Physical Chemistry Third Edition

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 con-

stant volume,dV0 for each infinitesimal step, so thatw0 for this step. The same is

true for the heating process. The only nonzero contribution towis 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 thatdwrevis not an exact differential.

EXAMPLE 2.8

a.Sincedwrevcorresponds to Eq. (B-19) of Appendix B withN0, show thatdwrevdoes

not satisfy the Euler reciprocity relation of Eq. (B-20) to be an exact differential.

b.Calculate the line integral ofdPfor 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,dV0 butdT0.

For the isothermal process,dV0 butdT0.

Solution

a. dwPextdV+ 0 PdV+0 (reversible processes)

(

∂P

∂T

)

V



(

∂ 0

∂V

)

T

 0

b.Let 5.000 L be calledV 1 and 10.00 L be calledV 2. Let 298.15 K be calledT 1 and 200.00 K

be calledT 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