Thepropertiesofgases
1A Theperfectgas
Answerstodiscussionquestions
D1A.1 Anequationofstateisanequationthatrelatesthevariablesthatde nethe stateofasystemtoeachother.Boyle,Charles,andAvogadroestablishedthese relationsforgasesatlowpressures(perfectgases)byappropriateexperiments.
Boyledeterminedhowvolumevarieswithpressure(V ∝ 1/ p),Charleshow volumevarieswithtemperature(V ∝ T ),andAvogadrohowvolumevaries withamountofgas(V ∝ n ).Combiningalloftheseproportionalitiesintoone gives
∝ nT p
Insertingtheconstantofproportionality, R ,yieldstheperfectgasequation
V = R nT p or pV = nRT
Solutionstoexercises
E1A.1(a) Fromtheinsidethefrontcovertheconversionbetweenpressureunitsis:1atm ≡ 101.325kPa ≡ 760Torr;1baris105 Paexactly.
(i) Apressureof108kPaisconvertedtoTorrasfollows 108kPa × 1atm 101.325kPa × 760Torr 1atm = 810Torr
(ii) Apressureof0.975baris0.975 × 105 Pa,whichisconvertedtoatmas follows 0.975 × 105 Pa × 1atm 101.325kPa = 0.962atm
E1A.2(a) eperfectgaslaw[1A.4–8], pV = nRT ,isrearrangedtogivethepressure, p = nRT /V eamount n isfoundbydividingthemassbythemolarmassof Xe,131.29gmol 1 p = n (131g) (131.29gmol 1 ) (8.2057 × 10 2 dm3 atmK 1 mol 1 ) × (298.15K) 1.0dm3 = 24.4atm
Sono ,thesamplewouldnotexertapressureof20atm,but24.4atm ifitwere aperfectgas.
E1A.3(a) Becausethetemperatureisconstant(isothermal)Boyle’slawapplies, pV = const. ereforetheproduct pV isthesamefortheinitialand nalstates
einitialvolumeis2.20dm3 greaterthanthe nalvolumeso Vi = 4.65+2.20 = 6.85dm3 . p i = Vf Vi × p f = 4.65dm3 6.85dm3 × (5.04bar) = 3.42bar
(i) einitialpressureis3.42bar
(ii) Becauseapressureof1atmisequivalentto1.01325bar,theinitialpressure expressedinatmis 1atm 1.01325bar × 3.40bar = 3.38atm
E1A.4(a) Ifthegasisassumedtobeperfect,theequationofstateis[1A.4–8], pV = nRT . Inthiscasethevolumeandamount(inmoles)ofthegasareconstant,soit followsthatthepressureisproportionaltothetemperature: p ∝ T eratio ofthe nalandinitialpressuresisthereforeequaltotheratioofthetemperatures: p f / p i = Tf / Ti . epressureindicatedonthegaugeisthatinexcess ofatmosphericpressure,thustheinitialpressureis24 + 14.7 = 38.7lbin 2 . Solvingforthe nalpressure p f (remembertouseabsolutetemperatures)gives p f = Tf Ti × p i = (35 + 273.15) K ( 5 + 273.15) K × (38.7lbin 2 ) = 44.4...lbin 2
epressureindicatedonthegaugeisthis nalpressure,minusatmospheric pressure:44.4... 14.7 = 30lbin 2 . isassumesthat(i)thegasisbehaving perfectlyand(ii)thatthetyreisrigid.
E1A.5(a) eperfectgaslaw pV = nRT isrearrangedtogivethepressure p = nRT V = n 255 × 10 3 g 20.18gmol 1 × (8.3145 × 10 2 dm3 barK 1 mol 1 ) × (122K) 3.00dm3 = 0.0427bar
E1A.6(a)
Notethechoiceof R tomatchtheunitsoftheproblem.Analternativeisto use R = 8.3154JK 1 mol 1 andadjusttheotherunitsaccordingly,togivea pressureinPa.
p = [(255 × 10 3 g)/(20.18gmol 1 )] × (8.3145JK 1 mol 1 ) × (122K) 3.00 × 10 3 m3
= 4.27 × 105 Pa
where1dm3 = 10 3 m3 hasbeenusedalongwith1J = 1kgm2 s 2 and1Pa = 1kgm 1 s 2
evapourisassumedtobeaperfectgas,sothegaslaw pV = nRT applies. e taskistousethisexpressiontorelatethemeasuredmassdensitytothemolar mass.
