD _{} PD _{} PE _{} PE _{}
(1)
Na _{NO3} Na _{NO3}
Similar to the derivation in the text (pages 102103), E
c c c c
K _{NO3} K _{NO3}
c c c c
Substituting the definitions of G and M gives:
c c K _{NO3} c c c c
FH
K
IF
H
G
K
IJ
F
HG
I
KJ
FH
IK
F
HG
I
KJ
F
HG
I
KJ
0 0 0 0 0 0 D D E E E D G G M M G M M G M ( ) ( ) (3) G G M whereSolutions Manual 41 3 0 0 3 1 K NO 3 0.01 mol 9.970 10 mol L 1 kg water
(using the mass density of water at 25 C: 0.997 g cm )
c D_{} c D u
D
and
Solving for G and M gives
c c Na 0 NO3 0 _{ mol L} E D _{0 01}_{.} 1 3 1 3 1 4.985 10 mol L 5.000 10 mol L G u M u (4)
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]
S E D RT x x s s s v ln (5) where _{s} 3 NO _{K} _{Na} a s w x (6) x is the mole fraction of the solvent (water) given by
3 1 ( ) xD xD D xD xD + Lys NO _{K} _{N} 1 ( ) s w xE xE xE _{} xE xE _{} xE
Because solutions are dilute, we expand the logarithmic terms in Eq. (5), making the approximation ln(1A) A : 3 Lys 3 NO _{K} _{Na} NO _{K} _{Na} ( ) ( RT x x x x x ) w x x E E D D _{º} S _{«} _{»} ¬ ¼ v E E D ª
Again, because solutions are very dilute,
i i i c c x c c
¦
i w i with 1 w wc  vTherefore, with these simplifying assumptions, the osmotic pressure is given by
Using the relationships with the original concentrations, we have S RT c[ _{NO}D _{} cD_{} cD _{} cE _{} cE_{} cE _{} cE ]
Solutions Manual 42 E 0 Because we obtain M (7)
The lysozyme concentration is
S RT c[( D_{} G M) ( c D_{} G) ( ) ( M c E_{} G M) ( c E_{} M) ( c )]
NO3 K _{NO3} Na Lys
0 0 0 0 c c c c NO3 K NO3 Na 0 0 0 0 D D E E S RT(2c0D_{} 2c0E_{} c0E 4G4 ) K Na Lys c_{Lys} g 1 L mol 1 L mol L 0 2 4 1 2 14 000 1 429 10 E ( / , ) _{.} _{u}
Substitution of values in Eq. (7) gives the osmotic pressure
) S u u u u u u u u u u ( . ( [( . ) ( . ) ( . ( . ) ( . ) ] 8314 51 298 2 9 97 10 2 0 01 1 429 10 4 4 985 10 4 5 000 10 3 4 3 3 Pa L mol K ) K) (mol L ) 1 1 1 354 Pa
17. Because only water can diffuse across the membrane, we apply directly Eq. (441) derived in the text: lna ln(x ) RT w w w w J Svpure (1)
where subscript w indicates water.
Since the aqueous solution in part D is dilute in the sense of Raoult’s law, . This reduces Eq. (1) to:
Jw  1 lnx ln( x x )(x x ) RT w w 1 2 2 A A A A pure Sv (2) or equivalently,
Solutions Manual 43 S RT x x w ( _{A} _{A} ) pure 2 v AtT = 300 K, v_{pure}_{w} 0.018069 mol L .1
Mole fractions and can be calculated from the dimerization constant and the ss balance on protein A: x_{A} x_{A2} ma K a a x x  105 2 2 2 2 A A A A (4) 3 1 18.015 18.069 cm mol 0.997 w w w M d v ( / , ) . 5 5 000 1 81 10 5 2 _{2} (1000 / 18.069) mol A mol water u A A _{x} _{x} _{ (5) }
Solving Eqs. (4) and (5) simultaneously gives 7 34 10 5 38 10 6 2 . .
Substituting these mole fractions in Eq. (3), we obtain x x A A u u 6 S u u u u . 0 01 56 bar 1756 Pa ( . ) ( ) ( . . ) ( . ) 0 0831451 300 5 38 10 7 34 10 0 0 069 7 6 6 18 18. a) From Eq. (445): S c RT M RTB c 2 2 2 * "
Ploting (with S in pascal and c_{2} in g L1_{) as a function of }_{c}
2, we obtain the protein’s
mo the intercept and the second virial osmotic coefficient from the slope. S / c_{2}
Solutions Manual 44 0 20 40 60 BSA Concentration, g L1 35 37.5 40 42.5 45 47.5 50 S /c2 , Pa L g 1
Intercept = RT/M_{2} = 35.25 Pa L g = 35.25 Pa m kg from which we obtain From a leastsquare fitting we obtain:
1 3 1 M_{2} RT 29 1 35 25 8 15 8 314 35 25 70 321 70 321 u . . ) ( . ) . . kg mol , g mol1 RTB*_{. Therefore, B}*_{ = 7.92u10}8_{ L mol g}2_{.}
cific v me is given by the ratio molecular volume/molecular mass. le is
(
Slope = 0.196 =
The protein’s spe olu The mass per partic
19 1 2 23 70, 321 1.17 10 g molecule 6.022 10 A M m N u u
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 10 4 2 95 10 24 25 3 1 . . u u _{ m molecule}
which corresponds to a molecular radius of 4.13u109_{ m or 4.13 nm.}
For the specific volume: 2 95 10 117 10 2 52 10 2 52 25 19 6 3 1 3 1 . . . . u u u _{ m g} _{cm g}
b) Comparison of this value with the nonsolvated value of 0.75 cm g , indicates that the par3 1 ticle is hydrated.
