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.

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    $\begingroup$ Can you think of how you would measure it if you knew the gases were different? $\endgroup$ Jul 24, 2018 at 0:05
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    $\begingroup$ 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. $\endgroup$
    – Juan Perez
    Jul 24, 2018 at 2:08
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    $\begingroup$ 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. $\endgroup$ Jul 24, 2018 at 2:32
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    $\begingroup$ 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. $\endgroup$ Jul 24, 2018 at 2:41
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    $\begingroup$ @Javi The work you do in pushing them into the pure gas cylinders is equal to the work the pure gases do in pushing a second piston at the other end of each pure gas cylinder against its surroundings. So the net work is zero. $\endgroup$ Jun 25, 2021 at 11:35

1 Answer 1


So not ideal gases, but the entropy of mixing two salts has been measured directly, and it matches your theoretical equation very well (with volume reparameterized to concentration):


N. J. Seeley, 1972, "Entropy of Mixing - An Electrochemical Measurement." J. Chem. Educ. 49(3):212


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