First,theamount n isexpressedasthemass m dividedbythemolarmass M to give pV = (m / M )RT ;divisionofbothsidesby V gives p = (m /V )( RT / M ). equantity (m /V ) isthemassdensity ρ ,so p = ρ RT / M ,whichrearranges to M = ρ RT / p;thisistherequiredrelationshipbetween M andthedensity.
M = ρ RT p = (3.710kgm 3 ) × (8.3145JK 1 mol 1 ) × ([500 + 273.15] K) 93.2 × 103 Pa = 0.255...kgmol 1
where1J = 1kgm2 s 2 and1Pa = 1kgm 1 s 2 havebeenused. emolarmass ofSis32.06gmol 1 ,sothenumberofSatomsinthemoleculescomprising thevapouris (0.255... × 103 gmol 1 )/(32.06gmol 1 ) = 7.98. eresultis expectedtobeaninteger,sotheformulaislikelytobeS8 .
E1A.7(a) evapourisassumedtobeaperfectgas,sothegaslaw pV = nRT applies;the taskistousethisexpressiontorelatethemeasureddatatothemass m . is isdonebyexpressingtheamount n as m / M ,where M isthethemolarmass. Withthissubstitutionitfollowsthat m = MPV / RT . epartialpressureofwatervapouris0.60timesthesaturatedvapourpressure m = MpV RT = (18.0158gmol 1 ) × (0.60 × 0.0356 × 105 Pa) × (400m3 ) (8.3145JK 1 mol 1 ) × ([27 + 273.15] K) = 6.2 × 103 g = 6.2kg
E1A.8(a) Consider1m3 ofair:themassofgasistherefore1.146kg.Ifperfectgasbehaviourisassumed,theamountinmolesisgivenby n = pV / RT
n = pV RT = (0.987 × 105 Pa) × (1m3 ) (8.3145JK 1 mol 1 ) × ([27 + 273.15] K) = 39.5...mol
(i) etotalamountinmolesis n = n O2 + n N2 . etotalmass m iscomputed fromtheamountsinmolesandthemolarmasses M as
m = n O2 × M O2 + n N2 × M N2
esetwoequationsaresolvedsimultaneouslyfor n O2 togivethefollowingexpression,whichisthenevaluatedusingthedatagiven
n O2 = m M N2 n M O2 M N2 = (1146g) (28.02gmol 1 ) × (39.5...mol) (32.00gmol 1 ) (28.02gmol 1 ) = 9.50...mol
emolefractionsaretherefore x O2 = n O2 n = 9.50...mol 39.5...mol = 0.240 x N2 = 1 x O2 = 0.760
epartialpressuresaregivenby p i = x i p tot
p O2 = x O2 p tot = 0.240(0.987bar) = 0.237bar
p N2 = x N2 p tot = 0.760(0.987bar) = 0.750bar
(ii) esimultaneousequationstobesolvedarenow n = n O2 + n N2 + n Ar m = n O2 M O2 + n N2 M N2 + n Ar M Ar
Becauseitisgiventhat x Ar = 0.01,itfollowsthat n Ar = n /100. etwo unknowns, n O2 and n N2 ,arefoundbysolvingtheseequationssimultaneouslytogive
n N2 = 100 m n ( M Ar + 99 M O2 ) 100( M N2 M O2 ) = 100 × (1146g) (39.5...mol) × [(39.95gmol 1 ) + 99 × (32.00gmol 1 )]
100 × [(28.02gmol 1 ) (32.00gmol 1 )] = 30.8...mol
From n = n O2 + n N2 + n Ar itfollowsthat
n O2 = n n Ar n N2 = (39.5...mol) 0.01 × (39.5...mol) (30.8...mol) = 8.31...mol
emolefractionsare
x N2 = n N2 n = 30.8...mol 39.5...mol = 0.780 x O2 = n O2 n = 8.31...mol 39.5...mol = 0.210
epartialpressuresare
p N2 = x N2 p tot = 0.780 × (0.987bar) = 0.770bar
p O2 = x O2 p tot = 0.210 × (0.987bar) = 0.207bar
E1A.9(a)
Note:the nalvaluesarequitesensitivetotheprecisionwithwhichtheintermediateresultsarecarriedforward.
evapourisassumedtobeaperfectgas,sothegaslaw pV = nRT applies. e taskistousethisexpressiontorelatethemeasuredmassdensitytothemolar mass.
