Supplemental Activities - gchem · Boyle’s law states that for a fixed amount of gas at a...
Transcript of Supplemental Activities - gchem · Boyle’s law states that for a fixed amount of gas at a...
SupplementalActivitiesModule:StatesofMatter
Section:GasLaws&GasMixtures–Key
GasLawsActivity1 ThepurposeofthisactivityistopracticeyourunderstandingofGasLaws.
1. Testyourself:howmanyofthegaslawequationscanyouwritedown?
Answersherewillvary.Herearesomeofthevalidanswers:
€
PV = k(or P1V1 = P2V2)
VT
= k
(orV1T2 =V2T1)
Vn
= k
(orV1n2 =V2n1)
PVT
= k
or P1V1T1
=P2V2T2
⎛
⎝ ⎜
⎞
⎠ ⎟
PV = nRT
2. Torr,mmHg,atm,barandPaareallunitsofgaspressure,whichistheratioofthecombinedforceofthegasparticleimpactsandthesurfaceareaofthegascontainer.
3. RobertBoylestudiedtherelationshipbetweenpressureandvolumeofafixedamountofgasataconstanttemperature.ExplainBoyle’slawwithwordsandanequation.
HereistheequationforBoyle’slaw:
€
PV = k(or P1V1 = P2V2)
wherePisthepressureofthegas,Visthevolumeofthegasandkisaconstant.
Boyle’slawstatesthatforafixedamountofgasataconstanttemperature,thepressureandvolumeofthatgasareinverselyproportional.
4. JacquesCharlesstudiedtherelationshipbetweentemperatureandvolumeofafixedamountofgasataconstantpressure.ExplainCharles’lawwithwordsandanequation.
HereistheequationforBoyle’slaw:
€
VT
= k
(orV1T2 =V2T1)
whereVisthevolumeofthegas,Tistheabsolutetemperatureofthegasandkisaconstant.
Charles’lawstatesthatforafixedamountofgasataconstanttemperature,thetemperatureandvolumeofthatgasaredirectlyproportional.
5. LordKelvindevelopedanabsolutetemperaturescalecalledtheKelvinscale.Itdefinesanabsolutezeropointatwhichsubstanceshavetheirminimumvalueofthermalenergyandatwhichpuresubstancesexistasperfectcrystals/lattices.
6. Thecombinedgaslawcombineswhichtwogaslaws?Explainthecombinedgaslawwithwordsandanequation.
ThecombinedgaslawcombinesCharles’lawandBoyle’slaw.
Hereistheequationforthecombinedgaslaw:
€
PVT
= k
or P1V1T1
=P2V2T2
⎛
⎝ ⎜
⎞
⎠ ⎟
wherePisthepressureofthegas,Visthevolumeofthegas,Tistheabsolutetemperatureofthegasandkisaconstant.
Thecombinedgaslawstatesthattheratiooftheproductofthepressureandvolumeofagaswithitsabsolutetemperatureisaconstant.
7. Boyle’slaw,Charles’lawandthecombinedgaslawcanbewrittenastwostatelawsaswellasonestatelaws.Whatismeantwhenwediscussthe“state”ofagassample?
Thestateofagassampleissimplythesetofvariablesthatdefinetheconditionsofthegasataparticularmoment.Sothestateofthegasmayincludethemeasurementsforthepressure,volume,temperatureandamountofthegas.
8. Assumingaconstantmolarquantityofgas,howcouldyouproducethefollowingeffects?
a. DecreasepressureYoucoulddecreasetemperatureorincreasevolume
b. DecreasevolumeYoucoulddecreasetemperatureorincreaseexternalpressure
c. IncreasepressureYoucouldincreasetemperatureordecreasevolume
d. IncreasevolumeYoucouldincreasetemperatureordecreaseexternalpressure
Activity2 Thepurposeofthisactivityistopracticeyourmasteryofthequantitativenatureofgaslaws.
1. Agasoccupies11.2litersat0.860atm.Whatisthepressurewhenthevolumeis15.0liters?Assumethatthetemperatureandamountofgasremainthesame.
Herewerecognizethatonlyvaluesforvolumeandpressurearegivenandaskedforintheproblem.Thetemperatureandamountofgasremainconstant.Therefore,wecanuseBoyle’sLawtosolvethisproblem.
