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

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    $\begingroup$ Please consider expanding your question to show that you are contributing original thought and to better illustrate the points of confusion you are having. If you don't, the question is likely to be closed. $\endgroup$ – Curt F. Feb 11 '15 at 5:55
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The Joule coefficient is $(\frac{\partial T}{\partial V})_U$

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

The Joule-Thompson Coefficient is $(\frac{\partial T}{\partial P})_H$

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.

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  • $\begingroup$ 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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