Gibbs paradox for a small numbers of particles

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greypilgrim
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Hi.

Trying to solve the Gibbs paradox for two identical volumes of ideal gas with ##N## particles each, I found the mixing entropy to be
$$\Delta S=2N \log(2)-\log((2N)!)+2\log(N!)\enspace .$$
The usual approach now uses Stirling's approximation to the order ##\log (n!)\approx n\log (n)-n## which indeed gives ##\Delta S=0##.

However, this is not zero for small ##N##. I assume this is because in the mixed case, fluctuations where the number of particles is different in the two volumes are still quite dominant for small number of particles, is this correct?

I used Mathematica to plot above function up to ##N=10^{10}## (it stops plotting after that), and it still looks to be monotonically increasing, yet very slowly and only up to about a value of 11. I somehow always assumed Stirling's approximation to be better for much smaller values. At which ##N## will above expression for ##\Delta S## start decreasing?
 
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A google of the subject says that the paradox is caused if a non-extensive form is used for the entropy equation. The problem is not because of Stirling's approximation. The google also says the Sackur-Tetrode expression for entropy is an extensive form that corrects this difficulty. The Wikipedia article appears to be a good one for this topic.
 
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I am using the (so-called) extensive form, the one that accounts for indistinguishability by dividing the number of microstates by the number of particle permutations ##N!##. That's what gave me the expression for ##\Delta S## in #1, which is only zero using Stirling's approximation. As I wrote in #1, numerically I get a monotonically increasing ##\Delta S## for up to ##N=10^{10}##, and my question is at what ##N## Stirling's approximation seems to "kick in", making this expression decrease.

The Sackur-Tetrode equation on the Wikipedia article on the Gibbs paradox is already simplified with Stirling's approximation, so it's not surprising that it leads to ##\Delta S=0## for any ##N##. This form only gets truly extensive with Stirling's approximation, which seems to be good only for ridiculously large ##N##.
As I stated in #1, it cannot be exactly extensive because there are microstates in the mixed case where the number of particles in the partial volumes differ from the case where the volumes are separated. Stirling's approximation sort of says that these fluctuations become negligible for large ##N##.

The non-extensive form that leads to the Gibbs paradox, where ##\Delta S## is even proportional to ##N##, contains no factorial, so there's no reason to use Stirling's approximation there.
 
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greypilgrim said:
I am using the (so-called) extensive form, the one that accounts for indistinguishability by dividing the number of microstates by the number of particle permutations ##N!##. That's what gave me the expression for ##\Delta S## in #1, which is only zero using Stirling's approximation. As I wrote in #1, numerically I get a monotonically increasing ##\Delta S## for up to ##N=10^{10}##, and my question is at what ##N## Stirling's approximation seems to "kick in", making this expression decrease.

The Sackur-Tetrode equation on the Wikipedia article on the Gibbs paradox is already simplified with Stirling's approximation, so it's not surprising that it leads to ##\Delta S=0## for any ##N##. This form only gets truly extensive with Stirling's approximation, which seems to be good only for ridiculously large ##N##.
As I stated in #1, it cannot be exactly extensive because there are microstates in the mixed case where the number of particles in the partial volumes differ from the case where the volumes are separated. Stirling's approximation sort of says that these fluctuations become negligible for large ##N##.

The non-extensive form that leads to the Gibbs paradox, where ##\Delta S## is even proportional to ##N##, contains no factorial, so there's no reason to use Stirling's approximation there.
One other suggestion is to use the more exact Stirling's approximation ## ln(N!)=N ln(N)-N +\frac{1}{2}ln(2 \pi N) ## to evaluate the expression. I'm not sure how exact the expression ## \Delta S=0 ## is supposed to be for the expressions that are being used.