- #1
greypilgrim
- 517
- 36
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?
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?