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Re: [Phys-l] Temp & Energy density



It seems that this particular question runs into a few problems (there may be many others which I don't notice):

i) "only a few gas particles" -- how is one to define 'energy density' or 'temperature' when there are so few particles that thermodynamic averages don't really apply? I think it was to give us an easier time during the compression phase, so that we could more easily play the role of
ii) Maxwell's Demon. If anything, the problem should be resolvable by reviewing the resolutions to problems where this guy appears;
iii) "free expansion" (also known as Joule expansion). This problem is a time-reversed "free expansion" in reverse. Callen gives (as problem 4.2-3) the following:

A monatomic ideal gas is permitted to expand by a free expansion from V to V+dV. The entropy change [in the problem] is dS = NR/V *dV [the problem can be done by showing that during a free expansion, dU = 0, and since we deal with a free expansion, PdV = 0, so that dU = cNRdT = 0 ==> dT = 0, so there's no temperature change. Also, dS = Cv*dT/T + P/T*dV ==> dS = P/T*dV, but for ideal gases, P/T = NR/V so that dS = NR/V*dV. ]

Callen goes on to say [forgive my strange formatting, as I've copied from an OCR'd text] that
"Whether this atypical (and infamous) 'continuous free expansion', process should be considered as quasi-static is a delicate point. On the positive side is the observation that the terminal states of the infinitesimal expansions can be spaced as closely as one wishes along the locus. On the negative side is the realization that the system necessarily passes through nonequilibrium states during each
expansion; the irreversibility of the microexpansions is essential and irreducible.
The fact that dS > 0 whereas dQ = 0 is inconsistent with the presumptive
applicability of the relation dQ = TdS to all quasi-static processes. We *define*
(by somewhat circular logic!) the continuous free expansion process as being
'essentially irreversible' and *non-quasi-static*. "

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________________________________
From: Paul Lulai <plulai@stanthony.k12.mn.us>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Monday, August 3, 2009 8:17:28 AM
Subject: [Phys-l] Temp & Energy density

Hello.
I am curious about the conditions for which energy density = temp of a
gas.
For this question I have a small cylinder if gas with only a few gas
particles.
Using work, if I squeeze a piston down when the few gas particles are
at the bottom of the cylinder, then I have not given them any extra
energy through a work process. Their temperature should remain constant.
Using energy density, this sneaky work can't be ignored. Energy
density would show a temp difference regardless of how the piston was
compressed.
This leads me to think the energy density solution includes some
assumptions that I do not know.
I appreciate any help available.
Thanks.



Sent from my iPod so I can blame Apple for my typos.

Paul Lulai
St. Anthony Village Senior High
Http://prettygoodphysics.wikispaces.com
US First RoboHuskie Team 2574

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