Clausius Clapeyron Equation Assumptions And Critical Thinking

"Clapeyron equation" and "Clapeyron's equation" redirect here. For a state equation, see ideal gas law.

The Clausius–Clapeyron relation, named after Rudolf Clausius[1] and Benoît Paul Émile Clapeyron,[2] is a way of characterizing a discontinuous phase transition between two phases of matter of a single constituent. On a pressure–temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of the tangents to this curve. Mathematically,

where is the slope of the tangent to the coexistence curve at any point, is the specific latent heat, is the temperature, is the specific volume change of the phase transition, and is the specific entropy change of the phase transition.

Derivations[edit]

Derivation from state postulate[edit]

Using the state postulate, take the specific entropy for a homogeneous substance to be a function of specific volume and temperature.[3]:508

The Clausius–Clapeyron relation characterizes behavior of a closed system during a phase change, during which temperature and pressure are constant by definition. Therefore,[3]:508

Using the appropriate Maxwell relation gives[3]:508

where is the pressure. Since pressure and temperature are constant, by definition the derivative of pressure with respect to temperature does not change.[4][5]:57, 62 & 671 Therefore, the partial derivative of specific entropy may be changed into a total derivative

and the total derivative of pressure with respect to temperature may be factored out when integrating from an initial phase to a final phase ,[3]:508 to obtain

where and are respectively the change in specific entropy and specific volume. Given that a phase change is an internally reversible process, and that our system is closed, the first law of thermodynamics holds

where is the internal energy of the system. Given constant pressure and temperature (during a phase change) and the definition of specific enthalpy, we obtain

Given constant pressure and temperature (during a phase change), we obtain[3]:508

Substituting the definition of specific latent heat gives

Substituting this result into the pressure derivative given above (), we obtain[3]:508[6]

This result (also known as the Clapeyron equation) equates the slope of the tangent to the coexistence curve, at any given point on the curve, to the function of the specific latent heat , the temperature , and the change in specific volume .

Derivation from Gibbs–Duhem relation[edit]

Suppose two phases, and , are in contact and at equilibrium with each other. Their chemical potentials are related by

Furthermore, along the coexistence curve,

One may therefore use the Gibbs–Duhem relation

(where is the specific entropy, is the specific volume, and is the molar mass) to obtain

Rearrangement gives

from which the derivation of the Clapeyron equation continues as in the previous section.

Ideal gas approximation at low temperatures[edit]

When the phase transition of a substance is between a gas phase and a condensed phase (liquid or solid), and occurs at temperatures much lower than the critical temperature of that substance, the specific volume of the gas phase greatly exceeds that of the condensed phase . Therefore, one may approximate

at low temperatures. If pressure is also low, the gas may be approximated by the ideal gas law, so that

where is the pressure, is the specific gas constant, and is the temperature. Substituting into the Clapeyron equation

we can obtain the Clausius–Clapeyron equation[3]:509

for low temperatures and pressures,[3]:509 where is the specific latent heat of the substance.

Let and be any two points along the coexistence curve between two phases and . In general, varies between any two such points, as a function of temperature. But if is constant,

or[5]:672

These last equations are useful because they relate equilibrium or saturation vapor pressure and temperature to the latent heat of the phase change, without requiring specific volume data.

Applications[edit]

Chemistry and chemical engineering[edit]

For transitions between a gas and a condensed phase with the approximations described above, the expression may be rewritten as

where is a constant. For a liquid-gas transition, is the specific latent heat (or specific enthalpy) of vaporization; for a solid-gas transition, is the specific latent heat of sublimation. If the latent heat is known, then knowledge of one point on the coexistence curve determines the rest of the curve. Conversely, the relationship between and is linear, and so linear regression is used to estimate the latent heat.

Meteorology and climatology[edit]

Atmosphericwater vapor drives many important meteorologic phenomena (notably precipitation), motivating interest in its dynamics. The Clausius–Clapeyron equation for water vapor under typical atmospheric conditions (near standard temperature and pressure) is

A typical phase diagram. The dotted green line gives the anomalous behavior of water. The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along phase boundaries.

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PROFESSOR: Let's do aoneminutereview,andthenmoveontotheClausius-Clapeyronequationandseehowfarwecangoonthat.So,remindingyouthen,whatyoulearnedlasttimeaboutphasetransitionsis,you drew aphasediagramintemperaturepressurespacehere.Andthenyouhadatriplepointsomewhere,which was auniquepoint.Thetriplepointtemperature,triplepointpressure.Thentherewasagas,solid,call itcoexistenceline.Whichkeptongoingadinfinitum.Andthea gasliquidcoexistentline.With acriticalpoint.Sothatallowedyoutogoaround.Youwantedtogofromliquidtogas,youcouldactuallygoaroundthiscriticalpoint, andnever actuallysee aphasetransition.Thentherewasasolidliquidcoexistence line.Whichusuallyhasanegativeslope,exceptforthetwomostimportantsubstancesonEarth,whicharewaterand silicon.Soforwaterandsilicon,wehavepositiveslope.H2Oandsilicon.Thisiswhythere'slifeonEarth.

