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Re: [Phys-l] T dS versus dQ

Yes it does the work mgh, but it produces significant KE of the piston. If mg is >> P A, then almost all of mgh goes into the piston KE. If you want to compress the gas, then (F + mg)/ A must be > = P at any instant. (Here F is an applied force to the piston and, again, we are assuming no energy passing through the cylinder wall.) The force of the piston acting on the gas cannot be greater than the force of the gas back on the piston.

It's like a car hitting a brick wall and driving through it. The wall didn't break because the force of the car on the wall was greater than the wall on the car. Their mutual equal and opposite force at some point exceeds the breaking strength of the wall.

Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bob Sciamanda [treborsci@verizon.net]
Sent: Friday, January 15, 2010 2:03 PM
To: Forum for Physics Educators
Subject: 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)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/



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