Here are a couple of ideas that may help answer the original
question. Executive summary:
1) Ensemble average!
2) Equipartition of *all* forms of energy!
In more detail:
1) Suppose we have a single particle in a tall, otherwise-
empty tube in a gravitational field. Most of the time, it
is a free particle, following a parabolic trajectory. Its
interactions with the side-walls are elastic and contribute
nothing to the story. Many times per minute, it interacts
_elastically_ with the bottom of the tube, reverses course,
and begins a new parabola.
Every few weeks, it interacts inelastically with the bottom
of the tube, and this is the process whereby it comes into
equilibrium with the temperature of the bottom (which is
connected to a heat bath).
Note the contrast:
-- If you prepare a thermal ensemble of such systems, you
will observe a thermal distribution of energies, and
observe equipartition of energy.
-- OTOH if you just look for a few minutes at one instance
of the system, i.e. one element of the ensemble, you will
not see a thermal distribution. In fact, rather than a
Boltzmann distribution of total energy, there will be a
delta-function distribution.
Let's be clear: The classical macroscopic variables such
as pressure, density, entropy, and temperature are _defined_
in terms of the ensemble average.
Operationally, if I use a liquid-in-glass thermometer to
measure the temperature of a swimming pool, I do not need
an ensemble of swimming pools, but that's only because for
a sufficiently-large system the time average is a good
approximation to the ensemble average. In any case, we
must recognize that this-or-that convenient operational
measure of temperature can never replace the definition of
temperature ... and for a small system (especially a single-
particle system) we need to be extra careful.
The fundamental laws of thermodynamics apply just fine to
small systems ... but many of your favorite approximations
apply only to "sufficiently large" systems.
2) Let's talk about equipartition of energy. In the aforementioned
ensemble, all of the following are true:
-- Boltzmann distribution of kinetic energy
-- Boltzmann distribution of potential energy
-- Boltzmann distribution of total energy.
Students may have a hard time with the last two items, because
again and again they have been fed the story that
☠ "Thermal energy" is random "kinetic energy" ☠
That story is nonsense! It has nothing to do with thermodynamics.
The only case in which it is close to true is for an ideal gas,
where the thermal energy is kinetic for the simple reason that
*all* the energy is kinetic. Here we are considering a tall
cylinder of gas in a gravitational field, in which case there is
a nontrivial potential energy, and the potential energy will be
Boltzmann-distributed.
It is an excellent, easy, informative homework exercise to
work out the general expression for a sample of gas in a
cylinder of height H in a gravitational field. I guarantee
that the potential energy is thermal energy, in the sense that
it contributes to the heat capacity. (It should go without
saying that you should check the correspondence limit, i.e.
the case of a not-very-tall cylinder, and show that you can
recover the elementary textbook formula for the heat capacity
of an ideal gas.)
Again: The ensemble has:
-- Boltzmann distribution of kinetic energy
-- Boltzmann distribution of potential energy
-- Boltzmann distribution of total energy.
Those three statements are consistent, since the exponential of
a sum is the product of exponentials. The probability distribution
of the KE is statistically independent of the distribution of PE,
so we have a product of probabilities.
In particular: On an ensemble-averaged basis, we have a Boltzmann
distribution of PE ... which for our single-particle gas means
we have a Boltzmann distribution of heights. At any particular
height, we have a Boltzmann distribution of KE.
It must be emphasized that these Boltzmann distributions are
correct for the ensemble but not for medium-short-term time
averages for a single element of the ensemble (that is, a single
instance of the one-particle system).
=========================
Bottom line: I'm happy now. I know why the right answer is right
*and* I finally figured out why the wrong answer is wrong.
*) The fundamental laws of thermodynamics apply to this system
just fine. This is exceedingly unsurprising.
*) There is no law that guarantees you can replace an ensemble
average by a medium-short-term time average ... and this system
is a poster child, demonstrating why the ensemble average is
fundamental whereas the medium-short-term average is not.
If anybody is not happy with this resolution, please re-ask the
question.