Physical Chemistry , 1st ed.

Example 18.8

Calculate the rotational partition function of gaseous NH 3 at 1000 K. The ro-

tational temperatures are 13.6 K, 13.6 K, and 8.92 K. The symmetry number

for ammonia is 3.

Solution

Since only two different rotational temperatures are given for ammonia, we

conclude that the molecule is a symmetric top. We can use equation 18.41

to determine qrot, and we must take care that the correct rotational temper-

ature goes in the correct term. Since 13.6 K is repeated twice, it is used for

r, A.We get

qrot

3

1/2



1

1

0

3

0

.6

0

K

K



1

8

0

.9

0

2

0

K

K



1/2

Solving:

qrot 460

Note how all of the units cancel, so that qrotis a pure number.

18.7 The Partition Function of a System

At this point, we have determined the complete partition function Qfor a mol-

ecule. It is

Qqtransqelectqvibqrotqnuc

and all of the parts ofQhave been evaluated mathematically. This molecular

partition function is strictly applicable to a single molecule. What if we have a

system that has many noninteracting molecules?

We start off by suggesting that the total energy of a system is the sum of the

individual types of energy a molecule can have: electronic, translational, vi-

brational, and so on. The total energy of the system is the sum of the energies

of the individual particles. Thus, due to the original definition of the partition

function, the overall partition function for the system is the productof the in-

dividual partition functions ofNmolecules:

QsysQ 1 Q 2 Q 3 QN

If the system is composed of only one kind of molecule, then all of the indi-

vidual Qivalues are the same, and we simply have Qbeing multiplied by itself

Ntimes. To write this another way, we have

QsysQN

However, this doesn’t account for the fact that the individual molecules in the

system are indistinguishable at the macroscopic level. Recall the examples of

the balls in boxes at the beginning of Chapter 17. We found that there were

fewer possible unique arrangements when the two balls were the same color.

Another way to consider this is that there will be fewer possible arrangements

if we cannot tell which gas molecules are which; that is, if they are indistin-

guishable. (We do know, however, that the molecules are the same compound.

We just can’t tell, say, one molecule of water apart from any other molecule of

water in our system.) Similarly, the above expression for Qsysis overvalued.

Statistics can show that the value is too high by a multiplicative factor ofN!

636 CHAPTER 18 More Statistical Thermodynamics