Editing Joule–Thomson effect (section)

==The Joule–Thomson (Kelvin) coefficient==
[[File:Joule-Thomson curves 2.svg|thumb|400px|Fig. 1 – Joule–Thomson coefficients for various gases at atmospheric pressure]]

The rate of change of temperature <math>T</math> with respect to pressure <math>P</math> in a Joule–Thomson process (that is, at constant enthalpy <math>H</math>) is the ''Joule–Thomson (Kelvin) coefficient'' <math>\mu_{\mathrm{JT}}</math>. This coefficient can be expressed in terms of the gas's specific volume <math>V</math>, its [[specific heat capacity#Heat capacity of compressible bodies|heat capacity at constant pressure]] <math>C_{\mathrm{p}}</math>, and its [[coefficient of thermal expansion]] <math>\alpha</math> as:<ref name=Perry/><ref name=Edmister/><ref>{{cite web|url=http://www.chem.arizona.edu/~salzmanr/480a/480ants/jadjte/jadjte.html|title=Joule Expansion|author=W.R. Salzman|publisher=Department of Chemistry, [[University of Arizona]]|access-date=23 July 2005|archive-url=https://web.archive.org/web/20120613235628/http://www.chem.arizona.edu/~salzmanr/480a/480ants/jadjte/jadjte.html|archive-date=13 June 2012|url-status=dead}}</ref>
:<math>\mu_{\mathrm{JT}} = \left( {\partial T \over \partial P} \right)_H = \frac V {C_{\mathrm{p}}}(\alpha T - 1)\,</math>

See the {{section link||Derivation of the Joule–Thomson coefficient}} below for the proof of this relation. The value of <math>\mu_{\mathrm{JT}}</math> is typically expressed in °C/[[bar (unit)|bar]] (SI units: [[kelvin|K]]/[[pascal (unit)|Pa]]) and depends on the type of gas and on the temperature and pressure of the gas before expansion. Its pressure dependence is usually only a few percent for pressures up to 100 bar.

All real gases have an ''inversion point'' at which the value of <math>\mu_{\mathrm{JT}}</math> changes sign. The temperature of this point, the ''Joule–Thomson inversion temperature'', depends on the pressure of the gas before expansion.

In a gas expansion the pressure decreases, so the sign of <math>\partial P</math> is negative by definition. With that in mind, the following table explains when the Joule–Thomson effect cools or warms a real gas:

{| class="wikitable" style="text-align:center;"
! If the gas temperature is
! then <math>\mu_\text{JT}</math> is
! since <math>\partial P</math> is
! thus <math>\partial T</math> must be
! so the gas
|-
| below the inversion temperature
| positive
| rowspan=2 | always negative
| negative
| cools
|-
| above the inversion temperature
| negative
| positive
| warms
|}

[[Helium]] and [[hydrogen]] are two gases whose Joule–Thomson inversion temperatures at a pressure of one [[atmosphere (unit)|atmosphere]] are very low (e.g., about 40 K, −233&nbsp;°C for helium). Thus, helium and hydrogen warm when expanded at constant enthalpy at typical room temperatures. On the other hand, [[nitrogen]] and [[oxygen]], the two most abundant gases in air, have inversion temperatures of 621 K (348&nbsp;°C) and 764 K (491&nbsp;°C) respectively: these gases can be cooled from room temperature by the Joule–Thomson effect.<ref name=Perry/><ref name=":Atkins1" />

For an ideal gas, <math>\mu_\text{JT}</math> is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.

=== Theoretical models ===
[[File:Dieterici gas inversion temperature plot.png|thumb|For the Dieterici gas, the relation between reduced pressure and reduced inversion temperature is <math>\tilde p = (8-\tilde T_I) e^{\frac 52 - \frac 4{8-\tilde T_I}}</math>. Reproduced from Fig. 17 of <ref name=":0" />.]]
For a [[Van der Waals equation|Van der Waals gas]], the coefficient is<ref>{{Cite web |date=2017-01-25 |title=10.3: The Joule-Thomson Experiment |url=https://phys.libretexts.org/Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book%3A_Heat_and_Thermodynamics_(Tatum)/10%3A_The_Joule_and_Joule-Thomson_Experiments/10.03%3A_The_Joule-Thomson_Experiment |access-date=2023-07-05 |website=Physics LibreTexts |language=en}}</ref><math display="block">\mu_\text{JT}=-\frac{V_m}{C_{p}} \frac{R T V_m^{2} b-2 a(V_m-b)^{2}}{R T V_m^{3}-2 a(V_m-b)^{2}}.</math>with inversion temperature <math>\frac{2a}{bR}\left(1 - \frac{b}{V_m}\right)^2</math>.


For the [[Real gas#Dieterici model|Dieterici gas]], the reduced inversion temperature is <math>\tilde T_I = 8 - 4/\tilde V_m</math>, and the relation between reduced pressure and reduced inversion temperature is <math>\tilde p = (8-\tilde T_I) e^{\frac 52 - \frac 4{8-\tilde T_I}}</math>. This is plotted on the right. The critical point falls inside the region where the gas cools on expansion. The outside region is where the gas warms on expansion.<ref name=":0">{{Cite book |last=Pippard |first=Alfred B. |title=Elements of classical thermodynamics: for advanced students of physics |date=1981 |publisher=Univ. Pr |isbn=978-0-521-09101-5 |edition=Repr |location=Cambridge |pages=74–77}}</ref>