- Joule–Thomson effect
In

physics , the**Joule–Thomson effect**or**Joule–Kelvin effect**describes thetemperature change of a gas or liquid when it is forced through avalve or porous plug while kept insulated so that no heat is exchanged with the environment.cite book |author=R. H. Perry, D. W. Green |title=Perry's Chemical Engineers' Handbook |publisher=McGraw-Hill Book Co. |year=1984 |isbn=0-07-049479-7] cite book |author=B.N. Roy |title=Fundamentals of Classical and Statistical Thermodynamics |publisher=Wiley |year=2002 |id=ISBN 0-470-84313-6] cite book|author=W. C. Edmister, B. I. Lee |title=Applied Hydrocarbon Thermodynamics|edition= 2nd edition |Volume=Vol. 1 |publisher=Gulf Publishing |year=1984 |isbn=0-87201-855-5] This procedure is called a "throttling process " or "Joule-Thomson process". [*cite book |author=F. Reif. |title=Fundamentals of Statistical and Thermal Physics | chapter=Chapter 5 – Simple applications of macroscopic thermodynamics |publisher=McGraw-Hill |year=1965 |isbn=07-051800-9*] At room temperature, all gases excepthydrogen ,helium andneon cool upon expansion by the Joule-Thomson process. [*cite book |author=A. W. Adamson|title=A textbook of Physical Chemistry |edition=1st Edition |chapter=Chapter 4 – Chemical thermodynamics. The First Law of Thermodynamics|publisher=Academic press|year=1973*] [*cite book |author=G.W. Castellan |title=Physical Chemistry |edition=2nd Edition |chapter=Chapter 7 – Energy and the First Law of Thermodynamics; Thermochemistry |publisher=Addison-Wesley |year=1971*]The effect is named for

James Prescott Joule andWilliam Thomson, 1st Baron Kelvin who discovered it in1852 following earlier work by Joule on "Joule expansion," in which a gas undergoes free expansion in avacuum .**Description**The "adiabatic" (no heat exchanged) expansion of a gas may be carried out in a number of ways. The change in temperature experienced by the gas during expansion depends on the initial and final pressure, but also on the manner in which the expansion is carried out.

*If the expansion process is reversible, meaning that the gas is inthermodynamic equilibrium at all times, it is called an "isentropic " expansion. In this scenario, the gas does positive work during the expansion, and its temperature decreases.

*In afree expansion , on the other hand, the gas does no work and absorbs no heat, so the internal energy is conserved. Expanded in this manner, the temperature of anideal gas would remain constant, but the temperature of a real gas may either increase or decrease, depending on the initial temperature and pressure.

*The method of expansion discussed in this article, in which a gas or liquid at pressure P_{1}flows into a region of lower pressure P_{2}via a valve or porous plug under steady state conditions and without change in kinetic energy, is called the Joule–Thomson process. During this process,enthalpy remains unchanged (see Appendix).Temperature change of either sign can occur during the Joule-Thomson process. Each real gas has a Joule–Thomson (Kelvin) inversion temperature above which expansion at constant enthalpy causes the temperature to rise, and below which such expansion causes cooling. This inversion temperature depends on pressure; for most gases at

atmospheric pressure , the inversion temperature is aboveroom temperature , so most gases can be cooled from room temperature by isenthalpic expansion.**Physical mechanism**As a gas expands, the average distance between

molecule s grows. Because of intermolecular attractiveforce s (see "Van der Waals force "), expansion causes an increase in thepotential energy of the gas. If no external work is extracted in the process and no heat is transferred, the total energy of the gas remains the same because of theconservation of energy . The increase in potential energy thus implies a decrease inkinetic energy and therefore in temperature.A second mechanism has the opposite effect. During gas molecule collisions, kinetic energy is temporarily converted into potential energy. As the average intermolecular distance increases, there is a drop in the number of collisions per time unit, which causes a decrease in average potential energy. Again, total energy is conserved, so this leads to an increase in kinetic energy (temperature). Below the Joule–Thomson inversion temperature, the former effect (work done internally against intermolecular attractive forces) dominates, and free expansion causes a decrease in temperature. Above the inversion temperature, gas molecules move faster and so collide more often, and the latter effect (reduced collisions causing a decrease in the average potential energy) dominates: Joule-Thomson expansion causes a temperature increase.

