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I have a question regarding the Gibbs Paradox.

Let us assume that the two partial volumes of the box are equal in size $V_1=V_2$, and that $N$ particles of a monatomic ideal gas are in each of the two chambers. All gases involved have the same particle mass and energy per particle before and after.

Now let's assume two cases:

1.both halves contain the same gas and this consists of distinguishable particles.

2.the halves contain different gases, which themselves consist of distinguishable particles.

If one calculates now the entropy change, is it not then actually the same calculation. The particles in case 1 are distinguishable, in case 2 they are different gases, but their particles among themselves are also distinguishable. In case 2, the particles cannot be more distinguishable than in case 1, can they?

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    $\begingroup$ Your question is vague: what do you mean "but their particles among themselves are also distinguishable"? Then you say "In case 2, the particles cannot be more distinguishable than in case 1, can they?" You made the particles distinguishable, didn't you? So what are you asking exactly? $\endgroup$
    – Themis
    Nov 7, 2022 at 23:18
  • $\begingroup$ Sorry for being so vague, I really just wanted to know if in case two, by being different gases, the particles are even more distinguishable from each other than if they are the same gases. I have to calculate the entropy change and I think all the time that this is the same calculation, the particles in case 1 are distinguishable among themselves and in case 2 they are different gases, but the particles among themselves are also distinguishable, so basically it is the same problem, or am I missing something? $\endgroup$
    – Lambda
    Nov 8, 2022 at 10:54
  • $\begingroup$ I still think you need to edit your question for clarity, but I added an answer below. $\endgroup$
    – Themis
    Nov 8, 2022 at 11:26

1 Answer 1

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Let's consider the reverse problem of the one you are considering:

  • In case I we have 8 balls and the task is to divide them into two equal piles.

  • In case II we have 4 balls of kind 1 and 4 balls of kind 2 mixed together and the task is to separate them in two piles with all balls of the same kind in the same pile.

Which task takes greater effort?

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Answer: The task that takes the greater effort is the one with the larger entropy, that's case II. Each time we pick one ball we must determine its kind because we are dealing with two distinguishable kinds. In case I we don't need to identify the type, hence less effort.

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