Measuring the entropy of mixing ideal gases

Suppose I have two compartments of volume V separated by a barrier. Each one has an ideal gas, pure, of N molecules, and with equal pressure and temperature. However, I don't know whether both gases are the same one or different ones.

I remove the barrier, and wait until equilibrium is reached.

If the gases were different, theory predicts the change of entropy is $\Delta S=-2Nk(\ln [V/2V])$
If they were the same, theory predicts it is $\Delta S = 0$, since I didn't actually mix anything, and there is no change in state.

Can I somehow experimentally measure this $\Delta S$?

Not predict it from theory, but calculate it from real actual thermodynamic measurements, and without being able to identify the gases, to see whether I obtain $0$ or another number.

If I can't, then it seems to me it's an empty concept.

• Can you think of how you would measure it if you knew the gases were different? – Chet Miller Jul 24 '18 at 0:05
• Not really, hence the question. I used calorimeters to find entropy changes but I think it won't help here. I can think as each ideal gas being blind to the other and see the mixing as merely an expansion against vacuum. Which is how I get the formula I described. But there is no measurement there. – Juan Perez Jul 24 '18 at 2:08
• Suppose you have the gas mixture in a cylinder (each gas at a different partial pressure), and, connected to the cylinder, are two semi-permeable membranes that permit each of the gases to pass, but not the other. So you can separate them into pure gases at pure gas pressures equal to their partial pressures in the mixture, by pushing them through the membranes into attached pure-gas cylinders. This requires no net work, no net heat, and no change in internal energy. You then impress each gas isothermally to a specified higher pressure. The latter requires removal of heat. – Chet Miller Jul 24 '18 at 2:32
• The heat transferred would determine the entropy change. This would be the reverse of the entropy change of mixing the gases. The final specified pressure for the pure gases would be the same as the mixture before the separation. – Chet Miller Jul 24 '18 at 2:41
• Agree with Chester's approach. Perhaps some sort of calorimetry could determine the quantity of heat transferred out of each gas. But shouldn't the equation be – Bob D Jul 28 '18 at 3:33