Re: [Phys-L] source of cooling for a propane tank

[John Denker <jsd@av8n.com>]

On 12/14/2016 09:14 AM, Robert Cohen asked:

> I would like to know the main reason why a propane tank cools when it
> is used.  When I look online, I typically find one of four
> explanations:

[slightly reformatted]

> 1a. Adiabatic cooling
> 1b. That is, the gas that remains inside is colder because it does work
> pushing the rest of the gas out.
>
> 2a. Evaporative cooling.
> 2b. That is, removing the gas lowers the pressure and the liquid
> propane boils as a result, leading to evaporative cooling.
>
> 3a. Joule-Thomson cooling 
> 3b. That is, the expanding gas must work against the attractive force
> between the propane molecules.
>
> 4a. Colder molecules "left behind"
> 4b. That is, the faster molecules are more likely to leave, and so
> the slower and cooler molecules remain behind.

Note:  As a pedagogical simplification, consider putting the tank in
contact with a thermostatic heater, so that it remains at constant
temperature.  Rather than observing the change in temperature,
observe the amount of energy that the heater must supply to hold
the temperature constant.

  Every so often, if desired, you can remove some energy to obtain
  the original overall behavior, but with a clearer picture of what
  is going on.

Item (2) is /usually/ dominant.  Think about removing one mole of
propane from a large, partly-full tank.  You draw product from the
top, i.e. from the gas phase, but the liquid immediately boils to
replace it.  The system remains on the SVP curve (saturated vapor
pressure) so at constant temperature the gas pressure and density
don't change at all.

   Exceptionally, at the end of the game, when the tank is devoid
   of liquid and only vapor remains, then sure, the gas will expand
   and cool, in accordance with item (1).

Normally, as long as you have coexistence of the two phases, item (2)
is a much better explanation.  Look at the plot of enthalpy versus
temperature to understand how the latent heat enters the picture.
Sure, there's some slope in the gas phase, but it's small potatoes
compared to the latent heat.  This graph is for water but the idea
applies to other systems:
  https://www.av8n.com/physics/img48/water-enthalpy.png

Item (3) is not responsive to the question.  It is perhaps the answer
to a different question, i.e. what happens to the gas downstream of
whatever valve or impedance you are using to keep the product from
leaving the tank all at once.

As for item (3b), if we were to lift it out of context, it would be
correct.  It is the explanation for (irrelevant) JT expansion, but
also the microscopic explanation for (relevant) latent heat of
evaporation.

Item (4) is 99% wrong.  In such situations it is a good approximation
and a valuable pedagogical simplification to assume that the whole
works remains near thermodynamic equilibrium throughout.  Therefore
the stuff that leaves is not any hotter than the stuff that stays
behind.

At the next level of detail:  You can't expand anything at strictly
zero rate, so we have to ask what happens then the rate is small
but nonzero.  Imagine a well-controlled expansion in front of a
moving piston.  The molecules that bounce off the moving piston
are to a small degree colder than the others.  (This makes contact
to our recent discussion about analyzing things in the CM frame.)

This is a small effect, small in proportion to the rate of expansion
... but integrated over the whole expansion, it is the microscopic
explanation for 100% of the cooling, so it can't really be ignored.
The calculation is straightforward at the level of introductory
physics;  it's the physics of bunting a baseball with a backwards
moving bat.  For compression it's the physics of hitting the ball
with a forward-moving bat.

  The same way of looking at it works for a photon gas (not just
  for particles).  The photons get Doppler shifted.  From there,
  it should come as no surprise that the quantum calculation is
  essentially the same.  You can dress it up as time-dependent
  perturbation theory, at the graduate level (or hard-core
  undergrad physics major level) ... but basically it's just
  a Doppler shift.

In any case, the point is, the particles aren't colder because
you «selected» them à la Maxwell's demon, but because you /made/
them colder, by direct mechanical action.

  There do exist selector-type devices, which is why item (4)
  is not quite 100% wrong, but that's nowhere near relevant to
  the question that was asked.