€
P1 = 0.860atmV1 =11.2LV2 =15.0LP2 = ?P1V1 = P2V2(0.860atm)(11.2L) = (P2)(15.0L)
P2 =(0.860atm)(11.2L)
(15.0L)P2 = 0.642atm
Weobservethatasthevolumeofthegasincreased,thepressureofthegasdecreased.ThisobservationmakessenseaccordingtoBoyle’sLaw,whichdescribestheinverselyproportionalrelationshipbetweenvolumeandpressureofafixedamountofgasataconstanttemperature.
2. You’reworkingonyourchemistryhomeworkwithafriend.Yourfriendconsidersaproblemdescribingagasthatiscooledfrom120˚Cto40˚C.Yourfriendassumesthatthevolumemusthavedecreasedbyafactorofthreefrom1.50Ldownto0.50L.Isyourfriendcorrect?Whyorwhynot?
No,yourfriendisnotcorrect.Itisimportanttorememberthatcalculationswiththegaslawsmustbedoneintheabsolutetemperaturescale.Ontheabsolutetemperaturescaleadecreasefrom120˚Cto40˚Ccorrespondstoadecreasefrom393Kto313K.Thevolumeofa
gasisdirectlyproportionaltotheabsolutetemperature(Kelvin)sothevolumewould’veactuallydecreasedfrom1.50Lonlydowntoabout1.19L.
3. Agasoccupies60.0mLat33˚C.Whatchangeinvolumedoesthisgasexperienceifitiscooleddownto5.00˚C?Assumethatthepressureandamountofgasremainconstant.
Herewerecognizethatonlyvaluesforvolumeandtemperaturearegivenandaskedforintheproblem.Thepressureandamountofgasremainconstant.Therefore,wecanuseCharles’Lawtosolvethisproblem.
€
V1 = 0.600LT1 = 33.0ÝC = 306.15KT2 = 5.00ÝC = 278.15KV2 = ?ΔV = ?V1T2 =V2T1(0.600L)(278.15K) = (V2)(306.15K)
V2 =(0.600L)(278.15K)
(306.15K)V2 = 0.545LΔV =V2 −V1 = 0.545L − 0.600L = −0.0549L
Weobservethatasthetemperatureofthegasdecreased,thevolumeofthegasdecreased.ThisobservationmakessenseaccordingtoCharles’Law,whichdescribesthedirectlyproportionalrelationshipbetweenvolumeandtemperatureofafixedamountofgasataconstantpressure.
4. A3250mLgassampleat24.5˚Chasapressureof1825mmHg.Youchangethetemperatureofthegas.Thenewvolumeis4250mLandthenewpressureis1.50atm.Whatisthenewtemperature?
Herewerecognizethatvaluesforvolume,pressuretemperaturearegivenandaskedforintheproblem.Therefore,wemustusetheCombinedGasLawtosolvethisproblem.Notethoughthatthetwopressurevaluesareindifferentunits.Wemustconvertoneofthemsothattheirunitsmatchinordertosolvetheproblem.Let’sconverttoatm.
€
P1 =1825mmHg × 1atm760mmHg
= 2.401atm
V1 = 3250mLT1 = 24.5ÝC = 297.65KP2 =1.50atmV2 = 4250mLT2 = ?P1V1T1
=P2V2T2
(2.401atm)(3250mL)(297.65K)
=(1.50atm)(4250mL)
T2
T2 =(1.50atm)(4250mL)(297.65K)
(2.401atm)(3250mL)T2 = 243.14K = −30.0ÝC
Activity3 ThepurposeofthisactivityistoinvestigateAvogadro’slawandstartdevelopingasmallparticlemodelofgases.
1. AmedeoAvogadrostudiedtherelationshipbetweenvolumeandamount/molesofagasataconstanttemperatureandataconstantpressure.ExplainAvogadro’slawwithwordsandanequation.
HereistheequationforAvogadro’slaw:
€
Vn
= k
(orV1n2 =V2n1)
whereVisthevolumeofthegas,nisthemolesofthegasandkisaconstant.
Avogadro’slawstatesthatforagasataconstanttemperatureandpressure,thevolumeofthatgasisdirectlyproportionaltotheamountofgaspresent.
2. Howdoesasmallparticlesimulatorillustratethemicroscopicbehaviorofgases?Inyourexplanation,describehowaregasesrepresentedandtheobservanceofthegaslaws.
