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

Question

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?

Answer 1

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.

Answer 2

Yes, there can be an entropy increase when mixing two gases, but in the case of mixing macroscopic gases it is very slight.

Let me be perfectly precise. An increase in entropy is equivalent to an irreversible process, and so in any sort of situation where the entropy increases we should be able to identify the corresponding irreversible behaviour. In the case of two equal volumes containing exactly equal numbers of indistinguishable particles, when we remove the partition between the volumes, the number of particles in each volume is able to fluctuate (only the sum is conserved). After we then re-insert the partition then we are no longer certain that each volume has an identical number of particles as we were at the start. This loss of knowledge is irreversible and by this mechanism the entropy is increased. For two large volumes the entropy increase is (if I recall correctly) only logarithmic in the number of particles, and so in most discussions it is neglected.

On the other hand if we consider the case of releasing individually-wrapped particles that you suggest, then the entropy increase is enormous as we would have to spend a great deal of effort to perfectly repartition the particles back into singly occupied spaces.

However, the above process is not the Gibbs paradox. The Gibbs paradox can refer to one of two things:

• erroneous predictions of giant mixing entropies from mistreating indistinguishable particles as distinguishable (and vice versa), as Lubos describes.
• a thought experiment where we take distinguishable particles and gradually diminish the differences between them until they are indistinguishable (at what level of distinction does the mixing entropy disappear).