Chemistry: It’s a gasgenchem1csustan.wdfiles.com/local--files/start/chapter6...Gas Density...
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Chemistry: It’s a gas Part IV Molar mass of a gas Density of a gas Dalton’s Law: high altitudes and scuba diving Kinetic Molecular Theory: fast gases are hot! I have a chemistry test on Friday, there is nothing down there that will make me panic. Warrior Chemistry Club Meeting and Tutoring: 5 pm Thursday, S139
Transcript of Chemistry: It’s a gasgenchem1csustan.wdfiles.com/local--files/start/chapter6...Gas 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
