Editing Joule–Thomson effect (section)

==Derivation of the Joule–Thomson coefficient==

It is difficult to think physically about what the Joule–Thomson coefficient, <math>\mu_{\mathrm{JT}}</math>, represents. Also, modern determinations of <math>\mu_{\mathrm{JT}}</math> do not use the original method used by Joule and Thomson, but instead measure a different, closely related quantity.<ref>{{cite book|title=Physical Chemistry|last=Atkins|first=Peter|publisher=W.H. Freeman and Co.|year=1997|isbn=978-0-7167-2871-9|location=New York|pages=[https://archive.org/details/physicalchemistr00atki/page/89 89–90]|edition=6th|url=https://archive.org/details/physicalchemistr00atki/page/89}}</ref> Thus, it is useful to derive relationships between <math>\mu_{\mathrm{JT}}</math> and other, more conveniently measured quantities, as described below.

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 [[triple product rule|cyclic rule]]; in terms of these three variables that rule may be written
:<math>\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.</math>

Each of the three partial derivatives in this expression has a specific meaning. The first is <math>\mu_{\mathrm{JT}}</math>, the second is the constant pressure [[heat capacity]], <math>C_{\mathrm{p}}</math>, defined by
:<math>C_{\mathrm{p}} = \left(\frac{\partial H}{\partial T}\right)_P </math>

and the third is the inverse of the ''isothermal Joule–Thomson coefficient'', <math>\mu_{\mathrm{T}}</math>, defined by
:<math>\mu_{\mathrm{T}} = \left(\frac{\partial H}{\partial P}\right)_T </math>.
This last quantity is more easily measured than <math>\mu_{\mathrm{JT}}</math> .<ref>{{cite journal|title=The pressure variation of the heat function as a direct measure of the van der Waals forces|last1=Keyes|first1=F.G.|date=1932|journal=Proc. Natl. Acad. Sci. U.S.A.|doi=10.1073/pnas.18.4.328|pmid=16587688|last2=Collins|first2=S.C.|volume=18|issue=4|pages=328–333|bibcode=1932PNAS...18..328K|pmc=1076221|doi-access=free}}</ref><ref>{{cite journal|title=A flow calorimeter for the measurement of the isothermal Joule–Thomson coefficient of gases at elevated temperatures and pressures. Results for nitrogen at temperatures up to 473 K and pressures up to 10 MPa and for carbon dioxide at temperatures up to 500 K and pressures up to 5 MPa|last1=Cusco|first1=L.|date=1995|journal= The Journal of Chemical Thermodynamics|doi=10.1006/jcht.1995.0073|last2=McBain|first2=S.E.|last3=Saville|first3=G.|volume=27|issue=7|pages=721–733}}</ref> Thus, the expression from the cyclic rule becomes
:<math>\mu_{\mathrm{JT}} = - \frac{\mu_{\mathrm{T}}} {C_p}.</math>

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

To proceed further, the starting point is the [[fundamental thermodynamic relation|fundamental equation of thermodynamics]] in terms of enthalpy; this is
:<math>\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P.</math>

Now "dividing through" by d''P'', while holding temperature constant, yields
:<math>\left(\frac{\partial H}{\partial P}\right)_T = T\left(\frac{\partial S}{\partial P}\right)_T + V</math>
The partial derivative on the left is the isothermal Joule–Thomson coefficient, <math>\mu_{\mathrm{T}}</math>, 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
:<math>\left(\frac{\partial S}{\partial P}\right)_T= -\left(\frac{\partial V}{\partial T}\right)_P= -V\alpha\,</math>
where ''α'' is the cubic [[coefficient of thermal expansion]]. Replacing these two partial derivatives yields
:<math>\mu_{\mathrm{T}} = - T V\alpha\ + V. </math>

This expression can now replace <math>\mu_{\mathrm{T}}</math> in the earlier equation for <math>\mu_{\mathrm{JT}}</math> to obtain:
:<math>\mu_{\mathrm{JT}} \equiv \left( \frac{\partial T}{\partial P} \right)_H = \frac V {C_{\mathrm{p}}} (\alpha T - 1).\,</math>

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 <math>\mu_{\mathrm{JT}}</math> 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 [[thermal expansion#Isobaric expansion in ideal gases|expansion in gases]]), the Joule–Thomson coefficient of an ideal gas is zero at all temperatures.<ref>{{cite book|title=Thermodynamics|url=https://archive.org/details/thermodynamicsin00call|url-access=registration|last=Callen|first=H.B|publisher=John Wiley and Sons|year=1960|location=New York|pages=[https://archive.org/details/thermodynamicsin00call/page/112 112–114]}}</ref>