c03 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come
THE HEAT CAPACITY OF AN IDEAL GAS 49To findCpwe notice that, sinceH=U+pV,wehaveCp−CV=(
∂H
∂T
)
p−
(
∂U
∂T
)
V=(
∂U
∂T
)
p+
(
∂pV
∂T)
p−
(
∂U
∂T
)
VFor an ideal gas, we recall the Joule experiment which shows that the energy is a
function ofTonlyU=f(T), hence
(
∂U
∂T)
p=
(
∂U
∂T
)
VandCp−CV=(
∂pV
∂T)
p=
(
∂RT
∂T
)
p=R
(
∂T
∂T
)
p=R
Now we see thatCp−CV=R= 8 .3JK−^1 mol−^1andCp∼= 12. 5 + 8. 3 ∼= 20 .8JK−^1 mol−^1A useful unitless parameter isγ=Cp/CV. From the thermodynamics we have
developed so far,γ=Cp/CV= 20. 8 / 12. 5 = 1 .66. Indeed, for He and Hg vapor,γ
is close to the ideal value. For more complicated molecules, however,γbegins to
fall off significantly. For ammonia vapor,γhas dropped to 1.3 and it appears to be
approaching 1.0 for the relatively complicated molecule diethyl ether.
The problem is that more complicated molecules can rotate and vibrate. The
NH 3 molecule resembles a pyramid with a triangular base. It can rotate with 3
degrees of rotational freedom in 3-space. The total number of degrees of freedom for
translational plus rotational motion is 3+ 3 =6; hence the molar kinetic energy is
Ukin==6(1/ 2 RT). From this, by the same reasoning as before, we getCV= 3 R
andCp=CV+R= 3 R+R= 4 R. TakingR=8.3 J K−^1 mol−^1 ,CV= 3 R∼= 24. 9
JK−^1 mol−^1 , andCp= 4 R= 33 .2. This leads toγ=33/25=1.33 (unitless) in good
agreement with the experimental value for ammonia, which is 1.31 forγ,butthe
results are not so good forCVandCp. Hydrogen is an intermediate case because it
normally has enough thermal energy to rotate end-over-end (inx,y2-space) but not
enough to spin on its axis.
Evidently, hydrogen and ammonia rotate but do not vibrate under the conditions
of Table 3.1. Most chemical bonds also stretch and bend. These motions yield extra