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Chapter 2 2.8 PERFECT GASES AND THE FIRST LAW
The First Law of Thermodynamics
Perfect Gases
An ideal gas was defined in Chapter 1 as a gas that obeys the equation of state PV
nRT. The molecular picture of an ideal gas is one with no intermolecular forces. If we
change the volume of an ideal gas while holding T constant, we change the average
distance between the molecules, but since intermolecular forces are zero, this distance
change will not affect the internal energy U. Also, the average translational kinetic
energy of the gas molecules is a function of T only (as is also true of the molecular
rotational and vibrational energies—see Sec. 2.11) and will not change with volume.
We therefore expect that, for an ideal gas, U will not change with V at constant T and
( U/ V) will be zero. However, we are not yet in a position to prove this thermody-
T
namically. To maintain the logical development of thermodynamics, we therefore now
define a perfect gas as one that obeys both the following equations:
PV nRT and 10U>0V2 0 perfect gas (2.66)*
T
An ideal gas is required to obey only PV nRT. Once we have postulated the second
law of thermodynamics, we shall prove that ( U/ V) 0 follows from PV nRT,
T
so there is in fact no distinction between an ideal gas and a perfect gas. Until then, we
shall maintain the distinction between the two.
For a closed system in equilibrium, the internal energy (and any other state func-
tion) can be expressed as a function of temperature and volume: U U(T, V).
However, (2.66) states that for a perfect gas U is independent of volume. Therefore U
of a perfect gas depends only on temperature:
U U1T2 perf. gas (2.67)*
Since U is independent of V for a perfect gas, the partial derivative ( U/ T) in
V
Eq. (2.53) for C becomes an ordinary derivative: C dU/dT and
V
V
dU C dT perf. gas (2.68)*
V
It follows from (2.67) and C dU/dT that C of a perfect gas depends only on T:
V
V
C C 1T2 perf. gas (2.69)*
V
V
For a perfect gas, H U PV U nRT. Hence (2.67) shows that H depends
only on T for a perfect gas. Using C ( H/ T) [Eq. (2.53)], we then have
P
P
H H1T2, C dH>dT, C C 1T2 perf. gas (2.70)*
P
P
P
Use of ( U/ V) 0 [Eq. (2.66)] in C C [( U/ V) P]( V/ T)
T P V T P
[Eq. (2.61)] gives
C C P10V>0T2 perf. gas (2.71)
P
P
V
From PV nRT, we get ( V/ T) nR/P. Hence for a perfect gas C C nR or
P P V
C P,m C V,m R perf. gas (2.72)*
We have m C ( U/ V) [Eq. (2.63)]. Since ( U/ V) 0 for a perfect gas,
J V T T
it follows that m 0 for a perfect gas. Also, m C ( H/ P) [Eq. (2.65)].
J JT P T
Since H depends only on T for a perfect gas, we have ( H/ P) 0 for such a gas,
T
and m 0 for a perfect gas. Surprisingly, as Fig. 2.9 shows, m for a real gas does
JT JT
not go to zero as P goes to zero. (See Prob. 8.37 for analysis of this fact.)
We now apply the first law to a perfect gas. For a reversible volume change,
dw PdV [Eq. (2.26)]. Also, (2.68) gives dU C dT for a perfect gas. For a fixed
V
amount of a perfect gas, the first law dU dq dw (closed system) becomes
dU C dT dq P dV perf. gas, rev. proc., P-V work only (2.73)
V