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).