1.14.38: Joule-Thomson Coefficient
An important property of a given gas is its Joule-Thomson coefficient [1-3]. These coefficients are important from two standpoints;
- intermolecular interaction, and
- liquefaction of gases.
A given closed system contains one mole of gaseous chemical substance \(\mathrm{j}\) at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\). The molar enthalpy of the gas \(\mathrm{H}_{\mathrm{j}}\) describes its molar enthalpy defined by equation (a).
\[\mathrm{H}_{\mathrm{j}}=\mathrm{H}_{\mathrm{j}}[\mathrm{T}, \mathrm{p}] \nonumber \]
Then,
\[\mathrm{dH}_{\mathrm{j}}=\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{p}} \, \mathrm{dT}+\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{p}}\right)_{\mathrm{T}} \, \mathrm{dp} \nonumber \]
Hence at constant enthalpy, \(\mathrm{H}\),
\[\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{p}} \, \mathrm{dT}=-\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{p}}\right)_{\mathrm{T}} \, \mathrm{dp} \nonumber \]
Or,
\[\left(\frac{\partial T}{\partial p}\right)_{H}=-\left(\frac{\partial H_{\mathrm{j}}}{\partial p}\right)_{T} \,\left(\frac{\partial T}{\partial H_{j}}\right)_{p} \nonumber \]
The Joule-Thomson coefficient for gas \(\mathrm{j}\), \(\mu_[\mathrm{j}}\) is defined by equation (e).
\[\mu_{\mathrm{j}}=\left(\frac{\partial T}{\partial p}\right)_{\mathrm{H}(\mathrm{j})} \nonumber \]
For all gases (except helium and hydrogen) at \(298 \mathrm{~K}\) and moderate pressures \(\mu_{\mathrm{j} > 0\). At room temperature and ambient pressure, \(\mu_{\mathrm{j}}\) is \(0.002 \mathrm{~K Pa}^{-1}\) for nitrogen and \(0.025 \mathrm{~K Pa}^{-1}\) for 2,2-dimethylpropane [3].
Further the isobaric heat capacity for chemical substance \(\mathrm{j}\) is defined by equation (f).
\[\mathrm{C}_{\mathrm{pj}}=\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{p}} \nonumber \]
Hence from equations (d), (e) and (f),
\[\mu_{\mathrm{j}}=-\frac{(\partial \mathrm{H} / \partial \mathrm{p})_{\mathrm{T}}}{\mathrm{C}_{\mathrm{pj}}} \nonumber \]
Then,
\[\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{p}}\right)_{\mathrm{T}}=-\mu_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}} \nonumber \]
Equation (h) marks an important stage in the analysis. For example, \(\mathrm{C}_{\mathrm{pj}} > 0\). From the definition of enthalpy \(\mathrm{H}_{\mathrm{j}}\),
\[\mathrm{U}_{\mathrm{j}}=\mathrm{H}_{\mathrm{j}}-\mathrm{p} \, \mathrm{V}_{\mathrm{j}} \nonumber \]
Equation (i) is differentiated with respect to \(\mathrm{V}_{\mathrm{j}} at fixed \(\mathrm{T}\). Thus,
\[
\[\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}}=\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}}-\mathrm{V}_{\mathrm{j}} \,\left(\frac{\partial \mathrm{p}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}}-\mathrm{p} \nonumber \]
Or,
\[\left(\frac{\partial U_{j}}{\partial V_{j}}\right)_{T}=\left(\frac{\partial p}{\partial V_{j}}\right)_{T} \,\left[\left(\frac{\partial H_{j}}{\partial V_{j}}\right)_{T} \,\left(\frac{\partial V_{j}}{\partial p}\right)_{T}-V_{j}\right]-p \nonumber \]
Then,
\[\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}}=\left(\frac{\partial \mathrm{p}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{T} \,\left[\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{p}}\right)_{\mathrm{T}}-\mathrm{V}_{\mathrm{j}}\right]-\mathrm{p} \nonumber \]
Using equation (h),
\[\left(\frac{\partial U_{j}}{\partial V_{j}}\right)_{T}=-\left(\frac{\partial p}{\partial V_{j}}\right)_{T} \,\left[\mu_{j} \, C_{p j}+V_{j}\right]-p \nonumber \]
An important application of equation (m) concerns the case where chemical substance \(\mathrm{j}\) is a perfect gas. In this case,
\[\mathrm{p} \, \mathrm{V}_{\mathrm{j}}=\mathrm{R} \, \mathrm{T} \nonumber \]
Or,
\[\mathrm{p}=\mathrm{R} \, \mathrm{T} \, \frac{1}{\mathrm{~V}_{\mathrm{j}}} \nonumber \]
Hence,
\[\left(\frac{\partial p}{\partial V_{j}}\right)_{T}=-R \, T \, \frac{1}{V_{j}^{2}}=-\frac{p}{V_{j}} \nonumber \]
Then from equation (m),
\[\left(\frac{\partial U_{j}}{\partial V_{j}}\right)_{T}=\frac{p}{V_{j}} \,\left[\mu_{j} \, C_{p j}+V_{j}\right]-p \nonumber \]
Or,
\[\left(\frac{\partial U_{j}}{\partial V_{j}}\right)_{T}=\frac{p \, \mu_{j} \, C_{p j}}{V_{j}} \nonumber \]
But
\[\operatorname{limit}(\mathrm{p} \rightarrow 0) \frac{\mathrm{p} \, \mu_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}}}{\mathrm{V}_{\mathrm{j}}}=0 \nonumber \]
Then
\[\operatorname{limit}(\mathrm{p} \rightarrow 0)\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}}=0 \nonumber \]
A definition of a perfect gas is that \(\left(\frac{\partial \mathrm{U}_{\mathrm{j}}}{\partial \mathrm{V}_{\mathrm{j}}}\right)_{\mathrm{T}}\) is zero. Then all real gases are perfect in the \(\operatorname{limit}(\mathrm{p} \rightarrow 0)\).
Footnotes
[1] James Prescott Joule(12818-1889) William Thomson (1824-1907); Later Lord Kelvin Some authors refer to the Joule-Thomson coefficient; e.g. E. B. Smith, Basic Chemical Thermodynamics, Clarendon Press, Oxford, 1982, 3rd. edn., page 119. Other authors refer to the Joule –Kelvin Effect; e.g. E. F. Caldin, Chemical Thermodynamics, Clarendon Press, Oxford, 1958,page 81. Other authors refer to either the Joule-Thomson or Joule-Kelvin Effect; e.g. M. H. Everdell, Introduction to Chemical Thermodynamics, English Universities Press, London 1965, page 57.
[2] M. L. McGlashan, Chemical Thermodynamics, Academic Press, London 1979, page 94.
[3] Benjamin Thompson (1753-1814); later Count von Rumford, married Lavoisier’s widow.