Joule Thomson expansion

  • #1
99
1
Wikipedia says,
Unlike a free expansion , in Joule Thomson expansion work is done causing the change in internal energy. Whether the internal energy increases or decreases is determined by whether work is done on or by the fluid; that is determined by the initial and final states of the expansion and properties of the fluid.

How we would determine work done by or on the gas in this expansion?

This is what I was thinking,

Work done before the porous plug = -(Pressure of gas before throttling)/(density of the gas before throttling)= W1

Work done after throttling= +(Pressure of gas after expansion)/(Density of the gas after throttling)= W2

Total Work(W)= W1+W2

If W >0 , then the gas cools down
W<0 , then the gas heats up

Am I going in right direction?

I have considered work done by system as positive in sign convention.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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Answers and Replies

  • #2
20,451
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Your assessment of the work done per unit mass of gas passing through the porous plug by its surroundings is correct (i.e., in line with the derivation of the open system, control volume, version of the first law of thermodynamics). And the Wiki statement about its effect on the internal energy per unit mass of gas is also correct (since the change in enthalpy per unit mass is zero). But I don't think that this necessarily says anything about the change in temperature, since the internal energy is a function of both temperature and pressure (i.e., the JT effect depends on the deviation of the gas from ideal gas behavior). This is because the deviations from ideal behavior of both Pv and U contribute to this.
 

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