# Joule-Thomson Coefficient

## Main Question or Discussion Point

When a real gas, as differentiated from an ideal gas, expands at constant enthalpy (i.e., no heat is transfered to or from the gas, and no external work is extracted), the gas will be either cooled or heated by the expansion. That change in gas temperature with the change in pressure is called the Joule-Thomson coefficient and is denoted by µ, defined as:

µ = (dT/dP) at constant enthalpy

The value of u depends on the specific gas, as well as the temperature and pressure of the gas before expansion. For all real gases, µ will equal zero at some point called the "inversion point". If the gas temperature is below its inversion point temperature, µ is positive ... and if the gas temperature is above its inversion point temperature, µ is negative. Also, dP is always negative when a gas expands. Thus:

If the gas temperature is below its inversion temperature:
-- µ is positive and dP is always negative​
-- hence, the gas cools since dT must be negative​

If the gas temperature is above its inversion temperature:
-- µ is negative and dP is always negative​
-- hence, the gas heats since dT must be positive​

"Perry's Chemical Engineers' Handbook" provides tabulations of µ versus temperature and pressure for a number of gases, as do many other reference books. For most gases at atmospheric pressure, the inversion temperature is fairly high (above room temperature), and so most gases at those temperature and pressure conditions are cooled by isenthalpic expansion.

Helium and hydrogen are two gases whose Joule-Thomson inversion temperatures at atmospheric pressure are very low (e.g., about −222 °C for helium). Thus, helium and hydrogen will warm when expanded at constant enthalpy at atmospheric pressure and typical room temperatures.

It should be noted that µ is always equal to zero for ideal gases (i.e., they will neither heat nor cool upon being expanded at constant enthalpy).

By contrast, when external work is extracted during the expansion of a gas (as when a high-pressure gas is expanded through a turboexpander), the expansion is isentropic (i.e., occurs at constant entropy) rather than isenthalpic as in a Joule-Thomson expansion. For an isentropic gas expansion, the gas temperature always cools and the temperature drop is more than would be achieved by an isenthalpic Joule-Thomson expansion.

Milt Beychok

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anorlunda
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Thanks for sharing. Very informative.

This was a necro post from 2006. It didn't ask any question, so it got no replies. But it does have 39K views from people who found it via Internet searches. It must be a really hot topic.

In 2019, we have PF Insights for articles in addition to threads for questions. Alas, I can't contact the OP. His email bounces. But I am bumping it in the Materials forum just to bring it to your attention.

Chestermiller
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@anorlunda There is a section on Joule-Thomson in practically every thermo book that covers essentially this same material.