University of Nebraska - LincolnDigitalCommons@University of
Nebraska - LincolnRobert Katz Publications Research Papers in
Physics and Astronomy1-1-1958Physics, Chapter 16: Kinetic Teory of
GasesHenry SematCity College of New YorkRobert KatzUniversity of
Nebraska - Lincoln, [email protected] this and
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-Lincoln.Semat, Henry and Katz, Robert, "Physics, Chapter 16:
Kinetic Teory of Gases" (1958). Robert Katz Publications. Paper
166.htp://digitalcommons.unl.edu/physicskatz/16616KineticTheoryof
Gases16-1 General GasLawThe behavior of a gas under various
condition8 of temperature and pressurehas already been studied in
80me detail. When the pressure of aconstantmassof gas is
nottoogreat, say less than about 2 atm, we find thatagasobeys the
followingrelationships:at constant temperature PV=
constant;atconstantvolume P= KT;atconstantpressure V=
K'T.(16-1)(16-2)(16-3)Thesethreeequationsarespecial casesof
asingleexperimental equationwhich gives the relationship between
the pressure P, the volume V,and theFig. 16-1 Twosteps inthe
derivationof the general gas law; anisothermal processfollowed
byaconstant-pressureprocess.absolutetemperature Tof a constant mass
of gas. Wemayderivethegeneral formof the gas law from the above
equations.Let us consider a gas containedina cylinder witha
closelyfittingpiston, asshowninFigure16-1. Theinitial
conditionofthegas maybe299300 KINETIC THEORY OF GASES
16-1ordescribedintermsofitsinitial pressurePi, itsinitial volumeVi,
anditsinitial temperatureTi . The gas is allowed to expand at
constant tempera-ture, saybykeepingthecylinder immersedin abath
ofmelting ice, untilits new pressure is Pf and its new volume is V2
Since the expansion was atconstant temperature, wefindfromEquation
(16-1)thatPiVi= Pf V2Now suppose that the gas is heated to a higher
temperature Tf , the volumebeing allowed to expand to a new
valueVj, but the pressure on the pistonbeingmaintainedat
thesamevaluePf throughout this process. Then,fromEquation(16-3)we
maywriteV2= Vf= /('TiTf'TiV2 =Vf -TfSubstituting forV2 into
thefirst of theabove equations, we findPiVi =
PfVf.TiTf(16.4)Equation(16-4)is one formof the generalgas law.
Since the initial state,described by the subscript i, and the final
state, described by the subscript!, are entirely arbitrary, the
only way in which the quantitieson the right-and left-hand sides of
the equationcanbe equal is foreach quantity tobeseparately equal to
the same constant. Thus we may rewrite the gas law as (16-5)wherec
isaconstant whosevaluedepends uponthemassof theenclosedgas.
Anyconvenient units maybe used for the pressure andvolume, butthe
temperatureTmustalwaysbe the absolute
temperature.IllustrativeExample. Agivenmassofair occupiesavolumeof
2,000cm3at27Cwhen itspressurecorresponds to the pressure at
thebaseof acolumn ofmercury75cmhigh. Theairiscompresseduntil
itsvolumeis1,200cm3, andits pressurecorrespondsto225cmofmercury.
