Chemistry:It’saGas

PartIICombinedGasLawAvogadro’sLawIdealGasLaw

Theozone(O3)layerprotectsusfromharmfulultravioletradiation.UVradiationcausesskincancerandcataracts.Chlorofluorocarbonsusedinairconditionersandhairspraycausetheozone layertobedepletedaroundthepolesoftheplanet.

Globalwarmingisduetomoleculestrappingheatinthetroposphere. Carbondioxide,waterandmethanegasesalltrapheatandwarmuptheplanet.

What is causing the planet to warm up?

A. Increasedcarbondioxideemissionsfromburninghydrocarbons.B. Increasedanimalproductdiets.C. Increasednumberofhumansontheplanet.D. Alloftheabove.E. Noneofthese, theearthiscoolingoff

Writedownonethingyoucandotodecreaseglobalwarming.

What is the relationship between atmospheric pressure and altitude?

Asaltitude increases,pressure:A. IncreasesB. DecreasesC. StaysthesameD. Thereisnorelationship

More gas molecules = more pressure

Pressure is a force applied to an area.More molecules create more collisions against the walls of the container.More collisions= more force.

Why are there more gas molecules at sea level than at the top of Mount Everest?A. Gravity pulls the molecules close to

earthB. The temperature is lower at sea levelC. Gas has less buoyancy at sea levelD. The temperature is lower at the top of

Mount Everest.E. None of these.

What is the relationship between the volume and the number of moles of a gas?

V=kn

Atconstanttemperature andpressure,asthenumberofmolesincreases:A.ThevolumedecreasesB.ThevolumeincreasesC.ThevolumestaysthesameD.Noneofthese

Avogadro’s Law:Ifpressureandtemperature areconstant,whatisthenewvolumeif1.2molesina2.4Lballoonhas5.3molesofgasadded?A. 11LB. 13LC. 10.6LD. 19LE. 0.077L

V1 =2.4LV2=Xn1 =1.2molesn2 =1.2+5.3=6.5moles

2.4L = X.1.2moles6.5moles

• WhatIstherelationshipbetweenpressureandtemperature?

• Atconstantvolumeandnumberofmolesofgas,pressureisdirectlyproportionaltotemperature.

• IfPincreases,Tincreases:P =kT

Amonton’s Law

CombinedGasLawAtconstantnumberofmolesofgas(anygas)P1V1=P2V2T1 T2

StandardtemperatureandpressureT=273K,P=1atm

WhatisthevolumeofagasatSTP,ifitoccupies25mLat750mmHgand25°C?

CombinedGasLaw P1V1=P2V2T1 T2

StandardtemperatureandpressureT=273K,P=1atmWhatisthevolumeofagasatSTP,ifitoccupies25mLat750mmHgand25°C?P1= P2=V1= V2=T1 =T2=

P1=0.98atm P2=1atmV1=25mL V2=?T1 =298K T2=273K

SolveforV2=(0.98atm)(25mL)(273) V2 =22.4mL(298k)(1atm)

GasLawsGasLaw Relationship

?Whatisconstant?

Whatequation?

Boyle’sLaw P∝ 1V

Charles’Law

Avogadro’sLaw

Amonton’sLaw

CombinedGasLaw

GaslawsLaw Relationship What is constant Equation neededBoyle’s P∝ 1

Vmoles, temperature P1V1 = P2V2

Charles’ V∝T pressure, moles V1 = V2T1 T2

Avogadro’s V∝n pressure, temperature V1 = V2n1 n2

Amonton’s P∝T volume, moles P1 = P2T1 T2

Combined Use this for converting to STP:273K and 1 atm

moles P1V1 = P2V2T1 T2

Which gas law is related to breathing?A. Amonton’s LawB. Charles’LawC. Boyle’sLawD. CombinedGasLawE. Avogadro’sLaw

Which of the following statements is true?A. As the diaphragm and rib muscles contract, volume

increases, pressure decreases and air flows in.B. As the diaphragm and rib muscles contract, volume

decreases, pressure decreases and air flows in.C. As the diaphragm and rib muscles contract, volume

decreases, pressure increases and air flows out.D. As the diaphragm and rib muscles contract, volume

increases, pressure decreases and air flows out.E. None of these.

When a 2.0 L weather balloon goes from sea level (1 atm, 23°C) to 10 km the pressure decreases to 0.495 atm and the temperature -18°C. What is the new volume?

A. 3.48LB. 0.85LC. 1.15LD. -2.81LE. Noneofthese

Usethecombinedgaslaw.

2.0Lx1atm =0.495atm xVolume296K 255K

255Kx2.0Lx1atm =Volume296Kx.495atm

Idealgaslaw:PV=nRTR=0.0821 Latm

moleKHowmanymolesofgasarepresentina3.2Lcontainerat30°Cand750mmHg?PV=nRTDividebothsidesbyRT:PV =nRT

A. 1.28moles.B.974moles.C.0.127molesD.0.14molesE.noneoftheseRemembertoconvertmmHgtoatm (760mmHg=1atm)and°CtoK,(273+°C=K)

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

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

PV = nRT PV = nRT

0.9921 atm x 0.127 L. = n0.0821 (Latm/moleK)371Kn = 0.00413 molesM = 0.495 g/.00413 molesM = 119.85 g/mole

GasDensity

• Findmoles,multiplybymolarmasstogetgrams(PV=nRT)• Density=grams/L