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Re: [Phys-l] T dS versus dQ



So that's the question: Is there any scenario in
which it is not OK to replace dQ by either dE or
T dS? If so, what is the scenario, and why is
dQ irreplaceable? Operationally, how do I measure
dQ in this scenario, and/or how do I calculate it?

I have an ideal gas in a thermally insulated cylinder and piston. I suddenly compress the piston. To be specific, suppose the piston's speed profile starts from zero, rapidly (practically stepwise) rises up to the speed of sound, then drops rapidly back to zero once the gas is compressed by dV (which is negative). In practice I might accomplish this by having a huge weight sitting on the piston which is at the top end of the cylinder and held in place by a pin. I pull out the pin and let the piston fall a distance dx until it slams into a stop.

I think we have T dS > 0 (because the process is certainly irreversible), dQ = 0, dE > 0 (because work -P dV was done on the gas) and so it doesn't look like dQ can be equal to either T dS or dE. I "computed" dQ by noting that the cylinder (including the piston) is thermally insulated (and I'm further helped that the process is so fast there isn't time for heat transfer even if it weren't insulated, noting that no thermal insulation is perfect in the real world).

Okay, fire away. -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/