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Re: dS=0



At 11:13 AM 1/7/00 -0700, Jim Green wrote:

The question on the page is "Why, if both Q-things and W-things are both
mechanical work, what then is the difference between W and Q if any?" The
answer is that Q-things change the entropy and W-things do not.

Hmmm. The way I use the symbols Q and W, the question is phrased in a way
that is highly likely to produce confusion, and the suggested answer just
increases the probability of confusion.

In many situations, it's bad luck to talk about Q-things as being distinct
from W-things.


To be specific, let's consider the case of a brick connected to a piston in
a cylinder full of ideal gas.

B
|
| | |
| | |
|=====|
| G |
| G |
| G |
|____G|


We can even imagine that the brick is oscillating up and down in response
to its weight, its inertia, and the gas-pressure on the piston. The
oscillation repeatedly transfers energy between the gas and the brick. We
arrange the following conditions:
a) the piston velocity is sufficiently small (timescale not too short),
b) frictional losses are sufficiently small, and
c) the walls of the cylinder are sufficiently well thermally insulated
(timescale not too long) ...
... so that we can invoke the adiabatic approximation.



The question arises, should the energy transfer be considered dQ or dW?
Answer: NO! Neither one.



A) Yes, energy can be transferred.
B) Yes, energy can be subdivided as dQ + dW.
*) But no, please don't crossbreed those two ideas.
In general, energy transfers cannot be subdivided.

We can escape from the mess by writing multiple equations:
dE(brick) = dQ(brick) + dW(brick)
dE(gas) = dQ(gas) + dW(gas)

where the "d" operator denotes a change in the properties of a *single*
subsystem, not a transfer from one subsystem to another.

We can also invoke the principle of conservation of energy,
dE(brick) + dE(gas) = 0

Now it turns out that dQ(brick) vanishes (because we are not coupling to
its internal degrees of freedom) and dW(gas) also vanishes (because it the
gas is ideal). Hence we can conclude
dQ(gas) = -dW(brick)

-------------

To reiterate, it is begging for trouble to ask whether the "process"
transfers dQ or dW. In fact the whole point of this process is that it
converts dQ(gas) to dW(brick) and back again.

==========================


Now, let's return to the question

what then is the difference between W and Q if any?"

It turns out that (depending on details of how we play the game) it might
be operationally difficult to tell whether a black box is internally doing
a W-thing or a Q-thing.

By way of illustration, contrast the gas cylinder with another cylinder
containing no gas but a spring instead.


B B
| |
| | | | | |
| | | | | |
|=====| |=====|
| G | | S |
| G | | S |
| G | | S |
|____G| |__S__|



If (big if) we stipulate that we will not look inside the cylinder, then
(with an artfully chosen spring) it could be arbitrarily difficult to tell
the difference between the two cylinders. Recall that we are assuming that
the timescale is short enough to disallow significant transfer of thermal
energy through the walls of the cylinder.

OTOH if we change the assumptions, making a major change in the insulation
and/or the timescale, we can observe the temperature of the contents of the
cylinder and it will be very different in the two cases.

The fact that the brick is undergoing a dW process tells you nothing about
what is going on in the cylinder. In general, W is not conserved. In
general, Q is not conserved. E is conserved. It's bad luck to talk about
the transfer of something that isn't conserved by the transfer-process.

(There are situations, such as in a well-made h&@t exchanger for example,
where Q is sufficiently well conserved that it makes sense to talk about
transferring it. But don't get tricked into thinking that that's the
general case.)