25.2 The Probability Distribution for a Dilute Gas 1053
For ln(N!)
ln(N!)≈
1
2
ln(2πN)+Nln(N)−N (25.2-20)
Stirling’s approximation is more nearly correct for larger values ofN. Because we
assume thatNjis a large number, the first term is much smaller than the other terms.
We omit it and write
ln(N!)≈Nln(N)−N (25.2-21)
You can see why we assumed that our system is very large. Not only did we require that
gjNjfor a typical energy level, we also need to assume thatNjis a large number
in order to use this approximation.
Exercise 25.10
Using a table of factorials or a calculator that computes factorials, find the percent error in the
results of the expression of Eq. (25.2-20) and that of Eq. (25.2-21) forN10,N60, and
N200. Find the difference between the results of Eq. (25.2-20) and of Eq. (25.2-21) for
N 1 × 109.
Substitution of Eqs. (25.2-18) and (25.2-21) into Eq. (25.2-17) gives a set of simul-
taneous equations (one equation for each value ofi):
∂
∂Ni
(∑
j
[Njln(gj)−Njln(Nj)+Nj+αNj−βNjεj]−αN+βE
)
0
The terms−αNandβEare independent of theN’s and have zero derivatives. Because
N 1 ,N 2 ,N 3 , and so on, are all independent variables and since all of theNvariables
except forNiare held fixed in the differentiation, only the term in the sum withji
has a nonzero derivative, and we have
ln(gi)−ln(Ni)− 1 + 1 +α−βεi0(i1, 2, 3,...) (25.2-22)
Fortunately, each equation contains only one variable. We solve Eq. (25.2-22) forNi:
Nigieαe−βεi (25.2-23)
The probability that a randomly selected molecule is found in energy leveljis equal
to the fraction of molecules in that energy level:
pj
Nj
N
1
N
gjeαe−βεj (probability of levelj) (25.2-24)
where we return to the use of the subscriptjto designate a molecule energy level. All
of the states of a level are equally probable since they have the same energy, so the
probability of a molecular state is
pi
Ni
N
1
N
eαe−βεi (probability of statei) (25.2-25)
where we now use the subscriptito designate a molecule state.