4.3 Solutions 279
4.43 (a) Use the relation
dU=Tds−PdV (1)
Here,
dV=0(∵V=constant) and
U=aV T^4 (2)
dU= 4 aV T^3 dT=Tds
(
ds
dT
)
V
= 4 aV T^2
IntegratingS=^43 aT^3 V
(b)F=U−TS=aV T^4 −
4
3
aT^4 V=−
1
3
aV T^4
p=−
(
∂F
∂V
)
T
=
1
3
aT^4 =
1
3
u
4.44 According to Dulong-Petit’s law the molar specific heats of all substances,
with a few exceptions like carbon, have values close to 6 cal/mol◦C−^1 .The
specific heat of Cu iskgK^387 − 1 =^0 gK.387J− 1 = 0 .0926cal/gK−^1. Therefore, the atomic
mass of Cu= 0. 09266 = 64 .79 amu.
4.3.3 StatisticalDistributions .............................
4.45 Probability for the rotational state to be found with quantum numberJis given
by the Boltzmann’s law.
P(E)∝(2J+1) exp[−J(J+1)^2 / 2 I 0 kT
whereI 0 is the moment of inertia of the molecule,kis Boltzmann’s constant,
andTthe Kelvin temperature. The two lowest states haveJ=0 andJ= 1
I 0 =M(r/2)^2 +M(r/2)^2 =
1
2
Mr^2 ,whereM=938 MeV/c^2
2 I 0 =Mr^2 = 938 ×(1. 05 × 10 −^10 )^2 /c^2
c= 197 .3MeV− 10 −^15 m
kT= 1. 38 × 10 −^23 ×
50
1. 6 × 10 −^13
= 43. 125 × 10 −^10
^2
2 I 0 kT
=
^2 c^2
Mc^2 r^2 kT
=
(197.3)^2 × 10 −^30
938 ×(1. 05 × 10 −^10 )^2 × 43. 125 × 10 −^10
= 0. 8728
ForJ= 1 ,
J(J+1)^2
2 kT
= 1 ×(1+1)× 0. 8728 = 1. 7457
ForJ= 0 ,P(E 0 )∝^1.^0
ForJ= 1 ,P(E 1 )∝(2× 1 +1) exp(− 1 .7457)= 0. 52
∴P(E 0 ):P(E 1 )::1:0. 52