Heat capacity ofa diatomic gas 77states tolevell,correspondingintuitivelytodifferent possibledirectionsfor the
angular momentum.
Armedwiththese results we can at once writedown the partitionfunctionfor
rotation
ZZZrot=∑
l= 0 , 1 ...( 2 l+ 1 )exp[−l(l+ 1 )/T] (7.8)Hence the problemisinprinciplesolved, evenifthe sum (7.8) mustbe evaluated
numericallyrather thanin terms ofknownfunctions.
Moreover, for all gases except hydrogen, there is a great simplification. When the
numbersfor the moments ofinertiaofa gas suchasO 2 orN 2 are substitutedinto
(7.7a), wefindthat thecharacteristic temperatureisabout 2 K (2.1 KforO 2 ,2.9K
forN 2 but 8 5 K forH 2 ). Since these other gases have liquefied (indeed solidified)
at suchlow temperatures,itisalways trueinthegaseous state thatT.Very
manyrotational states are therefore excited, and the sum (7.8) maybe replaced by
an integral in this ‘classical’ limit. We writey=l(l+ 1 )and treatyas a continuous
variable, so thatdy=( 2 l+ 1 )dl.Henceinthis approximation
ZZZrot=∫∞
0∫∫
exp(−y/T)dy=(−T/)[exp(−y/T)]∞ 0
=T/ (7.9)This simple result at once tells us the thermodynamic properties of these diatomic
gases. Thethermalpart ofthe totalpartitionfunctionZ(equation (7.3))isobtainedby
multiplyingZZZtrans(equation ( 6. 6 )) byZZZrot(equation (7.9)). As discussed in Chapter 3,
the ground state contributions from vibration and electronic excitation give rise to
zero-point terms only.SinceZ∝V(from the translationalcontribution only),it
remainstrue thatPV=NkkkBTas discussed in section 6.3.3. But the additional factor
ofTin the partition function from rotation gives an additionalNkkkBtoCV. And the
totalheat capacityis
CV=CV,trans+CV,rot=(
3
2
+ 1
)
NkkkB=
5
2
NkkkB (7.10)a result which follows fromZ ∝ T^5 /^2 , as the reader may verify (compare
section 6 .3.1). Hence also the ratioCCP/CV =^75 for thediatomic gas, a ratioin
goodagreement withexperiment.