Page 66 - Modern physical chemistry

3.14 The Maxwell Distribution Law 55

Substituting this energy and the given temperature into the Boltzmann distribution law tells
us that
=
In ~ _~ = _ 1.37 x 10- 22 J = -3.33 x 10-2
No kT (1.3807xl0-23JK-IX298.15K)

whence
~=0.967
No

and
N = (0.967X89) = 86.

3.14 The Maxwell Distribution Law

Before Boltzmann, Maxwell employed a symmetry argument to construct a distribu-
tion law for the translational energy of molecules in an ideal gas.
The fundamental idea was that collisions between gas molecules establish a statisti-
cal distribution. Thus, there is no preferred direction in absence of an externally imposed
field. Furthermore, each component of velocity of a molecule is independent of its orthog-
onal components.
Consider a pure ideal gas at equilibrium at temperature T and volume V. Let the prob-
ability that the rectangular components of the velocity of a molecule are between v." and
Vx + dvx , between Vy and Vy + dvy, and between V z and V z + dvz be.lt:v.,,) dvx·,.lt:vy) dvy.,
.It: v z ) dvz , respectively.
Since these are independent, the joint probability .It:v.,,, vy, vz) dvx dvy dvz equals the
product of the three probabilities:

f( vx, vy, Vz ) dvx dvy dvz = f( vx ) f( vy) f( vz ) dvx dvy dvz . [3.83]
Because of the assumed isotropy in space, the individual probabilities depend only on
the pertinent magnitudes and so on the pertinent squares. Thus, one may rewrite equa-
tion (3.83) in the form

2 2 2 2 2 2
4>( Vx + Vy + Vz ) dvx dvy dvz = 4>( Vx ) 4>( Vy ) 4>( Vz ) dvx dvy dvz . [3.84]

Equation (3.84) is satisfied when the independent variables in function 4> are the
powers of a common base. Without loss of generality, one may set

[3.85]

Since each molecule must have an x-component of velocity, we see that

[3.86]

Substituting form (3.85) into equation (3.86) and introducing the result from example
10.5 yields
112
AI: e-av.~ dvx = A ~ = 1, [3.87]
[
)