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Re: [Phys-l] T dS versus dQ

On 01/12/2010 06:33 PM, Bob Sciamanda wrote:
Equilibrium thermodynamics handles non-equilibrium (irreversible) processes,
if they connect well defined initial and final equilibrium states, by noting
1) The initial and final states are equilibrium states in which state
variables are well defined. and collectively define the system state.
2) One can then construct an alternate, reversible process consisting of a
series of equilibrium states which evolve from the initial to the final
state according to the "laws" of reversible, equilibrium thermodynamic
processes. Every state in this reversible process is an equilibrium state
and has well defined state variables.
3) One then uses the physics of equilibrium, reversible processes to
calculate, along the chosen, reversible path of equilibrium states, whatever
system state changes are of interest in the transition from the initial to
the final states.
4) Since "a state is a state is a state", these system changes are identical
to the system changes undergone by the originally considered irreversible

That's an innnnteresting approach, boldly stated.

Alas, if you actually follow that approach, you will
discover that in many cases it doesn't work.

A) Rumford's exciting Boring experiment (1798) is
irreversible. That's the main point of the exercise.
It cannot be approximated by any number of reversible

To say the same thing another way: In any reversible
step, entropy is a conserved quantity. In the real
world, entropy is not a conserved quantity.

B) When calculating the _stability_ of an equilibrium
situation, you need to consider the effect of perturbations
away from equilibrium. Again, that's the main point of
the exercise. If you assume that the perturbed state
is in equilibrium -- in particular, if you assume that
E can be written as a function of the same state variables
as the equilibrium state -- then the stability calculation
will get the wrong answer.

C) Et cetera. Need I go on?

Even something as central and basic as the Carnot heat engine
requires two heat baths, at temperature T1 and temperature T2.
Each heat bath is in equilibrium with itself ... but not with
the other.

My specific comment:
During the Carnot cycle the system under consideration is always in a well
defined equilibrium state. This in no way requires the two heat baths to be
in equilibrium with each other. What are you thinking of?

Are we perhaps quibbling over the definition of "system"?
I choose to draw my "system" boundary such that it
includes both heat sinks.

There is a well-known and easy-to-prove theorem that says
if *any* system is in equilibrium, then all parts of the
system must be at the same temperature. Clearly my system
is not in equilibrium. This point is so obvious that it
is often overlooked. I've always been known for my keen
grasp of the obvious.