Kinetic Th of Gases

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University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1-1958 Physics, Chapter 16: Kinetic eory of Gases Henry Semat City College of New York Robert Katz University of Nebraska - Lincoln, [email protected] Follow this and additional works at: hp://digitalcommons.unl.edu/physicskatz Part of the Physics Commons is Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Semat, Henry and Katz, Robert, "Physics, Chapter 16: Kinetic eory of Gases" (1958). Robert Katz Publications. Paper 166. hp://digitalcommons.unl.edu/physicskatz/166

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Kinetic Th of Gases

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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 additional works at: htp://digitalcommons.unl.edu/physicskatzPart of the Physics CommonsTis Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska -Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska -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'