CH353
Homework assignment 4 (
due Thursday March 25
)
1
(10). Three kilograms of
H
2
O at T=10
±
C are mixed with 1 kg of D
2
O at T=30
±
C and
form an ideal solution. Assuming that the C
p
value of D
2
O is the same as that of H
2
O (1
kcal kg
1
K
1
), calculate
∆
S
in this process, considering the separate steps of temperature
equilibration and isothermal mixing.
Step 1: For the equilibrium temperature T
f
,
(3
) 283
(1
) 303
(4
)
p
pp
f
Ck
g
K
g
K
g
T
+=
ii
, from which T
f
= 288K
(3
)ln(288/ 283)
)ln(288/303)
7.39 /
Ip
p
SCk
g
C
k
g
J
K
∆=
+
=
Step 2:
112 2
(l
n
l
n)
II
SR
n
x
x
x
x
−
+
x
1
= n
1
/(n
1
+n
2
)
x
2
= n
2
/(n
1
+n
2
)
where n
1
= 3kg/(0.018 kg/mol) = 166.67 mole and n
2
= 1kg/(0.02 kg/mol) = 50 mole,
all this gives us
∆
S
II
= 973.17 J/K
∆
S
total
=
∆
S
I
+
∆
S
= 980.56 J/K
2
(10).
In the class, we have calculated the change in the entropy
∆
S of
several
compounds when they are mixed to form an ideal solution at constant P and T. Consider
n
1
moles of compound 1 forming an ideal solution with n
2
moles of compound 2. Let
0
1
v
and
0
2
v
be the molar volumes of the pure compounds.
(a) What is the total volume of the mixture? Is it larger, smaller, or the same as the
total volume of the pure compounds? (Hint: use the following relationship: V =
12
,,
Tn n
G
P
∂
∂
).
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View Full DocumentNote: It was erroneously stated in this problem that
V = 
12
,,
Tn n
G
P
∂
∂
.
The sign, of
course, is “+”, as we all know. Using the wrong sign would not have changed the
answer in this problem.
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 Spring '08
 LIM
 Physical chemistry, Thermodynamics, pH, 1 kg, 1 K, RT ln, 3.51 g, 0 g

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