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I was wondering if gases, which are in the process of mixing, do any work on each other. I haven’t found many resources on this.

We all know that gases, which have a higher pressure than their surroundings, expand. For example, suppose a (approx. massless) piston divides an ideal gas and the ambient $\pu{1 atm}$ pressure. If we heat the ideal gas, the pressure rises, and the piston gets displaced so that the ideal gas can re-equilibriate to $\pu{1 atm}$ pressure, so outside pressure and inside pressure are the same.

But what is up with the case when two gases at different pressure mix? For example, an ideal gas is in a closed container at $p_1 (\lt p_2), V_1 (\gt V_2),$ and $\pu{273 K}$. Now this container is connected to a smaller one, inside is a different ideal gas at $V_2$, $p_2$, and $\pu{273 K}$. Now, of course the gases will mix with each other, to maximize entropy by mixing.

But does the higher pressure gas perform work against the lower pressure gas while expanding into the other container, like the first example I gave? Or is it like a free expansion (a.k.a no work done), because ideal gases don't occupy volume by themselves? And how would we calculate the work?

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  • $\begingroup$ "to maximize entropy by mixing" sounds like the gases have free-will, desires, and goals. $\endgroup$ Commented Apr 4, 2022 at 0:34
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    $\begingroup$ In ideal gas context, work is done/absorbed by expansion/compression of regions with different pressure. It is irrelevant if they are filled by different gases or the same gas. Spontaneous mixing of 2 different gases of the same pressure does not do any work, it just increases entropy and decreases Gibbs energy. $\endgroup$
    – Poutnik
    Commented Apr 4, 2022 at 8:21

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Consider what happens to just the gas at higher pressure, if allowed to expand to a lower pressure -- what happens to its temperature? Then consider the gas at lower pressure, and the converse.

As an example, consider two containers at 300 K, one with helium at 10,000 kPa (~100 bar), another, of the same volume, with He at 100 kPa (~1 bar). When a valve opens between them, the equalized pressure would be ~5,000.5 kPa, with far more gas (by mass) expanding than being compressed. Would you expect the final temperature to be lower or higher?

Of course, this is a highly simplified example. One would need to take in to account the heat capacity of differing gases, the volumes of the containers, the mass of gas in each, initial temperature of each and, to be closer to a "real life" answer, departure from ideal gas laws, and thermal storage in the containers themselves. Oh... and if the gas expand through a venturi or through a turboexpander.

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  • $\begingroup$ If I understood your example correctly, I would expect the final temperature to be lower. If we assume adiabatic conditions (dQ=0), only work is done, and as much more helium expands to the other side, it also needs to push against the 1 bar of helium on the other side, and needs to overcome more and more pressure as it builds up on the side that was originally 1 bar. How could we quantify this though? $\endgroup$
    – Mäßige
    Commented Apr 4, 2022 at 13:58
  • $\begingroup$ @Mäßige, one gas does not compress another. If gas A in container A expands to fill both container A and B, its partial pressure is the same, regardless of whether B originally contained nothing, or He under 100,000 kPa. Each molecule of gas does its own "pushing", and all molecules can fit between each other. Just consider each separately -- easily calculated. $\endgroup$ Commented Apr 4, 2022 at 23:33
  • $\begingroup$ Basically, if the partial pressure is the same regardless, then no work should be done, because the ideal gas molecules fit between each other and have no interactions? $\endgroup$
    – Mäßige
    Commented Apr 5, 2022 at 7:17
  • $\begingroup$ Is work done when a gas expands from one container to another? Could that move a piston, or turn a turboexpander, or change the temperature of the gas and its container? $\endgroup$ Commented Apr 5, 2022 at 20:51
  • $\begingroup$ No. As you put it: When a gas expands freely (that would mean against no outside pressure) to just fill another container, dW=0 and dQ=0 (adiabatic container), that means dU=0 and no change of temperature, just different pressure and volume. So no work done at all, if we consider each gas seperately while mixing, as per ideal gas law (molecules themselves occupy no volume). That means that generally no work is done when two ideal gases (with p₁, V₁, T₁; and p₂, V₂, T₂) are getting mixed. Correct? (May differ when dealing with real gases) $\endgroup$
    – Mäßige
    Commented Apr 6, 2022 at 16:06

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