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Re: dS=0

Jim,
I await your equations, but it seems that they are going to simply show
that a gas process for which dE = - PdV is adiabatic. This conclusion
follows directly from the first law: dE = dQ - PdV. (Similar results
apply to the generalized version dE = dQ + YdX.)

This is not universally true of *any* PdV process; only those for which dE
= - PdV (ie: dQ=0). For a finite ideal gas process the succession of
PdV's must follow a definite curve: PV^gamma = Constant, in order to be an
adiabatic process.

Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor

----- Original Message -----
From: Jim Green <JMGreen@SISNA.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, January 06, 2000 9:08 PM
Subject: Re: dS=0


Oh my!!! Bob is quite right. The sketch of the derivation is there but
the equations are not. . . .