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Physics 211 (Moore) Spring 2010
Problem Set #5 Solution
April 22, 2010
1. Reif 5.19
a.
The equation of state reads
(
p
+
av

2
)(
v

b
) =
RT
. We want to solve for the critical point in terms
of
a
and
b
. The easiest way to do this is to ﬁrst take a derivative with respect to
v
(treating
T
as a
constant) of both sides, and set
dp/dv
to zero. We ﬁnd that
(
v

b
)
2
v
3
=
RT
2
a
Next we cross multiply to clear out the fraction and take another derivative with respect to
v
. (This
works since we are only neglecting
d
2
p/dv
2
, which we’re setting to zero anyway.) This leads to a
simple formula for
v
which gives
v
c
= 3
b
. Plugging back into the above equation gives
RT
c
=
8
a
27
b
.
Inverting these formulae give
a
=
9
8
v
c
RT
c
b
=
v
c
3
b.
Plugging the formula for
a
and
b
back into the equation of state gives
p
c
=
3
8
RT
c
v
c
c.
The original equation of state can be written
±
p
0
+
a
p
c
v
2
c
v
0
2
²±
v
0

b
v
c
²
=
RT
0
T
c
p
c
v
c
Then plugging in the values for
a
and
b
and
p
c
gives
±
p
0
+
3
v
0
2
²±
v
0

1
3
²
=
8
3
T
0
2. Reif 5.24
The minimum amount of heat which the refrigerator rejects (while still, presumably, operating
reversibly) is given when the total entropy change of the system is zero. This means any entropy lost by the
ice must be gained by the reservoir, and this must furthermore be the
only
entropy gained. Clearly if the
reservoir has an inﬁnite heat capacity, the answer is
Q
=
mL
, so the problem is how to deal with the ﬁnite
heat capacity case.
Consider a generic moment in the heating process. The reservoir is at some
T > T
0
and some mass of
water has already been frozen. We consider freezing a small mass
dm
further. The icewater mix must lose
an entropy of
dS
ice
=

Ldm/T
0
in the freezing process. Assuming optimality, the reservoir must gain an
entropy
dS
r
= +
Ldm/T
0
. But
dS
r
=
CdT/T
and equating these and integrating leads immediately to
T
=
T
0
e
Lm/CT
0
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Problem Set #5 Solution
April 22, 2010
for the whole process. The heat gained by the reservoir is then
Q
=
CT
0
±
e
Lm/CT
0

1
²
In the limit of a large heat capacity,
Q
≈
CT
0
Lm
CT
0
+
1
2
³
Lm
CT
0
´
2
+
...
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 Spring '11
 abc
 Physics

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