First,theamount n isexpressedasthemass m dividedbythemolarmass M to give pV = (m / M )RT ;divisionofbothsidesby V gives p = (m /V )( RT / M ) equantity (m /V ) isthemassdensity ρ ,so p = ρ RT / M ,whichrearranges to M = ρ RT / p;thisistherequiredrelationshipbetween M andthedensity. M = ρ RT p = (1.23kgm 3 ) × (8.3145JK 1 mol 1 ) × (330K)
20.0 × 103 Pa = 0.169kgmol 1
erelationships1J = 1kgm2 s 2 and1Pa = 1kgm 1 s 2 havebeenused.
E1A.10(a) Charles’law[1A.3b–7]statesthat V ∝ T atconstant n and p,and p ∝ T at constant n and V .Fora xedamountthedensity ρ isproportionalto1/V ,soit followsthat1/ ρ ∝ T .Atabsolutezerothevolumegoestozero,sothedensity goestoin nityandhence1/ ρ goestozero. eapproachisthereforetoplot 1/ ρ againstthetemperature(in ○ C)andthenbyextrapolatingthestraightline ndthetemperatureatwhich1/ ρ = 0. eplotisshowninFig1.1.
θ /○ C ρ /(gdm 3 )(1/ ρ )/(g 1 dm3 )
851.877 0.5328 01.294 0.7728
1000.946 1.0571
edataareagood ttoastraightline,theequationofwhichis (1/ ρ )/(g 1 dm3 ) = 2.835 × 10 3 × (θ /○ C) + 0.7734
einterceptwith1/ ρ = 0isfoundbysolving 0 = 2.835 × 10 3 × (θ /○ C) + 0.7734
isgives θ = 273 ○ C astheestimateofabsolutezero.
E1A.11(a) (i) emolefractionsare x H2 = n H2 n H2 + n N2 = 2.0mol 2.0mol + 1.0mol = 2 3 x N2 = 1 x H2 = 1 3
(ii) epartialpressuresaregivenby p i = x i p tot . etotalpressureisgiven bytheperfectgaslaw: p tot = n tot RT /V
p H2 = x H2 p tot = 2 3 × (3.0mol) × (8.3145JK 1 mol 1 ) × (273.15K) 22.4 × 10 3 m3 = 2.0 × 105 Pa
p N2 = x N2 p tot = 1 3 × (3.0mol) × (8.3145JK 1 mol 1 ) × (273.15K) 22.4 × 10 3 m3 = 1.0 × 105 Pa
Expressedinatmospherestheseare2.0atmand1.0atm,respectively.
(iii) etotalpressureis
(3.0mol) × (8.3145JK 1 mol 1 ) × (273.15K) 22.4 × 10 3 m3 = 3.0 × 105 Pa or3.00atm.
Alternatively,notethat1molatSTPoccupiesavolumeof22.4dm3 ,whichis thestatedvolume.Asthereareatotalof3.0molpresentthe(total)pressure mustthereforebe3.0atm.
Solutionstoproblems
P1A.1 (a) eexpression ρ gh givesthepressureinPaifallthequantitiesarein SIunits,soitishelpfultoworkinPathroughout.Fromthefrontcover, 760Torrisexactly1atm,whichis1.01325×105 Pa. edensityof13.55gcm 3 isequivalentto13.55 × 103 kgm 3 .
p = p ex + ρ gh
= 1.01325 × 105 Pa + (13.55 × 103 kgm 3 ) × (9.806ms 2 )
× (10.0 × 10 2 m) = 1.15 × 105 Pa
Figure1.1
(b) ecalculationofthepressureinsidetheapparatusproceedsasin(a)
p = 1.01325 × 105 Pa + (0.9971 × 103 kgm 3 ) × (9.806ms 2 ) × (183.2 × 10 2 m) = 1.192... × 105 Pa
evalueof R isfoundbyrearrangingtheperfectgaslawto R = pV / nT
R = pV nT = (1.192... × 105 Pa) × (20.000 × 10 3 m3 ) [(1.485g)/(4.003gmol 1 )] × ([500 + 273.15] K)
= 8.315JK 1 mol 1
P1A.3 eperfectgaslaw pV = nRT impliesthat pVm = RT ,where Vm isthemolar volume(thevolumewhen n = 1).Itfollowsthat p = RT /Vm ,soaplotof p against T /Vm shouldbeastraightlinewithslope R . However,realgasesonlybecomeidealinthelimitofzeropressure,sowhatis neededisamethodofextrapolatingthedatatozeropressure.Oneapproachis torearrangetheperfectgaslawintotheform pVm / T = R andthentorealise thatthisimpliesthatforarealgasthequantity pVm / T willtendto R inthelimit ofzeropressure. erefore,theinterceptat p = 0ofaplotof pVm / T against p isanestimateof R .Fortheextrapolationofthelinebackto p = 0tobereliable, thedatapointsmustfallonareasonablestraightline. eplotisshownin Fig1.2. p/atm Vm /(dm3 mol 1 )( pVm / T )/(atmdm3 mol 1 K 1 )
0.75000029.8649 0.0820014
0.50000044.8090 0.0820227
0.25000089.6384 0.0820414
p/atm ( pV m / T )/( atmdm 3 mol 1 K 1 )
Figure1.2
edatafallonareasonablestraightline,theequationofwhichis ( pVm / T )/(atmdm3 mol 1 K 1 ) = 7.995 × 10 5 × ( p/atm) + 0.082062
eestimatefor R isthereforetheintercept,0.082062atmdm3 mol 1 K 1 . edataaregivento6 gures,buttheydonotfallonaverygoodstraightline sothevaluefor R hasbeenquotedtoonefewersigni cant gure.
