2A. Kinetic Molecular Theory Particles in an ideal gas…have no volume.have elastic collisions.are in constant, random, straight-line motion.don’t attract or repel each other.have an avg. KE directly related to Kelvin temperature.
3B. Real Gases Particles in a REAL gas… have their own volume attract each otherGas behavior is most ideal…at low pressuresat high temperaturesin nonpolar atoms/molecules
4C. Characteristics of Gases Gases 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
5C. Characteristics of Gases Gases can be compressed.no volume = lots of empty spaceGases undergo diffusion & effusion.random motion
6D. TemperatureAlways use absolute temperature (Kelvin) when working with gases.ºFºCK-45932212-273100273373K = ºC + 273
20C. Gay-Lussac’s LawThe pressure and absolute temperature (K) of a gas are directly relatedat constant mass & volumePT
21PV T V T P T PV = k P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1 D. Combined Gas LawPVTVTPTPV= kP1V1T1=P2V2T2P1V1T2 = P2V2T1
22E. Gas Law ProblemsA gas occupies 473 cm3 at 36°C. Find its volume at 94°C.CHARLES’ LAWGIVEN:V1 = 473 cm3T1 = 36°C = 309KV2 = ?T2 = 94°C = 367KTVWORK:P1V1T2 = P2V2T1(473 cm3)(367 K)=V2(309 K)V2 = 562 cm3
23E. Gas Law ProblemsA gas occupies 100. mL at kPa. Find its volume at 200. kPa.BOYLE’S LAWGIVEN:V1 = 100. mLP1 = 150. kPaV2 = ?P2 = 200. kPaPVWORK:P1V1T2 = P2V2T1(150.kPa)(100.mL)=(200.kPa)V2V2 = 75.0 mL
24E. Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3 A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.COMBINED GAS LAWGIVEN:V1 = 7.84 cm3P1 = 71.8 kPaT1 = 25°C = 298 KV2 = ?P2 = kPaT2 = 273 KP TVWORK:P1V1T2 = P2V2T1(71.8 kPa)(7.84 cm3)(273 K)=( kPa) V2 (298 K)V2 = 5.09 cm3
25E. Gas Law ProblemsA gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?GAY-LUSSAC’S LAWGIVEN:P1 = 765 torrT1 = 23°C = 296KP2 = 560. torrT2 = ?PTWORK:P1V1T2 = P2V2T1(765 torr)T2 = (560. torr)(309K)T2 = 226 K = -47°C
28A. Avogadro’s Principle Equal volumes of gases contain equal numbers of molesat constant temp & pressuretrue for any gasVn
29UNIVERSAL GAS CONSTANT B. Ideal Gas LawPVnTVnPVT= k= RUNIVERSAL GAS CONSTANTR= Latm/molKR=8.315 dm3kPa/molK
30UNIVERSAL GAS CONSTANT B. Ideal Gas LawPV=nRTUNIVERSAL GAS CONSTANTR= Latm/molKR=8.315 dm3kPa/molK
31B. Ideal Gas Law IDEAL GAS LAW P = ? atm n = 0.412 mol Calculate the pressure in atmospheres of mol of He at 16°C & occupying L.IDEAL GAS LAWGIVEN:P = ? atmn = molT = 16°C = 289 KV = 3.25 LR = Latm/molKWORK:PV = nRTP(3.25)=(0.412)(0.0821)(289)L mol Latm/molK KP = 3.01 atm
32B. Ideal Gas Law IDEAL GAS LAW V = ? n = 85 g T = 25°C = 298 K Find the volume of 85 g of O2 at 25°C and kPa.IDEAL GAS LAWGIVEN:V = ?n = 85 gT = 25°C = 298 KP = kPaR = dm3kPa/molKWORK:85 g 1 mol = 2.7 mol32.00 g= 2.7 molPV = nRT(104.5)V=(2.7) (8.315) (298)kPa mol dm3kPa/molK KV = 64 dm3
33IV. Gas Stoichiometry at Non-STP Conditions (p. 347-350) Ch. 10 & 11 - GasesIV. Gas Stoichiometry at Non-STP Conditions (p )
34A. Gas Stoichiometry Moles Liters of a Gas STP - use 22.4 L/mol Non-STP - use ideal gas lawNon-STP ProblemsGiven liters of gas?start with ideal gas lawLooking for liters of gas?start with stoichiometry conv.
35B. Gas Stoichiometry Problem What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?CaCO3 CaO CO25.25 g? L non-STPLooking for liters: Start with stoich and calculate moles of CO2.5.25 gCaCO31 molCaCO3100.09g1 molCO2CaCO3= 1.26 mol CO2Plug this into the Ideal Gas Law to find liters.
36B. Gas Stoichiometry Problem What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?GIVEN:P = 103 kPaV = ?n = 1.26 molT = 25°C = 298 KR = dm3kPa/molKWORK:PV = nRT(103 kPa)V =(1mol)(8.315dm3kPa/molK)(298K)V = 1.26 dm3 CO2
37B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C?4 Al O2 2 Al2O315.0 Lnon-STP? gGIVEN:P = 97.3 kPaV = 15.0 Ln = ?T = 21°C = 294 KR = dm3kPa/molKWORK:PV = nRT(97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K)n = mol O2Given liters: Start with Ideal Gas Law and calculate moles of O2.NEXT
38B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?4 Al O2 2 Al2O315.0Lnon-STP? gUse stoich to convert moles of O2 to grams Al2O3.0.597mol O22 mol Al2O33 mol O2g Al2O31 molAl2O3= 40.6 g Al2O3
40Ptotal = P1 + P2 + ... A. Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.Ptotal = P1 + PPatm = PH2 + PH2O
41A. Dalton’s LawHydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.GIVEN:PH2 = ?Ptotal = 94.4 kPaPH2O = 2.72 kPaWORK:Ptotal = PH2 + PH2O94.4 kPa = PH kPaPH2 = 91.7 kPaLook up water-vapor pressure on p.899 for 22.5°C.Sig Figs: Round to least number of decimal places.
42A. Dalton’s Law DALTON’S LAW GIVEN: Pgas = ? Ptotal = 742.0 torr A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas?The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.DALTON’S LAWGIVEN:Pgas = ?Ptotal = torrPH2O = 42.2 torrWORK:Ptotal = Pgas + PH2O742.0 torr = PH torrPgas = torrLook up water-vapor pressure on p.899 for 35.0°C.Sig Figs: Round to least number of decimal places.
43B. Graham’s Law Diffusion Spreading of gas molecules throughout a container until evenly distributed.EffusionPassing of gas molecules through a tiny opening in a container
44KE = ½mv2 B. Graham’s Law Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas.At the same temp & KE, heavier molecules move more slowly.Larger m smaller v because…KE = ½mv2
45B. Graham’s Law Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass.Ratio of gas A’s speed to gas B’s speed
46B. Graham’s LawDetermine the relative rate of diffusion for krypton and bromine.The first gas is “Gas A” and the second gas is “Gas B”.Relative rate mean find the ratio “vA/vB”.Kr diffuses times faster than Br2.
47Put the gas with the unknown speed as “Gas A”. B. Graham’s 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”.
48B. Graham’s 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.
49TEAM PRACTICE!Work the following problems in your book. Check your work using the answers provided in the margin.p. 324SAMPLE PROBLEM 10-6PRACTICE 1 & 2p. 355SAMPLE PROBLEM 11-10PRACTICE 1, 2, & 3