25.2 The Probability Distribution for a Dilute Gas 1049
and the most probable distribution resembled each other. Although we do not prove
this fact, the average distribution and the most probable distribution become identical
in the limit of an infinitely large system, which is called thethermodynamic limit.
LetW({N}) be the number of system microstates that correspond to a given
distribution{N}. The sum of theW’s for all distributions that correspond to the correct
values ofE,V, andNobeys
∑
{N}
W({N})Ω (25.2-5)
where the sum includes one term for each distinct distribution. By the second postulate
of statistical mechanics, all system microstates with the correct values ofE,V, andN
are equally probable. The distribution with the largest value ofWis therefore the most
probable distribution.
We now obtain an expression forWfor a given distribution. Consider energy level
numberjwith degeneracygj. LetNjbe the number of molecules occupying states in
levelj. We will assume that our system contains very many molecules. Since we have
already assumed that it is a dilute gas, we must assume that it is confined in an extremely
large volume. Equation (22.1-4) shows that when the dimensions of a rectangular
container become very large, the translational energy levels of a particle confined
in the container become very close together. The degeneracies of the translational
energy levels also become very large, and if necessary, we can group energy levels
of nearly equal energies together to get very large degeneracies. These assumptions
correspond to
Nj 1 (25.2-6)
and
gjNj (25.2-7)
whereNjis the number of molecules occupying molecule energy levelj, andgjis the
degeneracy of the level. The condition of Eq. (25.2-7) is calleddilute occupation.
We first assume that the molecules are fermions, so that no more than one molecule
can occupy each state. We need the number of ways to divide thegjstates of levelj
intoNjoccupied states andgj−Njunoccupied states. We denote this number bytj.
Counting up this number of ways is an elementary statistics problem.^3 Consider first
a set ofNballs andNboxes, with each box able to hold no more than one ball. The
first ball can go into any ofNboxes, the second ball can go into any ofN−1 boxes,
and so forth, until the final ball has only one empty box to go into. Since the choices
are independent, the number of ways of making all of the choices is the product of the
number of ways of making the individual choices:
Number of ways of choosingN(N−1)(N−2)···(3)(2)(1)N! (25.2-8)
whereN! stands forN factorial, the product of all of the integers starting withNand
ranging down to 1. We say thatN! is the number ofpermutationsofNobjects.
Now consider the number of ways to pickNoccupied states fromgpossible states,
whereg>N. This is analogous to having more boxes than balls. There aregchoices
(^3) J. E. Freund,Modern Elementary Statistics, 7th ed., Prentice-Hall, Englewood Cliffs, NJ, 1988,
p. 97ff.