P1A.5 Foraperfectgas pV = nRT whichcanberearrangedtogive p = nRT /V . e amountinmolesis n = m / M ,where M isthemolarmassand m isthemassof thegas. erefore p = (m / M )( RT /V ). equantity m /V isthemassdensity ρ ,andhence
p = ρ RT / M
Itfollowsthatforaperfectgas p/ ρ shouldbeaconstantatagiventemperature. Realgasesareexpectedtoapproachthisasthepressuregoestozero,soa suitableplotisof p/ ρ against p;theinterceptwhen p = 0givesthebestestimate of RT / M . eplotisshowninFig.1.3.
p/kPa ( p / ρ )/( kPakg 1 m 3 )
Figure1.3
edatafallonareasonablestraightline,theequationofwhichis ( p/ ρ )/(kPakg 1 m 3 ) = 0.04610 × ( p/kPa) + 53.96
P1A.7
einterceptis ( p/ ρ )lim p →0 ,whichisequalto RT / M . M = RT ( p/ ρ )lim p →0 = (8.3145JK 1 mol 1 ) × (298.15K) 53.96 × 103 Pakg 1 m3 = 4.594×10 2 kgmol 1
eestimateofthemolarmassistherefore45.94gmol 1 .
(a) Foraperfectgas pV = nRT soitfollowsthatforasampleatconstant volumeandtemperature, p 1 / T1 = p 2 / T2 .Ifthepressureincreasesby ∆ p foranincreaseintemperatureof∆T ,thenwith p 2 = p 1 + ∆ p and T2 = T1 + ∆T isfollowsthat
Foranincreaseby1.00K,∆T = 1.00Kandhence
= (6.69 × 103 Pa) × (1.00K) 273.16K = 24.5Pa
Anotherwayoflookingatthisistowritetherateofchangeofpressure withtemperatureas
p
T = p 1 T1 = 6.69 × 103 Pa 273.16K = 24.5...PaK 1
(b) Atemperatureof100.00 ○ Cisequivalenttoanincreaseintemperature fromthetriplepointby100.00 + 273.15 273.16 = 99.99K
p′ = ∆T ′ × ( ∆ p ∆T ) = (99.99K) × 6.69 × 103 Pa 273.16K = 2.44... × 103 Pa
e nalpressureistherefore6.69 + 2.44... = 9.14kPa
(c) Foraperfectgas∆ p/∆T isindependentofthetemperaturesoat100.0 ○ C a1.00Kriseintemperaturegivesapressureriseof24.5Pa ,justasin(a).
P1A.9 emolarmassofSO2 is32.06 + 2 × 16.00 = 64.06gmol 1 .Ifthegasisassumed tobeperfectthevolumeiscalculatedfrom pV = nRT
V = nRT p = n ( 200 × 106 g 64.06gmol 1 ) (8.3145JK 1 mol 1 ) × ([800 + 273.15] K) 1.01325 × 105 Pa
= 2.7 × 105 m3
Notetheconversionofthemassinttomassing;repeatingthecalculationfor 300tgivesavolumeof4.1 × 105 m3 . evolumeofgasisthereforebetween0.27km3 and0.41km3
P1A.11
Imagineacolumnoftheatmospherewithcrosssectionalarea A. epressure atanyheightisequaltotheforceactingdownonthatarea;thisforcearises fromthegravitationalattractiononthegasinthecolumn above thisheight–thatis,the‘weight’ofthegas.