Determinethetemperature ofthegas after it
hasbeencompressed.FromEquation(16-4) wehavePiVi=PfVf.Ti Tf16-3
KINl''l'lC THl'ORY OF GASl'S 301fromwhichThe pressure at the baseof
acolumn of mercuryh em high isgiven byP = hpg,wherep isthedensityof
themercury. Substitutingnumerical values, wehave75 emXpgX2,000
cm3225 emXpgX1,200 cm3,300.2 abs Tf1j = 540.4 abs.16-2 TheUniversal
GasConstant RThe constant cappearinginEquation(16-5) canbe
evaluatedfor anygiven mass of a gas. Let us designate the value
ofthisconstant foragrammolecular weight, or mole, of a gas
bythesymbol R. A grammolecularweight of anysubstance isan amount of
thatsubstance whose mass, expressedin grams, is numerically equal
tothe molecular weight of thesubstance. In thelimit of low
pressures, the value of Ris independent of the chemical natureof
the gas, so that Ris known as the universal gas constant. In the
eventthat n moles of gas arepresent ina container, Equation(16-5)
mayberewritten asPV=nRT. (16-6)The numerical value of the gas
constant R can be determined by notingthat 1 moleof anygasoccupies
avolume of 22.4 liters atapressureof 76cm of mercury at ooe;
putting these values into Equation (16-6), we get7 erg cal -2 liter
atmR=8.31X10 =1.99 1 oK=8.21X10 1 oKmole K mo e mo e16-3
KineticTheoryof GasesFromtheprecedingdiscussionwehaveseen that all
gasesexhibit similarthermal andmechanical properties, regardless
oftheir chemical composi-tion, aslong as their pressure is
sufficiently small. Thisbehavior isquiteunlikethat of
thesamesubstances inliquidor solidform, wherethesesubstances
exhibit widely different thermal and elastic properties. We areled
to infer that the molecules of a gas are sufficiently far apart so
that theyrarely interact with each other. The pressure of a gas
then results from thecollisions of the molecules of the gas with
the walls of thecontainer. Themovingmolecules
ofthegascompletelyfill everycontainerinwhichthegas is placed.We may
construct a theory of an ideal gas which is in good
agreementwiththeexperimental results
describedintheprecedingsections onthebasis of a few simple
assumptions. We shall assume that a gas is composedof molecules
that aresosmall that, toa first approximation, theymaybecon-sidered
as point masses. We assumefurther that the molecules do not302
KINETIC THEORY OF GASES16-3exert forceson each other except during
collisions. We shall furtherassumethat
themoleculesofthegasareperfectlyelastic, andthat thecontainer
ismade of perfectly elastic, rigid walls. This implies that
mechanical energyis conservedincollisions betweenmolecules. Ifthis
werenot thecase,wewouldexpect toobservethat thepressureofatankof
gaswoulddi-minish with time, as themolecules lostmechanicalenergy
in inelasticcol-lisions. Forthesakeof simplicityweshall assumethat
thegasisinacubical container of edge d andof volumeV=d3Fig. 16-2
Moleeules with equalvelocity components v1 near faceBeDE of
thecube..-J--:l----+---..",)DThe pressure exerted by the gas on the
walls of the container is due totheimpact of the molecules
onthewalls, and, whenin equilibrium, isequal
tothepressurethroughout thegas. Tocalculatethispressureletusassume
thattheimpact ofamoleculewithawall isanelasticimpact;that is, if a
moleculeis approachingthewall withavelocityvandmo-mentum mv, then
it will leave the wall with a velocity -v and a momentum-mv. The
change in momentum of the molecule produced by this impactwill thus
be- 2mv. To determine the pressure on the walls of the
container,let usfirst
calculatetheforceexertedbythemoleculesononeofthesixfacesofthecube,
saythefaceBCDEofFigure16-2, andthendividebyits area.Let us consider
those molecules which at some instant are very close tothis face.
Only those molecules whose velocities have components
perpen-dicular to this face, and directed toward it, will strike it
and rebound. Sup-poseweconsider asmallnumberofmoleculeswhichhave
thesame valueVI forthis velocity component. The number of these
molecules which willstrike this face during a small time interval
t:.t will be one half of the num-ber contained inasmall
volumeAt:.l, whereAis equalto the areaof theface of the cube and
t:.l =VI t:.t; the other half having a velocity componentof
magnitudeVI are moving away from the wall. If ni represents the
num-ber of molecules per unit volume which have a velocity
component ofmagnitudeVbthenthenumber strikingthisface of
thecubeintime t:.twillbe16-3 KINETIC THEORYOF GASES 303Since each
suchmolecule wiII have itsmomentum changed by- 2mvias a result of
this impact, the impulse imparted to the wall will be equal
andopposite to it, or +2mvl' The impulse F1 t:.t on the wall
produced by thesecollisions in timet:.t will then befromwhichThe
pressure on the wallproduced by the impactof these molecules isF1
2PI =A =nlmvlWe can now consider another group of molecules, n2 per
unit volume, whichhaveaslightly different velocitycomponent V2in
thisdirection;they willproduce anadditionalpressureP2given byP2= n
m v ~Inthisway, wecanbreakupthegasintodifferent
groupsofmolecules,eachgroupcontributingasimilar termtothepressureon
thisfaceof thecube. Thetotal pressureP duetoall thedifferent groups
of moleculeswilltherefore beof the formP=nl mvi + n m v ~ + n 3 m v
~ + ....Thisequationcanbesimplifiedbyintroducinganew
termcalledtheaverageofthesquaresofthecomponentsofthevelocitiesof
all themoleculesmovingperpendicularto faceAanddefinedby
theequationnlvi + n v ~ + n 3 v ~ +nin which n represents the total
number of molecules per unit volume. Sub-stituting thisvalueof v ~
intheequation forthepressure, wegetP= n m v ~ (16-7)Therewill
beasimilarexpressionforthepressureoneachofthesixfacesof thecube,
except thatthe factor v ~ will bereplacedby theappro-priateaverage
of thesquares of thecomponents of thevelocities of themolecules for
thatparticular face.The velocity v of anyone molecule may be in any
direction; itcan beresolved intothree mutually perpendicular
components Vx , Vy , Vz Themagnitude of v in terms of the
magnitudes of these components is given byv2=v; +v ~ +v;.304
KINETIC THEORY OF GASES 16-3There will be a similar equation for
the square of the velocity of each mole-culeof thegas intermsof
thesquares of its three mutually perpendicularcomponents.
