The point is that dS is an exact differential, but TdS is not:
Given any pair of well defined (initial and final) system states, the
integral of dS over any path (and with any valid constraints) connecting
those states results in the same numerical value. Not so for the integral
of TdS - its integral will be path (and constraint) dependent.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.velocity.net/~trebor/
trebor@velocity.net
----- Original Message -----
From: "Jim Green" <JMGreen@SISNA.COM>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Tuesday, August 03, 2004 5:48 PM
Subject: Re: Questionable interpretation of physics
| >
| >You can't write TdS as dQ or d(anything else). TdS is an
| >inexact differential, and cannot be written as the derivative
| >of Q or anything else. For details, see
| > http://www.av8n.com/physics/thermo-forms.htm
| >and references therein.
|
| Wait a minute John, I agree that it should be dbarQ as there is no exact
| differential for Q but not TdS???? I guess that I need more tutoring
here.
|
| And I do wish that there were a lower case Ð in the system for us physics
| people.
|
| Jim
|
|
| Jim Green
| mailto:JMGreen@sisna.com
| http://users.sisna.com/jmgreen