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February 26

Ideal gasses

Can Maxwell's demon "cheat" the ideal gas law by pulling a piston a little each time it can do so without allowing atom(s) to bounce off the piston while the piston's moving? Presumably this isn't a real cheat either, any more than the one where the demon sorts molecules with a gate. If a microscopic piston could accelerate fast enough at a lucky time it seems like it could outrun all gas in the cylinder (is the easiest gas xenon?), expand it many times and return to zero speed before any gas particle caught up but I don't know if significant frictional heating or breakage from g-forces would happen in the real world. Sagittarian Milky Way (talk) 00:38, 26 February 2022 (UTC)[reply]

Note that for pressure P:

P={\frac {F\cdot {\text{distance}}}{A\cdot {\text{distance}}}}={\frac {\text{Work}}{\text{Volume}}}={\frac {\text{Energy (J)}}{{\text{Volume }}({\text{m}}^{3})}}.

For an ideal gas, the energy present in the gas is therefore given by PV which, when conserved, is a constant, Simply increasing the volume reduces its pressure. So the short answer is no, the ideal gas law isn't "cheated". --Modocc (talk) 02:43, 26 February 2022 (UTC)[reply]

I never understood how the gas particles were supposed to know to slow down (average speed), instead of just the joules per cubic meter decreasing. Since temperature is joules per particle. But when Comet Holmes had a sudden mass dump the gas sphere unsurprisingly expanded the same number of miles per day, no slowdown. Sagittarian Milky Way (talk) 06:46, 26 February 2022 (UTC)[reply]

Maxwell's demon doesn't cheat the ideal gas law, but it does cheat the second law of thermodynamics. As your piston moves out while no gas particles collide with it, the gas does no work on the piston and the gas doesn't cool down. The pressure drops inversely proportional to the volume. Why it doesn't work in the real world is always tricky with Maxwell's demon. For your second question, in a normal cilinder with a piston moving out, the gas particles know to slow down by the Doppler effect when bouncing against a moving piston: when bouncing off a moving target, the energy of the particle changes in the reference frame in which the target moves. PiusImpavidus (talk) 09:48, 26 February 2022 (UTC)[reply]

Yes I knew Maxwell's demon "cheated" a different thermodynamics law. So it is 100% the effect of the bounce surface moving away as I suspected but it seemed like a paradox cause presenting it as a law made me think was an uncheatable law of volume. What if you used a very skinny carbon nanotube as a cylinder to trap millions of xenons and you kept pulling the plunger till one time you got lucky and it expanded without any cooling? Could that work? Not sure if anyone tried, there might be nothing to learn. Sagittarian Milky Way (talk) 17:51, 26 February 2022 (UTC).[reply]

If there is no contact with the piston no work is being done by the piston, either on it or by it. Thus the xenon temperature and energy nRT remain constant and the density and pressure of the xenon gas become reduced as it expands into the larger volume. -Modocc (talk) 19:32, 26 February 2022 (UTC)[reply]

With such a piston the gas will expand or contract while preserving its temperature. So, the entropy change will be

\Delta S=nR\ln {\frac {V}{V_{0}}}

, which can either positive or negative. However I suppose that the total entropy of the system including the piston and the mechanism that moves it will have its entropy always increased. Ruslik_Zero 19:52, 26 February 2022 (UTC)[reply]

@Sagittarian Milky Way: Suppose you succeed to move the piston away so that the chamber length gets doubled whilst avoiding collisions during piston's moves. Then each gas particle needs to travel a doubled distance between collisions with the piston. As particles' velocities did not change, the frequency of collisions with the piston got halved. As a result the pressure dropped to a half of its initial value:

PV={\frac {P_{initial}}{2}}\cdot (2V_{initial})=P_{initial}V_{initial}=\mathrm {const.}

The law doesn't seem "cheated". --CiaPan (talk) 20:45, 26 February 2022 (UTC)[reply]

Usually both temperature and pressure drops but you can trick it so only pressure drops, that seems like cheating. You're right the usual effect on temperature of compression alone or expansion alone must have a different name. Sagittarian Milky Way (talk) 22:47, 26 February 2022 (UTC)[reply]

See https://en.wikipedia.org/wiki/Adiabatic_process#Adiabatic_heating_and_cooling It gives the different regimes: "The adiabatic compression of a gas causes a rise in temperature of the gas. Adiabatic expansion against pressure, or a spring, causes a drop in temperature. In contrast, free expansion is an isothermal process for an ideal gas." -Modocc (talk) 23:43, 26 February 2022 (UTC)[reply]

This is why I should've stayed in school, I never had high school physics, biology or the late part of chemistry year. Sagittarian Milky Way (talk) 00:24, 27 February 2022 (UTC)[reply]

See Joule expansion catslash (talk) 17:16, 27 February 2022 (UTC)[reply]

Interesting article, thank you. Sagittarian Milky Way (talk) 21:21, 27 February 2022 (UTC)[reply]