Ifweaddthesquaresofthecomponent velocitiesinthexdirection and
divide this sum by the total number of molecules,we will getthe
average value of the square of this velocity component; it will
berepresented b y ~ Similarly, ~ and ~ will
representtheaveragesquaresofthevelocitiesintheyandzdirections,
respectively.
Byaddingtheseaveragesquaresofthethreevelocitycomponents, weget2
2+2+2v = vxvy v.,where v2is the averageof the squares of the
velocities of all the molecules.Since the velocities of the
molecules have all possible directions, the averagevalue of the
squares of the velocity in anyone direction should be the sameas in
anyother direction, orso thatIf we take the x
directionasperpendicular to the faceA,we canwritev2= 3 v ~so
thatEquation(16-7) becomes(16-8)(16-9)Recallingthat
thekineticenergyofamovingmoleculeisequal to !mv2,Equation(16-8) may
be written asP= in(!mv2).Since n is the number of molecules per
unit volume, we see that the pressureis numerically equal to two
thirdsthekineticenergyof the molecules in aunitvolumeof gas.Let
ussuppose thatNoisthetotal numberofmolecules inamoleofgas,
calledAvogadro'snumber, whichiscontainedina volume V. Thenthe
number ofmoleculesper unit
volumenisgivenbytheexpressionNon=V'Substituting forn into
Equation(16-8), we findPV= iNo X!mv2Equation (16-9)is a theoretical
resultobtained fromour hypotheses aboutanideal gas,
relatingthepressureandvolumeof 1 moleof anideal gas.16-3KINETIC
THEORY OF GASES305or if(16-10)If wecompare thisresultto
theexperimentalequationgiveninEquation(16-6), whichfor 1moleof gas
becomesPV=RT,wefindthetheoretical and experimentalresultstobe
inagreement ifRT= iNa !mv2,1:2 3 R ,-mv =- - 7.2 2NaIt is customary
to define a new constant k, called Boltzmann's constant,such
thatRk=-NaSince R is the gas constant per mole, and Na is the
number of molecules inamole of gas, theconstant k
maybedescribedasthegas constant permolecule. Intermsofkthe
preceding equationbecomes(16-11)that is, the mean kineticenergy of
translationof amolecule of gas is givenby ! theproduct of
Boltzmann'sconstant bytheabsolutetemperature.This equationgives us
somephysicalmeaningof temperature foranidealgas. For suchagas
thetemperature isassociated with thekineticenergyof the random
translational motion of the molecules of the gas. Accordingto
Equation (16-11)the average energy of each molecule, and therefore
thetotal internal energyof anideal gas, is associatedwithits
temperature.Thus the internal energy oj an ideal gas is a junction
oj its temperature only,and notof its pressure or volume.Inour
derivationofthegaslawintheformof Equation(16-9),
weusedtheword"molecules"todescribetheparticleswithwhichwe
weredealing. Thesemolecules weredescribed by
theconditionthattheyweresmall,relatively far apart, and perfectly
elastic. Thus thisequation mightbe used to describe the behavior of
any aggregate of particles whose physicaldimensions were small
compared to their averageseparation,provided thattheseparticles
wereelastic andrarelyinteracted witheachother. Theneutrons in
anuclear reactor satisfy theseconditions. If the neutrons
areinequilibriumwiththematerial of the reactor at a temperature T,
wespeakof themasthermalneutrons. Themeanvelocity
vofthesethermalneutrons may be obtained fromEquation (16-11). The
particles of acolloidalsuspension mayalso be thought of asthough
they were moleculesof an idealgas, and itis foundthatthesealsoobey
thegaslaws.Equation (16-9) incorporates another result called
Avogadro's hy-pothesis, first statedbyAvogadroin1811, that all
gasesoccnpyingequal306 KINETIC THEORY OF GASES 16-4volumes at
thesame temperatureand pressurecontain equal numbers of mole-cules.