**The Joule–Thomson (Kelvin) coefficient**The rate of change of temperature $T$ with respect to pressure $P$ in a Joule–Thomson process (that is, at constant enthalpy $H$) is the "Joule–Thomson (Kelvin) coefficient" $mu\_\{JT\}$. This coefficient can be expressed in terms of the gas's volume $V$, its heat capacity at constant pressure $C\_\{p\}$, and its

coefficient of thermal expansion $alpha$ as: [*[*]*http://www.chem.arizona.edu/~salzmanr/480a/480ants/jadjte/jadjte.html Joule Expansion*] (by W.R. Salzman, Department of Chemistry,University of Arizona ):$mu\_\{JT\}\; equiv\; left(\; \{partial\; T\; over\; partial\; P\}\; ight)\_H\; =\; frac\{V\}\{C\_\{pleft(alpha\; T\; -\; 1\; ight),$

See the Appendix for the proof of this relation. The value of $mu\_\{JT\}$ is typically expressed in °C/bar (SI units: K/Pa) and depends on the type of gas and on the temperature and pressure of the gas before expansion.

All real gases have an "inversion point" at which the value of $mu\_\{JT\}$ 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 $partial\; P$ is always negative. With that in mind, the following table explains when the Joule–Thomson effect cools or warms a real gas:

Helium andhydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of one atmosphere are very low (e.g., about 51 K (−222 °C) for helium). Thus, helium and hydrogen warm up when expanded at constant enthalpy at typical room temperatures. On the other handnitrogen andoxygen , the two most abundant gases in air, have inversion temperatures of 621 K (348 °C) and 764 K (491 °C) respectively: these gases can be cooled from room temperature by the Joule–Thomson effect.For an ideal gas, $mu\_\{JT\}$ is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.

**Applications**In practice, the Joule–Thomson effect is achieved by allowing the gas to expand through a throttling device (usually a

valve ) which must be very well insulated to prevent any heat transfer to or from the gas. No external work is extracted from the gas during the expansion (the gas must not be expanded through aturbine , for example).The effect is applied in the Linde technique as a standard process in the

petrochemical industry , where the cooling effect is used to liquefy gases, and also in manycryogenic applications (e.g. for the production of liquid oxygen, nitrogen, andargon ). Only when the Joule–Thomson coefficient for the given gas at the given temperature is greater than zero can the gas be liquefied at that temperature by the Linde cycle. In other words, a gas must be below its inversion temperature to be liquefied by the Linde cycle. For this reason, simple Linde cycle liquefiers cannot normally be used to liquefy helium, hydrogen, orneon .**Appendix****Proof that enthalpy remains constant in a Joule-Thomson process**In a Joule-Thomson process the

enthalpy remains constant. To prove this, the first step is to compute the net work done by the gas that moves through the plug. Suppose that the gas has a volume of V_{1}in the region at pressure P_{1}(region 1) and a volume of V_{2}when it appears in the region at pressure P_{2}(region 2). Then the work done on the gas by the rest of the gas in region 1 isP_{1}V_{1}. In region 2 the amount of work done by the gas is P_{2}V_{2}. So, the total work done by the gas is:$P\_2\; V\_2\; -\; P\_1\; V\_1,$

The change in internal energy plus the work done by the gas is, by the first law of thermodynamics, the total amount of heat absorbed by the gas (here it is assumed that there is no change in kinetic energy). In the Joule-Thompson process the gas is kept insulated, so no heat is absorbed. This means that

:$E\_2\; -\; E\_1\; +\; P\_2\; V\_2\; -\; P\_1\; V\_1\; =\; 0,$

where $E\_1$ and $E\_2$ denote the internal energy of the gas in regions 1 and 2, respectively.

