1000 Solved Problems in Modern Physics

260 4 Thermodynamics and Statistical Physics

E= 3 NkT/ 2 (10)

Combining (8), (9) and (10)

α=

m

2 kT

(11)

and A=N(α/π)^3 /^2 =N(m/ 2 πkT)^3 /^2 (12)

Using (11) and (12) in (5)

N(ν)dν= 4 πN(m/ 2 πkT)^3 /^2 ν^2 exp(−mν^2 / 2 kT)dν

4.2N(ν)dν=^4 πN(m/^2 πkT)^3 /^2 ν^2 exp(−mν^2 /^2 kT)dν (1)

PutE=^12 mν^2 ,dE=mνdν (2)

Use (2) in (1) and simplify to obtain

N(E)dE=

2 πNE^1 /^2

(πkT)^3 /^2

exp

(

−

E

kT

)

dE

4.3 The average speed

<ν>=

∫∞

0 νN(ν)dν

N

= 4 π

( m

2 πkT

) 3 / 2 ∫∞

0

ν^3 exp(−mν^2 / 2 kT)dν (1)

where we have used the Maxwellian distribution

Putα=

m

2 kT

(2)

so that

∫∞

0

ν^3 e−αν

2

dν=

1

2 α^2

(3)

Combining (1), (2) and (3)

<ν>=

(

8 kT

πm

) 1 / 2

=

√

8 RT

M

(4)

wheremis the mass of the molecule,Mis the molecular weight andRthe gas

constant.

4.4<ν^2 >=

∫∞

0 ν

(^2) N(ν)dν
N
= 4 π
( m
2 πkT

) 3 / 2 ∫∞

0

ν^4 exp(−mν^2 / 2 kT)dν

withα=

m

2 kT

andx=αν^2 ;dx= 2 ανdν

The integral,I=

∫∞

0

ν^4 e−αν

2

dν=

1

2 α^5 /^2

∫∞

0

x^3 /^2 e−xdx=

3

√

π

8 α^5 /^2

Therefore,<ν^2 >= 4 π

( m

2 πkT

) 3 / 2 3 √π

8

(m

2 kT

) 5 / 2 =

3 kT

m

<ν^2 >^1 /^2 =(3kT/m)^1 /^2