Instead of dS and TdS let's talk about things easier to "visualize", ie. dV
and PdV of a gas.
Both P and V are state variables; their numerical values are part of the
specification of any equilibrium state of the gas. Both dP and dV are exact
differentials. The integrals of dP and dV over any path (including
constraint specifications) connecting two equilibrium states are simply the
changes in those state functions: (P2-P1) and (V2-V1), regardless of
whatever process/path was used to get from state 1 to state 2. Once you are
in a given state, both P and V are specified, no matter what the previous
history.
On the other hand, the infinitesimal quantity PdV is not an exact
differential. Its integral between two states is not assured to be some
state-specified quantity like eg. (P2*V2 - P1*V1). The integral of PdV is
in fact the area under the path taken in (P, V) space, a patently path
dependent quantity.
In the same way the integral of dS is just S2-S1, and the integral of dT is
just T2-T1. But the integral of TdS is not determinable from only end state
quantities, it is path/process dependent. It is the area under the path
taken in (T,S) space.
As an aside, note that the quantity (P2*V2-P1*V1) is the integral of the
exact differential d(PV) which is equivalent to (PdV + VdP). This last
equivalence has an rather obvious graphical interpretation when you evaluate
it as the sum of two areas in P,V space.
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: Wednesday, August 04, 2004 2:35 PM
Subject: Re: Questionable interpretation of physics
| At 09:48 04 08 2004 , the following was received:
| >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, can you amplify this a bit? Maybe give an example?
|
| Jim
|
|
| Jim Green
| mailto:JMGreen@sisna.com
| http://users.sisna.com/jmgreen