Supposethattheheight h isincreasedbyd h . eforceonthearea A isreduced becauselessoftheatmosphereisnowbearingdownonthisarea.Speci cally, theforceisreducedbythatduetothegravitationalattractiononthegascontainedinacylinderofcross-sectionalarea A andheightd h .Ifthedensityof thegasis ρ ,themassofthegasinthecylinderis ρ × A d h andtheforcedueto gravityonthismassis ρ gA d h ,where g istheaccelerationduetofreefall. e changeinpressured p onincreasingtheheightbyd h isthisforcedividedby thearea,soitfollowsthat d p = ρ g d h
eminussignisneededbecausethepressuredecreasesastheheightincreases. edensityisrelatedtothepressurebystartingfromtheperfectgasequation, pV = nRT .Ifthemassofgasis m andthemolarmassis M ,itfollowsthat n = m / M andhence pV = (m / M )RT .Takingthevolumetotherightgives p = (m / MV )RT equantity m /V isthemassdensity ρ ,so p = ( ρ / M )RT ; thisisrearrangedtogiveanexpressionforthedensity: ρ = Mp/ RT isexpressionfor ρ issubstitutedintod p = ρ g d h togived p = ( Mp/ RT ) g d h . Divisionby p resultsinseparationofthevariables (1/ p) d p = ( M / RT ) g d h ele -handsideisintegratedbetween p 0 ,thepressureat h = 0and p,the pressureat h eright-handsideisintegratedbetween h = 0and h
p p
1
0 Mg RT d h [ln p] p p 0 = Mg RT [ h ] h 0 ln p p 0 = Mgh RT
eexponentialofeachsideistakentogive p = p 0 e h / H with H = RT Mg
Itisassumedthat g and T donotvarywith h .
(a) epressuredecreaseacrosssuchasmalldistancewillbeverysmallbecause h / H ≪ 1.Itisthereforeadmissibletoexpandtheexponentialand retainjustthe rsttwoterms:e x ≈ 1 + x p = p 0 (1 h / H )
isisrearrangedtogiveanexpressionforthepressuredecrease, p p 0
p p 0 = p 0 h / H
P1A.13
Ifitisassumedthat p 0 isoneatmosphereandthat H = 8km, p p 0 = p 0 h / H = (1.01325 × 105 Pa) × (15 × 10 2 m) 8 × 103 m = 2Pa (b) epressureat11kmiscalculatedusingthefullexpression p = p 0 e h / H = (1atm) × e (11km)/(8km) = 0.25atm
Imagineavolume V oftheatmosphere,attemperature T andpressure p tot . Iftheconcentrationofatracegasisexpressedas X partspertrillion(ppt),it meansthatifthatgaswerecon nedtoavolume X × 10 12 × V attemperature T iswouldexertapressure p tot .Fromtheperfectgaslawitfollowsthat n = pV / RT ,whichinthiscasegives n trace = p tot ( X × 10 12 × V ) RT
Takingthevolume V tothele givesthemolarconcentration, c trace
c trace = n trace V = X × 10 12 × p tot RT
Analternativewayoflookingatthisistonotethat,atagiventemperatureand pressure,thevolumeoccupiedbyagasisproportionaltotheamountinmoles. Sayingthatagasispresentat X pptimpliesthatthevolumeoccupiedbythe gasis X × 10 12 ofthewhole,andthereforethattheamountinmolesofthegas is X × 10 12 ofthetotalamountinmoles
n trace = ( X × 10 12 ) × n tot
isisrearrangedtogiveanexpressionforthemolefraction x trace
x trace = n trace n tot = X × 10 12 epartialpressureofthetracegasistherefore
p trace = x trace p tot = ( X × 10 12 ) × p tot
econcentrationis n trace /V = p trace / RT ,so
c trace = n trace V = X × 10 12 × p tot RT
(a) At10 ○ Cand1.0atm
c CCl3 F = X CCl3 F × 10 12 × p tot RT = 261 × 10 12 × (1.0atm) (8.2057 × 10 2 dm3 atmK 1 mol 1 ) × ([10 + 273.15] K) = 1.1 × 10 11 moldm 3
c CCl2 F2 = X CCl2 F2 × 10 12 × p tot RT = 509 × 10 12 × (1.0atm) (8.2057 × 10 2 dm3 atmK 1 mol 1 ) × ([10 + 273.15] K) = 2.2 × 10 11 moldm 3
(b) At200Kand0.050atm
c CCl3 F = X CCl3 F × 10 12 × p tot RT = 261 × 10 12 × (0.050atm) (8.2057 × 10 2 dm3 atmK 1 mol 1 ) × (200K)
= 8.0 × 10 13 moldm 3
c CCl2 F2 = X CCl2 F2 × 10 12 × p tot RT = 509 × 10 12 × (0.050atm) (8.2057 × 10 2 dm3 atmK 1 mol 1 ) × (200K)
= 1.6 × 10 12 moldm 3
1B Thekineticmodel
Answertodiscussionquestions
D1B.1 ethreeassumptionsonwhichthekineticmodelisbasedaregiveninSection1B.1onpage11.
