260
4 Thermodynamics and Statistical Physics
E = 3N kT/2
(10)
Combining (8), (9) and (10)
α =
m
2kT
(11)
and A = N (α/π)
3/2
= N (m/2πkT )
3/2
(12)
Using (11) and (12) in (5)
N (ν)dν = 4π N (m/2πkT )
3/2
ν
2 ex p(−mν
2
/2kT )dν
4.2 N (ν)dν = 4π N (m/2πkT )
3/2
ν
2 ex p(−mν
2
/2kT )dν
(1)
Put E =
1
2
mν
2
, dE = mνdν
(2)
Use (2) in (1) and simplify to obtain
N (E)dE =
2π N E
1/2
(πkT ) 3/2 exp
−
E
kT
dE
4.3 The average speed
< ν >=
∞
0 ν N (ν)dν
N
= 4π
m
2πkT
3/2 ∞
0
ν
3 exp(−mν
2
/2kT )dν
(1)
where we have used the Maxwellian distribution
Put α =
m
2kT
(2)
so that
∞
0
ν
3 e
−αν
2 dν =
1
2α 2
(3)
Combining (1), (2) and (3)
< ν >=
8kT
πm
1/2
=
8RT
M
(4)
where m is the mass of the molecule, M is the molecular weight and R the gas
constant.
4.4 < ν
2
>=
∞
0 ν
2 N (ν)dν
N
= 4π
m
2πkT
3/2 ∞
0
ν
4 exp(−mν
2
/2kT )dν
with α =
m
2kT
and x = αν
2 ; dx = 2ανdν
The integral, I =
∞
0
ν
4 e
−αν
2 dν =
1
2α 5/2
∞
0
x
3/2 e
−x dx =
3
√ π
8α 5/2
Therefore, < ν
2
>= 4π
m
2πkT
3/2 3
√
π
8
m
2kT
5/2 =
3kT
m
< ν
2
>
1/2
= (3kT /m)
1/2