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Consider the case of a wall dividing a box into sections 1 and 2, each of volume $V_0$.

Let be $X$ is an ideal gases. Section 1 contains $N$ particles of $X$ and section 2 contains $N$ particles of $X$.

The entropy of mixing (removing the divider) is $0$ in the above case. I understand that an argument for this is that it is a reversible process as we can just as well place the divider back and the system will be in an identical state to its initial condition (assume indistinguishable particles).

My question is that if entropy is a measure of the number of states a particle can take on, a state being a specification of all the values of the variables needed to completely describe the particles, i.e its position and momentum.

If we remove the divider, each particle can now take up twice as many positions and so should have twice as many available states to be in, this surely increases the entropy?

I can't seem to resolve the two different ideas.

Can someone clarify on this?

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You are surprised that $S(T, 2V, 2N) = 2S(T, V, N)$ but this formula can be obtained with statistical physics considering a mix of two identical gas

When you remove the barriers and the two gas are identicals you do not increase the number of unknowns configurations of the system, therefore, entropy does not increase. (This is a really intuitive view of the problem I don't know if you should consider it like that)

But when you remove the barriers with two differents gas, it is obvious that there is a change of the unknowns configuration of the system

Considering the last statement and the demonstrations for both discernable and non-discernable mix on the French Wikipedia you should be able to understand it

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  • $\begingroup$ "When you remove the barriers and the two gas are identicals you do not increase the number of unknowns configurations of the system". But consider particle 1 in one of half of the system before its divider is removed. It can only have positions in half the volume, after the divider is removed it can now take on two times as many positions. Isn't this an increase in the configurations it can take? I understand the logic used in the answer but I cant seem to resolve it with my above statement. $\endgroup$ – Vishal Jain Dec 10 '19 at 11:27
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    $\begingroup$ I believe the answer to your statement is in the indiscernibility, read the two full demonstrations (use google trad if needed) $\endgroup$ – Oka Dec 10 '19 at 13:12
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So when you mix the gases, say that the probability of any gas particle existing in the i'th state is $p_i$, so now what do you say is the probability of particles being in various states? $\Pi_ip_i$. Now entropy is defined as the average value of $-log(p_i)$, and as $p_i$ is multiplicative, $-log(p_i)$ is additive, i.e. entropy is additive

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