Didyouhear thestoryof why there'slifeonEarth?This isthe secret oflife.Withoutthis,thisiswhywe'rehere.Becauseofthisproperty.So,thereasonwhytheseslopescomefromthe,soI'lltellyouthesecretoflife.Butfirstletmeremindyouwhatthecoexistencecurveis.dp/dT.CoexistenceisdeltaSoverdeltaV,ordeltaHoverTdeltaV.Andthat'stheClapeyronequation.Now,theslopeofthecurvespointsusby,look atthissolidliquidline,deltaS,forsolidliquid.Ilooked at it uphere.It'sgoingtobeSliquidminusSsolid.Andyouknowthat'sgreaterthanzerobecausetheentropyintheliquidstateismuchbiggerthanit isin thesolid state.You have acrystal,entropy's verysmall.Liquidstate,muchmoredisorder.Got apositivesign here.Yes.Questions?

STUDENT: [INAUDIBLE]

PROFESSOR: So,thisisalmosteverything.Ithasthisslopehere.Whichisanegativeslope.Andwaterhasapositiveslope.Right?Isthatwrong?It'swrong.Let'sgothroughtheargument.All right.All right.Let'sgothroughthe argument.So,thenyouhavedeltaV,anddeltaVisVsolidminus Vliquid,this isper mole.Now,almosteverysubstancehasVsolidless thanVliquid.Sothisisnegative.Soalmosteverysubstance,you havedeltaSdividedbydeltaV isnegative.Anegativeslope.You'reright.SoIhave itbackwards.Idohaveitbackwards.OK.See,Iwasreallytryingtomakeit sothat mydrawingwasright.Whichiswhy Iinverted thesolid andliquidhere. So Ireallywantedtoberight,butIendedupbeingwrong.Liquidis lessthan solid.Wellthisiswater.Why do Ihave it backwardsin my notes. I've got tocorrectthat.Sothisismostsubstances,ispositive.OKso,andmostsubstances,OK?Exceptforwaterandsilicon.Becauseforwaterandsiliconthisisreversed.Becausethemolarvolumeofsolid forwaterisbiggerthanthatforliquid.So this isbigger thanhere.Thank you. OK.Thanksforcatchingit.

OK,sowhathappensifthemolarvolumefortheliquidisbiggerthanthemolarvolumeofthesolid.Thedensityofwaterwhichistheinverseofthemolevolume,sothedensity,themolardensity,is equaltooneoverthemolarvolume.Soyou've gotthe molardensityforliquid isthensmallerthanthemolar densityfor thesolid.It'soneover,right?Sothat'swhyicefloats.Becauseit'slessdensethanliquidwater.Sowhatdoyouthinkwouldhappeninthewinterificedidn'tfloat,intheCharlesRiver.Itwouldgotothebottom.It wouldfreeze,solidandyouwouldhavenofish.You wouldhavenolife. Itwouldfreezesolidandthatwouldbetheendoflife.Andyou know,fivehundredmillionyearsagotherewouldbenofish.Therewouldbenodinosaurs.Therewouldbenoanything,there wouldbe nohumans.That'sthesecret of life.Andthefactthatthisisalsothecase isveryusefulformakingsilicon.Forprocessingsilicon.Anotherthesecretoflife.Wherewould webewithoutsilicon?Ourcivilizationwould be,we'dstillbeusingstonesandthings,right?OK,so it'ssupersuperimportant.

Alright,sothisisaClapeyronstory.Whichweendedupgettingright.EventhoughIinsistedongettingitwrong.SowhathappensnowisMr.Clausiuscamearoundandherealizedthatyoucanmakesomeapproximationsthatareveryuseful.Sothefirstapproximationis,youcanrealizethatthemolarvolumeofthegasisalwaysmuchbiggerthanthevolumeofthesolidortheliquid.AndsoasaresultwheneveryouhavethesedeltaV's,forsublimation,ordeltaVforvaporization,whichisthevolumethegasminusthevolumeofthesolid.Orthevolumeofthegasminusthevolumeoftheliquid,youmightaswellignorethevolume oftheliquidorofthesolid.Andthiscanjust beequaltoroughlythevolumeofthegas.Sothisisthefirstapproximationhemade.So now, ifyougobacktothetotheClapeyronequationup here,withthisapproximationthendp/dT,thecoexistenceline,isdeltaHdividedbyTVgas.Forsublimation.Orvaporization.Youdon'thavethedeltaVdownthereany more.

