# Work done in mixing ideal gases

During our lecture on physical chemistry at university, we only talked about work in the context of volume work done by an ideal gas against a surrounding constant pressure.

(1)Let me explain. An ideal gas is located inside a container, one of the walls is a movable piston (massless). Initially, both outside pressure and gas pressure are at equilibrium, when we heat the gas, it expands against the piston, that means work done against the outside pressure, until equilibrium is reached again.

(2)But what happens if we let two ideal gases mix?

Suppose that inside an adiabatic container of distinct volume, an adiabatic wall is separating two portions of the container’s volume, each filled with one mole of ideal gas. One gas is in state 1 (p₁, T₁, V₁), the other in state 2 (p₂, T₂, V₂), and p₁>p₂; T₁>T₂; V₁>V₂.

If we remove the wall, surely the higher pressure gas in the bigger part of the container moves into the other part, moving the lower pressure gas. But is work done here?

While researching, I found an answer which indicated that work isn’t done here like in example (1) although there is a pressure difference; that is because the ideal gases fill no volume by themselves so the two gases are basically expanding freely doing no work at all.

Is that true? And if not, how can we calculate the work done here? I know, that two gases at the same pressure don’t do work on each other though (called free mixing).

Your adiabatic fixed volume system is an isolated system, i.e., it does not exchange mass or energy in the form of work or heat with its surroundings and so it does not undergo a change in internal energy per the first law.

That does not, however, mean work is not being done within the system. Work is done by the gas expanding, in which it loses energy, and done on the gas that is compressed, in which it gains an equal amount of energy, for a total internal energy change of zero.

Hope this helps.

• Ah that makes sense. So the work in total is zero, because the compressed gas gains work energy, and the expanding gas loses the same amount of work, and the energy is distributed again in the mixture, so dU=0 because of dQ=0 (adiabatic condition) and therefore dW=0 for the adiabatic system?
– 冰淇淋
Apr 9, 2022 at 13:46
• Correct. $W$ (and $Q$) depends on how you define the system and the surroundings. For example, if one of the gases was defined as the system and the other as the surroundings, then $W\ne 0$. Apr 9, 2022 at 16:49
• W in this definition would be not equal to zero in this case because the volume is expanded (wall is removed) and the pressure changes?
– 冰淇淋
Apr 9, 2022 at 21:22
• But if you are defining the contents of a rigid thermally insulated vessel and the surroundings as everything outside the vessel, $W=0$. But if you are defining the system as the higher pressure gas on one side of the piston inside the vessel as the system and the lower pressure gas on the other side of the piston as the surroundings then $W\ne0$. Are we on the same page now? Apr 9, 2022 at 21:30