4.3 Solutions 265
4.18 1eV=kT
T=1 eV
k=
1. 6 × 10 −^19 J
1. 38 × 10 −^23 J/K
= 11 ,594 K
4.19 (a) For a perfect gas at temperatureT, the kinetic energy from translation
motion
1
2
m<νx^2 >+1
2
m<νy^2 >+1
2
m<ν^2 z>=3
2
RT
N 0
(1)
whereRis the gas constant andN 0 is Avagadro’s number. The energy of
the 3 degrees of freedom of translation is therefore on the average equal to
3
2 RT/N^0 for each molecule. Using this result together with the principle of
the equipartition of energy, it is concluded that in a system at temperatureT
each degree of freedom contributes,^12 NR 0 Tto the total energy.
If each molecule hasndegrees of freedom, the total internal energyUof a
gram-molecule of a perfect gas at temperatureT,U=1
2
nRT (2)
The molecular heat at constant volumeCis equal to(∂U
∂T)
ν, and is therefore
given byCν=1
2
nR (3)
For a perfect gas
Cp−Cν=R (4)ThereforeCp=Cν+R=(n+2)R
2(5)
andγ=Cp
Cν= 1 +
2
n(6)
(b) For monatomic moleculen=3, for translation (rotation and vibration are
absent),γ= 1 .667.
For diatomic moleculen=5 (3 from translation and only 2 from rotation
as the rotation about an axis joining the centres of atoms does not contribute)
andγ= 1. 4
If vibration is included thenn=7 andγ= 1. 2864.20 According to Chapman and Enskog
K=
η
m[
5
2
dEt
dT+
dE′dT]
(1)
whereEtis the translational energy andE′
the energy of other types.
Ifβdenotes the number of degrees of freedom of the molecule due to causes
other than translation, the total number of degrees of freedom of the molecule
will be 3+β.