Ideal Gas: an unexpected consequence of indistinguishability and the $N!$ term

Question

I came across an issue which bugs me.

Considering an ideal gas of $N$ non-interacting particles on a $1D$ container of length $L$, its (configurational) entropy $S$ reads $$ S = kT \ln \Big(\frac{L^N}{N!} \Big) = kT N \Big( \ln (\frac{L}{N})+1 \Big)$$ the last step being justified by Stirling's approximation for large $N$. The configurational entropy is all there is in this system, no energy contribution for the free energy.

The particles are non-interacting so one would assume that the entropy would grow as the number of particles increases.

Yet that is not the case: $S$ is not monotonic, and $$ \frac{\partial S}{\partial N} = -kT \ln \Big( \frac{L}{N}\Big) $$ equals $0$ for $N = L$, whatever (positive) $T$.

So if one puts the box of side $L$ in contact with a particle reservoir, from which partciles can come and go without penalties, $N$ particles are required to minimise the free energy.

This is most surprising, and it comes from having included the $N!$ term in the denominator of the entropy expression. A rather significant change: without $N!$ term the entropy grows monotonically with $N$.

But how come? Where does this limit on the number of particles comes from, if they are non-interacting? I would expect, the more of them, the higher the entropy.

EDIT

following requests for clarification, I add details on howthe formula in question is derived. The partition function $Z_1$for an "energyless" particle (no interactions, no kinetic energy) in a 1D box of length $L$ is given by $$ Z_1 = \int_0^L \exp{-(\frac{0}{kT}}) \mathrm{d}x = L$$ As we have $N$ independent indistinguishable particles, the overall partition function $Z$ equals $$ Z = \frac{Z_1^N}{N!}$$ As there is no energy, the free energy equals $-TS$, and the relationship betwwem $Z$ and free is the final step.

Answer 1

Notice that you have the wrong formula. For a 1D ideal gas the 1-particle partition function is given by

$$Z_1 = \frac{1}{2\pi\hbar} \int dq dp \exp{\left(-\frac{1}{kT}\frac{p^2}{2m}\right)} = \frac{L}{\lambda}$$

where $\lambda = \sqrt{2\pi\hbar^2/mkT}$ is the thermal wavelength and $L$ is the length of the system. Therefore for $N$ indistiguishable particles, the partition function is given by: $$ Z = \frac{Z_1^N}{N!}$$ In the canonical ensemble, the entropy is given by $$S = \frac{\partial}{\partial T} (kT \log Z)$$ After some calculations, one gets the formula: $$S=k\left[\log{\frac{L^N}{\lambda^N N!}}+ \frac{N}{2} \right]$$ Applying Stirling's approximation, $\log{N!}\approx N\log{N}-N$ one obtains the correct formula: $$S=k N \left[\log{\frac{L}{\lambda N}}+ \frac{3}{2} \right]$$ And therefore its derivative: $$\left(\frac{\partial S}{\partial N}\right)_{T,V} = k \left[\log{\frac{L}{\lambda N}}+ \frac{1}{2} \right]>0$$ so the entropy does increase with the number of particles.

As to the other part of your question, that's not actually what it means to maximise entropy. Given a system with fixed $L,N$ and average energy (in the case of the canonical ensemble) the entropy is maximised by a certain distribution, and you get the formula that is the one above. You don't extremise with respect to $N$.