1000 Solved Problems in Modern Physics

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