Kinetic Theory of Gases - Physics and...

EP271KineticTheoryNotesM.P.Bradley

KineticTheoryofGasesSupplementaryLectureNotesforEP271MichaelPatrickBradley(Pleasenotethatthecalculationofpressureintermsofatomic/molecularmean‐squarevelocitycloselyfollowsthetreatmentgivenbyI.N.Levinein“PhysicalChemistry”3rd.ed.ThediscussionofmolecularenergydistributionsfollowsLevineandalsothenotesofProf.AdamBourassa)StatisticalMechanicsWehavelearnedfromourtextbookaboutentropyviaClausius’definitionoftheminimumentropychangedSassociatedwithanamountofheat

€

δQabsorbedbya

systemattemperatureT:

€

dS =δQT

Thisisadefinitionofentropyintermsofmacroscopicallymeasurableheattransferandtemperature,which“tames”theprocess‐dependentheattransfer

€

δQthroughtheuseofan“integratingfactor”,whichturnsouttobejust

€

1/T ,whereTisourpreviouslydefined“thermodynamictemperature”,normallymeasuredinKelvin.Whilethisdefinitionisperfectlyserviceableanduseful,andenablesustoputCarnot’sresultsforheatengineefficiencyonanicetheoreticalbasis,itdoesnotgiveinsightintothephysicalmeaningoftheentropyS.Forthat,wehavetogototheatomicpictureofmatter.Thestatisticalbehavioroflargecollectionsofatomsormoleculesgivesawayofcalculatingtheentropyandrelatedquantities.Thisapproachtounderstandingthermodynamicsfromthebottomupisreferredtoas“statisticalmechanics”.GasesAgasisanon‐condensed(i.e.“volume‐filling”)collectionofatomsormolecules(particles).Exceptforcollisionsbetweenparticles,theydon’treallyinteractwithoneanother.Wehavetalkedextensivelyaboutgasesinthiscoursebutuptillnowwehavenotreallyconsideredtheirmicroscopicbehaviourindetail.Howeversinceweknowthatgasesarecomposedonmanysmallfreelymovingatomsand/ormolecules,weshouldbeabletousethisinformationtohelpusgetabetterunderstandingofgasbehaviour.So,whatisagas?AgasisacollectionofNatomsormoleculesinanon‐condensed(i.e.volume‐filling)statewheretheyaremovingrelativelyquickly(typicalspeedsinexcessof300m/satroomtemperature),butinrandomdirections.Thegasparticles(i.e.atomsormolecules)aretypicallyseparatedbydistanceswhicharemuchlargerthantheparticlesize.Particlesinagasaretypicallynotstronglyinteractingandaremovingrandomlyinalldirections.

EP271KineticTheoryNotesM.P.Bradley

Afewbasicphysicspoints:Eachgasmoleculehasaposition(x,y,z)andavectorvelocitywithcomponents(vx,vy,vz)Eachgasmolecule’spositionevolves“ballistically”accordingtothesimplekinematicequations:

x(t)=x0+vxty(t)=y0+vytz(t)=z0+vzt

(Notethattheinitialposition(x0,y0,z0)oftheatom/moleculeattimet=0israndom.)Becausetherearenosignificantforcesbetweengasmolecules,normallytheatomic/molecularvelocitiesremainconstantuntiltheatomormoleculescollideswithawallofthechamberoranotheratomormolecule.EquipartitionofEnergy Toquantifythekineticenergyofanatomormoleculeweneedaveryimportantresultfromstatisticalmechanics/kinetictheoryofgasescalledthe“equipartitiontheorem”orthe“equipartitionofenergy”.Theequipartitiontheoremstatesthateach“degreeoffreedom”ofanatomormoleculehasanassociatedmeanenergyof

€

12 kBT wherekB=1.3806503x10‐23J/KisBoltzmann’sconstant.

