Joule-Thomson effect: why does a gas cool if it's below the inversion temperature?

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.

• Every mathematical model has its range of applicability. Check to ensure that you are not exceeding that range in your calculations. – David White May 29 '19 at 16:13

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 $$(\beta T-1)=\left(\frac{\partial \ln{z}}{\partial \ln{T}}\right)_P$$