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The Gibbs paradox

Consider an ideal gas of N particles in a container with a volume V. A partition separates the container into two sections with volumes tex2html_wrap_inline609 and tex2html_wrap_inline611 , respectively, such that tex2html_wrap_inline651 . Also, there are tex2html_wrap_inline653 particles in the volume tex2html_wrap_inline609 and tex2html_wrap_inline657 particles in the volume tex2html_wrap_inline611 . It is assumed that the number density is the same throughout the system


Figure 2:

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 tex2html_wrap_inline583 term).

The entropies tex2html_wrap_inline663 and tex2html_wrap_inline665 before the partition is removed are


and the total entropy is tex2html_wrap_inline667 .

After the partition is removed, the total entropy is


Thus, the difference tex2html_wrap_inline669 is


This contradicts our predicted result that tex2html_wrap_inline671 . Therefore, the classical expression must not be quite right.

Let us now restore the tex2html_wrap_inline583 . Using the Stirling approximation tex2html_wrap_inline675 , 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 tex2html_wrap_inline677 is constant, the terms appearing in the log are all 1 and, therefore, vanish. Hence, the change in entropy, tex2html_wrap_inline671 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.

Mark Tuckerman
Thu Feb 20 00:47:55 EST 2003