NowBoltzmann’sconstantisrelatedtoouroldfriendthemolargasconstant

€

R =8.314J/mol.Kviatherelation

€

R = NA kB whereNA=6.0221415x1023particles/mole.Sonowwehaveadirect,quantitativeconnectionbetweentemperatureTandmeanmolecularenergy—thisisaveryimportantresult.DegreesofFreedom(DOF)So—whatisamolecular“degreeoffreedom”?Wellinformally,amoleculardegreeoffreedom(DOF)isanyofthe“onedimensional”waysamoleculecanstoreenergy,e.g.becauseofavelocitycomponentsvx,vy,vzORarotationalmotionaboutthex,y,orzaxes,orinvibrationalmodesassociatedwithchemicalbonds.Formally,eachDOFcorrespondstoaquadraticterminthetotalatomicormolecularenergyexpression(theatomicormolecular“Hamiltonian”).Forasimplediatomicmoleculethisis

€

E = 12mCM vx

2 + vy2 + vz

2( ) + 12 Ixxωx

2 + 12 Iyyωy

2

EP271KineticTheoryNotesM.P.Bradley

wherewehaveignoredvibrationalmotionbecausethisnormallynotexcitedexceptathightemperatures.Sincethekineticenergyofanatomormoleculeisdueto3velocitycomponentswehave3degreesoffreedomassociatedwiththis,whichgivesusthefollowingresult:

€

εtr = 12mv

2 = 12m v 2 =

32kBT

RelationbetweenAverageAtomic/MolecularKineticEnergy&GasPressure:Let’simagineourgasatomsormoleculestobeconfinedinsidearectangularboxofdimensions(lx,ly,lz).Supposewelookatonewallofthebox(sayoneofthelxxlzwalls).Whatpressureisexertedonthatwall?Well,pressureisthemacroscopicmanifestationofthefactthatthewallisbeingbombardedbillionsandbillionsoftimespersecondbyfast‐movinggasparticles.Eachtimeagasparticlecollideswiththewallitchangesthey‐componentofitsmomentumfrompyto–py.Iftheparticle’smomentumbeforethecollisionwaspy=mvy,thenafterthecollisionitwillbe–mvy.

NowthegeneralformofNewton’s2ndlawis

€

F = d p

dt,i.e.forceisrelatedtothetime

rateofchangeofmomentum.Nowinthecaseofourgasmolecule,whenitcollideswiththewallandreversesitsy‐componentofmomentum,itdoessobecauseduringthedurationtimeofthecollisionΔt(whichmaybeveryshort,lessthanananosecond)thereisanaverageforceFyexertedonitbythewall.Sinceaction=reaction,byNewton’s3rdlaw,duringthecollisionthemoleculeexertsthesameforcebackonthewall,butactingintheoppositedirection.So,wecanfindthetotalforceonthewallbyaddinguptheforcesexertedbyallthemoleculeswhentheycollidewithit.The“translational”kineticenergyofagasatomormolecule(i.e.thepartassociatedwiththemolecule’smotion(=translation)isgivenby:

€

εtr = 12mv

2 = 12m(vx

2 + vy2 + vz

2)

€

Fy,i = may,i = mdpy,idt

=ddt(mvy,i)

2mvy,i =< Fy,i > Δt

Nowthecleverobservationinvolvedinthisderivationistorealizethattheaveragetimebetweencollisionscanberelatedtotheboxsizeasfollows:

EP271KineticTheoryNotesM.P.Bradley

€

Δt =2lyvyi.Thisthenenablesustowritetheaverageforceas

€

< Fy >= < Fy,i >i=1

N

∑ =mvy,i

2

ly=mly

vy,i2

i=1

N

∑i=1

N

∑

Nowtheatoms/moleculesofthegasDONOTallhavethesamespeed,soallthevyiaredifferent.However,wecancalculateandtalkaboutanaverageormeanvalueofthesquareofthemolecularspeed

€

< vy2 >≡

1N

vy,i2

i=1

N

∑

Withthisdefinitionthey‐componentoftheforceonthewallweselectedis

€

< Fy >=mNly

< vy2 > Now,thepressureexertedonthewallisjusttheforceonthe

walldividedbyitscross‐sectionalarea(recallthat1Pascal=1Newton/m2)sowe

gettheresultthat

€

P =mN < vy

2 >

lxlylz=mN < vy

2 >

V

Nowthereisnothingspecialaboutthey‐direction,wejusthadtopicksomethingtogetstarted,soonaverageitmustbetruethat

