Re: Where does the kenetic energy go during a Joules Thompson effect?

First, you have to realize that you are mixing different concepts.

Temperature is a fundamental -- basically defined by saying there is no 
net energy transfer between two things if they are at the same temperature.

Then there are concepts of energy.  One can start with the internal energy 
which in some sense measures how much energy the fluid has as a result of 
molecular motions and interactions.  Then there is *enthalpy*, which is 
the sum of the internal energy and PV, where P is the pressure and V is 
the volume.  This also has dimensions of energy, and is convenient for 
working with flow processes such as expansion through a valve.  The 
definition of a Joule-Thomson process is a change in pressure at constant 
enthalpy; the Joule-Thomson coefficient is the derivative of temperature 
with respect to pressure with enthalpy held constant (remember, you always 
have to say what you are holding constant with these thermodynamic 
derivatives!).

"Heat" is energy flowing from one place to another.  It really is not a 
relevant concept to the question you asked.

So, what happens when you expand a fluid at constant enthalpy?  As the 
pressure is lowered, the fluid expands, and this changes the interactions 
between the molecules (because on average they are farther apart).  For 
the moment, let's imagine the temperature is the same, and look at changes 
in the enthalpy.  An expansion will often raise the internal energy by 
lessening the attractive intermolecular forces.  The PV part of the 
enthalpy may go either direction depending on the conditions.  Suppose the 
net effect of this is for the enthalpy to go up.  We have already said 
that the actual conditions for Joule-Thomson expansion are constant 
enthalpy, so in order for the enthalpy to go back down to its original 
value the temperature must go down.  So that's the origin of Joule-Thomson 
cooling -- the total energy (measured by the enthalpy) stays the same, but 
the temperature has to go down to compensate for the increase in enthalpy 
due to expansion.  In terms of your title question, the kinetic energy 
(proportional to the temperature) goes into making the molecules be 
farther apart, which is a higher enthalpy state.

A major example of this is when the expansion of a liquid causes it to 
vaporize.  Since vapor is a much higher energy (and enthalpy) state than a 
liquid at the same temperature, the temperature is decreased by such an 
expansion so that the enthalpy remains constant.  This expansion is the 
key step in most refrigeration cycles.

A couple of notes.  First, for an ideal gas there are no intermolecular 
forces and the Joule-Thomson coefficient is zero.  Second, there are 
conditions where the balance of forces is such that the Joule-Thomson 
coefficient is negative, so you get heating upon an expansion at constant 
enthalpy.  Liquid water near room temperature is one case where this 
happens; I believe helium gas is another.

Finally, I'll mention that you should be able to find a more exhaustive 
discussion of the Joule-Thomson effect, complete with equations, in any 
college engineering thermodynamics textbook.

Allan Harvey, aharvey@boulder.nist.gov
"Don't blame the government for what I say, or vice-versa."