# Entropy of mixing

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?

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
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