Solutions Manual 45
c) Plotting (with S in pascal and c_{2} in g L1) as a function of c_{2} for the data at pH = 7.00, we obtain: S / c2 15 25 35 45 55 40 42.5 45 47.5 50 52.5 55 57.5 60 S /c2 , Pa L g 1 BSA Concentration, g L 1
Slope = 0.3317, which is steeper that at pH = 5.34, originating a larger second virial osmotic coefficient: B*_{= 1.338u10}7_{ L mol g}2_{ = 13.38u10}8_{ L mol g}2_{.}
s charged. The charged protein particles require counterions so electro e counterions form an ion atmosphere around a central protein particle and therefore this particles and its surrounding ion atmosphere have a larger excluded
volum e.
It the value of B*_{ at pH = 7.00 and that at pH = 5.37 for the }
uncharged m ontribution of the charge to B*_{:}
(13.38 7.92) u108_{ L mol g}2_{ = }
At pH = 7.00, the protein i neutrality is obtained. Th e than the uncharged particl
is the difference between olecule gives the c
1000 4 2 22 1 z M Um_{MX}
In this equation, we take the solu on mass density . Moreover, M_{2} = 70,321 g mol1_{, and m} MX 0.15 mol kg1: U1Uwater1g cm3 ti z2 8 _{)}2_{u}_{( . ) ( .}_{10} _{u} _{0 15 10}_{u} 3_{)} 4 5 46 10 70 321 1000 162 u u u  ( . ) ( ,
or Because pH is higher than the protein’s isoelectric point, the BSA must be negatively charged. Hence, z = 13.
9.
a) From Eq. (445) and according to the data: z r13. 1 S c c RT M RTB c c 0 2 0 *_{(} _{)} _{"}
Solutions Manual 46
Ploting (with S in pascal and in g L1_{) as a function of c} _{}_{ c}
0, we obtain the
molecular weight from the intercept and the second virial osmotic coefficient from the c_{2} S / (c c 0) solute’s slope. (cc_{0}), g L1 30 40 50 60 S /( cc0 ), P a L g 1 16.0 16.5 17.0 17.5 18.0
RT/M_{2} = 14.658 Pa L g1_{= 14.658 Pa m}3_{ kg}1_{from which we obtain}
From a leastsquare fitting we obtain: Intercept = 2 1 (298.15) (8.314) 14.659 14.659 169.109 kg mol RT M u 1 169,109 g mol
Slope = 0.053495 = RTB*_{. Therefore, }_{B}*_{ = 2.16}_{u10}8_{ L mol g}2_{.}
b) number of molecules in the aggrega parison of the molecular weight of the original ether (M = 390 g mol1_{) with that obtained in a): }
Number of molecules in the aggregate =
The te is obtained by com
169,109 390  434
Assuming that the colloidal particles arespherical, 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.,}
Prin opalan, 3rd. Ed.,
Marcel Dekker) that B*_{ is related to the excluded volume V}ex_{ through }
ciples of Colloid and Surface Chemistry, 1997, P.C. Hiemenz, R. Rajag
B N V M 2 _{2}2 A * ex
Solutions Manual 47
From a), B*_{ = 2.16u10}8_{ L mol g}2_{= 2.} _{0}5_{ m}3_{ mol kg}2_{ and M}
2 = 169,109 g mol1 =
169.109 kg mol1_{. The above equation gives V}ex_{ that is 4 times the actual volume of the particles}
(we rical). Calculation gives for the aggregate’s volume,
5.13u1025_{ m}3_{ (or 30.90 dm}3_{ mol}1_{) with a radius of 4.97u10}9_{ m or 5 nm.}
16u1 assume that the aggregates are sphe
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
5
1. Initial pressure P : moles of gas T a i  for n_{1} 1 at constant nd V: P V n B i 1 11 1 n RT_{1} V or P n RT V n RTB V i 1 1 2 11 2 Final pressure P_{f}: after addition of n_{2}moles of gas 2 at same T and V:
2 1 2 1 2 f _{2} (n n RT) (n n ) RTB P V V where Pressure change (n_{1}n_{2})2B n B_{1}2 _{11}2n n B_{1 2 12}n B_{2}2 _{22} : P ' 'P P P n RT n V n n B B RT V f i 2 (2 1 2 12 22 22) _{2} 12 Solving for B : 2 2 12 2 2 22 1 2 1 2 V P B n V n n RT § _{'} · ¨ ¸ ¨ ¸ © ¹ n B 49