€

<vx2>=< vy

2 >=< vz2 >

Nowsince

€

< v 2 >=< vx2 + vy

2 + vz2 >=

1N

(vx,i2 + vy,i

2 + vz,i2 )

i=1

N

∑ =1N

vx,i2

i=1

N

∑ +1N

vy,i2

1

N

∑ +1N

vz,i2

1

N

∑

Wehavetheresultthat

€

< v 2 >=< vx2 > + < vy

2 > + < vz2 >

Since

€

< vx2 >=< vy

2 >=< vz2 > wecanwrite

€

< vx2 >=< vy

2 >=< vz2 >=

13

< v 2 >

Sowegetfinallytheresultthat

€

P =mN < v 2 >

3Vforanidealgas(i.e.oneinwhichthe

moleculestravelballisticallybetweencollisionsandtherearenointermolecularforces).Wecangoabitfurtherhere.Theaveragetranslationalkineticenergyofagasatom

ormoleculeis

€

< εtr >= 12m < v 2 >⇒< v 2 >=

2 < εtr >m

Thereforewecanre‐writeour

expressionforthepressureofaclassicalidealgasintermsofthemeanatomicormolecularkineticenergy:

€

PV =23N < εtr >

Thisisourfirstreallyimportantkinetictheoryresult.YoucanseehowwehavebeenabletorelateamacroscopicallymeasurableparameterlikepressuretotheaveragebehaviourofalargenumberNofatomsormolecules.

EP271KineticTheoryNotesM.P.Bradley

KINETICTHEORYANDDISTRIBUTIONS OK,upuntilnowwehavejustsampledkinetictheoryinthemostbasicway:weacceptthatagasiscomposedonalargenumberNoratomsofmolecules,andwecouldevencalculateanicerelationshipbetweenpressureandthemean‐squarevelocityofthosemolecules(whichinturnwasrelatedtotheirmeankineticenergy):

€

P =mN < v 2 >

3V butsofarthatwasaboutasfaraswewereabletogo.Now,wecangofarther.TheMaxwell(orMaxwellBoltzmannMB)distribution:Sinceweknowthattheatomsormoleculesinagasdonotallhavethesame

velocities,weneedawayfordescribingthis.TheMaxwellBoltzmanndistributiongivestheanswer.TheprobabilityPthatagivenatomormoleculehasanxcomponentofvelocitybetweenvxandvx+dvxisgivenbyP=f(vx)dvx,where

€

f (vx ) =m

2πkBT

1/ 2

e−mvx

2

2kBT

Noticethatthishastheformofa“Gaussian”curveordistribution

€

P ~ e−vx2 / 2σ 2

wherethe“standarddeviation”

€

σ 2 = kBT /m SincethedefiniteintegralofaGaussianovertheentiredomainisgivenby

€

e−x 2

2σ 2 dx−∞

∞

∫ = 2πσ ,wecanseethatthevelocitydistributionisNormalized,i.e.

thedefiniteintegralfrom

€

vx = −∞ to

€

vx = +∞ isequalto1:

€

f (vx )dvx =1−∞

∞

∫ Thisisimportantbecausef(vx)isaprobabilitydistribution.Probabilitiesare

measuredfrom0to1,with0meaningnochancewhatsoever,and1meaningcompletecertainty.Sinceeverygasmoleculehassomevalueofthexcomponentofitsvelocitylyingbetween‐∞and+∞,theintegraloftheprobabilitydistributionovertheentirerangeofvelocitiesmustaddto1,andthenormalizationconditionguaranteesthat.Thereisnothingspecialaboutthex‐componentofvelocity(itisjustalabelwe

pickedforconvenience)sothesamedistributionappliestovyandvzMaxwellSpeedDistribution:Thevelocityofagivenmoleculeisavectorgivenbyitscomponents(vx,vy,vz),

eachofwhichisdistributedaccordingtotheabovedistributionlaw.Butinmanycasewearemoreinterestedinthespeedvofthemolecule,whichisgivenby

EP271KineticTheoryNotesM.P.Bradley

€

v = vx2 + vy

2 + vz2 .Thespeedisparticularimportantbecausethekineticenergyis

givenby

€

12mv

2 Nowitisimportanttorealizethattheprobabilitydistributionformolecularspeed

isdifferentfromtheprobabilitydistributionformolecularvelocitycomponents,eventhoughthespeedisafunctionofthosesamevelocitycomponents.Whyisthis?Well,theanswerissimple,onceyoulookattheproblemcorrectly.Basicallytheimportantthingisthattherearemanycombinationsofvelocitycomponents(vx,vy,vz)whichwillgivethesamefinalvalueofthespeed

€

v = vx2 + vy

2 + vz2 .

