Physical Chemistry , 1st ed.

n 2 objects in subsystem 2, and niobjects in the ith subsystem, then the num-

ber of ways of accomplishing this arrangement, labeled C, is given by the com-

bination formula

C (17.1)

where N! is “Nfactorial” or 1  2  3  4 Nand the ni! in the denomi-

nator is the product of all of the nivalues. The nivalues are called occupation

numbers.By definition, 0! 1 and 1! 1.

In our future consideration of atomic and molecular systems, the number

of subsystems Nand the occupation numbers will be very large. We would

then have to evaluate N! where Nis a very large number, on the order of

Avogadro’s number. However, it is easily shown on a calculator that N! gets

very large, very fast. (A point of calculator trivia is that 69! is just under 10^100 ,

which is the largest factorial that many calculators can evaluate.) We will need

to find some way of evaluating factorials of very large numbers.

There is a method of estimating the natural logarithmof factorials.Stirling’s

approximationsays that, for large N,

ln N! Nln NN (17.2)

This approximation will be useful when we apply combination statistics to col-

lections of molecules. To give you an example of how well it works, consider

the following table:

N ln N! Nln NN % error

30 74.66 72.04 3.51

100 363.74 360.52 0.885

5000 37,591 37,586 0.0133

Notice that the percentage error between ln N! and Nln NNgoes down as

Nincreases. Stirling’s approximation gets better as Nincreases, so it will be very

useful in considering macroscopic systems of moles of atoms and molecules.

Probabilities can also be used to determine average values of some variable.

Consider a variable uthat can have certain possible individual values uj.

Further, we will represent the probability that any particular value ujexists as

Pj. The average value of the variable u, which we designate u, is given by the

expression

possible

u (17.3)

We can illustrate the correctness of this equation with a simple example. A

class of seven students is given a quiz worth up to 10 points. The individual

scores are 7, 9, 9, 4, 2, 10, and 8. What is the average score on the quiz? One

way to determine an average is to add the individual scores, then divide by the

number of scores:

score

4

7

9

^7 ^9 ^9 ^4 ^2 ^10 ^8 ^7

7



values

j 1

ujPj





j

Pj

N!





m

i 1

ni!

588 CHAPTER 17 Statistical Thermodynamics: Introduction