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Source: The1stLawofThermodynamicsLesson4

I am looking for difference between adiabatic expansion and Joule-Thomson effect. Looking at the setup demonstrating Joule-Thomson effect, I find no essential difference between the set-up for the two.

What is the feature of the setup for Joule-Thomson effect that makes the process isenthalpic, as oppose to adiabatic expansion? (I am not looking for proof initial H= final H)

What is the use of the throttle/ porous plug? If it is the throttle that makes the process isenthalpic, what is the physical reason behind it?

From another source,

The gas is first compressed isothermally, then expanded isothermally (may be at different temperature as initial temperature)

I am not sure what it means. Does it mean that its temperature jumps abruptly from initial temperature to final temperature? It has only two temperature throughout the process? How is this possible? I expect the temperature to change continuously, because the volume of the gas changes throughout the process.


1 Answer 1


In the Joule-Thompson experiment, there is major viscous dissipation of mechanical energy (in this case pressure) to internal energy occurring within the pores of the plug. This is analogous to Poiseuille flow of a viscous fluid in a tube. If the tube is adiabatic, the fluid flowing through a tube experiences a temperature rise as a result of the viscous "heating." The work to push the fluid through the plug is totally dissipated and converted to increased internal energy of the gas. The reason that viscous effects are so important in flow through a plug is that the pores are very small. So flowing the gas through the plug is analogous to flowing the gas through an array of very tiny tubes in parallel.

Ordinarily when a gas expands adiabatically, it cools. But, for flow of an ideal gas through a porous plug, the adiabatic cooling is exactly offset by the viscous heating. So there is no temperature change. And, as a result, there is no enthalpy change.

  • $\begingroup$ Interesting explanation. Does this also explain why Joule-Thompson expansion is not reversible, while adiabatic is? Because there is nothing to "cool" the gas while it is being pushed through porous plug toward higher pressure? $\endgroup$
    – Pygmalion
    Commented Jul 7, 2021 at 20:08
  • $\begingroup$ Yes. But I think you meant "toward lower pressure." In fact, it is also the explanation of entropy generation in fluids experiencing irreversible mechanical deformations. $\endgroup$ Commented Jul 7, 2021 at 20:19
  • $\begingroup$ You are right, there is no Joule-Thomson compression. It is impossible for the gas at a lower pressure to go through porous plug to the higher pressure. So the process is completely impossible, and the Joule-Thompson expansion is inherently irreversible. [I was a bit confused because some web pages talk about Joule-Thomson expansion and compression] $\endgroup$
    – Pygmalion
    Commented Jul 7, 2021 at 20:29

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