1. egasconsistsofmoleculesinceaselessrandommotionobeyingthe lawsofclassicalmechanics.
2. esizeofthemoleculesisnegligible,inthesensethattheirdiameters aremuchsmallerthantheaveragedistancetravelledbetweencollisions; theyare‘point-like’.
3. emoleculesinteractonlythroughbriefelasticcollisions.
Anelasticcollisionisacollisioninwhichthetotaltranslationalkineticenergy ofthemoleculesisconserved.
D1B.3
Noneoftheseassumptionsisstrictlytrue;however,manyofthemaregoodapproximationsunderawiderangeofconditionsincludingconditionsofambient temperatureandpressure.Inparticular,
(a) Moleculesaresubjecttolawsofquantummechanics;however,forall butthelightestgasesatlowtemperatures,non-classicale ectsarenot important.
(b) Withincreasingpressure,theaveragedistancebetweenmoleculeswilldecrease,eventuallybecomingcomparabletothedimensionsofthemolecules themselves.
(c) Intermolecularinteractions,suchashydrogenbonding,andtheinteractionsofdipolemoments,operatewhenmoleculesareseparatedbysmall distances. erefore,asassumption(2)breaksdown,sodoesassumption (3),becausethemoleculesareo encloseenoughtogethertointeracteven whennotcolliding.
Foranobject(beitaspacecra oramolecule)toescapethegravitational eldoftheEarthitmustacquirekineticenergyequalinmagnitudetothe gravitationalpotentialenergytheobjectexperiencesatthesurfaceoftheEarth. egravitationalpotentialbetweentwoobjectswithmasses m 1 and m 2 when separatedbyadistance r is
V = Gm 1 m 2 r
where G isthe(universal)gravitationalconstant.Inthecaseofanobjectof mass m atthesurfaceoftheEarth,itturnsoutthatthegravitationalpotential isgivenby
V = GmM R
where M isthemassoftheEarthand R itsradius. isexpressionimpliesthat thepotentialatthesurfaceisthesameasifthemassoftheEarthwerelocalized atadistanceequaltoitsradius.
AsamassmovesawayfromthesurfaceoftheEarththepotentialenergyincreases(becomeslessnegative)andtendstozeroatlargedistances. ischange inpotentialenergymustallbeconvertedintokineticenergyifthemassisto escape.Amass m movingatspeed υ haskineticenergy 1 2 m υ 2 ;thisspeedwill bethe escapevelocity υ e when
equantityinthesquarerootisrelatedtotheaccelerationduetofreefall, g ,inthefollowingway.Amass m atthesurfaceoftheEarthexperiences agravitational force given GMm / R 2 (notethattheforcegoesas R 2 ). is forceacceleratesthemasstowardstheEarth,andcanbewritten mg . etwo expressionsfortheforceareequatedtogive
isexpressionfor GM / R issubstitutedintotheaboveexpressionfor υ e togive
e =
2GM R = √2 Rg
eescapevelocityisthereforeafunctionoftheradiusoftheEarthandthe accelerationduetofreefall.
eradiusoftheEarthis6.37 × 106 mand g = 9.81ms 2 sotheescapevelocity is1.11 × 104 ms 1 .Forcomparison,themeanspeedofHeat298Kis1300ms 1 andforN2 themeanspeedis475ms 1 .ForHe,onlyatomswithaspeedin excessofeighttimesthemeanspeedwillbeabletoescape,whereasforN2 the speedwillneedtobemorethantwentytimesthemeanspeed. efractionof moleculeswithspeedsmanytimesthemeanspeedissmall,andbecausethis fractiongoesase υ 2 itfallso rapidlyasthemultipleincreases.Atinyfraction ofHeatomswillbeabletoescape,butthefractionofheaviermoleculeswith su cientspeedtoescapewillbeutterlynegligible.
