Physical Chemistry II
Chem 402
Spring 2011
PS # 9: (Ch 9: 1, 5, 15, 20, 22, 27, 32)
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P9.1) Ratcliffe and Chao [Canadian Journal of Chemical Engineering 47, (1969), 148] obtained the
following tabulated results for the variation of the total pressure above a solution of isopropanol
( P1* = 1008 Torr) and n-decane ( P2* = 48.3 Torr) as a function of the mole fraction of the n-decane in
the solution and vapor phases. Using these data, calculate the activity coefficients for both components
using a Raoult’s law standard state.
x2
y2
942.6
0.1312
0.0243
909.6
0.2040
0.0300
883.3
0.2714
0.0342
868.4
0.3360
0.0362
830.2
0.4425
0.0411
786.8
0.5578
0.0451
758.7
0.6036
0.0489
(Torr)
a) for isopropanol P1*  1008 Torr
for n-decane
P2*  48.3 Torr
Using the relations Pi  yi Ptotal , ai  Pi / Pi * , and  i 
ai
, the calculated activities and activity coefficients
xi
are shown below.
x2
y2
a1
a2
1
2
942.6
0.1312
0.0243
0.912
0.474
1.05
3.61
909.6
0.2040
0.0300
0.875
0.565
1.10
2.77
883.3
0.2714
0.0342
0.846
0.625
1.16
2.30
868.4
0.3360
0.0362
0.831
0.651
1.25
1.94
830.2
0.4425
0.0411
0.790
0.706
1.42
1.60
786.8
0.5578
0.0451
0.745
0.735
1.69
1.32
758.7
0.6036
0.0489
0.716
0.768
1.81
1.27
P (Torr)
207
P9.5) The partial molar volumes of water and ethanol in a solution with xH 2O = 0.60 at 25°C are 17.0
and 57.0 cm3 mol–1, respectively. Calculate the volume change upon mixing sufficient ethanol with 2.00
mol of water to give this concentration. The densities of water and ethanol are 0.997 and 0.7893 g cm–3,
respectively, at this temperature.
V  nH 2O V H 2O  nEt V Et
V H 2O  17.0 cm3 mol-1 and V Et  57.0 cm3 mol-1
nH 2O  3.75 and xH 2O 
nH 2 O
nH 2O  nEt
 0.450
3.75 mol
 0.450; nEt  4.58
3.75 mol  nEt
V  nH2O V H2O  nEt V Et
The total mixed volume is given by
 3.75 mol 17.0 cm3 mol-1  4.58 mol  57.0 cm3 mol-1
 325 cm3
Vunmixed  nH2O
M H 2O
H O
2
 nEt
M Et
 Et
 3.75 mol 18.02 g mol-1 
1cm3
0.997 g
+ 4.58 mol  46.07 g mol-1 
1cm3
0.7893 g
 335 cm3
V  V  Vunmixed  325 cm3  335 cm3  10. cm3
P9.15) At 39.9°C, a solution of ethanol (x1 = 0.9006, P1* =130.4 Torr) and isooctane ( P2* = 43.9 Torr)
forms a vapor phase with y1 = 0.6667 at a total pressure of 185.9 Torr.
a. Calculate the activity and activity coefficient of each component.
b. Calculate the total pressure that the solution would have if it were ideal.
a) The activity and activity coefficient for ethanol are given by
208
a1 
y1Ptotal 0.6667  185.9 Torr

 0.9504
P1*
130.4 Torr
1 
a1 0.9504

 1.055
x1 0.9006
Similarly, the activity and activity coefficient for isooctane are given by
a2 
2 
1  y1  Ptotal
*
2
P

0.3333  185.9 Torr
 1.411
43.9 Torr
a2
1.411

 14.20
x2 1  0.9006
b) If the solution were ideal, Raoult’s law would apply.
PTotal  x1 P1*  x2 P2*
 0.9006  130.4 Torr  1  0.9006   43.9 Torr
 121.8 Torr
P9.20) The partial pressures of Br2 above a solution containing CCl4 as the solvent at 25°C are found to
have the values listed in the following table as a function of the mole fraction of Br2 in the solution [G.
N. Lewis and H. Storch, J. American Chemical Society 39 (1917), 2544]. Use these data and a graphical
method to determine the Henry’s law constant for Br2 in CCl4 at 25°C.
P (Torr)
P (Torr)
0.00394
1.52
0.0130
5.43
0.00420
1.60
0.0236
9.57
0.00599
2.39
0.0238
9.83
0.0102
4.27
0.0250
10.27
The best fit line in the plot is PBr2  Torr   413 xBr2  0.063 . Therefore, the Henry’s law constant in
terms of mole fraction is 413 Torr.
P9.22) The densities of pure water and ethanol are 997 and 789 kg m–3, respectively. The partial molar
volumes of ethanol and water are 55.2 and 17.8  10–3 L mol–1, respectively. Calculate the change in
volume relative to the pure components when 2.50 L of a solution with xethanol = 0.35 is prepared.
V  nH 2OVH 2O  nethanolVethanol
V
 xH 2OVH 2O  xethanolVethanol  0.65 17.8 103 L mol-1  0.35 55.2 10-3 L mol-1
ntotal
V
 0.0309 L mol-1
ntotal
ntotal 
xethanol
xH 2O
2.50 L
= 80.9 mol  nH 2O  nethanol
0.0309 L mol-1
n
0.350
; nH 2O  52.6 mol nethanol  28.3 mol
 ethanol 
0.650
nH 2O
Videal  nethanol
M ethanol
ethanol
 nH 2O
M H 2O
H O
2
46.07 10 kg mol-1
18.02 10-3 kg mol-1
+
52.6
mol
= 2.60 L

789 kg m -3
998 kg m -3
V  V  Videal  0.10 L
 28.3 mol 
-3
210
P9.27) A volume of 13.75 L of air is bubbled through liquid toluene at 298 K, thus reducing the mass of
toluene in the beaker by 5.95 g. Assuming that the air emerging from the beaker is saturated with
toluene, determine the vapor pressure of toluene at this temperature.
P
nRT

V
5.95 g 
1 mol
 0.08314 L bar K -1  298 K
-1
92.14 g mol
= 0.116 bar
13.75 L
P9.32) Calculate the activity and activity coefficient for CS2 at xCS2 = 0.573 using the data in Table 9.3
for both a Raoult’s law and a Henry’s law standard state.
R

aCS
2

R
CS2

H

aCS
2

H
CS2

PCS2
*
CS2
P
R
aCS
2
xCS2

420.3 Torr
 0.820
512.3 Torr

0.820
 1.43
0.573
PCS2
k H ,CS2
H
aCS
2
xCS2


420.3Torr
 0.209
2010Torr
0.209
 0.365
0.573
mols A   xc  0   mols Liquid   xd  xc  .