Asmallparticlesimulatorillustratesthemicroscopicbehaviorofgasesbyrepresentingthegasparticlesassmallspheresthatdonotinteractwithoneanotherotherthancollidingelastically.Thegaslawsareobeyedinthesmallparticlesimulator.Forexample,ifthepressureandnumberofparticlesareheldconstantwhilethetemperatureofthesampleis
cooled,thevolumeofthesamplelowers.Inthisway,thesmallparticlesimulatorillustratesCharles’law.
3. Imaginethatyouareanargonatominsideasealedballoonfilledwithonlyargongas.Describeyourbehaviorandhowyourbehavioraffectsthemacroscopicmeasurementsofpressureandvolume.Describewhathappensmacroscopicallyandmicroscopicallytothepressureandvolumeiftheballoonbeginstoleak.
Asanargonatom,youbouncearoundthevolumeoftheballoon.Yourimpactswiththesidesoftheballoon,incombinationwiththeimpactsofyourfellowargonatoms,createaninternalpressurethatisequaltotheexternalpressure.Thankstoyourcollisionswiththesidesoftheballoon,theballoonmaintainsaparticularvolume,whichhasbeendictatedbythepressureoutsidetheballoonandthenumberofatomsinsidetheballoon.Unfortunately,whenyourniceballoonspringsaleak,someofyourfellowargonatomsbegintoleavetheballoon.Therearefeweratomstomakethecollisionnecessarytomaintainaninternalpressurethatisequaltotheexternalpressure.Theballoonbeginstoshrinkandbydecreasingitsvolume,thesurfaceareainsidetheballoonisalsodecreasing.Therefore,evenwithfeweratoms,youandyourremainingcompatriotatomscancollidewiththewallswithenoughfrequencytomaintainequilibriumbetweentheexternalandinternalpressure.
IdealGasLaw
ACTIVITY1ThepurposeofthisactivityistopracticeyourunderstandingoftheIdealGasLaw
1. Howistheidealgaslawdifferentthantheothergaslawspreviouslydiscussed?
Theidealgaslawisaonestatelaw.Theidealgaslawonlyconsidersonesetofconditionsforagassample.Theotherlawsareeitheroneortwostateslaws(theyaremostoftenusedastwostatelaws).
2. Explaintheidealgaslawwithwordsandanequation.
Hereistheequationfortheidealgaslaw:
€
PV = nRT
wherePisthepressureofthegas,Visthevolumeofthegas,nisthemolesofthegas,RistheuniversalgasconstantandTistemperature.
IfwerearrangetheidealgaslawforRweseethattheratioofpressuretimesvolumeovermolestimestemperaturegivesaconstantvalueforanyidealgas.
€
R =PVnT
3. WritedowndifferentvaluesforRthatyoucouldusewhenpressureisinatmandvolumeisinliters,whenpressureisintorrandvolumeisinliters,whenpressureisinPaandvolumeisinm3,andwhenyouneedtocalculatejoules.
Risaconstant.Youcanmanipulatetheunitstobestaddresstheproblemathand.OryoucanconvertothervaluesintheproblemtomatchthevalueofRthatyouwanttouse.ThisproblemhereissimplyhighlightingthefactthatRcanhavemanydifferentunits:
• Pressureinatm,volumeinLàR=0.08206Latmmol–1K–1• PressureinmmHg,volumeinLàR=62.36Ltorrmol–1K–1• PressureinkPa,volumeincm3àR=8.314m3Pamol–1K–1• ForjoulesàR=8.314Jmol–1K–1
4. Whatdoweassumeaboutidealgases?Idealgasesareinfinitelysmall,hardspheresthatdonotinteractwitheachother.Theyareessentially"blind"toothergasmoleculesandwillbounceoffofeachotherjustastheywouldbounceofawall.
5. Youarescubadivinginalargefishtank.Whileyouareatthebottomofthetank,youreleaseaballoonfullofairandwatchitasitrisestothesurface.Whatdoyounoticeaboutthevolumeoftheballoon?Thepressureatthebottomofthetankisgreaterthanthepressureatthetopofthetank,soasitrises,younoticethatitsvolumeincreases.
ACTIVITY2ThepurposeofthisactivityistopracticeyourunderstandingofthequantitativesideoftheIdealGasLaw.
