Page 75 - Physical Chemistry

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Chapter 2 ( x/ y) ( y/ z) ( z/ x) 1 [Eq. (1.34)] to relate these partial derivatives.
x
y
z
The First Law of Thermodynamics Replacement of x, y, and z with T, U, and V gives
0T 0U 0V
a b a b a b 1
0U V 0V T 0T U
0U 0T 1 0V 1 0U 0T
a b ca b d ca b d a b a b
0V T 0U V 0T U 0T V 0V U
0U
a b C m J (2.63)
V
0V T
where ( z/ x) 1/( x/ z) , ( U/ T) C , and m ( T/ V) [Eqs. (1.32), (2.53),
U
J
y
V
V
y
and (2.62)] were used.
Joule’s 1843 experiment gave zero for m and hence zero for ( U/ V) . However,
T
J
his setup was so poor that his result was meaningless. The 1924 Keyes–Sears experi-
ment showed that ( U/ V) is small but definitely nonzero for gases. Because of
T
experimental difficulties, only a few rough measurements were made.
In 1853 Joule and William Thomson (in later life Lord Kelvin) did an experiment
similar to the Joule experiment but allowing far more accurate results to be obtained.
The Joule–Thomson experiment involves the slow throttling of a gas through a rigid,
porous plug. An idealized sketch of the experiment is shown in Fig. 2.7. The system
is enclosed in adiabatic walls. The left piston is held at a fixed pressure P . The right
1
piston is held at a fixed pressure P P . The partition B is porous but not greatly so.
2
1
This allows the gas to be slowly forced from one chamber to the other. Because the
throttling process is slow, pressure equilibrium is maintained in each chamber.
Essentially all the pressure drop from P to P occurs in the porous plug.
2
1
We want to calculate w, the work done on the gas in throttling it through the plug.
The overall process is irreversible since P exceeds P by a finite amount, and an
1
2
infinitesimal change in pressures cannot reverse the process. However, the pressure
drop occurs almost completely in the plug. The plug is rigid, and the gas does no work
on the plug, or vice versa. The exchange of work between system and surroundings
occurs solely at the two pistons. Since pressure equilibrium is maintained at each pis-
ton, we can use dw rev PdV to calculate the work at each piston. The left piston
does work w on the gas. We have dw P dV P dV, where we use subscripts
1
L
L
L
L and R for left and right. Let all the gas be throttled through. The initial and final vol-
umes of the left chamber are V and 0, so
1
1
w 0 P dV P 0 dV P 10 V 2 P V
1 1
1
1
1
L
V 1 V 1
The right piston does work dw on the gas. (w is negative, since the gas in the right
R R
V 2
chamber does positive work on the piston.) We have w P dV P V . The
R
2 2
2
0
work done on the gas is w w w P V P V .
2 2
1 1
R
L
Porous Plug Adiabatic Wall
B
P 1 P 2 P 1 P 2 P 1 P 2
P , V , P , V ,
1
2
1
2
T 1 P 1 P 2 T 2
Figure 2.7
The Joule–Thomson experiment. (a) (b) (c)