Solutionstoexercises
E1B.1(a) (i) emeanspeedisgivenby[1B.9–16], υ mean = (8 RT /π M )1/2 ,so υ mean ∝ √1/ M . eratioofthemeanspeedsthereforedependsontheratioof themolarmasses
υ mean,H2
mean,Hg = ( M Hg M H2 )
(ii) emeantranslationalkineticenergy ⟨ E k ⟩ isgivenby 1 2 m ⟨ υ 2 ⟩,where ⟨ υ 2 ⟩ isthemeansquarespeed,whichisgivenby[1B.7–15], ⟨ υ 2 ⟩ = 3 RT / M . emeantranslationalkineticenergyistherefore
emolarmass M isrelatedtothemass m ofonemoleculeby M = mN A , where N A isAvogadro’sconstant,andthegasconstantcanbewritten R = kN A ,hence
emeantranslationalkineticenergyisthereforeindependentofthe identityofthegas,andonlydependsonthetemperature:itisthesame forH2 andHg.
isresultisrelatedtotheprincipleofequipartitionofenergy:amolecule hasthreetranslationaldegreesoffreedom( x , y ,and z )eachofwhich contributes 1 2 kT totheaverageenergy.
E1B.2(a) ermsspeedisgivenby[1B.8–15], υ rms = (3 RT / M )1/2 . υ rms,H2 = ( 3 RT M H2 )1/2 = ( 3 × (8.3145JK 1 mol 1 ) × (293.15K) 2 × 1.0079 × 10 3 kgmol 1 )1/2 = 1.90kms 1
where1J = 1kgm2 s 2 hasbeenused.Notethatthemolarmassisinkgmol 1 . υ rms,O2 = ( 3 × (8.3145JK 1 mol 1 ) × (293.15K) 2 × 16.00 × 10 3 kgmol 1 )1/2 = 478ms 1
E1B.3(a) eMaxwell–Boltzmanndistributionofspeeds, f (υ ),isgivenby[1B.4–14]. efractionofmoleculeswithspeedsbetween υ 1 and υ 2 isgivenbytheintegral
Iftherange υ 2 υ 1 = δ υ issmall,theintegraliswell-approximatedby f (υ mid ) δ υ
where υ mid isthemid-pointofthevelocityrange: υ mid = 1 2 (υ 2 + υ 1 ).Inthis exercise υ mid = 205ms 1 and δ υ = 10ms 1 .
fraction = f (υ mid ) δ υ = 4π × ( M 2π RT )3/2 υ 2 mid exp ( M υ 2 mid 2 RT ) δ υ = 4π × ( 2 × 14.01 × 10 3 kgmol 1 2π × (8.3145JK 1 mol 1 ) × (400K) )3/2 × (205ms 1 )2 × exp ( (2 × 14.01 × 10 3 kgmol 1 ) × (205ms 1 )2 2 × (8.3145JK 1 mol 1 ) × (400K) ) × (10ms 1 ) = 6.87 × 10 3
where1J = 1kgm2 s 2 hasbeenused. us,0.687%ofmoleculeshavevelocitiesinthisrange.
E1B.4(a) emeanrelativespeedisgivenby[1B.11b–16], υ rel = (8 kT /π µ )1/2 ,where µ = m A m B /(m A + m A ) isthee ectivemass.Multiplyingtopandbottomof theexpressionfor υ rel by N A andusing N A k = R gives υ rel = (8 RT /π N A µ )1/2 inwhich N A µ isthemolare ectivemass.FortherelativemotionofN2 andH2 thise ectivemassis
N A µ = M N2 M H2 M N2 + M H2 = (2 × 14.01gmol 1 ) × (2 × 1.0079gmol 1 ) (2 × 14.01gmol 1 ) + (2 × 1.0079gmol 1 ) = 1.88...gmol 1
υ rel = ( 8 RT π N A µ )1/2 = ( 8 × (8.3145JK 1 mol 1 ) × (298.15K) π × (1.88... × 10 3 kgmol 1 ) )1/2 = 1832ms 1
evalueofthee ectivemass µ isdominatedbythemassofthelightermolecule, inthiscaseH2
E1B.5(a) emostprobablespeedisgivenby[1B.10–16], υ mp = (2 RT / M )1/2 ,themean speedisgivenby[1B.9–16], υ mean = (8 RT /π M )1/2 ,andthemeanrelative speedbetweentwomoleculesofthesamemassisgivenby[1B.11a–16], υ rel = √2 υ mean . M CO2 = 12.01 + 2 × 16.00 = 44.01gmol 1 .
