Consider two identical boxes of ideal gases, at the same temperature, volume, and pressure. I will assume that gas A and gas B are distinguishable, but do not react with each other (maybe this is a lie).

We remove a wall between the two boxes. The gases will tend to mix over time, since that tends to increase entropy.

Because entropy is increasing, Gibbs energy should be decreasing. I have read that "The Gibbs free energy change is the maximum amount of non-expansion work that can be extracted from a closed system."

However, in my example, where there is no change in enthalpy, I don't see how the work can be extracted from simply the change in entropy (the arrangement of the gases).

Clearly I have missed something, as I am new to these terms. I hope my question is not too naive.

  • $\begingroup$ Determining change in entropy and enthalpy during mixing of gases $\endgroup$
    – Farcher
    Jun 4, 2022 at 7:20
  • $\begingroup$ You need to devise a reversible process for going from two individual gases at T, P , and V to a mixture of the gases as T, P, and 2V (and partial pressures P/2). Hint: such a process might involve use of semipermeable membranes and three cylinders. $\endgroup$ Jun 4, 2022 at 10:47

1 Answer 1


Mixing of ideal gases is just two expansion processes into the same final volume. Replace the partition with one permeable to gas B but not gas A, and using a piston permeable to gas A but not gas B, collect the work from B reversibly expanding into the final volume. Then switch pistons to one permeable to gas B but not to gas A, remove the partition, and collect the reversible work from A expanding. No entropy is ever generated, and you extracted work, in accordance with the difference in Gibbs free energies.


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