I am still having a hard time figuring out what this discussion is
about.
There is only one right answer for the entropy of such a system.
Let's review some scenarios:
1) Helium on one side of the partition, helium on the other side,
pull out the partition
1a) For a macroscopic system, to an excellent approximation nothing
happens, so the entropy is extensive.
1b) For a submicroscopic system, i.e. one atom on each side, the
entropy is not even approximately extensive, as we know from
the Sackur-Tetrode equation. There is nothing the least bit
weird about this; the energy isn't extensive either. The
boundary effects cannot be neglected.
2) Helium on one side, argon on the other side, pull out the partition.
There is an irreversible increase in entropy. The entropy of
mixing. I do not know how to ask (let alone answer) the question
of whether "the" entropy is extensive; since the operation is
irreversible there is no good way to define "the" entropy in such
a way as to make the question meaningful.
3) Helium plus neon on one side, argon plus krypton on the other side,
pull out the partition.
Again there is an irreversible increase of entropy.
4) Helium on one side, nothing on the other side, pull out the
partition.
This is the familiar free expansion. There is an irreversible
increase of entropy. Nothing weird or paradoxical here.
The single-atom case is not much different from the many-atom case.
Prosaic irreversible expansion.
Supposedly (but not really) new:
5) N distinguishable particles on one side, M distinguishable particles
on the other side, pull out the partition.
It seems obvious to me that this is just N+M copies of case (4). Each
particle has *no* counterpart on the other side, so all we have is
irreversible free expansion. The particles don't care about each
other because they are distinguishable and ideal.
The single-particle probabilities are independent, so calculating the
(multi-particle) microstate probability is trivial. Just multiply.
Calculating the entropy is trivial. As always, for independent
probabilities, the entropy is additive. http://www.av8n.com/physics/thermo/entropy.html#eq-s-combined-independent
Note that this result does not require "reversible processes" or
a partition function or even thermal equilibrium. It is a direct
corollary of the definition of entropy.
Note that additive is not remotely the same as extensive.
Asking whether "the" entropy is extensive is still meaningless AFAICT,
for the reasons mentioned above.
6) Put two samples of gas side by side but don't mix them (as suggested
in a previous email).
In this case, if the partition is still there, putting them side by
side doesn't do anything. If the partition is removed and they don't
mix, you're violating the equation of motion. Either way I don't
understand the point of the question.
All in all ... what am I missing?
============
As for "classical thermodynamics" and "dQ/T" for the "equivalent
reversible process" I have two admittedly somewhat contradictory
things to say:
1a) I don't think there is any self-consistent theory of classical
thermodynamics.
1b) I don't think there is any such thing as dQ in general.
1c) It requires some contortions to talk about a reversible process
that is "equivalent" to an irreversible process.
1d) Classical thermodynamics is the hard way to do this problem. In
contrast, the stat mech approach is simple and straightforward.
2) OTOH if we close our eyes and grit our teeth and just turn the
crank on the classical thermodynamics formalism, it really should
give the same answer ... for the simple reason that entropy is a
state function and temperature is a state function.
Note that saying that classical thermodynamics works /in this case/
is not meant as an endorsement. I want a theory of thermodynamics
that works all the time, not just in selected cases.