Theacceptedvaluefor thenumberof moleculesinamoleof gasNoisNo =
6.023X1023molecules/gm molecular wt.As we have alreadyseen, 1 mole
of gas occupies avolumeof 22.4 liters atooe and at a pressure of 76
em of mercury. If we perform the
calculationindicatedinEquation(16-10)
tofindthenumericalvalueofBoltzmann'sconstant, we obtaink =
1.38X10-16erg;oK.16-4 WorkDone by aGasWheneveragasexpandsagainst
some external force, itdoesworkontheexternal agency; conversely,
wheneveragasiscompressedbytheactionof some outside force, work is
done on the gas. To calculate the work
done(a)(b)pppn,IAllII,IUF=PAvFig. 16-3 (a) Expansion of a gas at
constant pressure. (b) Graphical representation ofthe work done
!1Jrasan area on the PV diagram.byagas, consideragasenclosed
inacylinder withatight
fittingpiston.Thepistonmaybeconnectedtoamechanical deviceonwhichit
exertssome force. The forceFactingonthepistonowing to
thepressurePofthe gas is given byF= PA,in which A is the
cross-sectional area of the piston, as shown in Figure 16-3.Suppose
that the piston is pushed out a small distance /lS, while the
pressureof the gas remains essentially constant. The workMY doneby
the gas in16-4 WORK DONE BYA GAS 307'moving the piston is given
bytiJY= F tis = PAtis,or (16-12)Thus theworkdoneby anexpandinggas
at constant pressure is equal tothe product of the pressure by the
change in volume. No mechanical
workisdonebyagasunlessthereisachangeinthevolumeofthegas.
Thefirstlaw of thermodynamics,tiQ= tiU + tiJY,may be rewritten for
processes involving gases astiQ= tiU + PtiV.(15.6)(16-13)Thus,
inanyprocessinwhichthe volumeof thegas remains
constant,calledanisovolumicprocess, anyheatdelivered to thegas must
appear asinternal energyandisthereforeexhibitedasachangein
thetemperatureof thegas.Let us calculatetheworkdonebyanideal gas
whichexpandsiso-thermally,thatis, at constant temperature, froman
initial volumeVI to afinalvolume V2 From the gas law
therelationshipbetween the variablesof pressure, volume,and
temperature forone mole of gas may be stated asPV= RT.The work done
may be represented as an integral, from Equation (16-12), asJY
=fdJY=.V2 PdV,and, substituting for Pits value fromthegas
law,RTP=-,Vwe findat constant temperature.IvV2 dVJY = RT -v,
VRecalling thatfdX--; = In x + C,(16-14)where C is aconstant of
integration, we find thatV2JY= RTln-VIfor the workdonebyone
moleofanidealgas in an isothermal expansion308 KINETIC THEORY OF
GASES 16-516-5 MolarHeatCapacityof aGasFig. 16-4 Workdone bya mole
of agas in an isothermal expansion attemperature T from volume V 1
tovolumeV2When a quantity of heat t:.Q is deliveredtoa gas, it
maychange the internalenergy of thegas byanamount t:.UV and may
also result in the performanceof anamount ofexternalwork t:.Jr
bythegasupontheoutsideworld, inac-cordance with the first law of
ther-modynamics. If the volume of thegas is kept constant, all the
heat isconvertedinto internal energy. Sincethe internal
energyUofamoleofgas is a function of temperature only, we may
definethe molar heat capac-ityat constant volumeofamoleofgasCvasat
temperatureTfromaninitial volumeVI toafinal volumeV2
FromEquation(16-14) weseethat whenthegasexpands, that is,
whenV2isgreater than Vb the work done isP positive, asisconsistent
withthesignconventiondevelopedinSection15-6.Theworkdoneis shownas
the areaunder the isotherm inFigure16-4.C _ t:.Q_ t:.Uv - t:.T-
t:.T(at constant volume), (16-15)so thatthechange in internal
energyt:.U maybe expressedast:.U= Cv t:.T.
(16-16)FromEquation(16-13) thefirst
lawofthermodynamicsasappliedtoanideal gas maybe rewritten ast:.Q=
Cv t:.T +pt:.V. (16-17)Let us considerthechangeintemperatureof
amoleofgaswhenaquantityof heat t:.Q is delivered to thegaswhile
thepressure is heldcon-stant but the volume is permitted to change.
