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

Sorry, I'm not seeing this minor triumph.

It's not the work you do on the piston that determines the final Temperature and Pressure in the gas (compressed to the same volume in each case), it's the work the piston does on the gas. The force you apply to the piston accelerates the piston as well as compressing the gas. At some point you have to drastically reduce the force you are applying (and for a really massive piston you may have to pull instead of push) in order to stop the motion of the piston. The kinetic energy of the piston started at zero and ends at zero - the net work you did ended up in the gas. Sometimes the force you apply at each position of the piston is more than when you do it slowly, sometimes it must be less to decelerate the piston. The integral of your F dx appears to be the same either way - fast or slow.

Also, by your argument below, if the pressure is ill-defined, why must it be on average higher at each position than the slow motion case? That appears to be strictly a guess to get the result you want. Why not - on average - lower? Or the same? As you said, if the piston is suddenly slowed down the pressure of the gas on the piston might momentarily be zero.

I have seen lots of statements so far that the final temperature "must" be higher, and the final entropy "must" be greater, but no actual clear argument as to why this is so. I would love to see an argument that would make it as clear to me as it appears to be to the rest of you - take it as a little teaching challenge.

Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of John Mallinckrodt [ajm@csupomona.edu]
Sent: Wednesday, January 13, 2010 11:53 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

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
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