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In a slow adiabatic expansion, gas expands without allowing heat to leave the system. However, as the process is reversibly slow, no net entropy change comes out of the process.

However, I have fruitlessly been able to understand, fundamentally, why heat will try to leave the system in an expansion, and why the temperature of the gasses decrease in an expansion. In my mind, I can't distinguish an adiabatic expansion from an adiabatic free expansion -- the gas particles have more room to roam -- why does that mean their energy decreases?

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The key lies in the collisions between the gas particles and the walls of the container. During the collisions, the gas particles exert force on the walls but the walls are stationary - so no work is done, and the particles rebound with the same energy.

When the walls are moving (as is the case during expansion), the gas particles do work on the walls, so they rebound with a different energy. If the walls are moving outward, the work is positive, so the new energy is smaller; if the walls are moving inward, the work is negative, so the new energy is larger.

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  • $\begingroup$ This highlights a key confusion of mine. Why do the gas particles do work on the walls if they're moving? I would assume they would only do work on the walls if they're moving it themselves. If we pull a piston back, increasing the volume of a container, then we are causing the wall to move, not the gas particles. In my mind, they're just doing their thing as usual, but take more time to rebound off the moving wall now. Do you see where I'm confused? $\endgroup$ – sangstar Dec 2 '17 at 22:55
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    $\begingroup$ Analyzing the situation in term of work done on the walls is perfectly fine, but it requires paying attention to the finite time the interaction takes, and the details of inter-atomic interaction are something that many introductory treatments shy away from. In class I just have students toss bouncy rubber balls at a board help by a companion: they compare the case with the board static to those with it approaching and receding. $\endgroup$ – dmckee --- ex-moderator kitten Dec 2 '17 at 23:01
  • $\begingroup$ This post on Gravity Assist Analogy talks about balls bouncing off a moving paddle. $\endgroup$ – mmesser314 Dec 3 '17 at 0:21
  • $\begingroup$ @dmckee What did they find when tossing the rubber balls at a board when it was stationary or moving? Am I missing some key, non-obvious insight here that the gas particles will basically be getting a different rebound energy when the container is moving? Could it be because of the fact that, in the case of the container expanding, since the wall is moving it makes the effective speed of the gas particle less, so its rebound is with less speed than if it rebounded on a stationary wall? $\endgroup$ – sangstar Dec 3 '17 at 14:40
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    $\begingroup$ @sangstar In the rest frame of the wall, the rebound speed is the same as the initial speed. However, in a frame in which the wall is moving, this is no longer true. $\endgroup$ – J. Murray Dec 3 '17 at 16:35
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It is question of density really. When a fixed mass of gas expands in a container the density decreases. If you take look at the expression pv=mnc^2/3 you will get some idea of what is happening. Divide both sides of the equation and keep c^2 constant you will see that p=dc^2/3(d=density,c^2=average velocity squared). This means the number of molecules impacting on a container(and consequently the pressure) decreases as the volume increases. This is indicated in the expression p=dc^2/3. Even though each single molecules is rebounding at a connstant momentum the numbers rebounding simultaneously is decreasing if the volume is increasing. The total momentum in any direction is the sum of all the momenta of each molecule.

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  • $\begingroup$ You just given an explanation for why the pressure is lower, but not one for why the RMS speed of the molecules is smaller. And indeed, if you perform a isothermal expansion you can get the case of less pressure but the same speed distribution. And the difference in pressure between a isothermal expansion and a adiabatic expansion is exactly the reason the latter has even lower pressure than the former. $\endgroup$ – dmckee --- ex-moderator kitten Dec 29 '17 at 3:19

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