Download HO #15 Maxwell Distribution

Maxwell Distribution of Molecular Velocities
We want to derive a formula for the distribution of velocities of molecules in an
ideal gas. We make the following assumptions about the distribution. Let the velocity
vector be given by v = (v1 , v2 , v3 ) . Let f1 (v1 ) be the distribution of the component of the
velocity in the x-direction, let f2 (v2 ) be the distribution of the component of the velocity
in the y-direction, and let f3 (v3 ) be the distribution of the component of the velocity in
the z-direction.
Assumption 1. The distributions of the components of the velocity are the same so that if
the velocity is denoted by v = (v1 ,v2 ,v3 ) with components v1 ,v2 , and v3 , then
f (vi ) = f1 (v1 ) = f2 (v2 ) = f3 (v3 ) .
Assumption 2. The distributions of the components of the velocity are independent so
that f (v1 ) f (v2 ) f (v3 )dv1dv2 dv3 is the probability of the velocity being in the region
[v1 , v1 + dv1 ] × [v2 , v2 + dv2 ] × [v3 , v3 + dv3 ] .
Assumption 3. The distribution of the total velocity only depends on its magnitude so that
f (v1 ) f (v2 ) f (v3 )dv1dv2 dv3 = ϕ (v12 + v22 + v32 )dv1dv2 dv3 .
From these assumptions one can then derive the precise form of the functions f
and ϕ . We proceed to do this. All we need is the last equation to start.
f (v1 ) f (v2 ) f (v3 ) = ϕ (v12 + v22 + v32 )
Take this equation and differentiate both sides with respect to v1 .
df
dϕ (u) ∂ u dϕ (u)
f (v2 ) f (v3 ) =
=
2v1
dv1
du ∂ v1
du
Then divide both sides by ϕ .
df
f (v2 ) f (v3 ) dϕ (u) 2v1
dv1
= du
ϕ
ϕ
df
dϕ (u)
f (v2 ) f (v3 )
2v1
dv1
du
=
f (v1 ) f (v2 ) f (v3 )
ϕ
df
dϕ (u)
dv1
= du
2v1 f (v1 )
ϕ
In this form it is clear that the expression on the left is only a function of v1 alone.
The expression on the right is a function of v1 , v2 , and v3 . This implies that these
expressions must be constant. Call the constant A. Thus, we get the following equation.
df
dv1
= 2Av1
f (v1 )
If we integrate both sides of this equation we get an expression for f.
ln( f (v)) = Av 2 + B
2
f (v) = e B e Av = Ce Av
2
∞
In order that f be a probability distribution, we must have
∫
f (v)dv = 1 and this is
−∞
impossible if A ≥ 0 . Assume that A = − µ 2 . Then C is determined by the next equation.
∞
∫ Ce
− µ 2 v2
dv = 1
−∞
∞
Using the fact that
∫e
− x2
dx = π , one can easily see that
−∞
µ
thus that we must have C =
. This gives us the form of f.
π
f (v) =
∞
∫e
−∞
− µ 2 v2
dv =
π
and
µ
µ − µ 2 v2
e
π
This is a normal or Gaussian distribution with mean zero and standard deviation
1
.
2µ
We also have the form for the function ϕ .
2
ϕ (v12 + v22 + v32 ) =
µ3
π
3
exp(− µ 2 (v12 + v22 + v32 ))
Using the ideal gas law one can show that
µ2 =
m
2kT
where m is the mass of the individual molecule of gas, T is the absolute temperature, and
k is Boltzmann’s constant. The velocity distribution varies only with temperature and the
mass of the molecule of gas.
To apply this to a specific gas, let us consider oxygen. The atomic weight of
oxygen is 16. The molecular weight of oxygen is 32 since two oxygen atoms bind
together in ordinary gaseous form. This gives us a value of
µ2 =
m
= 7.096 × 10 −6 kg J
2kT
using the standard temperature for T. This gives a value of
187.69 m s = 419.9 mph
as the standard deviation of the distribution of the velocity.
(
1
exp − 12
2πσ
( ))
x−µ 2
σ
The Normal Distribution
For oxygen at standard temperature the root mean square speed of oxygen
3p
ρ
where p is standard atmospheric pressure and ρ is the density of our substance, in this
case oxygen, at standard temperature and pressure.
molecules is determined to be 461 m s by a different derivation. It is given by
3