I am trying to understand the following commentary I found in Wikipedia about this paradox:

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?

  4. Furthermore the following is also said in the wikipedia article about this:

    "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?

  • $\begingroup$ 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. $\endgroup$
    – Mauricio
    Commented Nov 24, 2021 at 19:19
  • $\begingroup$ en.wikipedia.org/wiki/Intensive_and_extensive_properties $\endgroup$
    – Mauricio
    Commented Nov 24, 2021 at 19:23
  • $\begingroup$ @Mauricio yes, but why, where does the paradox lies ? $\endgroup$
    – imbAF
    Commented Nov 24, 2021 at 19:29
  • $\begingroup$ 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. $\endgroup$
    – Mauricio
    Commented Nov 24, 2021 at 19:43
  • $\begingroup$ see this analysis by Jaynes: damtp.cam.ac.uk/user/tong/statphys/jaynes.pdf $\endgroup$
    – Quillo
    Commented Sep 23, 2022 at 17:56

3 Answers 3


Distinguishability means a scheme to extract energy

On one side of the barrier, 238UF6, and on the other, 235UF6. You lower the barrier and it still is all more or less uranium hexafluoride gas. What has changed?

Well, you could have some very special channel, akin to the mitochondrial F0-F1 ATPase but working in gas phase, to extract energy out of a gradient of UF6, and that's the same as you could get with a piston. And you could have such two variants of this channel that they only pass U-235 or U-238 containing variants of the gas. In which case, letting the gas out and expand from one chamber to another yields a lot of energy, and remember that the partial pressures of the two "different" gases don't affect each others' equilibria. So there is real, usable work on each side, in the same amount as can be gotten from moving a piston, i.e. RT x ln (2V/V) x mole of particles, or kB T ln 2 per particle. In a reversible process the thermal energy would stay the same but the entropy would increase by kB ln 2 per particle, so the temperature would decrease.

If there is no effort to harvest the energy and the two sides are just left to mix, then although there are no overall pressure changes, the partial pressure is averaged between the starting value and 0 on each side. The entropy increases by kB ln 2 and the thermal energy increases in proportion, so there is no change in temperature. This extra thermal energy is the same energy that could hypothetically have pushed a piston.

Now suppose you have Maxwell's demon come along and swap all the particles back where they started. That gives one bit of data for each particle with entropy -kB ln 2 and requires an energy input of kB T ln 2, which is the same as the hypothetical piston energy again. The information - which half of the chamber each particle is on - contains the same energy per kelvin as can be got by mixing them. It is simply the opposite of the entropy, compressing the particle by identifying it into one of two sets rather than moving it there.

But what if you can't come up with a channel that only passes one isotope of uranium? Well, then you might decide to consider the particles are "practically indistinguishable". The increase in entropy represents the loss of a free energy you could never have accessed anyway, so it doesn't matter. The temperature stays constant, but the thermal energy increases with the entropy, but only at the cost of this unusable free energy.

But you didn't have to mix the particles to do that! All you had to do was say, I don't want to KNOW what isotope is on what side of the chamber for any of these particles, and assuming the same is true of your channels, now you've erased one bit of information for every particle, increasing the entropy by kB ln 2 without ever having to open up a hole for them to mix. With the same effects, as far as you know.

  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.


The Gibbs paradox is of historical value only and relates to the thinking that led Gibbs to guess the presence of the factorial $N!$ in the canonical partition function:

$$Q = \frac{1}{h^{3N}N!} \int e^{-E/k_B T} d\mathbf{r}^N d\mathbf{p}^N$$

Without the $N!$ the log of the partition function is not extensive, as it should, neither is entropy or any other extensive property obtained from the partition function.

To produce the paradox one must misapply the partition function without the $N!$. From an educational standpoint it doesn't make sense to ask someone who tries to understand statistical mechanics to first apply an incorrect equation so we can show them what the correct equation is.

My advice to anyone who wants to learn more about this topic is to first try to understand the proper application of statistical mechanics and in particular the notion of homogeneity when we talk about extensive and intensive properties. Only then can one try exploring the many wrong results one obtains when the $N!$ is omitted.


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