# Entropy and multiplicity of a gas when adding/removing wall in container

1. Oct 9, 2008

Consider an isolated system of two ideal, identical gases in thermal equilibrium. Gas A is occupying a volume A, separated by a wall from volume B, where gas B resides. They are in thermal contact. The volumes are the same, so are the pressures and temperatures, and it follows that the number of molecules on each side of the wall is the same.

We know that the multiplicity of each gas (at constant T) is proportional to V^N, where V is the volume and N is the number of particles.

Now, without disturbing the system in any other way, imagine that we remove the separating wall. Ideally, let's say the wall "disappears".

Suddenly, each gas has doubled its availiable volume, so the multiplicity increases. From the multiplicity, we can find the entropy as:

S = k ln(m); where k=Boltzmanns constant, m = multiplicity.

So the increase in entropy for the entire system is:

/\S = 2 * k ln[(2V)^N / V^N] = 2k ln(2^N); where V is half of the total volume

/\S is obviously positive.

Now, if we reintroduce the wall, each gas is again enclosed in volume V. If we consider the particles to be indistinguishable, keep temperature and all other parameters fixed, the multiplicity is now scaled down. We get a decrease in entropy. (Not allowed, ofcourse, since dS >= dQ/T)

Must be something wrong with our reasoning, so let's say that the change in entropy upon reintroducing the wall is 0. Then we would have invented an entropy-making machine for ourselves. Just take the wall out to increase the entropy and put it back again. Repeat for even more entropy and go on as long as you like!

Not reasonable, right?

Seems to me that the suspect here is the "indistiguishable particles" step.

Any ideas?

2. Oct 9, 2008

### atyy

3. Oct 11, 2008

### oasis2007

Atyy will you be able to help with grad course hw in statistical mechanics?