The above equation then implies that:

$H\_1\; =\; H\_2,$

where $H\_1$ and $H\_2$ denote the enthalpy of the gas in regions 1 and 2, respectively.

**Derivation of the Joule-Thomson (Kelvin) coefficent**A derivation of the formula :$mu\_\{JT\}\; equiv\; left(\; frac\{partial\; T\}\{partial\; P\}\; ight)\_H\; =\; frac\{V\}\{C\_\{pleft(alpha\; T\; -\; 1\; ight),$

for the Joule–Thomson (Kelvin) coefficient.

The partial derivative of T with respect to P at constant H can be computed by expressing the differential of the enthalpy dH in terms of dT and dP, and equating the resulting expression to zero and solving for the ratio of dT and dP.

It follows from the

fundamental thermodynamic relation that the differential of the enthalpy is given by::$dH\; =\; T\; dS\; +\; V\; dP,$ (here, $S$ is the

entropy of the gas).Expressing dS in terms of dT and dP gives:

:$dH\; =\; Tleft(frac\{partial\; S\}\{partial\; T\}\; ight)\_\{P\}dT\; +\; left\; [V+Tleft(frac\{partial\; S\}\{partial\; P\}\; ight)\_\{T\}\; ight]\; dP,$

Using

:$C\_\{P\}=\; Tleft(frac\{partial\; S\}\{partial\; T\}\; ight)\_\{P\},$ (see "Specific heat capacity"), we can write:

:$dH\; =\; C\_\{P\}dT\; +\; left\; [V+Tleft(frac\{partial\; S\}\{partial\; P\}\; ight)\_\{T\}\; ight]\; dP,$

The remaining partial derivative of S can be expressed in terms of the coefficient of thermal expansion via a

Maxwell relation as follows. From the fundamental thermodynamic relation, it follows that the differential of theGibbs energy is given by::$dG\; =\; -S\; dT\; +\; V\; dP,$

The

symmetry of partial derivatives of G with respect to T and P implies that::$left(frac\{partial\; S\}\{partial\; P\}\; ight)\_\{T\}=\; -left(frac\{partial\; V\}\{partial\; T\}\; ight)\_\{P\}=\; -Valpha,$

where $alpha$ is the coefficient of thermal expansion. Using this relation, the differential of H can be expressed as

:$dH\; =\; C\_\{P\}dT\; +\; Vleft(1-Talpha\; ight)\; dP,$

Equating dH to zero and solving for dT/dP then gives:

$left(\; frac\{partial\; T\}\{partial\; P\}\; ight)\_H\; =\; frac\{V\}\{C\_\{pleft(alpha\; T\; -\; 1\; ight),$

**See also***

Critical temperature

*Ideal gas

*Enthalpy andIsenthalpic

*Refrigeration

*Reversible process (thermodynamics) **References****Bibliography***cite book | author=Mark W. Zemansky | title=Heat and Thermodynamics; an intermediate textbook| publisher=McGraw-Hill | year=1968 | id= LCCN 67026891, "p."182, 335

*cite book | author=Daniel V. Schroeder|title=An Introduction to Thermal Physics|publisher=Addison Wesley Longman|year=2000| id=ISBN 0-201-38027-7 , "p."142

*cite book | author=Charles Kittel and Herbert Kroemer|title= Thermal Physics|publisher=W.H. Freeman and Co.|year=1980|id=ISBN 0-7167-1088-9**External links*** [

*http://scienceworld.wolfram.com/physics/Joule-ThomsonProcess.html Joule-Thomson process*] from Eric Weisstein's World of Physics

* [*http://scienceworld.wolfram.com/physics/Joule-ThomsonCoefficient.html Joule-Thomson coefficient*] from Eric Weisstein's World of Physics

* [*http://www.britannica.com/eb/article?tocId=9044025&query=Joule-Thomson%20effect&ct= Joule-Thomson effect*] from the truncated free online version of the Encyclopedia Britannica.

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