Joule–Thomson effect 113469 222132610 2008-06-27T18:37:46Z Rracecarr 1284233 /* Derivation of the Joule-Thomson (Kelvin) coefficent */ write out In [[physics]], the '''Joule–Thomson effect''' or '''Joule–Kelvin effect''' describes the [[temperature]] change of a gas or liquid when it is forced through a valve or porous plug while kept insulated so that no heat is exchanged with the environment.<ref name=Perry>{{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}}</ref><ref name=Roy>{{cite book |author=B.N. Roy |title=Fundamentals of Classical and Statistical Thermodynamics |publisher=Wiley |year=2002 |id=ISBN 0-470-84313-6}}</ref><ref name=Edmister>{{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}}</ref> This procedure is called a ''[[throttling process|throttling process]]'' or ''Joule-Thomson process''.<ref>{{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}}</ref> At room temperature, all gases except hydrogen and helium cool upon expansion by the Joule-Thomson process.<ref>{{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}}</ref> The effect is named for [[James Prescott Joule]] and [[William Thomson, 1st Baron Kelvin]] who discovered it in [[1852]] following earlier work by Joule on ''Joule expansion,'' in which a gas undergoes free expansion in a vacuum. ==Description== The ''[[adiabatic process|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 process (thermodynamics)|reversible]], meaning that the gas is in [[thermodynamic equilibrium]] at all times, it is called an ''[[isentropic]]'' expansion. In this scenario, the gas does positive [[mechanical work|work]] during the expansion, and its temperature decreases. *In a [[free expansion]], on the other hand, the gas does no work and absorbs no heat, so the internal energy is conserved. Expanded in this manner, though the temperature of an [[ideal gas]] would remain constant, 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<sub>1</sub> flows into a region of lower pressure P<sub>2</sub> 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|Appendix]]). Temperature change of either sign can occur during the Joule-Thomson process. Each real gas has a [[inversion temperature|Joule–Thomson (Kelvin) inversion temperature]]<ref name=Roy/> 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 above [[room 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 attractive [[force]]s (see ''[[Van der Waals force]]''), expansion causes an increase in the [[potential 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 the [[conservation of energy]]. The increase in potential energy thus implies a decrease in [[kinetic 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 <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_{JT}</math>. This coefficient can be expressed in terms of the gas's volume <math>V</math>, its [[heat capacity#Heat capacity of compressible bodies|heat capacity at constant pressure]] <math>C_{p}</math>, and its [[coefficient of thermal expansion]] <math>\alpha</math> as:<ref name=Perry/><ref name=Edmister/><ref>[http://www.chem.arizona.edu/~salzmanr/480a/480ants/jadjte/jadjte.html Joule Expansion] (by W.R. Salzman, Department of Chemistry, [[University of Arizona]])</ref> :<math>\mu_{JT} \equiv \left( {\partial T \over \partial P} \right)_H = \frac{V}{C_{p}}\left(\alpha T - 1\right)\,</math> See the [[#Derivation of the Joule-Thomson (Kelvin) coefficent|Appendix]] for the proof of this relation. The value of <math>\mu_{JT}</math> is typically expressed in [[Celcius|°C]]/[[bar (unit)|bar]] (SI units: [[Kelvin (unit)|K]]/[[Pascal (unit)|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 <math>\mu_{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 always negative. With that in mind, the following table explains when the Joule–Thomson effect cools or warms a real gas: {| class="wikitable" !If the gas temperature is!!then <math>\mu_{JT}</math> is!!since <math>\partial P</math> is!!thus <math>\partial T</math> must be!!so the gas |- |align=center|below the inversion temperature||align=center|positive||always negative||align=center|negative||align=center|cools |- |align=center|above the inversion temperature||align=center|negative||always negative||align=center|positive||align=center|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 51 K (−222 °C) for helium). Thus, helium and hydrogen warm up 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 °C) and 764 K (491 °C) respectively: these gases can be cooled from room temperature by the Joule–Thomson effect.