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Re: [Phys-l] T dS versus dQ

On 01/12/2010 11:37 AM, Carl Mungan wrote:

I have an ideal gas in a thermally insulated cylinder and piston. I
suddenly compress the piston. To be specific, suppose the piston's
speed profile starts from zero, rapidly (practically stepwise) rises
up to the speed of sound, then drops rapidly back to zero once the
gas is compressed by dV (which is negative). In practice I might
accomplish this by having a huge weight sitting on the piston which
is at the top end of the cylinder and held in place by a pin. I pull
out the pin and let the piston fall a distance dx until it slams into
a stop.

I think we have T dS > 0 (because the process is certainly
irreversible), dQ = 0, dE > 0 (because work -P dV was done on the
gas) and so it doesn't look like dQ can be equal to either T dS or
dE. I "computed" dQ by noting that the cylinder (including the
piston) is thermally insulated (and I'm further helped that the
process is so fast there isn't time for heat transfer even if it
weren't insulated, noting that no thermal insulation is perfect in
the real world).

OK, that's exactly the sort of scenario I would like to
discuss. The description seems clear. Let's continue
the analysis.

Note that AFAICT we all understand the physics here.
We are only discussing details of terminology and
concepts used to _model_ the physics.

If I understand the scenario correctly, we know the
amount mechanical work done by the outside agent, since
we observe the F dot dx or equivalently the P dV.

We know that
dE = - P dV + 0 [1]
since the cylinder is thermally insulated and there are
no other relevant contributions to dE.

Therefore *if* you are sure that
dE = dQ - P dV [2]
then you can conclude that dQ=0. Alas I do not consider
it axiomatic or obvious that equation [2] applies to
this scenario.

The way I look at it, *if* we think E can be expressed
as a function of S and V alone, then we can write
dE = T dS - P dV [3]
but ... in this scenario E is not a function of S and
V alone. If you insist that (S,V) is your state vector,
then E is not a function of state during the time that
the piston is provoking sound waves, shocks, et cetera
in the fluid. The analysis crashes and burns at this
point. We cannot proceed in this direction.

If you add more variables (more dimensions) to the
state vector, then you need more terms on the RHS of
equation [3] and equation [2]. This makes it clear
that we cannot safely conclude that dQ=0 on the basis
of equation [2].


Here's another angle on the analysis: This points up,
yet again, the multiple inconsistent definitions of
"heat". Some people write
dE = T dS - P dV
where the second term is clearly "work" in agreement
with the definition inherited from non-thermal mechanics,
and the other term presumably must be "heat". Note
that both T dS and P dV are functions of state, depending
only on the interior properties of the system.

Meanwhile, the suggestion above that
dQ = heat = 0
is obviously inconsistent with the idea that
T dS = heat > 0.

The only way I can make even a little bit of sense out
of this is to switch definitions and interpret the
statement dQ = heat = 0 as referring to "heat flow"
across the boundary ... something very different
from the "heat production" that occurs far from the
boundary, deep in the interior in this example.

Not coincidentally, this points up the multiple inconsistent
meanings of "adiabatic".

a) Sometimes "adiabatic" means isentropic.
b) Sometimes "adiabatic" means thermally insulated.

These are not the same. A dissipative system such as
today's scenario is the canonical example of something
that is thermally insulated but not isentropic.

My recommendation: do not say "dQ=0" as a synonym
for adiabatic. For that matter, don't say "adiabatic"
either. If you mean thermally insulated, say "thermally
insulated". If you mean isentropic, say "isentropic".
If you mean T dS, say "T dS".

So, unless I am misunderstanding this example, it
does not give us a reason to use dQ, but rather
gives us yet another reason for avoiding dQ.