Gibbs Paradox - why should the change in entropy be zero?

The Gibbs paradox deals with the fact that for an ideal gas with $N$ molecules in a volume $V$ seperated by a diaphragm into two subvolumes $V_1,V_2$ with $N_1,N_2$ particles in each subvolume, removing the diaphragm gives a nonzero change in entropy, but the change should be zero.

I don't understand why (conceptually) the change of entropy in this situation is supposed to be zero. Why isn't it positive - after all, removing the diaphragm gives the particles more freedom and thus increases the 'disorder' of the system and with entropy being a measure of this 'disorder' it should too increase. Conversely, if I put more and more diaphragms into the container, I could potentially isolate each particle in its own subvolume and leading the system to be very ordered, so the entropy should be very small.

What is wrong with this way of thinking?

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The paradox is: with such a big jump in entropy, you would see a significant change in your system - like a sudden temperature drop. Obviously, this doesn't happen, hence the paradox. –  gigacyan Jun 12 at 15:57

The entropy change should be zero – and essentially is zero, in the correct theory that takes the indistinguishability into account – because the thin membrane doesn't materially change the system and carries a tiny entropy by itself. The first reason is enough: the removal of the membrane is a reversible process – one may add the membrane back – so the entropy has to be zero. An entropy can't increase during a reversible process because it would decrease when the process is reversed – and that would violate the second law of thermodynamics.

In other words, the self-evident reversibility of the unphysical membrane means that $\delta S = \delta Q/T$ where $\delta Q$ is the heat flowing to the system – but it's clearly zero.

The paradox is removed when the indinguishability of the particles is appreciated. The calculable entropy change is zero, as expected. In some sense, we are implicitly assuming that the molecules are indistinguishable everywhere above. If the molecules carried some passports, they could have a Canadian and American passport in the volume $V_1,V_2$, respectively, which would be a very special state (none of the molecules is abroad) while the number of states would be increased because each molecule may be either in its own country/volume or abroad. This is indeed why the wrong classical calculation claims that the entropy would increase.

However, this prediction may be extracted even if the initial total volume $V_1+V_2$ is actually perfectly mixed before the membrane is added.

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