The thermal energy deliveredto the gas mustnowbeused todo external
work as well as tochange theinternal energy of the gas. The rise in
temperature of the gas will thereforebe less than in the case where
the volume of the gas is kept constant. Themolar heat capacityat
constant pressureCp might beexpressedinsuchunits as calories per
mole per degree, andmaybe defined throughtheequationt:.QCp = - (at
constant pressure),t:.T(16.18)16-5 MOLAR REA'!' CAPACITY OF A GAS
309orand fromEquation(16-17) wefind
Cp= From thegas law foramole of gaswehavePV=RT,RV= pl'.At
constant pressure both Rand P are constant, so that a change of
volume Visrelatedtoachangeintemperature
throughtheequationR=Substituting thisresult into the preceding
equation for Cp, wefindthatCp=Cv + R. (16-19)Themolarheat
capacityof agasat constant pressureis always greaterthan the molar
heat capacity atconstantvolumebythegas constant R.The internal
energy of amonatomicgas, such as helium, is entirely intheformof
kinetic energy of translation of the randommotions of theatoms of
the gas. From Section16-3 this internal energy may be stated
asU=NoX!mv2=!RT.(16-20)SinceR is aconstant, we findthatthechange in
internalenergy asso-ciated withachange in temperature is given
by=!R Thus the molar heat capacity of amonatomic gas at constant
volume maybe found by substituting the preceding result into
Equation (16-15), to findCv =!R (monatomicgas).
(16-21a)Substituting thisresult into Equation(16-19), we obtainCp =
iR (monatomicgas). (16-21b)It is customary to designate theratioof
the specificheat at constant pres-sure to the specific heat at
constant volume by the letter'Y(gamma). ThusCp'Y = - .
(16.22)CvSubstitutingfromEquations (16-21) into(16-22),
wefindthevalueof 'Yforamonatomic gas to be'Y= 1 (monatomic gas).310
KINETIC '.rHEORY OF GASES 16-5Inourdevelopment
ofthekinetictheoryofanidealgasweassumedthat
amoleculecouldbeconsideredasapoint massandshowed that theaverage
kinetic energy of translation per molecule is ikT (Equation
16-11).Therearethreeindependent directions of motionof translation,
saythex, y, and z directions. We say that the molecule has three
degrees of freedom;that is, threecoordinates are necessary to
specify the position ofthe mole-cule at any instant, one for each
degree of freedom. Since there is no reasonfor preferring one
direction rather thananother, we postulate the principleof
equipartitionof energy, that each degree of freedomshouldhave
thesame amount of energy. Referring to Equation 16-11, the amount
ofenergytobeassociatedwitheachdegreeof freedompermoleculeis
!kT.Theinternal energyof amoleof amonatomicgaswill thenbe iRT,
asgiven by Equation16-20.The idea of degreesof freedomcanbe
extended to diatomicand poly-tomicgases. It canbe shown thatthe
valueof 'Y canbe expressedasf+27=--'fwhere f is the number of
degrees of freedom per molecule. For a monatomic3+2 5gas we find
that 'Y = -- =- =1.67, in agreement with measured3 3values as shown
in Table 16-1.We mayextendthese ideas toa diatomic molecule whichwe
mayimagine to be two pointmassesafixeddistance apart; the
linejoining thetwoatomsistheaxis of themolecule.
Ifthediatomicmoleculeis con-sidered as a rigid body, then it will
have three degrees of freedomowing tothetranslational
motionoftheentiremolecule, plusacertainnumberofdegrees of
freedomowingtothe rotational motion of themolecule. Aglance
atTable16-1shows that 'Y = 1.4 fordiatomicgas, indicating thatf =
5; thustheremust betwoadditional degreesoffreedomofrotation.These
would correspondtorotations about twomutuallyperpendicularaxes
inaplane at right angles to thelinejoining thetwoatoms.Thus if
atordinary temperatures the molecules of adiatomicgas maybethought
of as rigid, and possessing no vibrational energy, the meanenergy
of each molecule must be ~ k T The total internal energy of a
moleof such agas is given byU=NoX ~ k T = ~ R T
(diatomicgas).(16-23)Fromthisexpressionwefindthemolal' heat
capacitiesofadiatomicgasto beCv= ~ R Cp = iR (diatomicgas),
(16-24)16-6 ADIABATIC PROCESSES 311andtheratioof the specific heats
l' isl' = i- (diatomicgas). (16-25)These results are in rather
remarkable agreement with experiment, asshownin Table 16-1.TABLE
16-1 THE MOLAR HEAT CAPACITY AT CONSTANT VOLUME,AND THE RATIO OF
SPECIFIC HEATS FOR SEVERAL GASESAtoms perCv*l'GasMoleculeTheory
Experiment Theory ExperimentArgon 1 2.98 2.98 1.67 1.67Helium 1
2.98 2.98 1.67 1.66Oxygen 2 4.97 5.04 1.40 1.40Nitrogen 2 4.97 4.93
1.40 1.40Carbonmonoxide 2 4.97 4.94 1.40 1.40* The units of Cv are
given as calories per mole per degreecentigrade.At high
temperatures the classical theory of the specific heats of gasesis
nolonger adequate todescribetheexperimentaldeterminations.
