Chemistry 131, Fall 2007
Solutions to Homework No. 8
Problem 1.
See Levine, problem 9.33.
(a) True. Since
x
i
<
1 for each component
i
in solution and
since ln
x
i
<
0 when
x
i
<
1, then Δ
mix
G
=
RT
∑
i
n
i
ln
x
i
<
0, at constant
T
and
P
(Levine eq. 9.44).
(b) True.
The formation of a solution is an irreversible
process, for which Δ
S
univ
>
0 (Levine eq. 3.39). Since
Δ
G
=

Δ
S
univ
/T
at constant
T
and
P
(Levine eq. 4.21),
then Δ
mix
G <
0 for the formation of any solution.
(c) False. For an ideal solution, Δ
mix
H
= 0 (Levine eq. 9.47)
and Δ
mix
S
=

R
∑
i
n
i
ln
x
i
(Levine eq. 9.46). Therefore,
Δ
mix
S
6
= Δ
mix
H/T
.
(d) False. The condition for phase equilibrium (Levine eq.
4.88) pertains to each individual component
i
in the system,
e.g., a solution, not to the system as a whole. Therefore, for
equilibrium between a solution and its vapor, the chemical
potential of each substance in the solution must equal (or
be less than, if the substance is nonvolatile) the chemical
potential of the vapor of that substance.
(e) False. The relation between
x
v
i
and
x
l
i
for an ideal solution
is given by Raoult’s law,
x
v
i
P
=
x
l
i
P
*
i
(Levine eq. 9.52).
Therefore, in general
x
v
i
6
=
x
l
i
.
(f) True (Levine eq. 9.55). Follows from
V
i
=
±
∂μ
i
∂P
²
T,n
j
(Levine eq. 9.31) and from
μ
i
=
μ
*
i
+
RT
ln
x
i
(Levine eq.
9.41).
Problem 3.
See Levine, problem 9.40.
V
=
X
i
n
i
V
i
(Levine eq. 9.16)
=
X
i
n
i
V
*
m
,i
(Levine eq. 9.55)
=
X
i
V
i
(Levine eq. 1.4)
=
X
i
w
i
ρ
i
(Levine eq. 1.2)
,
w
=
X
i
w
i
,
ρ
=
w
V
=
∑
i
w
i
∑
i
w
i
/ρ
i
,
where
w
i
and
w
represent the masses of component
i
and of
the whole solution, respectively. Note that equal masses of
benzene and toluene are in used in this problem, so the last
expression simpliﬁes to:
ρ
= 2
³
X
i
1
ρ
i
!

1
=
2
´
(0
.
8790 g
/
cm
3
)

1
+ (0
.
8668 g
/
cm
3
)

1
µ

1
=
0
.
8729 g
/
cm
3
.
Problem 5.