4.3 Solutions
259
of mass these two collisions appear to be equivalent so that c
= c. We can
then write
f (ν 1 ) f (ν 2 ) = f (ν 3 ) f (ν 4 )
or ln f (ν 1 ) + ln f (ν 2 ) = ln f (ν 3 ) + ln f (ν 4 )
( 1 )
Since kinetic energy is conserved
ν
2
1 + ν
2
2 = ν
2
3 + ν
2
4
(2)
Equations (1) and (2) are satisfied if
ln f (ν) ∝ ν
2
(3)
or f (ν) = A exp(−αν
2 )
( 4 )
where A and α are constants. The negative sign is essential to ensure that no
molecule can have infinite energy.
Let N (ν)dν be the number of molecules per unit volume with speeds ν to
+dν, irrespective of direction. As the velocity distribution is assumed to be
spherically symmetrical, N (ν)dν is equal to the number of velocity vectors
whose tips end up in the volume of the shell defined by the radii ν and +dν,
so that
N (ν)dν = 4πν
2 f (ν)dν
(5)
Using (4) in (5)
N (ν)dν = 4π Aν
2 ex p(−αν
2 )
( 6 )
We can now determine A and α. If N is the total number of molecules per
unit volume,
N =
∞
0
N (ν)dν
(7)
Using (6) in (7)
N = 4π A
∞
0
ν
2 exp(−αν
2 ) dν = 4π A(1/4)(π/α
3 )
1/2
or N = A(π/α)
3/2
(8)
If E is the total kinetic energy of the molecules per unit volume
E =
1
2
m
∞
0
ν
2 N (ν)dν =
4π Am
2
∞
0
ν
4 exp(−αν
2 )dν
or E = (3m A/4)(π
3
/α
5 )
1/2
(9)
where gamma functions have been used for the evaluation of the two integrals.
Further,