1. GivethedifferentpressureandtemperatureconditionsforbothSTPandSATPforgases.
STP:1atmand0˚C(273.15K) SATP:1barand25˚C(298.15K)
2. Whatdoestheterm“molarvolume”mean?WhatisthevalueformolarvolumeofanidealgasatSTP?
Molarvolumeisthevolumeoccupiedbyonemole.ForidealgasesatSTP,themolarvolumevalueis22.4L.
3. Calculatethemolesofgaspresentina910mLsampleat38˚Cand650Torr.
€
PV = nRTP = 650TorrV = 910mL = 0.910LR = 62.36 L ⋅Torr
mol⋅K
T = 38ÝC = 311.15Kn = ?
n =PVRT
=(650Torr)(0.910L)
(62.36 L ⋅Torrmol⋅K )(311.15K)
n = 3.05 ×10−2mol
4. Howmanyatomsofargongasareina137mLcontainerifthepressureinthecontaineris8.80X10-5mmHgandthetemperatureis794K?
€
PV = nRTP = 8.80 ×10−5mmHgV =137mL = 0.137L
R = 62.36 L ⋅mmHgmol⋅K
T = 794Kn = ?PV = nRT
n =PVRT
=(8.80 ×10−5mmHg)(0.137L)
(62.36 L ⋅mmHgmol ⋅K )(794K)
n = 2.43 ×10−10mol
2.43 ×10−10mol Ar ×6.022 ×1023atoms Ar
1mol Ar=1.47 ×1014atoms Ar
5. Whatisthetemperatureof.75molesofargonina18Lcontainerwithapressureof790Torr?
K
KmolatmLmol
LatmnRPVT
nRTPVmoln
TR
LV
atmtorratmTorrP
nRTPV
KmolatmL
4.304082.075.0
18*04.1
75.0?082.0
18
04.17601790
=
⋅
⋅×
==
=
=
=
=
=
=×=
=
⋅⋅
6. Isitpossiblefor1moleofairinanadult'slungstobeatSTP?Explainandprovebyusetheidealgaslaw.
Noit'snotpossible,youwoulddie.1moleofanidealgasatSTPhasavolumeof22.4L.Thiswouldbewaytoomuchforyourlungs,whichholdabout6L.PV=nRTAtSTP:1atmand273K𝑉 = 𝑛𝑅𝑇/𝑃
𝑉 =1 𝑚𝑜𝑙𝑒 0.082 𝐿𝑎𝑡𝑚𝐾𝑚𝑜𝑙 273𝐾
1 𝑎𝑡𝑚
𝑉 = 22.386 𝐿
GasDensity
ACTIVITY1Thepurposeofthisactivityistodemonstrateathoroughunderstandingofthedifferencebetweennumberdensityandmassdensity
1. Theratioofthemassofasubstanceandthevolumethatthemassoccupiesiswhatisconsideredtobethedensityofthatsubstance.
2. Massdensityistheratioofmassandvolumewhilenumberdensityistheratioofmolesormoleculesandvolume.
3. Assomegivenpressureandtemperature,howcouldtwo500mLclosedcontainersfilledonlywithidealgas(es)havethesamenumberdensitybutdifferentmassdensity?Provideatheoreticalexample.
Twocontainerswiththesamenumberdensityimpliesthattheyeachcontainthesamenumberofparticlesperunitvolume.Theimplicationisproventruewhenweconsidertheidealgaslaw:thetemperature,volumeandpressureareidenticalbetweenthetwocontainersandsothenumberofmolesineachcontainerisidenticalaswell(PV=nRTwhereRisaconstant).However,ifonecontainerwerefilledwithaheaviergas,itwouldhaveagreatermassdensitythanthecontainerwithalightergaseventhoughtheyhavethesamenumberofmolesperunitvolume.Forexample,argongasisaboutfourtimesheavierthanneongas.Ifone500mLcontainerheld2molesofargonandtheotherheld2molesofneon,theargonwouldhaveamassdensityofaboutfourtimesthatofthehelium.
4. Considertwoballoonseachfilledwiththesameamountofheliumgas.BothballoonsareatSATP.Then,youtakeoneballoonandcarefullyplaceitintoanopencontainerofliquidnitrogen(77K).Theotherballoonremainsunchanged.Drawbothballoonsintheirinitialandfinalstatesandincludevaluesfornumberandmassdensityineachdrawing.
