Joule kelvin expansion, thermodynamics

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Homework Statement




for a jk expansion, I know dU=0. I think this is because dQ=0 as its isolated and dW = 0 as there is no work done against the surroundings.

but is dW 0? does the gas do work against itself?

I have also seen that the temperature change of an ideal gas = 0. how can this be?
 
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Yes, dU = dQ = dW = 0. However, there is mass flow energy that must be accounted for. The Joule Kelvin (also called Joule Thomson) is a throttling process which usually reduces to an isenthalpic process.
 
I think I may have confused a jk expansion with a free expansion of gas where the gas is confined in volume V then is allowed to expand into a vacuum.

could you please explain further about this "flow energy"?
and for an ideal gas, I don't see how change in T =0 if there is a change in P and V
 

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