The first attempt to determine whether (∂U/∂V)T,nvanishes for real gases was made by Joule around 1843. His apparatus is schematically depicted in Figure 2.5. A sample of a gas (the system) was placed in one side of the apparatus and the other side of the apparatus was evacuated. Mechanical vacuum pumps were not yet invented, so a water aspirator was used. The system was allowed to equilibrate and the initial temperature of the apparatus was measured. The stopcock was then opened and the gas expanded irreversibly into the vacuum.
Because the surroundings were not affected during the expansion into a vacuum, wwas equal to zero. The gas expanded rapidly so there was little opportunity for heat to be transferred to or from the surroundings, and q therefore vanished to a good approximation. From the first law,∆U was equal to zero. If a change in tem- perature of the gas occurred, heat would be transferred to or from the surround- ings after the expansion was complete, and the final temperature of the surroundings
Chamber originally containing gas Chamber originally evacuated Valves
Gas inlet
Thermometer
To vacuum pump
Water bath
Figure 2.5 The Apparatus for the Joule Experiment (Schematic).
would differ from the initial temperature. If the heat capacity of the apparatus and the heat capacity of the gas are known, the change in temperature of the gas could be calculated.
The Joule experiment was carried out several times with various volumes for the second chamber. The ratio∆T /∆V would be determined for each experiment and extrapolated to zero value of∆V, where∆V is the final volume of the gas minus its initial volume. This extrapolation is equivalent to taking the mathematical limit, so the result is a partial derivative, called theJoule coefficientand denoted byµJ:
µJ lim
∆V→0
∆T
∆V
∂T
∂V
U,n
(2.4-13) The Joule coefficient is related to (∂U/∂V)T,nby use of the cycle rule, Eq. (B-15), and the reciprocal identity, Eq. (B-8):
∂U
∂V
T,n
− ∂T
∂V
U,n
∂U
∂T
V,n
−µJCV (2.4-14)
Exercise 2.10 Verify Eq. (2.4-14).
Joule was unsuccessful in his attempt to measure the Joule coefficient because the changes in temperature that occurred were too small to be measured by his ther- mometers, even though he used pressures up to 22 atm. Later versions of the exper- iment with better apparatus have given nonzero values of (∂U/∂V)T,n for real gases.
There are better ways than the Joule experiment to determine values of (∂U/∂V)T,n, and we will discuss them in Chapter 4. Once values forCVand for (∂U/∂V)T,n are obtained,∆Ucan be calculated for any process that begins with one equilibrium state and ends with another equilibrium state.
E X A M P L E 2.15
If the virial equation of state, Eq. (1.3-3), is truncated at the second term it can be shown that
∂U
∂V
T,n
∂Um
∂Vm
T,n RT2
Vm2 dB2
dT (2.4-15)
whereR is the gas constant and where Vm is the molar volume. The derivation of this equation is found in Example 4.8. For argon gas at 298.15 K,B2is approximately equal to
−15.8 cm3mol−1anddB2/dTis approximately equal to 0.20 cm3mol−1K−1. Assume that CV,m3R/2.
a. Find∆U,q, andwfor a reversible isothermal expansion of 1.000 mol of argon at 298.15 K from a volume of 2.000 L to a volume of 20.00 L. Compare with values obtained assuming ideal gas behavior.
b. Find the value of the Joule coefficient for 1.000 mol of argon at 298.15 K and a volume of 20.000 L.
Solution
Since our system consists of 1.000 mol, we calculate the change in the molar energy, UmU/n. We also calculate the work done per mole.
a. ∆Um
c ∂Um
∂Vm
T,n dVm
Vm2 Vm1
RT2B2 Vm2
dVm −RT2B2 1
Vm2− 1 Vm1
−(8.3145 J K−1mol−1298.15 K)2(0.20×10−6m3mol−1K−1)
×
1
0.0200 m3mol−1− 1 0.00200 m3mol−1
66.5 J mol−1≈67 J mol−1
w −
c
PdVm −RT Vm2
Vm1
1 Vm− B2
Vm2
dVm
w −RT
ln Vm2
Vm1
−B2 1
Vm2− 1 Vm1
−(8.3145 J K−1mol−1)(298.15 K)
×
ln 20.0 L
2.00 L
−
−15.8 cm3mol−1
1
20000 cm3mol−1− 1 2000 cm3mol−1
−
2479 J mol−1
2.303−7.11×10−3
−5690 J mol−1
q∆U−w67 J mol−1−(−5690 J mol−1)5757 J mol−1
Compare these values with those obtained if ideal behavior is assumed: ∆U0, w −5708 J mol−1,q5708 J mol−1.
b.
∂U
∂V
T,n
(8.3145 J K−1mol−1)(298.15 K)2
⎛
⎜⎝0.20×10−6m3mol−1K−1
0.02000 m3mol−1 2
⎞
⎟⎠
370 J m−3 µJ −
∂U
∂V
T,n 1
CV − 370 J m−3
12.17 J K−1 −30 K m−3
Exercise 2.11
a. Find∆U,q, andwfor an irreversible isothermal expansion at 298.15 K of 1.000 mol of argon with the same initial and final molar volumes as in the previous example but with a constant external pressure of 1.000 atm. Assume thatP(transmitted)Pextand assume the same equation of state as in the previous example. Compare your results with the values obtained in the previous example.
b. Find the change in temperature if 1.000 mol of argon initially at 298.15 K is expanded adia- batically (without any transfer of heat) into a vacuum so that its volume changes from 2.000 L to 20.00 L. Since it expands into a vacuum the surroundings are not affected.
400
300
200
100
0
T/K
V/L 5
0 10 15 20
Step 2: Constant-volume heating Step 1: Isothermal expansion
(2 L, 298.15 K) (20 L, 298.15 K)
(20 L, 373.15 K)
Figure 2.6 The Curve Representing the Path for the∆ULine Integral.
A change in internal energy for a nonisothermal process can be calculated by carrying out a line integral ofdU.
E X A M P L E 2.16
Calculate∆U for a process that takes 1.000 mol of argon fromT 298.15 K andV 2.000 L toT 373.15 K andV 20.000 L. Does the result depend on whether the process is reversible?
Solution
SinceUis a state function, we can choose any path with the same initial state and final state to calculate∆U. We choose to integrate along the reversible path shown in Figure 2.6, consisting of an isothermal expansion and a constant-volume change in temperature. The first step of the path is that of the previous example, so∆U1, the change in energy for that part, is equal to 67 J mol−1. For the second step
∆U2 T2
T1
CVdT CV(T2−T1)nCV,m(T2−T1) (1.000 mol)
3 2
R(T2−T1)
12.472 J K−1
(75.0 K)935 J
∆U∆U1+∆U267 J+935 J1002 J
BecauseUis a state function, the result does not depend on whether the actual process is reversible, so long as the initial and final states are metastable or equilibrium states.
Exercise 2.12
a. Calculateqandwfor the reversible process that follows the path used in the solution of the previous example.
b. Calculateqandwfor the reversible process corresponding to the following path. Step 1: The system is heated from 298.15 K to 373.15 K at a constant volume of 2.000 L; step 2: It is then expanded isothermally to a volume of 20.000 L.
c. Comment on the difference between theqandwvalues for parts a and b. What is the value of∆Ufor the process of part b?