I. Physical Properties

Gases

A. Kinetic Molecular TheoryParticles in an ideal gashave no volume.have elastic collisions. are in constant, random, straight-line motion.dont attract or repel each other.have an avg. KE directly related to Kelvin temperature.

B. Real GasesParticles in a REAL gashave their own volumeattract each otherGas behavior is most idealat low pressuresat high temperaturesin nonpolar atoms/molecules

C. Characteristics of GasesGases expand to fill any container.random motion, no attractionGases are fluids (like liquids).no attractionGases have very low densities.no volume = lots of empty space

C. Characteristics of GasesGases can be compressed.no volume = lots of empty spaceGases undergo diffusion & effusion.random motion

D. TemperatureAlways use absolute temperature (Kelvin) when working with gases.

E. PressureWhich shoes create the most pressure?

E. PressureBarometermeasures atmospheric pressure

E. PressureManometermeasures contained gas pressure

E. PressureKEY UNITS AT SEA LEVEL101.325 kPa (kilopascal)1 atm 760 mm Hg760 torr 14.7 psi

F. STPSTP

II. The Gas LawsBOYLESCHARLESGAY-LUSSAC

A. Boyles LawPV = k

Volume

(mL)

Pressure

(torr)

PV

(mLtorr)

10.0

760.0

7.60 x 103

20.0

379.6

7.59 x 103

30.0

253.2

7.60 x 103

40.0

191.0

7.64 x 103

A. Boyles LawThe pressure and volume of a gas are inversely related at constant mass & tempPV = k

A. Boyles Law

B. Charles Law

Volume

(mL)

Temperature (K)

V/T

(mL/K)

40.0

273.2

0.146

44.0

298.2

0.148

47.7

323.2

0.148

51.3

348.2

0.147

B. Charles LawThe volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

B. Charles Law

C. Gay-Lussacs Law

Temperature (K)

Pressure

(torr)

P/T

(torr/K)

248

691.6

2.79

273

760.0

2.78

298

828.4

2.78

373

1,041.2

2.79

C. Gay-Lussacs LawThe pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

= kPVPTV TPVTD. Combined Gas LawP1V1T2 = P2V2T1

GIVEN:V1 = 473 cm3T1 = 36C = 309KV2 = ?T2 = 94C = 367KWORK:P1V1T2 = P2V2T1E. Gas Law ProblemsA gas occupies 473 cm3 at 36C. Find its volume at 94C. CHARLES LAWTV(473 cm3)(367 K)=V2(309 K)V2 = 562 cm3

GIVEN:V1 = 100. mLP1 = 150. kPaV2 = ?P2 = 200. kPaWORK:P1V1T2 = P2V2T1E. Gas Law ProblemsA gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLES LAWPV(150.kPa)(100.mL)=(200.kPa)V2V2 = 75.0 mL

GIVEN:V1 = 7.84 cm3P1 = 71.8 kPaT1 = 25C = 298 KV2 = ?P2 = 101.325 kPaT2 = 273 KWORK:P1V1T2 = P2V2T1(71.8 kPa)(7.84 cm3)(273 K)=(101.325 kPa) V2 (298 K)V2 = 5.09 cm3

E. Gas Law ProblemsA gas occupies 7.84 cm3 at 71.8 kPa & 25C. Find its volume at STP. P T VCOMBINED GAS LAW

GIVEN:P1 = 765 torrT1 = 23C = 296KP2 = 560. torrT2 = ?WORK:P1V1T2 = P2V2T1E. Gas Law ProblemsA gas pressure is 765 torr at 23C. At what temperature will the pressure be 560. torr? GAY-LUSSACS LAWPT(765 torr)T2 = (560. torr)(309K)T2 = 226 K = -47C

III. Ideal Gas Law(p. 334-335, 340-346)Ch. 10 & 11 - Gases

A. Avogadros Principle

Gas

Volume

(mL)

Mass

(g)

Moles, n

V/n

(L/mol)

O2

100.0

0.122

3.81 ( 10-3

26.2

N2

100.0

0.110

3.93 ( 10-3

25.5

CO2

100.0

0.176

4.00 ( 10-3

25.0

A. Avogadros PrincipleEqual volumes of gases contain equal numbers of molesat constant temp & pressuretrue for any gas

PVTB. Ideal Gas LawVnPVnTUNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.315 dm3kPa/molK= R

B. Ideal Gas LawUNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.315 dm3kPa/molKPV=nRT

GIVEN:P = ? atmn = 0.412 molT = 16C = 289 KV = 3.25 LR = 0.0821Latm/molKWORK:PV = nRTP(3.25)=(0.412)(0.0821)(289) L mol Latm/molK KP = 3.01 atm

