Re: [Phys-l] T dS versus dQ

[curtis osterhoudt <flutzpah@yahoo.com>]

Note that my use of "non-adiabatic" in the previous email is quite useless, and probably even so useless as to be dangerous. I apologize.

 /************************************
Down with categorical imperative!
flutzpah@yahoo.com
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________________________________
From: curtis osterhoudt <flutzpah@yahoo.com>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Tue, January 12, 2010 11:51:07 AM
Subject: Re: [Phys-l] T dS versus dQ

  Is P well-defined enough in such a non-adiabatic process to claim that the work done is P*dV?

/************************************
Down with categorical imperative!
flutzpah@yahoo.com
************************************/




________________________________
From: Carl Mungan <mungan@usna.edu>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Tue, January 12, 2010 11:37:50 AM
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 into 
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
mailto:mungan@usna.edu    http://usna.edu/Users/physics/mungan/
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