User:Retired Pchem Prof/JouleThomsonDerivation

January 28, 2016

article Joule-Thompson effect

Derivation of the Joule–Thomson coefficient
It is difficult to think physically about what the Joule–Thomson coefficient, $$\mu_{\mathrm{JT}}$$, represents. Also, modern determinations of $$\mu_{\mathrm{JT}}$$ do not use the original method used by Joule and Thomson, but instead measure a different, closely related quantity. Thus, it is useful to derive relationships between $$\mu_{\mathrm{JT}}$$ and other, more convenient quantities. That is the purpose of this section.

The first step in obtaining these results is to note that the Joule–Thomson coefficient involves the three variables T, P, and H. A useful result is immediately obtained by applying the cyclic rule; in terms of these three variables that rule may be written


 * $$\left(\frac{\partial T}{\partial P}\right)_H\left(\frac{\partial H}{\partial T}\right)_P\left(\frac{\partial P}{\partial H}\right)_T = -1.$$

Each of the three partial derivatives in this expression has a specific meaning. The first is $$\mu_{\mathrm{JT}}$$, the second is the constant pressure heat capacity, $$C_{\mathrm{p}}$$, defined by


 * $$C_{\mathrm{p}} = \left(\frac{\partial H}{\partial T}\right)_P $$

and the third is the inverse of the isothermal Joule–Thomson coefficient, $$\mu_{\mathrm{T}}$$, defined by


 * $$\mu_{\mathrm{T}} = \left(\frac{\partial H}{\partial P}\right)_T $$.

This last quantity is more easily measured than $$\mu_{\mathrm{JT}}$$. Thus, the expression from the cyclic rule becomes


 * $$\mu_{\mathrm{JT}} = - \frac{\mu_{\mathrm{T}}} {C_p}$$.

This equation can be used to obtain Joule-Thompson coefficients from the more easily measured isothermal Joule–Thomson coefficient. It is used in the following to obtain a mathematical expression for the Joule-Thompson coefficient in terms of the volumetric properties of a fluid.

To proceed further, the starting point is the fundamental equation of thermodynamics in terms of enthalpy; this is


 * $$\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P.$$

Now "dividing through" by dP, while holding temperature constant, yields


 * $$\left(\frac{\partial H}{\partial P}\right)_{T} = T\left(\frac{\partial S}{\partial P}\right)_{T} + V$$

The partial derivative on the left is the isothermal Joule-Thomson coefficient, $$\mu_{\mathrm{T}}$$, and the one on the right can be expressed in terms of the coefficient of thermal expansion via a Maxwell relation. The appropriate relation is
 * $$\left(\frac{\partial S}{\partial P}\right)_{T}= -\left(\frac{\partial V}{\partial T}\right)_{P}= -V\alpha\,$$

where α is the cubic coefficient of thermal expansion. Replacing these two partial derivatives yields


 * $$\mu_{\mathrm{T}} = - T V\alpha\ + V $$.

This expression can now replace $$\mu_{\mathrm{T}}$$ in the earlier equation for $$\mu_{\mathrm{JT}}$$ to obtain


 * $$\mu_{\mathrm{JT}} \equiv \left( \frac{\partial T}{\partial P} \right)_H = \frac{V}{C_{\mathrm{p}}}\left(\alpha T - 1\right)\,$$.

This provides an expression for the Joule–Thomson coefficient in terms of the commonly available properties heat capacity, molar volume, and thermal expansion coefficient. It shows that the Joule–Thomson inversion temperature, at which $$\mu_{\mathrm{JT}}$$ is zero, occurs when the coefficient of thermal expansion is equal to the inverse of the temperature. Since this is true at all temperatures for ideal gases (see expansion in gases), the Joule–Thomson coefficient of an ideal gas is zero at all temperatures.