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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?

sangstar
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2 Answers2

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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.

J. Murray
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  • 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? – sangstar Dec 02 '17 at 22:55
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    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. – dmckee --- ex-moderator kitten Dec 02 '17 at 23:01
  • This post on Gravity Assist Analogy talks about balls bouncing off a moving paddle. – mmesser314 Dec 03 '17 at 00:21
  • @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? – sangstar Dec 03 '17 at 14:40
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    @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. – J. Murray Dec 03 '17 at 16:35
  • No way. So the change in momentum is due to the relatively reduced speed of the gas particles? Since they rebound with less speed they lost energy, and thus do work? – sangstar Dec 03 '17 at 16:54
  • That sounds like the tail wagging the dog - The gas particles exert force on the walls over a nonzero distance (because the walls are moving), so they do work. This corresponds to a loss of energy. The fact that the particles' impact with the wall is not the "primary force" causing the wall to move is irrelevant. – J. Murray Dec 03 '17 at 17:01
  • @J.Murray Then why then, do gas particles gain energy upon a volume compression? – sangstar Dec 04 '17 at 14:35
  • And my mind has serious inhibitions accepting the fact that supplying a force that isn't sufficient to move something when it is moving is now doing work when its movement is entirely due to a different force. In my mind, the work done is totally due to the piston being pulled by me and the gas particles rebounding off the moving wall, are not doing any work to the piston as they are not moving the walls - they never have had the force required to. – sangstar Dec 04 '17 at 14:40
  • When the walls are moving inward, then the rebound speed is greater, so the energy increases. In this case, the particles do negative work on the wall. As for your inhibitions, I'm not sure what to tell you. $W = \int F \ dx$. This case is basically the same as that of a 1D elastic collision between a small mass $m$ and an enormous mass $M$- the latter playing the role of the wall. Calculate the work done and energy transfer in that case, then take the limit as $M\rightarrow \infty$. – J. Murray Dec 04 '17 at 15:11
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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.

user177923
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  • 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. – dmckee --- ex-moderator kitten Dec 29 '17 at 03:19