A differential that is not exact is called aninexact differential. The differential
duM(x,y)dx+N(x,y)dy (2.1-20)
is an inexact differential if M andN are not the appropriate partial derivatives of the same function. The line integral of an inexact differential depends on the path of integration as well as on the initial point and the final point. We will show thatdwis an inexact differential by showing that two processes with the same initial and final states can correspond to different amounts of work done on the system.
E X A M P L E 2.7
Consider a reversible process with the same initial and final states as the process of Example 2.2, but with a different path. Calculate the work done on the ideal gas system of Example 2.2 if it is reversibly cooled at constant volume of 5.000 L from 298.15 K
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,dV 0 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−1mol−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.
E X A M P L E 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,dV 0 butdT 0.
For the isothermal process,dV 0 butdT 0.
Solution
a. dwPextdV +0PdV +0 (reversible processes) ∂P
∂T
V
∂0
∂V
T 0
b. Let 5.000 L be calledV1and 10.00 L be calledV2. Let 298.15 K be calledT1and 200.00 K be calledT2.
dP ∂P
∂T
V dT+
∂P
∂V
T
dV nR
V dT− nRT V2 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
dP nR V1
T2 T1
dT−nRT2 V2
V1
1 V2 + nR
V2 T1
T2
dT
nR V1
(T2−T1)+nRT2 1
V2− 1 V1
+ nR
V2
(T1−T2) (1.2516 mol)(8.3145 J K−1mol−1)
−98.15 K 0.00500 m3
+(1.2516 mol)(8.3145 J K−1mol−1)(200.00 K) 1
0.0100 m3− 1 0.0050 m3
+(1.2516 mol)(8.3145 J K−1mol−1)
98.15 K 0.01000 m3
−2.0428×105J m3−2.0813×105J m3+1.0214×105J m3 −3.103×105J m33.062 atm
This value is the same as in the previous example, showing that the integral ofdP is path-independent so far as these two paths are concerned.
Exercise 2.4
a. Calculate the amount of work done on the surroundings if the isothermal expansion of Example 2.2 is carried out at a constant transmitted pressure of 1.000 atm instead of rever- sibly, but with the same initial and final states as in Example 2.2. Why is less work done on the surroundings in the irreversible process than in the reversible process?
b. What is the change in the pressure of the system for the irreversible process?
P R O B L E M S
Section 2.1: Work and the State of a System
2.1 Calculate the work done on the surroundings if 1.000 mol of neon (assumed ideal) is heated from 0.0◦C to 250.0◦C at a constant pressure of 1.00 atm.
2.2 Calculate the work done on the surroundings if 100.0 g of water freezes at 0.0◦C and a constant pressure of 1.00 atm. The density of ice is 0.916 g cm−3and that of liquid water is 1.00 g cm−3.
2.3 Calculate the work done on 100.0 g of benzene if it is pressurized reversibly from 1.00 atm to 50.00 atm at a constant temperature of 293.15 K.
2.4 Calculate the work done on the surroundings if 1.000 kg of water is heated from 25.0◦C to 100.0◦C at a constant pressure of 1.00 atm.
2.5 a. The tension force for a spring that obeys Hooke’s law is given by
τ −k(x−x0)
wherexis the length of the spring,x0is its equilibrium length, andkis a constant called the spring constant.
Obtain a formula for the work done on the spring if its length is changed reversibly fromx0toxat constant volume.
b. Show that the force can be derived from a potential energy
V 1
2k(x−x0)2
and that the work done on the spring in part a is equal to the change in the potential energy.
2.6 a. Obtain a formula for the work done in reversibly and isothermally compressing 1.000 mol of a gas from a volumeV1to a volumeV2if the gas obeys the Redlich–Kwong equation of state.
b. Using the formula from part a, find the work done in reversibly compressing 1.000 mol of carbon dioxide from 10.00 L to 5.000 L at 298.15 K. Compare with the result obtained by assuming that the gas is ideal.
c. Using the formula from part a, calculate the work done on the surroundings if 1.000 mol of carbon dioxide expands isothermally but irreversibly from 5.000 L to 10.00 L at an external pressure of 1.000 atm. Compare with the result obtained by assuming that the gas is ideal.
2.7 a. Obtain a formula for the work done in reversibly and isothermally compressing 1.000 mol of a van der Waals gas from a volumeV1to a volumeV2.
b. Using the formula from part a, find the work done in reversibly compressing 1.000 mol of carbon dioxide
from 10.00 L to 5.000 L at 298.15 K. Compare with the result obtained by assuming that the gas is ideal.
c. Using the formula from part a, calculate the work done on the surroundings if 1.000 mol of carbon dioxide expands isothermally but irreversibly from 5.000 L to 10.00 L at an external pressure of 1.000 atm. Compare with the result obtained by assuming that the gas is ideal.
2.8 Test the following differentials for exactness:
a. T2dT+TVdV b. xexydx+yexydy c. xydx+12x2dy 2.9 Carry out the line integral:
c
nR
V dT−nRT V2 dV
on the path from (300.0 K, 10.0 L) to (300.0 K, 20.0 L) to (400.0 K, 20.0 L). Taken1.000 mol.
2.2 Heat
Joseph Black was the first to distinguish between the quantity of heat and the
“intensity” of heat (temperature) and to recognize latent heat absorbed or given off in phase transitions. However, Black believed in the caloric theory of heat, which incorrectly asserted that heat was an “imponderable” fluid called “caloric.”
This incorrect theory was not fully discredited until several decades after Black’s death.
Joseph Black, 1728–1799, was a Scottish chemist who discovered carbon dioxide (“fixed air”) by heating calcium carbonate.