Ifbothballoonshavethesameamountofgasandarebothatstandardambienttemperatureandpressure,thentheymusthavethesamevolumeinitially.Therefore,theirnumberdensitieswillbeidenticalinitially.Furthermore,becausethegasesinbothballoonsarethesame,theywillhaveidenticalmassdensitiesinitially.Inthefinalstate,thefirstballoonwillhavethesamevaluesfornumberandmassdensitybecauseitisunchanged.Thesecondballoonhoweverexperiencesatemperatureandvolumedecrease(Charles’law).Let’scalculatethenumberandmassdensitiesforeachballoonineachstate:
€
Initial Number Density
Balloon One and Balloon Two :PV = nRTnV
=PRT
= numberdensity
P =1.0barR = 0.08314 L ⋅bar
mol⋅K
T = 298.15K
nV
=PRT
=(1.0bar)
(0.08314 L ⋅barmol⋅K )(298.15K)
= 0.0403 molL
nV≈ 4.0 ×10−2 mol
L
Noticethatwedidnotneedtocalculatevolumeornumberofmolestodeterminethenumberdensityofthegas.Knowingthatmassandmolesarerelatedthroughthemolar
massofthesample,wecansimplymultiplythenumberdensitybythemolarmassofthegastocalculatemassdensity:
€
Initial Mass Density
Balloon One and Balloon Two :PV = nRTnV
=PRT
mass = n × molarmass
massdensity =nV× molarmass
molarmass = 4.0 gmol
massdensity = 0.0403 molL × 4.0 g
mol = 0.161 gLmassdensity ≈1.6 ×10−1 g
L
Inthefinalstate,balloononewillhavethesamevaluesforitsnumberdensityanditsmassdensitybecausenothingwaschanged.Thesecondballoon(intheliquidnitrogen)iswherewewillfindnewvalues:
€
Final Number Density
Balloon One :nV≈ 4.0 ×10−2 mol
L
Balloon Two :PV = nRTnV
=PRT
= numberdensity
P =1.0barR = 0.08314 L ⋅bar
mol⋅K
T = 77K
nV
=PRT
=(1.0bar)
(0.08314 L ⋅barmol⋅K )(77K)
= 0.156 molL
nV≈1.6 ×10−1 molL
€
Final Mass Density
Balloon One :
massdensity ≈1.6 ×10−1 gL
Balloon Two :PV = nRTnV
=PRT
mass = n × molarmass
massdensity =nV× molarmass
molarmass = 4.0 gmol
massdensity = 0.156 molL × 4.0 g
mol = 0.625 gL
massdensity ≈ 6.2 ×10−1 gL
Duetothedecreaseintemperature,balloontwodecreasedinvolume.However,thisballoonstillcontainedthesameamountofgasasinitsinitialstate.Therefore,weobserveanincreaseinbothitsnumberdensityanditsmassdensity.Noticethatbecausetheliquidnitrogencontainerisopen,thepressureremainsthesameasintheinitialstate.
5. Anadult'slungscanholdabout6L.Whatmassofaircananadultholdatapressureof102kPa?Normalbodytemperatureis37°Candairisabout20%oxygenand80%nitrogen.(101,325Pa=1atm)
Airis20%O2and80%N2,oxygenandnitrogenarediatomicmoleculesOxygen:32g/molNitrogen:28g/molOxygencontributes6.4g/molofairandnitrogencontributes22.4g/molofair.So,airhasatotalof28.8g/mol.(1atm/101,325Pa)(102000Pa)=1.007atm37°C+273=310K
airofgramsmolgmol
molK
molkLatm
LatmRTPVn
nRTPV
__85.68.28238.0
238.0310082.0
6007.1
=×
=×
×==
=
ACTIVITY2Thepurposeofthisactivityispracticeyourunderstandingusinggasdatatocomputemolarmassofgas
1. Useyourknowledgeoftheidealgaslawtowritetwoequationstosolveformolarmassofanidealgas–onewheremassisavariableandtheotherwheredensityisavariable.
