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The Joule-Thomson coefficient is given by $$\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_{H} = \frac{V}{C_{P}}(\beta T - 1),$$ where $\beta$ is the coefficient of thermal expansion. If the inversion temperature is defined by $T_{inv} = \frac{1}{\beta}$, then why is $\mu_{JT} > 0$ if $T < T_{inv}$, as stated by Wikipedia Joule-Thomson effect? I really don't get this, this should be basic math ? This results in not comprehending why a gas cools if it's below inversion temperature and vice versa.

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  • $\begingroup$ Every mathematical model has its range of applicability. Check to ensure that you are not exceeding that range in your calculations. $\endgroup$ – David White May 29 '19 at 16:13
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If you look up the coefficient of thermal expansion of air, you will find that it decreases with increasing absolute temperature. In fact, it decreases rapidly enough that even the product $\beta T$ decreases with increasing temperature. For air at room temperature and 1 bar, for example, the product is about 1.01 while, at 200 C, the product is about 0.99.

In terms of the compressibility factor z, the product of $\beta$ and T is given by: $$\beta T=1+\left(\frac{\partial \ln{z}}{\partial \ln{T}}\right)_P$$So $$(\beta T-1)=\left(\frac{\partial \ln{z}}{\partial \ln{T}}\right)_P$$

Get yourself a plot of z vs reduced temperature and reduced pressure and note how the right hand side of this equation decreases with increasing reduced temperature. It is positive at low reduced temperatures, and reaches a value of zero at a reduced temperature of about 4.

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