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The Joule Thomson coefficient for various gases can be found in textbooks, e.g. have I found that hydrogen has $\mu_{jt}=-0,024735$ K/bar and an inversiontemperatur of around 200 K.

Not having the background to understand its derivation: Is $\mu_{jt}$ constant for a given gas or is it a function of some parameter (apart from that it changes sign around the inversion temperature)?

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It is not a constant for a given gas. A value given in a table must be for some specific pressure and temperature. To see this, look at a diagram showing isenthalps (curves of constant enthalpy). They are not straight lines! However they are pretty straight in the low pressure region when the temperature is comfortably below the inversion temperature. I expect the value given in a table is probably the value at low pressure at some chosen temperature below the inversion temperature.

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  • $\begingroup$ This agrees with my own conclusion below, although I have not seen an actual isenthalp diagram. Thanks! $\endgroup$
    – Irenaius
    Commented Sep 28, 2020 at 11:52
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Defintion of the Joule-Thomson Effect:

$\left(\frac{\delta T}{\delta P}\right)_H=:\mu=\frac{V}{C_p}(\alpha T-1)$

with the general thermal expansion coefficient (hydrogen behaves almost like an ideal gas in this respect)

$\alpha=\frac{1}{V}\left(\frac{\delta V}{\delta T}\right)_P$

and the heat capacity at constant pressure

$C_p=T\left(\frac{\delta S}{\delta T}\right)_P$.

Thus function parameters are temperature, pressure and volume.

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  • $\begingroup$ "nothing gas/material specific" seems to suggest the coefficient does not depend on the fluid, but in fact it does depend. Also, both $\alpha$ and $C_p$ are in general not constant for any given fluid but depend on the state. I hope you will either correct or clarify this answer. $\endgroup$
    – Andrew Steane
    Commented Sep 28, 2020 at 13:21
  • $\begingroup$ It is not meant to mean that it does not depend on the fluid, which it sure does. And, both $\alpha$ and $C_p$ are functions with the above described parameters - is that what you mean by state? And you are totally right, this bit was misleading if not simply false. $\endgroup$
    – Irenaius
    Commented Sep 28, 2020 at 13:45

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