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# Re: [Phys-l] T dS versus dQ

I appreciate and value John Denker's endorsement of my post of 1/17 at 17:10 and, in return, heartily endorse the two very illuminating examples he provided in his post of 1/17 at 18:10. While I'm appreciating folks, I'd also like to acknowledge Brian's kind words.

It's interesting to me that the analysis of the rapid versus slow compression of a thermally insulated gas can cause so much confusion and I think the reason has to do with the fact that it can't be easily done by simply talking about work and energy (and it can't be done *quantitatively,* period, without knowing about the process and the boundary conditions in excruciating detail.) By considering things like the potential energy loss of a heavy piston as it falls or the kinetic energy added to and then delivered by the piston, it can be tempting to suppose incorrectly that the rate of the process is irrelevant to the final state.

Nevertheless, the analysis is almost trivial if one makes us of the idea of entropy and the physics of dissipation. The reasoning proceeds as follows:

1. Any rapid motion of the piston will necessarily result in a departure from thermal equilibrium for the gas.

2. The return to thermal equilibrium is accomplished via dissipation.

3. Dissipation necessarily produces new entropy within the gas.

4. There's no way to reduce the entropy in the gas because of the thermal insulation.

5. As a result the entire process leaves the gas with more entropy than it would have had if the process had been quasi-static.

6. The only way for the gas to have more entropy at a given volume is for it to have more energy.

7. The only way for the gas to have more energy is to have done more work on it.

The conclusion follows in complete generality for ANY process--jerky, oscillatory, smooth, or otherwise--that ever once leaves a thermally insulated system out of equilibrium on the way between thermal equilibria at specified initial and final mechanical boundary conditions.

John Mallinckrodt
Cal Poly Pomona