Chronology | Current Month | Current Thread | Current Date |
[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |
4. Somehow the term "reversible" showed up in the discussion. It does not appear in the question, so I couldn't figure out what relevance it had. 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. If the process is reversible, the system must be near thermodynamic equilibrium at all points in the process, but that is not a necessary condition in this 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?
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