PHYSICS 218
SOLUTION TO HW 10
Created: November 25, 2004
23:26pm
Last updated: December 9, 2004
1:01am
1.
Schroeder 3.32
(a) In moving the piston through a distance of Δ
x
= 1 mm, the work done on the system is
W
=
F
Δ
x
= 2000
×
10

3
= 2 J.
(b) We recall that a fast compression is an adiabatic process. Therefore, no heat is added to the gas.
(c) Assuming that all the work done on the system is converted into the internal energy of the gas, the
increase of internal energy is therefore Δ
U
= 2 J.
(d) Calculating the change in entropy appears to be complicated because all of our familiar thermody
namic variables change in the process. That is,
P
f
, V
f
, U
f
, T
f
are all different from
P
i
, V
i
, U
i
, T
i
. As
is always the case, there are several ways to calculate the change in entropy. Schroeder suggests
that you use the thermodynamic identity d
U
=
T
d
S

P
d
V
to calculate the change in entropy. In
this case, we want to look at the process in the
U

V
plane as outlined on pp. 111112 of Schroeder.
Our process is one in which
U
i
< U
f
and
V
i
> V
f
, so the two can be represented by two points in a
figure like Fig. (3.15), except that – contrary to the figure –
V
f
is below
V
i
in our case. We envision
replacing the actual process by two quasistatic processes, in which Step 1 is an increase in
U
from
U
i
to
U
f
with
V
held constant (as in Fig. 3.15) and Step 2 is a compression or step downward from
V
i
to
V
f
with
U
held constant at
U
i
(just opposite to Step 2 in Fig. 3.15). Since
V
is constant in
Step 1, we have d
U
=
T
d
S
and
Δ
S
1
=
d
U
T
=
C
V
d
T
T
=
C
V
ln
T
f
T
i
.
Now,
T
f
=
T
i
+ Δ
U/C
V
, so ln(
T
f
/T
i
) = ln[1 + Δ
U/
(
C
V
T
i
)]
≈
Δ
U/
(
C
V
T
i
) and Δ
S
1
≈
Δ
U/T
i
=
W/T
i
.
During Step 2, we have
T
d
S
=
P
d
V
since
U
(ergo
T
) is constant.
Hence, we have
Δ
S
2
= (1
/T
)
P
d
V
, so
Δ
S
2
=
NkT
T
d
V
V
=
Nk
ln
V
f
V
i
=
Nk
ln 1 +
Δ
V
V
i
≈
Nk
Δ
V
V
i
.
(Note Δ
V
<
0!).
Since
Nk/V
i
=
P
i
/T
i
, we have Δ
S
2
≈
P
i
Δ
V/T
i
.
Putting these together we
find that the total change in entropy Δ
S
is, Δ
S
= Δ
S
1
+ Δ
S
2
≈
(
W
+
P
i
Δ
V
)
/T
i
. Then, Δ
V
=

A
Δ
x
=

10

2
m
2
×
10

3
m =

10

5
m
3
, so
P
i
Δ
V
≈ 
1 J. Finally, Δ
S
≈
(2 J

1 J)
/
(300 K) =
3
.
33
×
10

3
J/K. This is positive as it should be since the process that is got us from
i
to
f
is sudden
– anything but quasistatic. It is possible to calculate Δ
S
in the more familiar
P

V
plane. In this
case you also need to define a third point in order to get from
i
to
f
along a quasistatic path. You
could use a quasistatic isothermal or a quasistatic adiabatic process along with a quasistatic process
at constant volume. You could also use both an isothermal and an adiabatic process. Since any of
these choices require calculating
P
and
V
at this third point, they turn out to be more complicated
analytically than the calculation in the
U

V
plane where quasistatic paths with one variable held
constant are simple.
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 Fall '04
 POLLACK
 Thermodynamics, Magnetism, Energy, Work, Entropy, Adiabatic process, Schroeder

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