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Re: [Phys-l] T dS versus dQ



On 01/12/2010 12:11 PM, LaMontagne, Bob wrote:
Since you specified a thermally insulated cylinder and piston, it
doesn't really seem to matter if you do the compression quickly or
slowly - the final temperature will be the same.

I disagree.

Since a reversible
adiabat will get you from the same initial state and to the same
final state, it appears dS = 0 as well as dq = 0. Is the supersonic
nature of the proposed compression an implication of a change in the
values of Cp and Cv so the slow reversible compression is not
equivalent?

I guess that's one way of saying it.

For that matter, the motion doesn't even need to be
supersonic; I reckon Carl mentioned that just by
way of emphasis. Any reasonably _sudden_ motion of
the piston will launch sound waves which will sooner
or later become thermalized, raising the temperature.

To say the same thing yet another way, the force
required for a sudden movement will be greater than
the force required for a gradual movement. This
means the F dot dx (aka P dV) will be larger than
you might have expected.

This falls in the same category as yesterday's
questions about stirring. Sudden motions of the
piston "stir" the fluid.