Re: [Phys-l] PV question

[Carl Mungan <mungan@usna.edu>]

> 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
--
Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/