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3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR T? 53

To within experimental accuracy, Joule found that dT surroundings = 0 . Because the
water bath and the system are in thermal equilibrium, dT = dT surroundings = 0 . With
this observation, Joule concluded that dq = 0 . Therefore, Equation (3.22) becomes
0U
a b dV = 0 (3.23)
0V T
Because dV Z 0 , Joule concluded that (0U>0V) = 0 . Joule’s experiment was not defin-
T
itive because the experimental sensitivity was limited, as shown in Example Problem 3.4.

EXAMPLE PROBLEM 3.4
In Joule’s experiment to determine (0U>0V) T , the heat capacities of the gas and the
water bath surroundings were related by C surroundings >C system L 1000 . If the precision
with which the temperature of the surroundings could be measured is ;0.006°C , what
is the minimum detectable change in the temperature of the gas?
Solution
View the experimental apparatus as two interacting systems in a rigid adiabatic enclosure.
The first is the volume within vessels A and B, and the second is the water bath and the
vessels. Because the two interacting systems are isolated from the rest of the universe,

q = C water bath ¢T water bath + C gas ¢T gas = 0
C water bath
¢T gas =- ¢T water bath =-1000 * (;0.006°C) =< 6°C
C gas
In this calculation, ¢T gas is the temperature change that the expanded gas undergoes
to reach thermal equilibrium with the water bath, which is the negative of the tempera-
ture change during the expansion.
Because the minimum detectable value of ¢T gas is rather large, this apparatus is
clearly not suited for measuring small changes in the temperature of the gas in an
expansion.

More sensitive experiments were carried out by Joule in collaboration with William
Thomson (Lord Kelvin). These experiments, which are discussed in Section 3.8,
demonstrate that (0U>0V) T is small, but nonzero for real gases.
Example Problem 3.3 has shown that (0U>0V) = 0 for an ideal gas. We next calcu-
T
V m, f
late (0U>0V) T and ¢U = 1 V m, i (0U>0V) dV m for a real gas, in which the van der
T
T
Waals equation of state is used to describe the gas, as illustrated in Example Problem 3.5.
EXAMPLE PROBLEM 3.5
For a gas described by the van der Waals equation of state,
2
P = nRT>(V - nb) - an >V 2 . Use this equation to complete these tasks:
.
a. Calculate (0U>0V) T using (0U>0V) = T(0P>0T) - P
T
V
b. Derive an expression for the change in internal energy, ¢U = 1 V i V f (0U>0V) dV ,
T
T
in compressing a van der Waals gas from an initial molar volume V to a final
i
molar volume V at constant temperature.
f
Solution
2
nRT n a
0c - d
0P £ V - nb V 2 ≥ nRT
a. Ta b - P = T - P = - P
0T V 0T V V - nb
2
2
nRT nRT n a n a
= - + =
V - nb V - nb V 2 V 2