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52 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy

Does the Internal Energy Depend More
3.3 Strongly on V or T?

Chapter 2 demonstrated that U is a function of T alone for an ideal gas. However, this
statement is not true for real gases, liquids, and solids for which the change in U with V
must be considered. In this section, we ask if the temperature or the volume depend-
ence of U is most important in determining ¢U for a process of interest. To answer this
question, systems consisting of an ideal gas, a real gas, a liquid, and a solid are consid-
ered separately. Example Problem 3.3 shows that Equation (3.19) leads to a simple
result for a system consisting of an ideal gas.

EXAMPLE PROBLEM 3.3

Evaluate (0U>0V) T for an ideal gas and modify Equation (3.20) accordingly for the
specific case of an ideal gas.

Solution
0U 0P 0[nRT>V] nRT
a b = Ta b - P = Ta b - P = - P = 0
0V T 0T V 0T V V
Therefore, dU = C dT , showing that for an ideal gas, U is a function of T only.
V

Example Problem 3.3 shows that U is only a function of T for an ideal gas.
Specifically, U is not a function of V. This result is understandable in terms of the
potential function of Figure 1.10. Because ideal gas molecules do not attract or repel
one another, no energy is required to change their average distance of separation
(increase or decrease V):

T f
¢U = C (T) dT (3.21)
V
3
T i
Recall that because U is only a function of T, Equation (3.21) holds for an ideal gas
even if V is not constant.
Next consider the variation of U with T and V for a real gas. The experimental
determination of (0U>0V) T was carried out by James Joule using an apparatus con-
sisting of two glass flasks separated by a stopcock, all of which were immersed in a
water bath. An idealized view of the experiment is shown in Figure 3.3. As a valve
between the volumes is opened, a gas initially in volume A expands to completely
Thermometer
fill the volume A + B. In interpreting the results of this experiment, it is important to
understand where the boundary between the system and surroundings lies. Here, the
decision was made to place the system boundary so that it includes all the gas.
Initially, the boundary lies totally within V , but it moves during the expansion so
A
A B that it continues to include all gas molecules. With this choice, the volume of the
P P i P 0
system changes from V before the expansion to V + V after the expansion has
A
A
B
taken place.
Water bath The first law of thermodynamics [Equation (3.13)] states that
FIGURE 3.3 0U 0U
dq - P external dV = a b dT + a b dV
Schematic depiction of the Joule experi- 0T V 0V T
ment to determine (0U>0V) T . Two spheri-
cal vessels, A and B, are separated by a However, all the gas is contained in the system; therefore, P external = 0 because a vac-
valve. Both vessels are immersed in a uum cannot exert a pressure. Therefore Equation (3.13) becomes
water bath, the temperature of which is
monitored. The initial pressure in each dq = a 0U b dT + a 0U b dV (3.22)
vessel is indicated. 0T V 0V T