Consider an ideal gas of *N* particles in a container with a volume *V*.
A partition separates the container into two sections with volumes
and , respectively, such that . Also, there
are particles in the volume and particles in the
volume . It is assumed that the number density is the same throughout
the system

If the partition is now removed, what should happen to the total entropy? Since the particles are identical, the total entropy should not increase as the partition is removed because the two states cannot be differentiated due to the indistinguishability of the particles. Let us analyze this thought experiment using the classical expression entropy derived above (i.e., we leave off the term).

The entropies and before the partition is removed are

and the total entropy is .

After the partition is removed, the total entropy is

Thus, the difference is

This contradicts our predicted result that . Therefore, the classical expression must not be quite right.

Let us now restore the . Using the Stirling approximation , the entropy can be written as

which is known as the *Sackur-Tetrode* equation.
Using this expression for the entropy, the difference now becomes

However, since the density is
constant, the terms appearing in the log are all 1 and, therefore, vanish.
Hence, the change in entropy, as expected. Thus, it seems that the 1/*N*! term
is absolutely necessary to resolve the paradox. This means that only
a correct quantum mechanical treatment of the ideal gas gives rise to
a consistent entropy.

Thu Feb 20 00:47:55 EST 2003