Re: [Phys-l] PV question

[Leigh Palmer <palmer@sfu.ca>]

Carl asks:

> Leigh wrote:
>
> > It is 
> > implicit in the question that in this process the state of >the system is always 
> > sufficiently near *mechanical* equilibrium that its pressure >is uniform 
> > throughout. Note that this does not imply the stronger >constraint that it be 
> > near *thermodynamic* throughout the process, though that is >the usual case.
>
> Okay I'll bite: Can you construct a specific example (preferably involving some actual setup) where the gas can follow the process described and *not* be in (or better: nearly in, which I agree with John D is my meaning of the term "quasistatic") thermodynamic equilibrium? Certainly for say an ideal monatomic gas in the usual cylinder arrangement, no such example is possible because N,V,P all have well-defined (near-equilibrium) values throughout the process and hence so does evey state variable. So I conclude you must be thinking of a different kind of example. Please provide the details. -Carl

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