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Re: [Phys-l] T dS versus dQ

Bob LaMontagne wrote:
My claim was that the final temperatures would be the same if the process proposed would be done quickly or slowly.

Bob, let me try to make more explicit what others have already said:

If you simply drop the weighted piston, it does the work mgh, all of which goes to increase the gas temperature.
But if you want to very slowly lower the weight, you will have to hold back with a varying upward force against the weight of the "falling" piston throughout the process.
The downward force of the piston upon the gas is now a varying force, increasing slowly to the final value mg, during the compression process.
The area under this variable F(x) curve is less than mgh, so that less work is done on the gas to go towards a temperature increase.

Hope this helps.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)