Gibbs paradox - Aetherwisp

The Gibbs paradox is a paradox regarding a non-quantum derivation of entropy that does not consider the indistinguishability of particles. The paradox revolves around how the entropy as derived from classical ensembles is intensive, despite entropy by definition needing to be extensive. It is resolved by taking the indistinguishability of particles into account when counting the number of states.

Description#

Stating that entropy is intensive is, in a way, the same as saying that the entropy of a system is not equal to the sum of its parts, which makes little sense. Gibbs sought to describe this phenomenon by taking two ideal gases, mixing them and seeing the results.

Consider two ideal gases, with $N_{1}$ and $N_{2}$ particles and $V_{1}$ and $V_{2}$ volumes, initially separated from each other. They have the same temperature $T$ . Let's call the total occupied volume $V=V_{1}+V_{2}$ . From the entropy of a classical ideal gas

S_{i}=k_{B}N_{i}\left[ \frac{3}{2}\ln\left( \frac{2\pi m_{i}k_{B}T}{h^{2}} \right)+\ln V_{i}+ \frac{3}{2} \right]

we can figure out the entropy of mixing as the difference between the total final entropy

S_{T}=\sum_{i=1}^{2} k_{B}N_{i}\left[ \frac{3}{2}\ln\left( \frac{2\pi m_{i}k_{B}T}{h^{2}} \right)+\ln V+ \frac{3}{2} \right]

(note that here we use $V=V_{1}+V_{2}$ as opposed to $V_{i}$ ) and the sum of individual entropies:

\Delta S=S_{T}-\sum_{i=1}^{2} S_{i}>0\tag{1}

If the gas density is the same for both gases, this simplifies greatly to

\frac{\Delta S}{k_{B}}=N_{1}\ln \frac{V}{V_{1}}+N_{2}\ln \frac{V}{V_{2}}>0\tag{2}

So far so... good, actually. This isn't illogical. The issues start to arise when we mix a gas with itself (i.e. the gases are the same). In this case, $m_{1}=m_{2}=m$ and the total entropy is

S_{T}=k_{B}N\left[ \frac{3}{2}\ln\left( \frac{2\pi mk_{B}T}{h^{2}} \right)+\ln V+ \frac{3}{2} \right]

The only difference from the previous formula is that, everything else being equal, the sum just amounts to $N_{1}+N_{2}=N$ , hence why $N$ lost the subscript. It is otherwise identical to before. Because of this, the entropy of mixing will be given by $(1)$ and $(2)$ . However, this is nonsense, and the reason for that is that mixing a gas with itself is perfectly reversible, so long as the original temperature and density are the same: removing the partitioning wall keeping the gases separate effectively does nothing because the gases are already in the same starting condition. Thus, putting the wall back fully reverses the process. And yet the entropy increases, despite nothing happening? Reversible transformations are supposed to have $\Delta S=0$ , so this can't be. This is the essence of the paradox.

Let's be more precise: something obviously does happen when you remove the partitioning wall. Atoms of gas 1 move into the volume of gas 2 and viceversa. But we know for a fact that entropy can't increase in this process, so by necessity this exchanging of atoms must do nothing with respect to entropy. In fact, this is the crux of the matter: we cannot distinguish between atoms, and not for lack of trying. It is simply an innate property of quantum particles that we need to accept and deal with. This indistinguishability, as it is known, is the true culprit of the paradox, and our lack of consideration for it in this classical proof is the cause of the incident. If we accept that we cannot distinguish particles from each other (and also not track their motion, see Uncertainty principle), the start state and end state of mixing two gases made of the same particles are identical, and so how could entropy change?

Correction#

The paradox is corrected by removing the overcounting of indistinguishable states. This is done by using so called correct Boltzmann counting, which simply involves dividing the number of states by $N!$ . Using this new way of counting, the entropy of an ideal gas comes out to be

\frac{S}{k_{B}N}=\ln\left[ \frac{V}{N}\left( \frac{4\pi m}{3h^{2}} \frac{U}{N} \right)^{3/2} \right]+ \frac{5}{2}

This is known as the Sackur-Tetrode equation and it was proven correct experimentally.

For a derivation of why specifically $N!$ , see Identical particles > Correct Boltzmann counting.