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Re: TdS is not dQ or d(anything)



Consider an isolated gaseous system divided into two subsystems by a
rigid, adiabatic wall. Suppose the two subsystems are in "private"
equilibrium at different temperatures. Each subsystem has a definite
entropy S1 and S2. The entropy of the composite system is then S = S1 +
S2. Now, if the adiabatic wall is "dissolved" an equilibrium ensues and
the new system entropy is S > S1 + S2. Note that since for this gas there
exists a definite function S(E,V,N) for equilibrium states, the final
equilibrium S is determined and calculable even before the mixing occurs.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "Pentcho Valev" <pvalev@BAS.BG>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, May 13, 2003 10:10 AM
Subject: Re: TdS is not dQ or d(anything)


| Bob Sciamanda wrote:
|
| > | . . .This is the origin of one of the lines
| > | of confusion in thermodynamics. The definition dS = dQrev/T gives
the
| > | impression that, at least for an ideal gas, the entropy is a state
| > function
| > | only if the system undergoes a reversible process. On the other
hand,
| > the
| > | nature of the ideal gas as reflected by eq. /1/ guarantees that the
| > entropy,
| > | if defined by dS = dQ/T, will also be a state function provided the
| > system is
| > | in equilibrium all along. . . . From: "Pentcho Valev"
<pvalev@BAS.BG>
| >
| > Pentcho,
| > I think this passage points out the heart of your confusion. Even
though
| > S (along with other state variables) is a state function, this does
not
| > preclude the system undergoing processes in which the state is not
even
| > defined, because the system is in dis-equilibrium. "S is a state
| > function" means that whenever the system is in a defined, equilibrium
| > state, the value of S is determined by that state.
|
| This was the original Gibbs idea (he never applied the concept of
entropy to
| non-equilibrium states) but then someone applied the concept of entropy
to
| reactions distant from equilibrium. When textbooks authors write DeltaG
< 0
| they in fact say: "The reaction is distant from equilibrium, approaches
| equilibrium, the entropy of the universe increases, the free energy of
the
| system decreases." Is the entropy defined or is it not defined for this
| particular non-equilibrium state? If it is not defined, are textbooks
authors
| right to write DeltaG < 0? Note that here we don't have the classical
| situation in which the process starts and ends in equilibrium. Rather,
for a
| chemical reaction, the initial state is the most distant from
equilibrium.
|
| Pentcho