Is there a difference between the Joule coefficient and the Joule–Thomson coefficient? Also, I am having a hard time understanding what they mean.


The Joule coefficient is $$\left(\frac{\partial T}{\partial V}\right)_U$$

Experimentally, Joule attempted to measure this value by expanding gas into an evacuated, insulated container, thus ensuring $U$ is constant.

The Joule–Thomson coefficient is $$\left(\frac{\partial T}{\partial P}\right)_H$$

Experimentally, this is realized by expanding a flow of gas in an insulted pipe from a high pressure upstream region to a lower pressure downstream region, the two regions being seperated by a porous frit.

For more information see THE JOULE AND JOULE-THOMSON EXPERIMENTS by Dr. J. B. Tatum, University of Victoria.

  • $\begingroup$ So, the Joule coefficient is a ratio of the change in temperature to the change in volume with constant internal energy, and the Joule-Thomson coefficient is a ratio of the change in temperature to the change in pressure under constant enthalpy? In the Joule experiment, the gas does no work, and the temperature doesn't change? $\endgroup$ – user4696 Feb 11 '15 at 15:19
  • $\begingroup$ This almost correct, only your statement "the temperature doesn't change" is incorrect. Joule didn't detect a temperature change, because his equipment wasn't good enough, but for a real gas (not an ideal gas) there is a temperature change. $\endgroup$ – DavePhD Feb 11 '15 at 15:22

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