Chemistry: It’s a gasgenchem1csustan.wdfiles.com/local--files/start/chapter6...Gas Density •Density = grams/L •Find moles, multiply by molar mass to get grams What is the density

Chemistry:It’sagas

PartIVMolarmassofagasDensityofagas

Dalton’sLaw:highaltitudesandscubadivingKineticMolecularTheory:fastgasesarehot!

IhaveachemistrytestonFriday,thereisnothingdowntherethatwillmakemepanic.

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

MolarMassofaGasAnexperimentshowsthata0.495gsampleofanunknowngasoccupies127mLat98°Cand754torr pressure.Calculatethemolarmassofthegas(hint:M =mass/numberofmoles).

V =127mL=0.127LP =754torr =0.9921atmT =98°C=371Km=0.495gR=0.0821(Latm/moleK)M =?

PV = nRT PV = nRT

0.9921 atm x 0.127 L = n0.0821 (Latm/moleK)371Kn = 0.00414 molesM = 0.495 g/0.00414 molesM = 120.38 g/mole

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

GasDensity•Density=grams/L•Findmoles,multiplybymolarmasstogetgrams

Whatisthedensityofoxygengasat25°Cand1atm?Hint:Pickanyvolume,1.0Listheeasiest!

n=PV =1atmx1.0L=0.0409molesRT0.0821(Latm/moleK)x298K

0.0409molesx32g/mole=1.31g1.0L

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

Whatisthedensityofcarbonmonoxidegasat25°Cand1atm?Assumethevolumeis1.0L.

n=1atmx1.0L =0.0409moles0.0821(Latm/moleK)298K

0.0409molesx28g/moleCO=1.145g

1.145g/1.0L =1.145g/L

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

Dalton’sLawandMolefractions

ΧA =molefractionA=molesofgasA/totalmolesofgasAir:χO2=0.2χN20.8pO2=χO2PTPT =pO2 +pN2

Ifascubadiverisatadepthwherethepressureis5atm,whatshouldthemolefractionofoxygenbetohaveapartialpressureof0.21atm?

0.21atm=X5.0atmX=0.042

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

Thepartialpressureofagasis________

A. thepressurethegasexertswhenpure.B. thetotalpressureofamixtureofgases.C. thepressureduetoagasinamixture.D. thesameasthevaporpressureofthegas.E. thepressureexertedbyonemoleculeofthegas.

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

Whatisthepartialpressureofoxygengasifitsmolefraction(X)is0.210andtheatmosphericpressureis190mmHg?

pO2=χO2PT

A. 0.84atmB. 39.9atmC. 0.0525atmD. 19.04atm

pO2 = 0.210(190mmHg/760mmHg/atm)

Inamixtureofgases,thegaswiththelargestmolefraction(X)willhavethe:A. largestnumberofmoleculespresent.B. highestkineticenergy.C. smallestnumberofmoleculespresentD. smallestmolarmass.E. largestmolarmass.

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

KineticMolecularTheoryofGasesAssumesthatgasmolecules:

1. Havetinyvolumescomparedwiththecollectivevolumetheyoccupy2. Moveconstantlyandrandomly3. Haveaveragekineticenergythatisproportionaltoabsolute

temperature(absolute=tempinKelvin,nomotionatOK,outer(empty)space=2.6K,OKhasnotbeenobtained.)

4. Engageinelasticcollisionswithwallsofcontainerandothergasmolecules

5. Actindependentlyofothergasmolecules

AverageKineticEnergy

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• Averagekineticenergy:KEavg =½m(urms)2• urms =theroot-mean-squarespeedofthemolecules

R=8.314J/molKT=KelvinJ=kgm2/sec2M =kg/mole

DiffusionandEffusion

• Graham’sLawofEffusionorDiffusion:Theeffusion/diffusionrateofanygasisinverselyproportionaltothesquarerootofitsmolarmass.

• Effusion– Theprocessbywhichagasescapesfromacontainerthroughatinyholeintoaregionoflowerpressure

• Diffusion– Thespreadofonesubstancethroughanother

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WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

DiffusionandEffusion (cont. 1)

Graham’sLawderivesadiffusionoreffusionrateoftwogases:

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A. O2<He<NO<CO2

B. He<O2<CO2<NOC. He<NO<O2<CO2

D. CO2<O2 <NO<HeE. Noneofthese

Listthefollowinggases,whichareatthesametemperature,intheorderofincreasingratesofdiffusion:O2,He,NO,CO2

Whatisthemolarmassofeachgas?

Whatistherelationshipbetweenmolarmassandrootmeanspeedofagas?

He=4g/moleNO=30g/moleO2 =32g/moleCO2 =44g/mole

Calculatethemolarmassofagasifequalvolumesofoxygengasandanunknowngastake3.25minand8.41min,respectively, toeffusethroughasmallholeatconstantpressureandtemperature.

Oxygen(x):M =32g/moleRate=1/3.25min(rate=1/time)

Othergas(y):M =unknownRate=1/8.41min

1/3.25= 8.41 = M1/8.41 3.25 0.032 kg/mole

Squarebothsides:2.592=M

0.032 0.032 x 6.69 = .214 kg/mole = 214 g/mole

Whatisunimportantwhenusingtheidealgaslaw?

A. Thechemicalidentityofthegassample.B. Thetemperatureofthegassample.C. Theamountofgas.D. Thepressureofthegassample.E. Thevolumeofthecontainerholdingthegassample.

WarriorChemistryClubMeetingandTutoring:5pmThursday,S139

Whichofthefollowingreactionswillresultinanincreaseintotalpressure?

A. 2H2(g)+O2(g)à 2H2O(l)B. 2HI(g)à H2(g)+I2(g)C. CH4(g)+2O2(g)àCO2(g)+2H2O(g)D. 2N2O(g)à2N2(g)+O2(g)E. noneofthese

Moregasmolecules=morepressure

When is the next Warrior Chemistry Club meeting with tutoring?A. Tuesdayat5pminS139B. Thursdayat5pminS139C. Fridayat9aminS139D. Wednesday at5pminS139E. Noneofthese