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.