TheMaxwellBoltzmanndistributionformolecularspeedsisgivenby

€

g(v) =m

2πkBT

3 / 2

4πv 2e−mv 2

2kBT .Noticethatishassomesimilaritiestothevelocity

componentdistributionsbutitisnotthesame!ApplicationsoftheMaxwellDistributions:Calculationoftheaverage/meanmolecularspeed:

€

v = v = vg(v)dv−∞

∞

∫ .Whydoesthisformulawork?Well,ifwethinkaboutitsimply,werememberthattheprobabilityofgettingspeedvisgivenby

€

P(v) ~ g(v)dv .Thentheweightedaveragespeed(ormathematicalexpectation)isgivenbythesum

€

vP(v)∑ (whereIhavebeenratherrelaxedaboutthediscretization,butthebasicideaiscorrect).SinceinthelimitofverymanyparticlesNwecantakethesumoverintoanintegral,wethengettheformulaquotedabove.Nowletusapplyit:

€

v = 4π m2πkBT

3 / 2

e−mv2 / 2kBTv 3dv

0

∞

∫

Noticethattheintegralextendsfrom0to∞withnonegativevaluesinthe

integrationdomainbecausethespeedisapositivedefinitequantity.Wecanevaluatethisintegralusingastandardresult:

€

x 2n+1e−ax2

dx =n!2an+10

∞

∫ whichisvalidfora>0andn=0,1,2,3….

ApplyingthisstandardGaussianintegraltoourcalculationof<v>weget(usingx

v,a=m/2kBT,n=1)

EP271KineticTheoryNotesM.P.Bradley

€

v = 4π m2πkBT

3 / 21!

2 m2kBT( )

2

= 4π m2πkBT

3 / 24kB

2T 2

2m2

=2 2π

kBTm

1/ 2

Thusweendupfinallywith

€

v =8πkBTm

Wecanseethattheaveragemolecularspeedinagasincreasesasthe

temperatureTgoesup(as

€

T1/ 2 ,anddecreasesforheavieratomsormoleculesas

€

m−1/ 2).Thisseemsintuitivelyreasonable.Mean‐SquareSpeed:Wecouldcalculatethemeanofthesquareofthespeed(normallycalledthe

“meansquarespeed”)inthesamemannerasabove:

€

v 2 = v 2g(v)dv0

∞

∫ .Asbeforethisisjustaweightedaverageormathematicalexpectationvalueforthesquaredspeedv2,wheretheprobabilityistheweightfactor.Thiscalculationisnottoodifficultbutitisalittlebitinvolved(infactIaskedyoutocarryoutasoneofthehomeworkproblems).Howeverthereisanotherwaytoapproachthisproblemwhichismuchsimpler,usingourearlierresultontheEquipartitionofEnergytheorem.Recalltheequipartitionofenergytheorem:foreverymolecularoratomicdegree

offreedom(basicallycorrespondingtoeveryquadratictermintheenergy),thereisonaverage1/2kBTofassociatedenergy.Fortranslationalkineticenergyofanatomormolecule,thekineticenergyis

€

εtr = 12mv

2 = 12mvx

2 + 12mvy

2 + 12mvz

2 .Sincetherearethreequadratictermsintheenergyexpression(correspondingtothe3translationaldirectionsofmotionx,y,z),theaverageormeankineticenergymustbe3/2kBT,i.e.

€

εtr = 12mv

2 = 12m v 2 = 3

2 kBT .Wecanrearrangethistoimmediatelygetour

desiredresultfor<v2>:

€

v 2 = 3 kBTm

Wesometimesliketotalkaboutthe“root‐mean‐square”speedor“RMS”speed

€

vRMS .Thisissimplygivenby

€

vRMS = v 2 =3kBTm

.NotethatthisisNOTthesame

asthemeanspeed(infactitisalittlelarger),butitdoeshavethesameunits.MostProbableAtomicorMolecularSpeed:Nowwehavecalculated2differentcharacteristicspeeds(themeanspeed

€

v andtheRMSspeed

€

vRMS )forgasatomsormoleculesofmassmattemperatureT.But

EP271KineticTheoryNotesM.P.Bradley

thereisstillanothercharacteristicmolecularspeedwhichmaybeuseful—the“mostprobable”speed

€

vMP .Thisisthemostlikelyspeedtoobserveifwepickasinglemoleculeatrandomandmeasureitsspeed.Itisgivensimplybythevaluewhichmaximizesthespeedprobabilitydistribution

€

g v( ) ,whichwefindbysettingthefirst

derivativeto0,i.e.

