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 kinetic energy is positive. But for ${\displaystyle T<T_{\text{inv}}}$, expansion causes temperature to decrease because the work of attractive intermolecular forces dominates, giving a negative change in average molecular speed, and therefore kinetic energy. ^[1]

The inversion temperature of air

The coefficients of the approximation of the air with the Van der Waals gas equation,

${\displaystyle P={\frac {n\,R\,T}{V-n\,b}}-a\,{\frac {n^{2}}{V^{2}}},}$

are

${\displaystyle a:=1.368\,{\frac {{\text{m}}^{6}\cdot {\text{bar}}}{{\text{kmol}}^{2}}},}$

${\displaystyle b:=0.0367\,{\frac {{\text{m}}^{3}}{\text{kmol}}}.}$

The inversion temperature is associated to the states of the gas where the Joule-Thomson coefficient ${\displaystyle \mu _{JT}\equiv \left({\frac {\partial T}{\partial P}}\right)_{H}}$ is equal to zero. A symbolic calculation of the Joule-Thomson coefficient on the ${\displaystyle P}$ versus ${\displaystyle T}$ plane is tedious task, but it can be carried out by a math software with the steps described below (on the example of the syntax of Maple). For convenience, the subscripts in the names of the expressions denote the state functions the named expression depends on.

The expression of the pressure of the Van der Waals gas is:

${\displaystyle P_{T,V}:={\frac {n\,R\,T}{V-n\,b}}-a\,{\frac {n^{2}}{V^{2}}}.}$

The internal energy of the Van der Waals gas is:

${\displaystyle U_{T,V}:={\frac {3}{2}}\,n\,R\,T-a\,{\frac {n^{2}}{V}},}$

therefore the enthalpy ${\displaystyle H=U+P\,V}$ of the gas is:

${\displaystyle H_{T,P,V}:={\frac {3}{2}}\,n\,R\,T-a\,{\frac {n^{2}}{V}}+P\,V.}$

The volume can be expressed from ${\displaystyle H_{T,P,V}}$ by solving the equation ${\displaystyle H_{T,P,V}=H}$ for ${\displaystyle V}$ as follows:

${\displaystyle V_{T,P,H}:={\rm {solve}}\left(H_{T,P,V}=H,V\right).}$

An equation binding the three state functions ${\displaystyle \{T,P,H\}}$ can be obtained by substituting ${\displaystyle V_{T,P,H}}$ for ${\displaystyle V}$ in the expression of the pressure ${\displaystyle P_{T,V}}$:

${\displaystyle P_{T,P,H}:=\left.P_{T,V}{\begin{matrix}{}\\{}\end{matrix}}\right|_{V=V_{T,P,H}},}$

which can be then solved for ${\displaystyle T}$:

${\displaystyle T_{P,H}:={\text{solve}}\left(P_{T,P,H}=P,T\right),}$

whence the Joule-Thomson coefficient ${\displaystyle \mu \equiv \left({\frac {\partial T}{\partial P}}\right)_{H}}$ can be calculated as follows:

${\displaystyle \mu _{P,H}:={\frac {\partial }{\partial P}}\,T_{P,H}.}$

The Joule-Thomson coefficient of the air as a function of the pressure and temperature

The Maple code for the calculation of the Joule-Thomson coefficient of the air as a function of the pressure and the temperature

So far, we expressed the Joule-Thomson coefficient in terms of the pressure and the enthalpy. In order to express the Joule-Thomson coefficient in terms of ${\displaystyle (P,T)}$, we first obtain the state equation of the volume ${\displaystyle V=V(T,P)}$ by solving equation ${\displaystyle P_{T,V}=P}$ for ${\displaystyle V}$,

${\displaystyle V_{T,P}:={\text{solve}}\left(P_{T,V}=P,V\right),}$

which is then substituted in the expression ${\displaystyle H_{T,P,V}}$ for ${\displaystyle V}$ in order to obtain the state equation ${\displaystyle H=H(T,P)}$ for the enthalpy:

${\displaystyle H_{T,P}:=\left.H_{T,P,V}{\begin{matrix}{}\\{}\end{matrix}}\right|_{V=V_{T,P}}.}$

Finally, the substitution of ${\displaystyle H_{T,P}}$ for ${\displaystyle H}$ in ${\displaystyle \mu _{P,H}}$ gives us the searched Joule-Thomson coefficient expressed in terms of the temperature and the pressure:

${\displaystyle \mu _{P,T}:=\left.\mu _{P,H}{\begin{matrix}{}\\{}\end{matrix}}\right|_{H=H_{T,P}}.}$

The obtained expression is extremely cumbersome, but the result can be plotted. The annexed plot shows the values of the Joule-Thomson coefficient in ${\displaystyle {\frac {\text{K}}{\text{bar}}}}$ on the pressure and temperature plane. The positive values of the Joule-Thomson coefficient correspond to the cooling effect because in the expression of ${\displaystyle \mu _{JT}\equiv \left({\frac {\partial T}{\partial P}}\right)_{H}}$ the denominator ${\displaystyle \partial P}$, corresponding to the drop of the pressure during the throttling process, is negative. The surface of the plot, where the values of the Joule-Thomson coefficient are positive and the throttling of the air results in the cooling of the gas, is marked in blue. Outside the range of the blue surface, the throttling causes heating. The contour separating the blue surface from the red surface corresponds to the inversion temperature of (the Van der Waals gas model of) the air.

The Maple code implementing the described procedure is provided in the annexed figure.

See also

• Critical point (thermodynamics)
• Phase transition
• Joule-Thomson effect

References

1. Charles Kittel and Herbert Kroemer (1980). Thermal Physics (2nd ed.). W.H. Freeman. ISBN   0-7167-1088-9.

External links

• Thermodynamic Concepts and Processes (Chapter 2) (part of the Statistical and Thermal Physics (STP) Curriculum Development Project at Clark University)