Addi-tional rotational andvibrational modesareexcitedat
hightemperaturesina manner whichis best describedbythe
quantumtheory of specificheats;this isbeyondthe scope ofthis
book.16-6 AdiabaticProcessesAnadiabaticprocessisoneinwhichnoheat
entersorleavesthesystem.In theformof anequationflQ=O.If we applythe
first lawof thermodynamics, mthe formof Equation(16-17),
toadiabaticprocesses, wefindflQ=0=Cv flT + PflV.ThusflVflT=
-p--.CvThe general gas law describes the behavior of a gas under
all circumstances.Thus foramole of gas we havePV=RT.IfP, V,
andTarepermittedsmall variations, wefind, ontakingdiffer-entials,
thatPfl V + V flP= R flT.312 KINETIC THEORY OF GASES 16-6If
wereplace thegas constant RbyR= Cp - CyfromEquation (16-19), and
substitute the value of t:,.T into the aboveequation, we
findPt:,.VPt:,.V + Vt:,.P=-(Cp - Cy ) -- =-("I - 1) Pt:,.V,Cyfor,
fromEquation (16-22), "I = CpjCy . Ontransposing, and
dividingtheaboveequation by theproduct PV, we findt:,.P t:,.Vp+'Y
y= O.Inthelimit ofsmallincrementswemayreplace t:,. byd, and
integratetofindorIn P +"I In V = constant,PV'Y= constant.
(16-26)The value of the constant is determined by the quantity of
gas present, sothat Equation(16-26)
maybeusedtodescribetherelationshipbetweenthe pressure and the
volume of any quantity of gas undergoing an adiabaticchange. If a
gas originally at pressure PI, volumeVI, and temperature TIis
compressed adiabatically, as in an insulated cylinder,
toanewpressurePz, the new volume of thegasVz may be found
fromEquation (16-26)bywriting(16-27)The final temperature of the
gas Tz may then be obtained from the gas lawPI VI PzVz--=--,TITzin
whichallquantities except Tz are nowknown.IllustrativeExample. A
massof gasoccupiesavolumeof8 liters at apres-sureof1 atm
andatemperatureof3000abs. Itiscompressedadiabaticallytoa volume of
2 liters. Determine (a) the final pressure and (b) the final
temperature,assuming it to be an ideal gas whose value of "I =
1.5.(a) Thefinal pressureof the gascan bedeterminedwiththe aidof
Equation(10-27) thus,so thatfromwhich16-7 'l'HE MAXWELL
DISTRIBUTION FUNCTION 313(b) Thefinal temperaturecanbefoundwith
theaidof thegeneralgaslawthus,so thatorT2= TlP2V2,PlVlT 3000 b 8
atmX212 = as,1 atmX81T2= 6000abs.16-7 TheMaxwell
DistributionFunctionIn Section16-3 weshowed how the properties of
agascouldbe accountedfor onthebasisofaverysimpleset
ofhypothesesabout thenatureofagas. Weassumed that
thegaswasmadeupofmanymoleculesinrapidmotion, and that the molecules
were sufficiently far apart so that the forcesN3SpeedFig. 16-5 The
Maxwellian distribution of molecular speeds. Relative numbers
ofmolecules having speeds in a unit speed interval at various
speeds are shown as ordinate,while speeds in unitsof the
mostprobablespeedare shownasabscissa.thatone molecule exertedon
another wereof minor importance andcouldbeneglected. We assumedthat
the molecules were perfectlyelastic sothat there was no loss in
mechanical energy in collisions between moleculesof the gas and the
wallsof the container. On the basisof such arguments,wecould
accountforthegaslaw, andwewereable toshow that the
tem-peratureofagaswasdirectlyrelatedtotheaveragevalueofthekineticenergyof
its molecules.Whenagasisinequilibriumat anabsolutetemperatureT,
thedis-tributionofvelocitiesofthemoleculesofthegasisgivenbyFigure16-5,called
the Maxwelliandistribution, according to the theory
firstdevelopedby Maxwell (1831-1879). This theory has now been well
verified by experi-ment as actuallydescribingthebehavior of gas
molecules. Infact, theMaxwelliandistributionmaybe takenas the
meaningof thetemperature314 KINETIC THEORY OF GASES 16-7of a gas,
for, if a collectionof gas molecules has a
