# Inversion temperature

The inversion temperature in thermodynamics and cryogenics is the critical temperature below which a non-ideal gas (all gases in reality) that is expanding at constant enthalpy will experience a temperature decrease, and above which will experience a temperature increase. This temperature change is known as the Joule-Thomson effect, and is exploited in the liquefaction of gases. Inversion temperature depends on the nature of the gas.

For a van der Waals gas we can calculate the enthalpy ${\displaystyle H}$ using statistical mechanics as

${\displaystyle H={\frac {5}{2}}Nk_{\mathrm {B} }T+{\frac {N^{2}}{V}}(bk_{\mathrm {B} }T-2a)}$,

where ${\displaystyle N}$ is the number of molecules, ${\displaystyle V}$ is volume, ${\displaystyle T}$ is temperature (in the Kelvin scale), ${\displaystyle k_{\mathrm {B} }}$ is Boltzmann's constant, and ${\displaystyle a}$ and ${\displaystyle b}$ are constants depending on intermolecular forces and molecular volume, respectively.

From this equation, we note that if we keep enthalpy constant and increase volume, temperature must change depending on the sign of ${\displaystyle bk_{\mathrm {B} }T-2a}$. Therefore, our inversion temperature is given where the sign flips at zero, or

${\displaystyle T_{\text{inv}}={\frac {2a}{bk_{\mathrm {B} }}}={\frac {27}{4}}T_{\mathrm {c} }}$,

where ${\displaystyle T_{\mathrm {c} }}$ is the critical temperature of the substance. So for ${\displaystyle T>T_{\text{inv}}}$, an expansion at constant enthalpy increases temperature as the work done by the repulsive interactions of the gas is dominant, and so the change in energy is negative. But for ${\displaystyle T, expansion causes temperature to decrease because the work of attractive intermolecular forces dominates, giving a positive change in energy.[1]