Re: [Phys-l] Gibbs paradox (redux)

[John Denker <jsd@av8n.com>]

On 06/24/2011 06:41 PM, Carl Mungan wrote:
> Okay, one more time on this topic. Maybe now I'm getting 
> somewhere. We'll see.
>
> Consider the system to be a glass of milk at room temperature, 
> ie. a colloid.

OK, that's a nice incisive question.
 
> (1) Would you agree that this system could be described as an 
> ideal gas of distinguishable particles?

Sure, why not.

> (2) If so, what is a formula for its entropy? (Consider only 
> the configurational entropy of the translational states of the 
> suspended fat globules.) 

> Sorry, replace "configurational entropy of the translational 
> states" with "translational entropy" [above]. In 
> other words, ignore any conformational changes of the 
> globules, entropy of the water solvent, etc. Treat the 
> globules as hard spheres moving around in water.

Gack, the language is confusing ... but I think I understand the
intent.

There is one clarification that I must insist on:  I am happy to 
go along with holding the "internal" coordinates of each particle
constant, so that we can ignore *changes* in the internal coordinates,
as stated ... but we cannot ignore these coordinates entirely.  It
turns out that they make a constant contribution to the entropy, 
but the constant is not zero.

> Write your answer in terms of V,U,N 
> where U = 1.5 NkT. If V, U, and N are all doubled, does S 
> double?

No, S does not double.  Not even close.

The explanation is too long and detailed for email.  I wrote it up
at
  http://www.av8n.com/physics/thermo/z-particles.html#sec-z-ideal-gas
wherein the bottom line is
  http://www.av8n.com/physics/thermo/z-particles.html#eq-s/v-snow

Note that what Carl calls "milk" I call "snow" (since snowflakes are
proverbially all different).

=====

In brief:  It all comes down to doing the delabeling properly.

For a nondegenerate gas of absolutely identical particles, the usual
factor of N! is correct.

If the particles are not all identical, N! is not correct.  However,
as always, you can get the right answer by doing the wrong calculation
and then adding a fudge factor at the end.  It seems a lot of people
stick in the factor of N! even when they shouldn't, and then fudge
the result by adding some sort of "spin entropy" term.

My point is that if you do the counting properly, a simple and principled
calculation gives you the correct expression for _the_ honest-to-goodness
entropy ... and you don't need to add any ad-hoc spin-entropy term, because 
that is automatically included, as it should be.

Bottom line:  spin entropy is real.  Entropy of mixing is real.
Identical-particle effects are crucial to thermodynamics, even 
in many situations where you might have expected the classical 
approximation to work.