€

dg(v)dv vMP

= 0 .ForthemomentIwillleavethiscalculationforyou

asanexercise.Theresultthatwegetis

€

vMP = 2 kBTm

.Youcanverifythatthisisa

maximumbyapplyingthesecondderivativetest,althoughtheshapeofthedistributiong(v)makesthismore‐or‐lessobvious.SummaryofCharacteristicMolecularSpeeds:OK,sowehavenowcalculated3differentcharacteristicspeedsforgasatomsor

moleculesofmassm,attemperatureT.Wecansummarizetheseresultsinatable:

MostProbableAtomic/MolecularSpeed

€

vMP = 2 kBTm

≈1.41 kBT /m

MeanAtomic/MolecularSpeed

€

vMP =8πkBTm

≈1.60 kBT /m

RMSSpeed

€

vMP = 3 kBTm

≈1.73 kBT /m

Nowwecanseethatthespeedshavethefollowinghierarchyofmagnitudes:

€

vMP < v < vRMS .Also,sincethespeeddistributionfunctiong(v)fallsoffforspeedshigherthanthemostprobablyspeedvMPwealsohavethefollowingresult:

€

g(vMP ) > g( v ) > g vRMS( )Soisthephysicalsignificanceofallofthis?Well,wecanseeprettyclearlythat

althoughaboxorchambercontaininganatomicormoleculargascontainsatomsormoleculewithawiderangeofspeeds,movingrandomlyinalldirectionsandcollidingofthewallsandoffeachother,wecannonethelessmakesomeprettydefinitestatementsabouttheaverageormeanvaluesofmanyatomic/molecularquantities,suchasthespeedandenergy,aswellasresultsforthepressureexertedbythegasonthewallsofthechamber.Thisispowerofthestatisticalmechanicsapproach(recallthatthekinetictheoryofgasesistheearlieststatisticalmechanics‐basedtheory).Ourdiscussionsofspeedsshowedusthatthereareseveralcharacteristicspeeds

wecancalculate,allofwhicharecloseinvalueandallofwhichincreaseasthetemperatureincreases.Thisisthebasisforthecommonstatementthatthetemperatureisameasureoftheaveragemolecularmotion,andthatinwarmerbodies,atomsandmoleculearemovingfaster.Thecharacteristicenergyassociated

EP271KineticTheoryNotesM.P.Bradley

witheachdegreeoffreedomiskBT,andthecharacteristicmolecularspeedsallhave

theform(numericalfactor)x

€

kBTm

.

Thismakessenseonthebasisofunitsandisalsoconsistentwiththeformulafor

thespeedofsoundinagas:

€

csound = γkBTm

where

€

γ =cpcvisthespecificheatratio

(Moran,Shapiroetal.callit“k”butthiscanbeconfusedwithBoltzmann’sconstantwhichcomesupeverywhereinstatisticalmechanicssoIprefertousetheGreeklettergammahere).Whyshouldthespeedofsoundberelatedtothecharacteristicatomic/molecularspeedsinagas?Wellsoundwavesadvancebetweenalternatecompressionsandrarefactionsbycollisionsbetweengasmolecules,soitmakessensethatthespeedwithwhichsoundcouldpropagatewouldbeofthesameorderofmagnitudeasthecharacteristicatomicormolecularspeeds.Atomic/MolecularEnergyDistributionSofarwehaveusedtheatomic/molecularMaxwell‐Boltzmannspeeddistribution

tocalculatesomecharacteristicspeedsforanidealgasofatomsormoleculesofmassmattemperatureT.Howeverwearealsooftenveryinterestedintheenergydistributionfortheseparticles.Thiswecaneasilyderive.Thekineticenergyisgivenby

€

ε = 12mv

2 .Whatwearetryingtocalculateisessentiallyachange‐of‐variablesfromatomic/molecularspeedvtoenergyE.Sincethereisaone‐to‐onemappingfromspeedtoenergy,thisisastraightforwardprocess.Byourdefinitionofprobabilitydistributionswemusthave

€

g(v)dv = g(ε)dε .Re‐arrangingweget

€

g(v) = g(E)dE /dv .Nowwecaneasilycalculate

€

dE /dv = mv ,sowegetfinally

€

gE (E) = g(v) /mv Fromthisweget,finally

€

gE (E) = 2π

1kBT( )

3 / 2Ee−E / kBT

Youcanverifythefactthatthisisaproperlynormalizeddistributionandthatit

givesameanenergywhichobeystheequipartitionfunction,ifyoulike(Iwillleavethisasanexerciseforyou,eagerstudent).