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*From*: John Denker <jsd@av8n.com>*Date*: Tue, 12 Jan 2010 12:30:45 -0700

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.

Comments?

**Follow-Ups**:**Re: [Phys-l] T dS versus dQ***From:*"Bob Sciamanda" <treborsci@verizon.net>

**References**:**[Phys-l] Thermodynamics question***From:*"LaMontagne, Bob" <RLAMONT@providence.edu>

**[Phys-l] T dS versus dQ***From:*John Denker <jsd@av8n.com>

**Re: [Phys-l] T dS versus dQ***From:*Carl Mungan <mungan@usna.edu>

**Re: [Phys-l] T dS versus dQ***From:*John Denker <jsd@av8n.com>

**Re: [Phys-l] T dS versus dQ***From:*Carl Mungan <mungan@usna.edu>

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