velocitydistributionwhichdiffersfromFigure16-5, thenwe maysay
thatthegashas not yetreachedthermal equilibriumandtherefore does
not have a
well-definedtemperature.Theaverageofthevelocitycomponentsofthemoleculesofagasinaparticular
directionmust bezero. Ifthiswerenot so,
thegasanditscontainerwouldbeintranslational motion. However,
theaveragevalueof the squares of the molecular velocities is
notzero, and is given by Equa-tion(16-11). FromFigure16-5weseethat
molecules whosespeedsaremore than three times the most probable
speed are extremely rare. Never-theless, thereare some molecules in
thegaswhichhaveverylarge speeds,for
thedistributioncurveapproachesthehorizontal axis asymptotically.We
mustalso note that, atagiventemperature, themoleculesof agasoflow
molecular weight are in more rapid motion than the molecules of a
gasof high molecular weight. This has the interesting consequence
thathydrogenandheliumaresteadilydiffusingout
oftheearth'satmosphere,for, at the temperature of theouter air,
some of these lighter molecules aremoving sufficiently rapidly to
attain the escape velocity of 11 km/secnecessary foraprojectile
toescape thegravitational pull oftheearth.If we call the energy
required to disrupt a chemical molecule its bindingenergy, we see
that, as the temperature of agas is raised, agreater propor-tionof
gasmoleculesmayhavekineticenergiesgreater thanthebindingenergy, so
that a molecule may be decomposed as a result of energy
transferduringacollision. Thusmoleculeswhicharestable at
ordinarytempera-turesmust
havebindingenergieswhicharelargecomparedtothemeankineticenergyof
amolecule at roomtemperature,
asgivenbyEquation(16-11).Itisinterestingthat moderntheories of
thestructureof atomsandmolecules have provided a justification of
abasic assumption of the kinetictheoryof gases. According to
thequantumtheory, molecules existonly incertainquantumstates,
eachhavingafixedamount ofenergy.
Thesearesometimescalledenergylevels.
Themoleculenormallyexistsinitsstateoflowest energy,
calleditsgroundstate, andcanonlyabsorbenergyinacollision with
another molecule in exactly theright amount toraise it
toastateofhigherenergy, calledanexcitedstate, ortodisrupt it
completely.In general, it is very unlikely that the colliding
molecules will have just theright amount of energyfor excitation,
sothat thecollisionsbetweenthemolecules of a gas result in no
absorption of energy by the molecules. Thekineticenergyis
conservedinthecollisionratherthanbeingtransferredto internal
excitation energy ofone molecule. The collisions
arethereforeperfectly elastic.PROBLEMS 315Problems16-1. A closed
vessel contains dry air at 25C and 76 cm of mercury pressure.Its
temperature is raised to 100C. Determine the pressure of the air,
neglectingthechange in volumeof thecontainer.16-2.
Amassofoxygenoccupies avolumeof1 liter atapressureof76cmof mercury
when its temperature is40C. The gas is allowed to expand until
itsvolume is1.5 liters and its pressure is 80 cm of mercury. (a)
Determine its finaltemperature. (b) Determine thenumberof molesof
oxygenin thesystem.16-3. Derivethegeneral gas lawfromEquations
(16-1) to(16-3) bycon-sideringthat the gas is taken fromTiPi Vi bya
constant-volume process toT 2Pf Vi,
andthencebyaconstant-pressureprocesstoTfPf Vf'16-4. Acertain
gashasadensityof0.001gm/cm3whenitstemperatureis50Candits pressure
is 4atm. What pressure will be neededtochange thedensityof thegas
to0.002 gm/cm3when itstemperatureis100C?16-5. Anautomobiletirehas
avolumeof 1,000in.3andcontains air at agauge pressure of
24Ib/in.2whenthe temperature is ODC. ,"Vhat will be
thegaugepressureof theairin thetireswhen
itstemperaturerisesto27DCanditsvolume increases to1,020 in.3?16-6.