€
molarmass =mass × RT
PV
molarmass =ρRTP
2. Giventhata2.16gramsampleofgasoccupies0.43Latapressureof1.2atmatatemperatureof298K.Calculatethemolarmassofthegas.
molarmass = mass×RTPV
mass = 2.16gR = 0.08206 L⋅atm
mol⋅K
T = 298KP =1.2atmV = 0.43L
molarmass = (2.16g)(0.08206L⋅atmmol⋅K )(298K )
(1.2atm)(0.43L)molarmass =102.4 g
mol
3. Agashasamolarmassof100.0g/mol.At25˚C,1.40molesofthegasexertsapressureof380torr.Whatisthedensityofthegasundertheseconditions?
€
PV = nRT
n =PVRT
n =mass
molarmass
denisty = ρ =massvolume
mass = denisty × volume = ρV
n =ρV
molarmassρV
molarmass=PVRT
molarmass × PV = ρVRT
ρ =molarmass × P
RTmolarmass =100.0 g
mol
P = 380torrR = 62.36 L ⋅torr
mol⋅K
T = 25ÝC = 298.15K
ρ =(100.0 g
mol )(380torr)(62.36 L ⋅torr
mol⋅K )(298.15K)
ρ = 2.04 gL
4. Theunknowngassampleat137°Cand0.989bar,hasadensityof0.698g/L.Calculatethemolarmassofthisunknowngas.
molarmass = ρRTP
ρ = 0.698 gL
R = 0.08314 L⋅barmol⋅K
T =137˚C = 410.15KP = 0.989bar
molarmass = (0.698gL )(0.08314 L⋅bar
mol⋅K )(410.15K )(0.989bar)
molarmass = 24.1 gmol
5. Agasexertsapressureof1.12atmina4Lcontaineratº19C.Youknowthedensityofthegasis1.5g/L.Whatisthemolecule?𝑃𝑉 = 𝑛𝑅𝑇
(1.12 𝑎𝑡𝑚)(4𝐿) = 𝑛 (0.082𝐿𝑎𝑡𝑚𝐾 𝑚𝑜𝑙
)(19 + 273) = .187 (0.187𝑔/4𝐿)(𝑥 𝑔/𝑚𝑜𝑙) = 1.5 𝑔/𝐿 𝑥 = 32𝑔/𝑚𝑜𝑙ProbablyO2
GasMixturesActivity1 Thepurposeofthisactivityisforyoucheckyourunderstandingofgasmixtures.
1. Explaintherelationshipbetweenthepartialpressureofagasinamixtureandthetotalpressureofthemixturewithwordsandanequation.
Hereistheequationforthisrelationship:
€
Pi = XiPTotal
wherePiisthepartialpressureofagasinamixture,Xiismolefractionofagasinamixture,andPTotalisthetotalpressureofthegasmixture.
Therelationshipstatesthatthepartialpressureofagasinamixtureisdirectlyproportionaltoitsconcentrationinthemixture(molefraction).
2. ExplainDalton’sLawofpartialpressurewithwordsandanequation.
Hereistheequationforthisrelationship:
€
PTotal = Pii=1
n
∑
ORPTotal = P1 + P2 + ...+ Pn
wherePiisthepartialpressureofagasinamixture,nrepresentsthenumberofindividualgasesinthemixtureandPTotalisthetotalpressureofthegasmixture.
Dalton’sLawofpartialpressurestatesthatthetotalpressureofagasmixtureisthesumofallthepartialpressuresofeachgaswithinthemixture.
3. Amolefractionisconsideredameasureofconcentration.Explainwhythisistrue.
Molefractionistheratioofthemolesofanindividualsubstancetothetotalnumberofmoles.Thisprovidesuswithavaluethatexpressestherelativepresenceofthatsubstanceinthemixture.Inthisway,molefractionisameasureofconcentration.
4. TrueorFalse?Itisimpossibletoactuallymeasuretheindividualpressuresofaparticulargaswithinamixtureofgases.Explainyouranswer.
True.Apressuresensorcannotdifferentiatebetweendifferenttypesofmoleculesthatcollidewithitinsideacontainer.Allcollisionsaresimplyfeltascollisions.SoweneedDalton’sLawofpartialpressurestohelpuscalculatepartialpressures.
Activity2 1. InthetroposhereofTitan,Saturn’slargestmoon,theatmosphericpressureisabout1.5bar.
Atthispointintheatmospherethereisapproximately4.9%methane(thevastmajorityoftheatmosphereisnitrogen)bynumber.CalculatethepartialpressureofmethaneinthestratosphereofTitan.
