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# Re: [Phys-l] T dS versus dQ

Jeffrey Schnick wrote:

John M: I took Bob's words to mean that the final temperature resulting from a fast compression would be the same as the final temperature resulting from a slow compression (not the same as the initial temperature). I disagree with Bob on this point in that the greater the speed of the pistion the greater the local pressure at the face of the piston. The relative velocity between the piston and the gas molecules is greater when the piston is closing in on them and hence the momentum transfer to the piston is greater with each collision and you have a greater collision rate; hence the greater pressure. Thus the piston does more work on the gas when the piston is moving faster so the final internal energy of the thermaly insulated gas must be greater and the final temperature of the gas must be greater.

Right. I prefer, however, to think about the work done in terms of force (and displacement of the piston) rather than pressure (and volume change of the gas), because the pressure in the gas is ill- defined. Indeed, the pressure that you want to use in this calculation can only be determined finding the force applied to the gas by the piston and then dividing by the piston area. Note also that one can at least imagine constructing a compression procedure during which that "pressure" is sometimes close to zero even when the piston is moving toward the gas. For instance, give a big push on the piston, then pull it back very quickly, and then immediately begin "compressing the gas" again. The point is that the big inward push and retreat creates a bit of a void that the piston moves through during the initial phase of the subsequent compression. It is a minor triumph of thermodynamics that we know that the net work done on the gas will, in ANY event be larger than that done in a reversible (isentropic in this case) compression precisely BECAUSE of the irreversible nature of the process.

John Mallinckrodt
Cal Poly Pomona