Now a door in the container wall is opened to allow the gas particles to mix between the containers. No macroscopic changes occur, as the system is in equilibrium. The entropy of the gas in the two-container system can be easily calculated, but if the equation is not extensive, the entropy would not be $$2S$$. In fact, the non-extensive entropy quantity defined and studied by Gibbs would predict additional entropy. Closing the door then reduces the entropy again to $$2S$$, in supposed violation of the Second Law of Thermodynamics.

I understand the reason why the entropy doesn't change (macroscopic at least). The state the system was before the removal of the wall is no different then that after the removal,simply because we cannot tell whether a particle of the left or right side moved in the other direction.

But the commentary I highlighted is what confuses me, therefore I have some questions about it:

First I tried to calculate the entropy of the system when we remove the thin wall, in the simple case where the 2 chambers have same gas, same nr. of particles, same $$T$$, $$P$$, $$V$$. I got as a results:

$$\Delta S =2nR\,\ln2$$ and since there is no exchange of heat with the environment or work done by the system or from the system, the entropy of the environment doesn't change. And as a consequence the total entropy (system + environment) > 0, which implies an irreversible process. I hope my understanding is correct up to this point.

1. Then the text proceeds with he following segment : " ... but if the equation is not extensive, the entropy would not be $$2S$$." What does it mean? What is the meaning of an extensive equation?

2. "In fact, the non-extensive entropy quantity defined and studied by Gibbs would predict additional entropy." This I don't understand at all.

3. " Closing the door then reduces the entropy again to $$2S$$, in supposed violation of the Second Law of Thermodynamics." Is this a violation of the 2nd law because the entropy which was $$S =2nR\ln2 + 2S$$ is reduced to $$2S$$. How would I be able to calculate this reduced entropy?

"Suppose the two volumes are separated by a barrier in the beginning. Removing or reinserting the wall is reversible, but the entropy increases when the barrier is removed by the amount $$\Delta S =2nR\ln2 >0$$ which is in contradiction to thermodynamics if you re-insert the barrier. This is the Gibbs paradox."

What is reversible here? I have never considered the removal or addition of a barrier as a process, and I cannot also understand how it is qualified as a reversible, process I guess? The only thing that I do remember is when we study a gas in one side of a volume and we remove the barrier, in this case the expansion of the gas in the whole volume is an irreversible process, but never has anything been said regarding the removal of the barrier.

I hope someone can help me understand what this paradox is about and how entropy changes, which violates the 2nd law?

Edit: And why would (supposedly) entropy change two $$2S$$ ($$S$$ for each sub system) when we reinstall the barrier and not to something else, something not symmetrical?

• Gibbs paradox: You start with two containers with $S$ on each, you open the door and now have a $S'=2S+\Delta S$, if you get $\Delta S =0$ then everything is ok, but if you somehow get $\Delta S\neq 0$ there is something paradoxically wrong in your derivation. Nov 24, 2021 at 19:19
• en.wikipedia.org/wiki/Intensive_and_extensive_properties Nov 24, 2021 at 19:23
• @Mauricio yes, but why, where does the paradox lies ? Nov 24, 2021 at 19:29
• Classical mechanics usually considers particles to be "followable", and do to that you get $\Delta S \neq 0$, which means that in principle you could extract work from such a system. Nov 24, 2021 at 19:43

1. For a definition of extensive and intensive properties see Wikipedia's article. Entropy is an extensive quantity like volume or mass. If you have two containers with entropies $$S_1$$ and $$S_2$$, then the total entropy is $$S_1+S_2$$.

2. Intuitively, in Gibbs' thought experiment, one could expect that nothing should happen when you open the barrier between two identical containers with entropy $$S$$ each ($$2S$$ at the beginning, $$2S$$ at the end). However, originally statistical mechanics predicted $$2S+2\ln2\;nR$$ after opening the barrier. If the boxes are identical and it is the same gas why should entropy increase by opening the box?

3. I am not so sure about this one. Should the entropy in each side be $$S$$ or $$S+\ln2\; nR$$ after opening and closing the barrier? If it is the former as claimed, then you indeed increase and decrease the entropy by opening back and forth the barrier (which makes no sense due to the 2nd law of thermodynamics).

4. "Reversible" here just means that the barrier can be closed and opened. However this simple process seems to modify the entropy of the total system.

Last question: I am also concerned on why should the gas go back to $$2S$$ when the barrier is closed again. It certainly has to be symmetrical because of the symmetries of the problem but I see no argument to say why it should not be $$2S + 2nR \ln2$$.

The whole problem is calling for a solution. If the gases are the same in each box there is no good argument for the entropy to go up. An increase of entropy implies that you could somehow do some work.

The microscopic resolution is that if, on one hand, particles that compose the gas are indeed identical (and we know that is the case because of quantum mechanics) you should add a factor in your equation ($$1/N!$$) in order to make the entropy extensive. If you account for that factor you get $$2S$$ (barrier closed) and $$2S$$ (barrier open), which is the expected behaviour.

On the other hand, if the gases are different (example: two different isotopes), then there is no reason to expect $$2S$$ after opening the barrier. As you open the barrier, the two gases mix and the mixing process is irreversible. Entropy rises and now even if you close the barrier the entropy in each container has been raised.