Determine the pressure of 4.032gmof hydrogen which occupies avolume
of 16.8 liters at a temperature of OC. The molecular weight
of-hydrogenis2.016.16-7. Determinetheaveragevalue of thekinetic
energyof the moleculesofagas (a) at ODCand(b) at 100C.16-8. (a)
What is the mass of a hydrogen molecule? (b) Determine
theaveragevelocityof amoleculeof hydrogen at27DC.16-9.
Themoleculesof acertaingashaveamassof5X10-24gm.
Whatisthenumberofmoleculespercubiccentimeterof thisgas
whenitspressureis106dynes/cm2anditstemperatureis27DC?16-10.
Calculatetheworkdoneincompressingonemoleofoxygenfromavolumeof22.4litersat
ODCand1 atmpressureto16.8litersat thesametem-perature.16-11. A
cylinder contains amoleof hydrogen atODC and76 cm of
mercurypressure. Calculate the amount of heat requiredto raise the
temperature of thishydrogen to50DC(a)keeping the pressureconstant,
and(b)keeping thevolumeconstant. (c) Whatisthevolumeof
thehydrogenwhen atODC?16-12. Acylindercontains32gmofoxygen at
ODCand76cmofmercurypressure. Calculate the amount of heat
requiredto raise the temperatureof thismassofoxygento80DC(a)
keepingthepressureconstant and(b) keepingthevolume constant. (c)
Howmuch mechanical workis done bythe oxygenineachcase?16-13. A
massofamonatomicgasoccupiesavolumeof 400cm3atatem-peratureof
l7Candapressureof 76cmof mercury. Thegasis
compressedadiabaticallyuntil its pressure is 90cmof mercury.
Determine (a) the finalvolumeof thegasand(b) thefinal temperatureof
thegas.16-14. A mass of a diatomic gas occupies a volume of 6
liters at a temperature316 KINETIC THEORY OF GASESof 27C and75cmof
mercurypressure. The gasexpandsadiabaticallyuntil
itsvolumeis8liters. What isthefinal temperatureofthegas?16-15.
Amole of gasat atmospheric pressure andOCis
compressediso-thermallyuntil its pressureis 2atm. Howmuchmechanical
workis doneonthegasduring thisoperation?16-16. Tengrams of
oxygenare heatedat constant atmospheric pressurefrom27C to127C. (a)
Howmuchheat isdeliveredtotheoxygen? (b) Whatfractionoftheheat
isusedtoraisetheinternalenergyof theoxygen?16-17. An air bubbleof
volume20cm3 is at thebottom of alake 40 mdeepwhere the temperature
is 4C. The bubble rises to the surface where the temper-ature is
20C. Assumingthat the temperature of the bubble is the same
asthatof thesurroundingwater, whatisitsvolumejust as
itreachesthesurface?16-18. Anideal gasfor which l' =
1.5isenclosedina cylinderof volume1 m3 under apressure of 3 atm.
The gas is expanded adiabatically to apressureof1 atm. Find(a)
thefinal volumeand(b) thefinal temperatureof thegasifits initial
temperaturewas20C.16-19. Amass of 1.3kgof oxygenofmolecular weight
32isenclosedinacylinderof volume1 m3 atapressureof 105nt/m2
andatemperatureof20C.Fromthesedatafindthe universal gas constant
Rassumingoxygentobeanideal gas.16-20. Agas ofmass mandmolecular
weight Mundergoesanisothermalexpansionfromaninitial pressurePI
andvolumeV I toafinal pressureP 2andvolumeV 2 while attemperatureT.
Find(a) theworkdonebythegasinthisexpansion, (b) theheat
flowtothegas, and(c) thechange ininternal energyofthegasin termsof
these symbols.16-21. A piece of putty is placedin avisewith
insulating jaws. A constantforceof100 nt is applied through
adistance of 2 cm. The putty isfoundnot tohaveits volume
changedinthis process. What is the changein theinternalenergy of
theputty?16-22. Prove that TV'Y-I= constantfor an
adiabaticprocess.16-23. Showthat the workdonebyagas
inanadiabaticexpansionfrominitialconditionsPi, Vitofinal
conditionsP1> Vf is givenby}f' = Pi Vi (V}-'Y_ V ~ - Y .1-
l'