Joule–Thomson Effect

1. Jun 13, 2013

sliorbra

Hello,

I have a problem to understand properly this effect. I understand that during the process the change in the internal energy of the gas is equal to the net work that have been done on it.

So far everything is fine. My problem is with the physical explanation of the phenomenon.
Lets assume for now that the attraction forces are the dominant one.
According to the explanation I always hear,when the gas expands, its potential energy increases, and because it doesn't have any other energy source [since the process assumed to be adiabatic], the source to the increase in the potential energy is his kinetic energy. Therefore, the temperature of the gas decreases from T1[the temperature in the initial region] to T2[the temperature in the final region].

BUT, in the first step of the process work is done on the gas, what increases his internal energy, and just then occurs expansion. So the situation where the gas uses his own energy to expand occurs just after some work was done on him.

So, why does the temperature of the gas can't be increased in the first step by the the work that is done on it, and then reduce due to the expansion, in such manner that the total temperature change is positive?

I feel that all the explanations I've read so far ignore the first step of the process...

2. Jun 13, 2013

Yanick

The Joule-Thomson coefficient ((dT/dP)H) can indeed have positive or negative values, depending on the gas or the pressures used in the two chambers. From Levine, 6th Edition, values of μJT can have values of +3 to -0.1 K/atm.

From the same book, the point of a JT expansion is to do it with the same gas at different pressures and build an isoenthalpic curve which can then be used to determine the sign and magnitude of μJT. Then you can use your information to figure out how to do real things, such as making liquid nitrogen, which is, I believe, done via a JT expansion.