PV= nRT KMT P= nRT – a(n/V)2 V-nb Gases Stoich. urms= (3RT/MM)1/2 Units mm, torr, atm, Pa, kPa, N/cm2 A.P. Chem. Ch. 5 Gases Pressure: Force Area CaPtot= Pa Dalton’s Law Ptot= Pa+ Pb… Boyle’s Law PV=k Charles’ Law V/T=k Avog. Law V/n=k Graham’s Law Ra/Rb = (MMb/MMa)1/2 Stoich. Devices Barometer At STP Manometer 0.0821 L atm/mol K 8.31 J/ mol K 22.4 L = 1 mole PV= nRT The way we choose to view it Non- STP P = dRT/MM Derivation Due to Avogadro’s Hypothesis.` KMT P = 2 nNA1/2mu2 3V P = 2 nKEper mol urms= (3RT/MM)1/2 Dimensionless Pts. The Way It Really Is. 3V KEper mol = (3/2)RT In constant motion Colliding 100% elast. Particle interactions Creating pressure P= nRT – a(n/V)2 w/o influence In such a way that Temp is dir. Prop. To average KE. V-nb Volumes of particles

Problems to emphasize in your HW: Format of the Test Option 1 – The marathon problem (1 system; many parts) 70 points Option 2 – The regular test (single parts; many systems) 70 points 13 multiple choice [~20 minutes] 2 free response [~25 minutes] Problems to emphasize in your HW: 43, 61, 67, 83, 85

The average kinetic energy of the gas molecules is Container (3.0 L) A B C Gas Methane Ethane Butane Formula CH4 C2H6 C4H10 Molar Mass (g/mol) 16 30. 58 Temp. (ºC) 27 Pressure (atm) 2.0 4.0 The average kinetic energy of the gas molecules is greatest in container A greatest in container B greatest in container C the same for all three containers How would you calculate the average KE per mole of gas? Avg. KEmol = 1.5 (8.31 J/mol K) 300.K Avg. KEmol = 3740 J/mol

The average velocity of the gas molecules is greatest in container A Container (3.0 L) A B C Gas Methane Ethane Butane Formula CH4 C2H6 C4H10 Molar Mass (g/mol) 16 30. 58 Temp. (ºC) 27 Pressure (atm) 2.0 4.0 The average velocity of the gas molecules is greatest in container A greatest in container B greatest in container C the same for all three containers How would you calculate the velocity of the methane gas? urms = [3(8.31J/mol K)(300. K)/0.016 kg]½ urms = 680 m/s

The number of gas molecules is greatest in container A Container (3.0 L) A B C Gas Methane Ethane Butane Formula CH4 C2H6 C4H10 Molar Mass (g/mol) 16 30. 58 Temp. (ºC) 27 Pressure (atm) 2.0 4.0 The number of gas molecules is greatest in container A greatest in container B greatest in container C the same for all three containers How would you calculate the number of particles of ethane? n = (4.0 atm)(3.0L) (0.0821 L atm/mol K)(300. K) n = 0.49 mol

The density of the gas is greatest in container A Container (3.0 L) A B C Gas Methane Ethane Butane Formula CH4 C2H6 C4H10 Molar Mass (g/mol) 16 30. 58 Temp. (ºC) 27 Pressure (atm) 2.0 4.0 The density of the gas is greatest in container A greatest in container B greatest in container C the same for all three containers How would you calculate the density of the ethane gas? d = (4.0 atm)(30. g/mol) (0.0821 L atm/mol K)(300. K) d = 4.9 g/L

greatest in container A greatest in container B Container (3.0 L) A B C Gas Methane Ethane Butane Formula CH4 C2H6 C4H10 Molar Mass (g/mol) 16 30. 58 Temp. (ºC) 27 Pressure (atm) 2.0 4.0 If the containers were opened simultaneously the diffusion rates of the gas molecules out of the containers through air would be greatest in container A greatest in container B greatest in container C the same for all three containers Calculate how much faster the methane diffuses compared to the ethane. (ratemeth/rateeth) = (30.g/mol/16g/mol)½ (ratemeth/rateeth) = 1.4 (times faster)

Consider the production of water at 450. K and 1.00 atm: 2 H2(g) + O2(g) --- > 2 H2O(g) If 300.0 mL of hydrogen is mixed with 550.0 mL of oxygen, determine the volume of the reaction mixture when the reaction is complete. You Have: 300.0 mL of H2 550.0 mL of O2 You Need: Twice as many milliliters of H2 than O2 (from balanced reaction equation) Consequence: H2 is the L.R. Since H2 is the L.R. then 300.0 mL of it is used and only 150 mL of the O2 is used. (2:1 mole ratio can be viewed as 2:1 volume ration as well due to Gay-Lussac’s Law) Also, 300.0 mL of H2O will be produced. The reaction mixture remaining will be 700 mL in volume from the 300.0mL of water made and the 400.0 mL of O2 unused.

√ √ UH2 MMI2 RateH2 timeI2 UI2 MMH2 RateI2 timeH2 52 s 253.8g amount H2/timeH2 √ UH2 MMI2 RateH2 timeI2 = = = UI2 MMH2 RateI2 timeH2 amount I2/timeI2 52 s 253.8g √ = timeH2= 4.6 s timeH2 2.02 g