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Re: [Phys-l] T dS versus dQ

On 01/17/2010 06:10 PM, John Mallinckrodt wrote a really nice
analysis of the situation ... more clear and much more complete
than anything I would have come up with. Thanks.

All four of the cases he considered need to be taken to
heart. However, I suspect that case 3 is the make-or-break
case, the one most likely to surprise and/or confuse people.
BC asked how we might demonstrate this operationally. If
I may be allowed to throw in my $.02, I will suggest a two-
part experiment that demonstrates that when JM says "we
don't know ..." it means we really, really don't know.

CASE 3: The energy input is specified and the gas is ideal:

1. We know the work done. (Because the energy input was entirely the
result of work.)

2. We know the final temperature. (Because we know the final energy.)

3. We don't know the final entropy. (Because we don't know the final

4. We don't know if stirring occurred. (Because we don't know if the
entropy increased.)

So, here's the setup: We start with a sample of pure
ideal gradient-free nonrelativistic nondegenerate monatomic
or at least polytropic gas in a thermally-insulated
cylinder in a known equilibrium state. There is a piston
with weights arranged so that there is initially mechanical
as well as thermal equilibrium. The plan is to add a
specified amount of energy.

In subcase 3a, we wiggle the piston at high frequency by
means of a sound transducer. This adds a little bit of
energy to the system. Sooner or later, presumably rather
soon, all of the sound is dissipated. The sound energy
causes the gas to expand. The gas expands at constant
pressure as determined by the weights. In so doing, it
does some work against the weights. This macroscopic
mechanical energy-output is strictly less than the sonic
energy-input. We keep doing this, little by little, until
the total delta E (sound in minus macroscopic mechanical
work out) is equal to the specified delta E required by
the statement of the problem.

In subcase 3b, we drive the piston down using a jack
screw. The weights have now become irrelevant. The
gas is compressed along a contour of constant entropy,
because we are doing everything gently and reversibly.
The volume goes down, the pressure goes up, and the
energy goes up. We keep doing this until we have added
the required amount of energy.

In the two subcases, the final energies are the same
and the final temperatures are the same. But the
entropies, volumes, and pressures are wildly different.

We remark that the product P(final) times V(final) is
the same in both subcases, because (for any polytropic
gas) the product is related to the energy. Not equal
to, but related to.

To summarize the final states:

3a relative to 3b

energy =
temperature =
entropy >
volume >
pressure <
PV product =

So, this is an operational way of reaching the result
that we previously reached in numerous other ways: it
is perfectly possible to have two states with the same
energy but different entropy.


To possibly make Bob S. happy, I remark that there are
other experiments, including completely reversible
experiments, that will also take us from the given
initial state to either of these two final states ...
or take us directly from 3a to/from 3b. But those
require different apparatus, and IMHO are not as useful
in helping us understand the physics of dissipation.
Anybody who wants to analyze this angle is welcome to
do so. The starting point is to observe that in any
equilibrium state, E,T,S,P, and V are functions of