<ref name=Perry/> For an ideal gas, <math>\mu_{JT}</math> 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 a [[turbine]], for example). The effect is applied in the [[Carl von Linde|Linde technique]] as a standard process in the [[petrochemical industry]], where the cooling effect is used to liquefy gases, and also in many [[cryogenic]] applications (e.g. for the production of liquid oxygen, nitrogen, and [[argon]]). 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, or [[neon]]. ==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<sub>1</sub> in the region at pressure P<sub>1</sub> (region 1) and a volume of V<sub>2</sub> when it appears in the region at pressure P<sub>2</sub> (region 2). Then the work done on the gas by the rest of the gas in region 1 is P<sub>1</sub> V<sub>1</sub>. In region 2 the amount of work done by the gas is P<sub>2</sub> V<sub>2</sub>. So, the total work done by the gas is :<math>P_2 V_2 - P_1 V_1\,</math> The change in internal energy plus the work done by the 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 :<math>E_2 - E_1 + P_2 V_2 - P_1 V_1 = 0\,</math> where <math>E_1</math> and <math>E_2</math> denote the internal energy of the gas in regions 1 and 2, respectively. The above equation then implies that: <math>H_1 = H_2\,</math> where <math>H_1</math> and <math>H_2</math> denote the enthalpy of the gas in regions 1 and 2, respectively. ===Derivation of the Joule-Thomson (Kelvin) coefficent=== A derivation of the formula :<math>\mu_{JT} \equiv \left( \frac{\partial T}{\partial P} \right)_H = \frac{V}{C_{p}}\left(\alpha T - 1\right)\,</math> 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: :<math>dH = T dS + V dP\,</math> (here, <math>S</math> is the [[entropy]] of the gas). Expressing dS in terms of dT and dP gives: :<math>dH = T\left(\frac{\partial S}{\partial T}\right)_{P}dT + \left[V+T\left(\frac{\partial S}{\partial P}\right)_{T}\right] dP\,</math> Using :<math>C_{P}= T\left(\frac{\partial S}{\partial T}\right)_{P}\,</math> (see ''[[Specific heat capacity#Heat capacity of compressible bodies|Specific heat capacity]]''), we can write: :<math>dH = C_{P}dT + \left[V+T\left(\frac{\partial S}{\partial P}\right)_{T}\right] dP\,</math> 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 the [[Gibbs energy]] is given by: :<math>dG = -S dT + V dP\,</math> The [[symmetry of partial derivatives]] of G with respect to T and P implies that: :<math>\left(\frac{\partial S}{\partial P}\right)_{T}= -\left(\frac{\partial V}{\partial T}\right)_{P}= -V\alpha\,</math> where <math>\alpha</math> is the coefficient of thermal expansion. Using this relation, the differential of H can be expressed as :<math>dH = C_{P}dT + V\left(1-T\alpha\right) dP\,</math> Equating dH to zero and solving for dT/dP then gives: <math>\left( \frac{\partial T}{\partial P} \right)_H = \frac{V}{C_{p}}\left(\alpha T - 1\right)\,</math> == See also == * [[Critical temperature]] * [[Ideal gas]] * [[Enthalpy]] and [[Isenthalpic]] * [[Refrigeration]] * [[Reversible process (thermodynamics)]] == References == {{reflist}} == 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. <!-- This link no longer works: * [http://www.nd.edu/~ed/Joule_Thomson/joule_thomson.htm Joule-Thomson effect module] from the University of Notre Dame--> [[Category:Thermodynamics]] [[Category:HVAC]] [[Category:Chemical engineering]] [[Category:Cryogenics]] [[Category:Hydrogen physics]] [[bg:Ефект на Джаул-Томсън]] [[cs:Joule-Thomsonův jev]] [[de:Joule-Thomson-Effekt]] [[es:Efecto Joule-Thomson]] [[fr:Effet Joule-Thomson]] [[it:Effetto Joule-Thomson]] [[ja:ジュール=トムソン効果]] [[pl:Efekt Joule'a Thomsona]] [[ru:Эффект Джоуля — Томсона]] [[uk:Ефект Джоуля-Томсона]] [[zh:焦耳-湯姆生效應]]