2.1 Work and the State of a System
49
to 200.00 K, then reversibly expanded from 5.000 L to 10.00 L at a constant temperature
of 200.0 K, and then reversibly heated at a constant volume of 10.00 L from 200.0 K to
298.15 K.
Solution
The three parts of the process must be integrated separately. In the cooling process at constant volume, dV 0 for each infinitesimal step, so that w 0 for this step. The same is
true for the heating process. The only nonzero contribution to w is from the isothermal
expansion:
w −(50.00 g)
1 mol
39.948 g
(8.3145 J K −1 mol −1 )(200.00 K) ln
10.00 L
5.000 L
−1443 J
which is not equal to the amount of work in Example 2.2.
Since a single case of path dependence is sufficient to show that a differential is
inexact, this example shows that dw rev is not an exact differential.
E X A M P L E 2.8
a. Since dw rev corresponds to Eq. (B-19) of Appendix B with N 0, show that dw rev does
not satisfy the Euler reciprocity relation of Eq. (B-20) to be an exact differential.
b. Calculate the line integral of dP for the process of Example 2.2. Show that the integral is
path-independent for the two paths of Example 2.2 and Example 2.7. For the second path
the integral will have to be done in three sections, but in each section only one term of the
line integral will be nonzero. For the heating and cooling processes, dV 0 but dT 0.
For the isothermal process, dV 0 but dT 0.
Solution
a.
dw P ext dV + 0 PdV + 0 (reversible processes)
∂P
∂T
V
∂0
∂V
T
0
b. Let 5.000 L be called V 1 and 10.00 L be called V 2 . Let 298.15 K be called T 1 and 200.00 K
be called T 2 .
dP
∂P
∂T
V
dT +
∂P
∂V
T
dV
nR
V
dT −
nRT
V 2 dV
We carry out the three integration steps. For the first step, the second term vanishes.
For the second step, the first term vanishes, and for the third step, the second term
vanishes.
n (50.00 g)
1 mol
39.948 g
1.2516 mol
49
to 200.00 K, then reversibly expanded from 5.000 L to 10.00 L at a constant temperature
of 200.0 K, and then reversibly heated at a constant volume of 10.00 L from 200.0 K to
298.15 K.
Solution
The three parts of the process must be integrated separately. In the cooling process at constant volume, dV 0 for each infinitesimal step, so that w 0 for this step. The same is
true for the heating process. The only nonzero contribution to w is from the isothermal
expansion:
w −(50.00 g)
1 mol
39.948 g
(8.3145 J K −1 mol −1 )(200.00 K) ln
10.00 L
5.000 L
−1443 J
which is not equal to the amount of work in Example 2.2.
Since a single case of path dependence is sufficient to show that a differential is
inexact, this example shows that dw rev is not an exact differential.
E X A M P L E 2.8
a. Since dw rev corresponds to Eq. (B-19) of Appendix B with N 0, show that dw rev does
not satisfy the Euler reciprocity relation of Eq. (B-20) to be an exact differential.
b. Calculate the line integral of dP for the process of Example 2.2. Show that the integral is
path-independent for the two paths of Example 2.2 and Example 2.7. For the second path
the integral will have to be done in three sections, but in each section only one term of the
line integral will be nonzero. For the heating and cooling processes, dV 0 but dT 0.
For the isothermal process, dV 0 but dT 0.
Solution
a.
dw P ext dV + 0 PdV + 0 (reversible processes)
∂P
∂T
V
∂0
∂V
T
0
b. Let 5.000 L be called V 1 and 10.00 L be called V 2 . Let 298.15 K be called T 1 and 200.00 K
be called T 2 .
dP
∂P
∂T
V
dT +
∂P
∂V
T
dV
nR
V
dT −
nRT
V 2 dV
We carry out the three integration steps. For the first step, the second term vanishes.
For the second step, the first term vanishes, and for the third step, the second term
vanishes.
n (50.00 g)
1 mol
39.948 g
1.2516 mol