Thenthenextthingthat,nextassumptionthatherealizedyoucanmakewas,well,allthesegasescould,they'relikeidealgases.Sowecan,insteadofhavingthevolumeofthegashere,we canuse theidealgaslaw.Thisisall,let'sdoallthispermole.SoVisequaltoRToverp.Youcanplugthisbackinhere.AndthenyouendupwiththeClausius-Clapeyronequation.Sothisisapproximationnow.WeputinRToverp forhere.WehavepdeltaH,wherethisiseithersublimationorvaporizationdividedbyRTsquared.dp/dT,sublimationorvaporization,andthis istheClausius-Clapeyronequation.Twoimportantapproximationsthatgoinhere.Andit'snotvalidforsolids orliquid.Youhaveto havea gasintherebecauseoftheidealgasfor theapproximationthatgoesinhere.

Andonceyouhavethat,youcanintegratebothsides.You'vegottoputthetemperaturesononeside.Putthepressuresontheotherside.I'll go ahead andcover this uphere.So you haveoneover pdp/dTisequaltodeltaH overRTsquared.Oneoverpdp/dTthat'sjustlikedlogp.Andthat'sanotherformthatyou'regoing toseeoftenfortheClausius-Clapeyronapproximationisdlogp /dTisequaltodeltaHoverRTsquared.AndI'mfreelydroppingthesublimationand thevaporizationbecauseweknowthat'swhatImeanhere.It'sonlyvalidforthosetwolines.Sothat'sanotherformthatyou'llsee.Andthat'snotdThere,that'sdlogp,dlogpdp.Yeah,that'sright,d logp / dT.And sonowyoucantakethedTtotheothersideandintegrate.Frompoint onetopoint twodlogp,isfrompoint onetopoint two,deltaHoverRTsquareddT.Andyoumaketheusualapproximation,whichwe'vemadebefore.ThatdeltaHisroughlyindependentoftemperature, orveryslowlychangingwithtemperature.Andthat'sfineaslongasyou'reinanarrowtemperaturerange.WhichisusuallythecasewhenyouhavetheseClausius-Clapeyronequationproblems.So youcantakethedeltaHoutoftheintegral,theRout oftheintegral.Andyoucanintegratebothsides.Togiveyoulogp2minuslogp1,isdelta HtimesR,dividedbyR, deltaHdividedbyR.Delta H overRtimesT2minusT1overT1T2.

Itlooksalotlikeandthetemperaturedependenceequilibriumconstant.Andinfact,butit'snot,right?Andsomepeopleusethatequationhereinsteadofequilibriumconstant.Temperaturedependence.That's,cangiverisetoproblems.Becauseobviouslythey'renotthesameequation.Butroughlyhasthesameform.So,anotherwaythatyou'llalsofindthiswrittenisveryoftenp1andT1willbesomereference.Pressuresandtemperature,maybeonebar,let'ssay.T1,298degreesKelvin.That's yourreferencepointandyouwant tofindoutthepressuretemperaturedependenceinanequation.Soifyourearrangeyourequationssop1,T1arenowsomereferencepoint.Theybecomenumbers.Soyouhavelogofareferencepoint,T1here isanumber.SoyouoftenseethisequationrewrittenthenaslogpisequaltodeltaHoverRT.AndIfeellikeI'vegotaminussignmissingsomewhere.Here.Ohyeah,whenyouintegratethis,theoneoverTsquared,there'saminussign.Oh,ItookcareofthatbydoingT2minusT1here.I thinkthere'sstillaminussignproblemsomewhere.I thinkthere'saminussignproblem.Let mecheckup thenoteshere.MinusoneoverT1minusT2overT1T2.Where'smyminussign?T2 minus T1.OK,soIgotminusT1, minusT2 over T1T2.T2 minus T1.Thisisfine,right.There'snominussignproblem.Plusaconstant.OK,soyou'lloftenseeitwrittenlikethat.

Andthatgivesyou arelationshipbetweenthepressureandthetemperaturethen,forasubstancewhereallthereferencepointinformation iscontainedin thisconstant,C.WhichiswheretheT1'sandthep'scomefrom.OK,anyquestions?We'lldoalittleexamplehereofhowthiscouldbeused.

