The document discusses how the temperature distribution in a column of gas under gravity may be affected by intermolecular attraction between gas particles in a van der Waals gas. It shows that when making a mean field approximation to account for intermolecular attraction, and assuming the gas is in equilibrium with Maxwell-Boltzmann velocity distributions at different heights, the temperature will remain constant with height. This is true no matter the dependence of potential energy on height. Equations derived by Walton that seemed to imply a temperature gradient with height actually assume non-equilibrium conditions that do not apply in this case.

1. Discussion of the temperature distribution in a column of gas under gravity,
including intermolecular attraction
A van der Waals gas differs from the simplest ideal gas by having particles with finite volume, and
which can also attract one another.
In this discussion, I will neglect the effect of finite size (can come back to this later!), and will take the
attraction as a mean value to the centre-of-mass of the gas (mean field approximation). The further
a molecule is from the centre-of-mass of the gas, the higher its potential energy is.
Can this effect the temperature distribution of the gas?
Well, some relevant equations for a van der Waals gas are
http://en.wikipedia.org/wiki/Joule-Thomson_effect
where temperature can be seen to change with pressure (the subscript 'H' refers to the case where no heat
or work is exchanged with the surroundings), and the barometric formula
where pressure can be seen to drop with height. Putting these two equations together one may expect for
a van der Waals gas under gravity that the temperature will drop with height even when no work or heat is
exchanged with the surroundings.
This can be investigated using the derivation of Walton
<http://www.tandfonline.com/doi/abs/10.1080/00107516908220108#preview>. Walton’s maths is for an
ideal gas with no intermolecular attraction under gravity, but can easily be generalised to include the mean-
field case under discussion.

3. First question is where this expression comes from?
This is easy enough. The 1D Maxwell-Boltzmann
distribution is a 'probability density function'. The
word 'probability density' is used as it is the integral
that is needed to give the probability, hence the 'dw'
at the end. The prefactor n2 is needed, as the MB
distribution is normalised to unity, whereas we want
to know the actual number of molecules to be found
in any given state, because this varies with height.
The number 'n2' is the number of particles per unit
volume in region 2.
Note that Walton uses 'w' for velocity whereas we
might be more used to 'v', but the maths is obviously
the same.

4. The next issue is to remove any reference to
'h'
One sees in this equation that the integral has
a lower limit of sqrt(2gh). However, this
assumes a specific dependence of potential
energy, PE, with height. For gravity this is of
course PE=mgh.
We can generalise the equation by simply
saying that to go from Region 1 to Region 2,
the particle must start with KE>=PE.
This means the lower velocity must be
½ m.w^2 = PE
w=sqrt(2.PE/m)
Filling that in gives the equations on the next
slide.

6. Similarly for the second page, the same
substitution can be made, giving the equations on
the right.
NOTE:
At this stage, none of the equations have a
reference for 'h', nor any particular dependence of
PE with h. As far as these equations are
concerned, the equations hold whether
PE=m.g.h
PE=m.g.h^2
PE=m.g.cos(h)
It doesn't matter what the dependence is, so long
as the gas is described by a MB distribution at
each position, it will be isothermal with height
under gravity.
Furthermore, because the gas is in contact with
walls of a container (or ground), the gas
molecules will be thrown off the surface with a MB
distribution. Hence the conditions will be met and
the gas will certainly be isothermal.

7. How to reconcile this with the first slide?
What was shown on the previous page is that so long as one has a MB distribution at different parts of
the system and the system is in equilibrium, then no matter what the PE function is, the gas will be
isothermal. This includes the case for a van der Waals gas under gravity in the mean-field
approximation.
But the equations on slide one seemed to imply a temperature difference, where did this go?
Well, the equation for dT/dP on slide one is mostly used for the case of a gas escaping through a
nozzle, where the gas flows from a region of high pressure to a region of lower pressure. This is a
non-equilibrium condition, and so the analysis of Walton doesn't apply. However, we are only
interested in the equilibrium condition, and have shown that this will be isothermal!
Similarly for the case of vertical convection, the initial gas velocity distribution is not MB (there will be a
vertival ‘offset’ velocity), hence the gas can cool, but convection requires a heat source/sink, and so is
not an equilibrium state of a closed system.
Notes:
Things omitted in the above analysis are
1, the finite size of the molecules
2, the detailed interparticle forces on a pair-by-pair basis