Re: TdS is not dQ or d(anything)

[Pentcho Valev <pvalev@BAS.BG>]

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