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Re: [Phys-l] T dS versus dQ

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. The higher temperature
(compared to that resulting from the corresponding reversible
compression) is evidence of the production of entropy. It would seem
that, without giving students the impression that entropy is conserved,
one could compute how much entropy was generated by the fast compression
by calculating how much entropy would have to be reversably transferred
(from an infinite collection of infinite heat reservoirs) into the gas
after it had been compressed reversably to the same final volume, in
order to bring the gas up to the final temperature resulting from the
rapid compression.

-----Original Message-----
From: [mailto:phys-l-] On Behalf Of LaMontagne, Bob
Sent: Tuesday, January 12, 2010 2:12 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

Since you specified a thermally insulated cylinder and piston, it
doesn't really seem to matter if you do the compression quickly or
slowly - the final temperature will be the same. Since a reversible
adiabat will get you from the same initial state and to the same final
state, it appears dS = 0 as well as dq = 0. Is the supersonic nature
the proposed compression an implication of a change in the values of
and Cv so the slow reversible compression is not equivalent?

Bob at PC

-----Original Message-----
From: [mailto:phys-l-] On Behalf Of Carl Mungan
Sent: Tuesday, January 12, 2010 1:38 PM
To: Forum for Physics Educators
Subject: 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
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
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