B. Ideal Gas LawCalculate the pressure in atmospheres of 0.412 mol of He at 16C & occupying 3.25 L. IDEAL GAS LAW

GIVEN:V = ?n = 85 gT = 25C = 298 KP = 104.5 kPaR = 8.315 dm3kPa/molKB. Ideal Gas LawFind the volume of 85 g of O2 at 25C and 104.5 kPa. = 2.7 molPV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK KV = 64 dm3IDEAL GAS LAW

IV. Gas Stoichiometry at Non-STP Conditions(p. 347-350)Ch. 10 & 11 - Gases

A. Gas StoichiometryMoles Liters of a GasSTP - use 22.4 L/mol Non-STP - use ideal gas law

Non-STP ProblemsGiven liters of gas? start with ideal gas lawLooking for liters of gas?start with stoichiometry conv.

1 molCaCO3100.09g CaCO3B. Gas Stoichiometry ProblemWhat volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25C?5.25 gCaCO3= 1.26 mol CO2CaCO3 CaO + CO2 1 molCO21 molCaCO35.25 g? L non-STPLooking for liters: Start with stoich and calculate moles of CO2.Plug this into the Ideal Gas Law to find liters.

WORK:PV = nRT(103 kPa)V =(1mol)(8.315dm3kPa/molK)(298K)V = 1.26 dm3 CO2

B. Gas Stoichiometry ProblemWhat volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25C?GIVEN:P = 103 kPaV = ?n = 1.26 molT = 25C = 298 KR = 8.315 dm3kPa/molK

WORK:PV = nRT(97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K)n = 0.597 mol O2B. Gas Stoichiometry ProblemHow many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21C?GIVEN:P = 97.3 kPaV = 15.0 Ln = ?T = 21C = 294 KR = 8.315 dm3kPa/molK4 Al + 3 O2 2 Al2O3 15.0 L non-STP? gGiven liters: Start with Ideal Gas Law and calculate moles of O2.NEXT

2 mol Al2O33 mol O2B. Gas Stoichiometry ProblemHow many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21C?0.597mol O2= 40.6 g Al2O34 Al + 3 O2 2 Al2O3 101.96 g Al2O31 molAl2O315.0Lnon-STP? gUse stoich to convert moles of O2 to grams Al2O3.

V. Two More Laws(p. 322-325, 351-355)Ch. 10 & 11 - Gases

A. Daltons LawThe total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

GIVEN:PH2 = ?Ptotal = 94.4 kPaPH2O = 2.72 kPaWORK:Ptotal = PH2 + PH2O 94.4 kPa = PH2 + 2.72 kPaPH2 = 91.7 kPaA. Daltons LawHydrogen gas is collected over water at 22.5C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.Look up water-vapor pressure on p.899 for 22.5C.Sig Figs: Round to least number of decimal places.The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

GIVEN:Pgas = ?Ptotal = 742.0 torrPH2O = 42.2 torrWORK:Ptotal = Pgas + PH2O 742.0 torr = PH2 + 42.2 torrPgas = 699.8 torrA. Daltons LawA gas is collected over water at a temp of 35.0C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? DALTONS LAWLook up water-vapor pressure on p.899 for 35.0C.Sig Figs: Round to least number of decimal places.The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the gas and water vapor.

B. Grahams LawDiffusionSpreading of gas molecules throughout a container until evenly distributed.EffusionPassing of gas molecules through a tiny opening in a container

B. Grahams LawSpeed of diffusion/effusionKinetic energy is determined by the temperature of the gas.At the same temp & KE, heavier molecules move more slowly.Larger m smaller v becauseKE = mv2

B. Grahams LawGrahams LawRate of diffusion of a gas is inversely related to the square root of its molar mass.

B. Grahams LawDetermine the relative rate of diffusion for krypton and bromine.Kr diffuses 1.381 times faster than Br2.The first gas is Gas A and the second gas is Gas B. Relative rate mean find the ratio vA/vB.

B. Grahams LawA molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Put the gas with the unknown speed as Gas A.

B. Grahams LawAn unknown gas diffuses 4.0 times faster than O2. Find its molar mass.The first gas is Gas A and the second gas is Gas B. The ratio vA/vB is 4.0.Square both sides to get rid of the square root sign.

TEAM PRACTICE!Work the following problems in your book. Check your work using the answers provided in the margin.p. 324 SAMPLE PROBLEM 10-6PRACTICE 1 & 2p. 355 SAMPLE PROBLEM 11-10PRACTICE 1, 2, & 3