Solet'sswitchtoanexamplenow.Let'sgototheexample,let'sfirstdotheexamplewhichisattheveryendofthenotes.SothisisanexampleofRDX.RDXisafamousexplosive,asI'msureyouknow.Andit'splasticexplosive.It'sgotameltingpointof204degreesCelsius.481degreesKelvin.Andit'saproblemfor,soit'saverypowerfulexplosiveandyoucan'tseeitin X-rays.Soit'saproblematairports.You've gotbeabletofind ifsomebody'scarryinga littlepiece ofRDX in theirluggage.Andso,thequestionbecomeswhatdoyouneedtoknowtodetectit?Youwant tofindthevaporpressure.You'vegottahave,andthemachinewhichbasicallyissensitiveenoughtodetectmoleculesofRDXthatare invaporatroomtemperature.Sothewaythatyoudothatis,youdo anexperiment.Whereyoumeasurethepressure, thevaporpressure,asafunctionoftemperature.Androomtemperature, itturnsoutifit'snotvolatileatall,if thevaporpressure isvery,reallytiny.Andsoinordertogetaccuratenumbers,youactuallydoyourmeasurementsatsignificantlyhighertemperaturesthanroomtemperature.Theydoitclosetothemeltingpoint,soyou'reactuallylookingatthesublimationofRDX.Youdoit,let'ssay,at400degreesKelvin,wherethemeltingpointisat481degreesKelvin. Soslightlybelowthemeltingpointyou'relookingatsublimation.

Andyouplot.LogpversusoneoverT.AndaccordingtoClausius-Clapeyron,thatshouldgiveyou astraightline.Andin fact,that'sexactlywhatyouseewhenyoutakeRDXand dothatexperiment.Youendupwith abunchofdatapoints.That'sfallnicelyon thisstraight line.Andthenyoucanextrapolatetoroomtemperature.Somewherehere,let'ssay,300degreesKelvin.Andfindoutwhatlog p is,at300 degreesKelvin.Andwhatyoufindisthat,andthat'sthegraphthat'sonthelastpageofthelecturenotes,thatthevaporpressureofRDXatroomtemperature is10totheminus11bar,That'stenpartspertrillion.SoittellsyouthatifyouwanttodetectRDXwithasniffermachineattheairport,thatmachinebetterbeabletotellyou,findonemoleculeofRDXoutofatenthofatrillionothermolecules,basically.Which isareallyhardthingtodo.Sothisgivesyou adesignruleforthesepiecesofequipment.Andwhythey'resoexpensive.

OK,beforewedothenextexample,let'ssee ifanyquestionsonClausius-Clapeyron.Yeah.

STUDENT: [INAUDIBLE]

PROFESSOR: So,passomeexponentialhere?Youknow,Idon'trecallseeingitasanexponentialform.Butthatdoesn'tmeanit'snotusedtheexponentialform.Thisisaneasywaytodoit,becausethenyou'vegotalinearrelationship.Andgenerallyweliketoseestraightlines.Sothisgivesyou anicestraight line.

OK,sonowlet'sdoaslightlymorecomplicatedexample.Whichis,somostofthetimeyoudon'thavepurematerial.Youhaveamixtureofsomesort.Forinstance,ifIhaveaglassofwateronthetableatroomtemperature.Abovetheglassofwater,I haveair,there'smyH2Ohere.AndthenI'vegotsomeair.Abovetheglassofwater.Andtheair'sinerttothewater.It'snotreactingwith thewater.Soit'slikeaninertgassittingontopofthewater.ThenIwant toaskthequestion,whatisthevaporpressureofthewater?Inthepresenceofthisinertgas,the air.It'snotreallythesameastheproblemthatwe'vebeenlookingatuphere.Right?Becausethisgasliquidcoexistenceline,thisdiagram,andthisClapeyronequation,isalldoneforapuresubstance.Andatroomtemperature,atroomtemperature,we'resittingsquarelyin theliquidphase.We'renotonthecoexistenceline.Atroomtemperatureandonebarpressure.Soinsteadofhavingthissystemhere,IlookedatthesystemwhereIhadthewater,H2O,withnothingontop.With noairontop.Exceptthecylinder.Now,I'dbettermakethesurfaceofmywaterastraightline,otherwiseI'mgoingtogetintrouble.There'sthewatersittingrighthere.AndIputacylinderwithonebarpressureontop.Mycylinder'sgoingtositsquarelyonthe stumpofthesurfaceofthewater.This isnotgoing tobeanywatervaporatonebarpressureonthatcylinder.Atonebar,we'rewayupintheliquidphase.Iwouldhavetodecreasethepressureonthatcylinderdownto,

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