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I have a question about free expansion of ideal gas.

Firstly, we will take the entire (isolated) container to be our thermodynamic system. We know that the gas in the left-part of the container has a temperature $T'$. We know that the first-law of thermodynamics must hold $\Delta U = Q + W$, but because the system is isolated there is no heat-transfer meaning $Q = 0$ and beacause the work can't be done by gas on vacuum in the right-side of the container $W = 0 \Rightarrow \Delta U = 0$. If $U$ doesn't change the temperature must stay the same. $$T = T'$$

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And now the part I do not understand.

If we take for our thermodynamic system just the part of the gas that will end up in the right-side of the container. Then we can say that the gas on the left will push the marked gas to the right side. Marked gas will receive the work by the gas from the left, and gas from the left will transfer work to the right part of the gas. Because $\Delta U = 0$ the temperature on the left and the right side of container will not be the same. $$T' \ne T_1 \ne T_2$$ How is that possible, we just concluded form the example above the the temperature shouldn't change?!?!?!

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2 Answers 2

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At equilibrium, the temperatures must be the same, as you show in your first paragraph. This doesn't preclude temporary fluctuations while the system is out of equilibrium.

I agree with your reasoning that the gas within the blue region tends to heat up temporarily as work is done on it to push it into the empty side. (This aspect is discussed by Baker in "An easy to perform but often counterintuitive demonstration of gas expansion.") Correspondingly, the gas outside the blue region tends to cool down temporarily as it does that work. Over time, however, heat transfer causes the temperature throughout the container to equilibrate at the original temperature. Make sense?

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  • $\begingroup$ It also depends on whether the channel between the two chambers is large enough to allow the heat transfer to occur within a reasonable amount of time for the chambers to equilibrate. $\endgroup$
    – Chet Miller
    Commented Mar 19, 2022 at 12:56
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Then we can say that the gas on the left will push the marked gas to the right side

No. Part of the definition of an ideal gas is that there are no interparticle interactions. Neither portion is "in the way" of the other and they cannot push on each other. Instead the gas that ends up (initially) in the second chamber is just the particles that happened to hit the door.

The particles are just bouncing of the walls (which we assume are solid and do not move), so there is no work done on the gas.

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  • $\begingroup$ Climate and atmospheric scientists will be quite surprised to hear that regions of air cannot do work on each other. $\endgroup$
    – Chemomechanics
    Commented Mar 18, 2022 at 20:12
  • $\begingroup$ @Chemomechanics, I hope climate scientists are not assuming the atmosphere is an ideal gas. But the question is asking about one. $\endgroup$
    – BowlOfRed
    Commented Mar 18, 2022 at 20:13
  • $\begingroup$ The compressibility factor of air at room temperature is 0.999, so they're welcome to make that assumption in my book when it's similarly warranted. $\endgroup$
    – Chemomechanics
    Commented Mar 18, 2022 at 20:21
  • $\begingroup$ An ideal gas is a specific model. Yes, air approaches that for many situations. But for the experiment above, the lack of interparticle interactions is significant. For example you can't get en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect on an ideal gas, only a real one. $\endgroup$
    – BowlOfRed
    Commented Mar 18, 2022 at 20:26
  • $\begingroup$ So you would maintain that a compressed ideal gas does not tend to cool down as it expands into an ideal gas atmosphere? $\endgroup$
    – Chemomechanics
    Commented Mar 18, 2022 at 20:35

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