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# Re: [Phys-l] PV question

• From: Carl Mungan <mungan@usna.edu>
• Date: Tue, 26 Jan 2010 13:51:06 -0500

Certainly the "usual" system would not behave in this manner, being in a state of mechanical but not thermodynamic equilibrium. The specific system I had in mind is the usual system with a gimmick. Place a second heat-incapacious thermally-insulating piston in the middle, and put different gases, say one monatomic and the other diatomic, on the two sides of that piston. One can now carry out reversible processes on this system that leave the two sides at different temperatures. Consider an adiabatic compression of the system from an initial state of thermodynamic equilibrium, for example. Which gas gets warmer?

Leigh

The monatomic gas. Defining t to be the ratio of the final and initial temperature of a gas, then t_monatomic = t_diatomic^1.4. Okay I accept that example of yours as a valid demonstration of a system with a single well-defined V and P but not T (but only because T has two piecewise uniform values).

But I don't see how this demonstrates your previous statement: "If the process is reversible, the system must be near thermodynamic equilibrium...."

Let's accept your two-piston system as not remaining in thermodynamic equilibrium for the sake of continuing the discussion. (At least not the whole system, although piecemeal the parts of the system certainly are. So John D might quibble here. Anyhow, let's just go on.) But it still certainly looks to me like your compression IS reversible. We did a slow, dissipation-free adiabatic compression. I can undo it with a slow, dissipation-free adiabatic expansion in such a fashion that I also undo all changes in the environment. That has to fit any reasonable definition of reversible, right? If so, your statement doesn't stand up.... -Carl
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