# 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

## 1 Answer

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.