υ mp = ( 2 RT M )1/2 = ( 2 × (8.3145JK 1 mol 1 ) × (293.15K) 44.01 × 10 3 kgmol 1 )1/2 = 333ms 1
υ mean = ( 8 RT π M )1/2 = ( 8 × (8.3145JK 1 mol 1 ) × (293.15K) π × (44.01 × 10 3 kgmol 1 ) )1/2 = 376ms 1 υ rel = √2 υ mean = √2 × (376ms 1 ) = 531ms 1
E1B.6(a) ecollisionfrequencyisgivenby[1B.12b–17], z = συ rel p/ kT ,withtherelative speedfortwomoleculesofthesametypegivenby[1B.11a–16], υ rel = √2 υ mean . emeanspeedisgivenby[1B.9–16], υ mean = (8 RT /π M )1/2 .Fromthe Resourcesection thecollisioncross-section σ is0.27nm2 z = συ rel p kT = σ p kT × √2 × ( 8 RT π M )1/2 = (0.27 × 10 18 m2 ) × (1.01325 × 105 Pa) (1.3806 × 10 23 JK 1 ) × (298.15K) × √2 × ( 8 × (8.3145JK 1 mol 1 ) × (298.15K) π × (2 × 1.0079 × 10 3 kgmol 1 ) )1/2 = 1.7 × 1010 s 1
where1J = 1kgm2 s 2 and1Pa = 1kgm 1 s 2 havebeenused.Notethe conversionofthecollisioncross-section σ tom2 :1nm2 = (1 × 10 9 )2 m2 = 1 × 10 18 m2
E1B.7(a) emeanspeedisgivenby[1B.9–16], υ mean = (8 RT /π M )1/2 ecollision frequencyisgivenby[1B.12b–17], z = συ rel p/ kT ,withtherelativespeedfor twomoleculesofthesametypegivenby[1B.11a–16], υ rel = √2 υ mean . emean freepathisgivenby[1B.14–18], λ = kT / σ p
(i) emeanspeediscalculatedas
υ mean = ( 8 RT π M )1/2 = ( 8 × (8.3145JK 1 mol 1 ) × (298.15K) π × (2 × 14.01 × 10 3 kgmol 1 ) )1/2 = 475ms 1
(ii) ecollisioncross-section σ iscalculatedfromthecollisiondiameter d as σ = π d 2 = π × (395 × 10 9 m)2 = 4.90... × 10 19 m2 .Withthisvalue themeanfreepathiscalculatedas
λ = kT σ p = (1.3806 × 10 23 JK 1 ) × (298.15K) (4.90... × 10 19 m2 ) × (1.01325 × 105 Pa) = 82.9×10 9 m = 82.9nm
where1J = 1kgm2 s 2 and1Pa = 1kgm 1 s 2 havebeenused.
(iii) ecollisionrateiscalculatedas
z = συ rel p kT = σ p kT × √2 × ( 8 RT π M )1/2 = (4.90... × 10 19 m2 ) × (1.01325 × 105 Pa) (1.3806 × 10 23 JK 1 ) × (298.15K) × √2 × ( 8 × (8.3145JK 1 mol 1 ) × (298.15K) π × (2 × 14.01 × 10 3 kgmol 1 ) )1/2 = 8.10 × 109 s 1
Analternativeforthecalculationof z istouse[1B.13–18], λ = υ rel /z , rearrangedto z = υ rel / λ
2 × (475ms 1 ) 82.9 × 10 9 m = 8.10 × 109 s 1
E1B.8(a) econtainerisassumedtobesphericalwithradius r andhencevolume V = 4 3 π r 3 isvolumeisexpressedintermsthetherequireddiameter d = 2r as V = 1 6 π d 3 .Rearrangementofthisexpressiongives d d = ( 6V π )1/3 = ( 6 × 100cm3 π )1/3 = 5.75...cm
emeanfreepathisgivenby[1B.14–18], λ = kT / σ p isisrearrangedto givethepressure p with λ equaltothediameterofthevessel p = kT σ d = (1.3806 × 10 23 JK 1 ) × (298.15K) (0.36 × 10 18 m2 ) × (5.75... × 10 2 m) = 0.20Pa
Notetheconversionofthediameterfromcmtom.
E1B.9(a) emeanfreepathisgivenby[1B.14–18], λ = kT / σ p. λ = kT σ p = (1.3806 × 10 23 JK 1 ) × (217K) (0.43 × 10 18 m2 ) × (0.05 × 1.01325 × 105 Pa) = 1.4 × 10 6 m=1.4µm
Solutionstoproblems
P1B.1 Arotatingslotted-discapparatusconsistsofaseriesofdisksallmountedona commonaxle(sha ).Eachdischasanarrowradialslotcutintoit,andtheslots onsuccessivediscsaredisplacedfromoneanotherbyacertainangle. ediscs arethenspunataconstantangularspeed.
