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Re: [Phys-l] T dS versus dQ

I don't consider the method I outlined for determining how much entropy
was generated to model what happens. It is just a calculational
technique that takes advantage of the fact that state variables depend
only on the state of the system, not on how it came to be in that state.
It is not a model.

-----Original Message-----
From: [mailto:phys-l-] On Behalf Of John Denker
Sent: Wednesday, January 13, 2010 12:09 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

On 01/13/2010 08:22 AM, Jeffrey Schnick wrote:
... It would seem
that, without giving students the impression that entropy is
one could compute how much entropy was generated by the fast
by calculating how much entropy would have to be reversably
(from an infinite collection of infinite heat reservoirs) into the
after it had been compressed reversably to the same final volume, in
order to bring the gas up to the final temperature resulting from
rapid compression.

Yes, you can do that calculation if you want to, as
some kind of consistency check, but you have to do it
carefully, and you shouldn't leave it out there as
the only way (or even the primary way) of thinking
about the problem. The straightforward approach --
treating the irreversible process as an irreversible
process -- is easier and safer.

Among the many problems with the "fake entropy reservoir"
approach is that it further muddies the already-muddy
notions of heat and work. We have a situation where
the real physics involves no flow of entropy across
the boundary of the system, but the model requires
entropy to flow in from the fake entropy reservoir.
Does this fake flow count as heat? The question is
unanswerable. The "yes" answer makes the model grossly
unfaithful to the real physics, while the "no" answer
makes the model grossly inconsistent with any known
definitions of heat and work and/or inconsistent with
the basic conservation laws.

On 01/13/2010 09:53 AM, John Mallinckrodt wrote:

... 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.

Quite so. Thanks for clarifying this point.

I have been tacitly assuming all along that everybody intended
P dV to be synonymous with F dx in this situation. I hope this
assumption didn't cause too much confusion.

Forum for Physics Educators