Re: dS=0

[Jim Green <JMGreen@SISNA.COM>]

No, Bob.  While they are working on my web page (as we speak, happily),
please let me clarify what is being discussed here:

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.  [Or better
that to the extent that an action changes entropy that portion of the total
work could be included in Q -- but I think that this is a dark alley] I
show this to be true for the classical piston and cylinder from _stats_.  I
wish I had another example of a W-like thing to show that statistically
dS=0 in that case too

I certainly is not fair to take a system where I say deltaQ=0 and show from
thermo that dS=0 -- that would be a tautology.

The question before us is "What is Q?"

The abominable traditional definition is the Q is something like "energy
transfer due to heat flow or due to a temperature difference"  and I
apologize to the list for even saying such a foul thing!

Neither is "energy transferred" nor does "heat flow".  Nor is a temperature
difference required for any of this.

Energy is a system property and can be changed -- increased or decreased
--  but it does not move.  The _system_ can move -- energetic _particles_
can move -- but the property (energy) can't move per se.  The ONLY way to
change the energy level of a system is to do WORK on it.

Remember the pervasive work/energy principle!!!!

Just as the property of color doesn't move -- it can be changed but "blue"
does not move -- only blue _things_ can move.

And, Q (sometimes called "heat") doesn't move either -- it is DONE the same
as work, W, is done.  Q is a kind of WORK.  It doesn't move.  Work doesn't
move.

If anyone thinks either "heat" or "energy" can move, tell us please what it
is that is moving without using the words "heat" or "energy" or their
equivalent - or invoking some unintelligible QM or GR argument -- this is a
classical discussion.

BTW in this regard the property of a system called "charge" doesn't move
either -- only charged _particles_ move - if we are being careful with our
language and trying to make sense of difficult subjects.

If one wants to use language where a _property_ moves independent from it's
system, one must carefully re-define the words before beginning -- and
hence re-define all the previous usage and equations as well.

In the present case one is quite mistaken to call "Q" "heat" AND say it
moves or is inside the system.  Call "heat" "thermal energy" if you will,
but then you must re-write the First Law.

And not nearly a year has passed. (:-)

Jim

At 02:24 07 01 2000 -0500, you wrote:
> Jim,
> I await your equations, but it seems that they are going to simply show
> that a gas process for which dE = - PdV is adiabatic.  This conclusion
> follows directly from the first law: dE = dQ - PdV.  (Similar results
> apply to the generalized version dE = dQ + YdX.)
>
> This is not universally true of *any* PdV process; only those for which dE
> = - PdV (ie: dQ=0).  For a finite ideal gas process the succession of
> PdV's must follow a definite curve: PV^gamma = Constant, in order to be an
> adiabatic process.
>
> Bob
>
> Bob Sciamanda (W3NLV)
> Physics, Edinboro Univ of PA (em)
> trebor@velocity.net

Jim Green
mailto:JMGreen@sisna.com
http://users.sisna.com/jmgreen