Theproblemstatestheatmosphereis4.9%methane“bynumber.”Wecaninterpret“bynumber”tomean4.9%ofthemolecules(andthereforemoles)atthispointinTitan’satmospherearemethane.Thismakesthecalculationofthemolefractionofmethanefairlystraightforward.Ifwehavesay100molesampleofTitan’satmosphere,then4.9molesofthismixturearemethane.Thismakesthemolefractionofmethane0.049.
€
PCH4 = XCH4Ptotal
XCH4=nCH4ntotal
=4.9mol100mol
= 0.049
PCH4 = (0.049)(1.5bar)PCH4 = 0.0735bar
2. YouaretheTitanexpertofyourlabgroupandyoudecidetorecreatetheatmosphereinthestratosphereofTitaninacontainerwithafixedvolumeof7.00L.YoudutifullyremindyourlabpartnersthatthestratosphereofTitanis,bynumber,98.4%nitrogengas,1.40%methanegasandtherestishydrogengas.Finally,youremindthemtomakesureallcalculationsareforatotalpressureof1.50barwhenthemixtureisatatemperature–179˚CinordertobestmimictheconditionsonTitan.Luckily,youalreadycreatedafreezerthatmaintainsaperfecttemperatureof–179˚CintowhichyouwillplaceyourmodelTitanatmosphere.Drawasmallparticlemodelofwhatisgoingoninthecontainerwhenthemixtureisatafinalpressureof1.5barandfinaltemperatureof–179˚C.Determinethepressureofeachgasinthecontainerandcomputehowmanygramsofeachgasyouwillplaceintothiscontainer(afterfirstevacuatingitofcourse).
Firstweneedtoknowhowmanytotalmolesofgaswillbepresentwiththegiven“Titan”-likeconditionsof1.5barand–179˚C:
€
Total Moles in Mixture :PV = nRTP =1.5barV = 7.00LR = 0.08314 L ⋅bar
mol⋅K
T = −179ÝC = 94.15KPV = nRT
ntotal =PVRT
ntotal =(1.5bar)(7.00L)
(0.08314 L ⋅barmol⋅K )(94.15K)
=1.341402mol
Soourmini-Titanstratospherecontains1.341402molesofgastotalin7.00Lat–179˚Ctoachieveapressureof1.5bar.
Now,weknowthepercentagebreakdownofthethreegasespresentinthismixture.Wealsoknowthepercentagesare“bynumber”meaningthesepercentagespertaintothemolarcompositionandnotthemasscomposition.Wecandeterminethenumberofmolesofeachtypeofgas:
€
Moles of each gas :nx = (percentage) × ntotal
nN2 = 0.984 ×1.341402mol =1.319939molnCH4 = 0.014 ×1.341402mol = 0.01877963molnH2
= 0.002 ×1.341402mol = 0.002682804mol
Thepartialpressuresofeachgascouldbecalculatedintwoways.Wecouldgothroughthetroubleofcalculatingthemolefractionsofeachgasandthencalculatepartialpressure.OR,wecouldrecognizethatthepercentagesbynumberareequivalenttothemolefractionofeachgas!Sowecansimplyusethepercentagesgivenintheoriginalproblemtocalculatepartialpressure:
€
Ptotal =1.5barPN2 = (percentageN2 ) × PtotalPN2 = (0.984)(1.5bar)PN2 =1.476bar
PCH4 = (percentageCH4 ) × PtotalPCH4 = (0.014)(1.5bar)PCH4 = 0.021bar
PH2= (percentageH2
) × PtotalPH2
= (0.002)(1.5bar)PH2
= 0.003bar
Finally,wecancalculatethegramsneededtobeaddedtothecontainerbymultiplyingthenumberofmolesbytheirmolarmasses:
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Mass of each gas :mass = n × molarmass
massN2 =1.319939mol N2 ×28g N2
1mol N2= 36.95831g N2
massCH4 = 0.01877963mol CH4 ×16g CH4
1mol CH4= 0.3004740g CH4
massH2= 0.002682804mol ×
2g H2
1mol H2= 0.005365608g H2
massN2 ≈ 37.0g N2
massCH4 ≈ 3.00 ×10−1g CH4
massH2≈ 5.37 ×10−3g H2
Here’sasmallparticlerepresentationofthe7.00Lmini-Titanatmosphereat–179˚Cand1.5barpressure.