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I see the second law of thermodynamics as an observed law, not an imposed law: it is broken sometimes on a larger scale and broken often on a molecular scale.

So, if a box contains only one gas molecule, will that molecule experience the same range of speeds as the many molecules of a larger sample?

I.E. Is the molecule moderated by its impacts with the solid walls of the container? does its Maxwell-Boltzmann curve gets narrower & higher?

If it is moderated, then consider the following.

Zeolites are porous minerals, used, for example, to filter isobutane from butane. (https://en.m.wikipedia.org/wiki/Zeolite)

Consider a thin horizontal slab of zeolite bisecting a container.

The bottom half of the horizontal slab is soaked with a solvent.

Above & below the slab is a gas that's weakly soluble in that solvent.

If, within the upper half of the slab, the solid walls of the tiny zeolite chambers moderate the gas then the free gas would have fewer slow (& fewer fast) molecules than normal.

Slow moving free gas molecules must be more soluble than fast moving free gas molecules.

So the free gas, within the zeolite, facing the upper surface of the solvent, must be less soluble than the free gas, outside the zeolite, facing the lower surface of the solvent.

Hence, more free gas leaves the upper surface of the solvent than dissolves back into it. Hence the solvent, there, becomes depleted.

Whereas the free gas facing the lower face of the solvent would be un-moderated and so would dissolve into the body of depleted liquid.

Hence there would be, overall, an upward movement of gas through the slab, resulting in a higher pressure in the upper half of the container than the lower half.

If that pressure difference were used (turbine taking power away) then the whole container would drop in temperature slowing & halting the process.

But if the container were put say, in a stream, then the process would stabilise & continue.

I’ve had this idea for decades (it’s like Maxwell’s demon but more convincing). I highly doubt it will work, but I don’t know why. Can anyone tell me.

Many thanks, David Porter.

Obviously physics is not my field. I am a philosopher.

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  • $\begingroup$ Considerations like these are standard for chemical processes. Nothing weird will happen; the system will find an equilibrium so that the rate of the gas coming out of the solvent is the same rate as the gas absorbed by the solvent, on both sides of the solvent, which can be different if the situation requires it. It will not be able to act as a Maxwell's demon. The zeolite wall has a temperature, and the gas molecules that bounce off that will equilibriate with that, ending up with the same kinetic energy distribution before and after collisions. $\endgroup$
    – naturallyInconsistent
    Commented Nov 25, 2023 at 7:30
  • $\begingroup$ Hmm. There's no chemical process. Nevertheless, it won't work. I wonder if the gas is moderated though, and if there's a potential idea there? $\endgroup$
    – David Porter
    Commented Nov 25, 2023 at 7:35
  • $\begingroup$ Is the Maxwell Boltzmann curve for the zeolite, though very much lower than for the gas, also a narrower curve; narrower because it's a solid, bound: less kinetic energy, more electrical(?) energy? $\endgroup$
    – David Porter
    Commented Nov 25, 2023 at 7:50
  • $\begingroup$ I suppose the answer may be "Yes, but it would make no difference". Ah well. All hopes on nuclear fusion. $\endgroup$
    – David Porter
    Commented Nov 25, 2023 at 7:58
  • $\begingroup$ I am trying to tell you that these physical considerations are necessary to get chemical reactions correct, even if there is no chemical process in your particular situation. No, the zeolite would have to be modelled by a different distribution, related to the Maxwell-Boltzmann. It does not matter, because thermal equilibrium requires that the different distributions be mathematically compatible with each other. $\endgroup$
    – naturallyInconsistent
    Commented Nov 25, 2023 at 7:59

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