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?

• 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? Commented Nov 7, 2022 at 23:18
• 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? Commented Nov 8, 2022 at 10:54
• I still think you need to edit your question for clarity, but I added an answer below. Commented Nov 8, 2022 at 11:26