1 / 9

Derivation

Units. A.P. Chem. Ch. 5. mm, torr, atm, Pa, kPa, N/cm 2. Gases. Pressure: Force Area. C a P tot = P a. Dalton’s Law P tot = P a + P b…. Boyle’s Law PV=k. Charles’ Law V/T=k. Avog. Law V/n=k. Graham’s Law R a /R b = (MM b /MM a ) 1/2. Stoich. Devices. Barometer.

dustinn
Download Presentation

Derivation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Units A.P. Chem. Ch. 5 mm, torr, atm, Pa, kPa, N/cm2 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 0.0821 L atm/mol K 8.31 J/ mol K Manometer 22.4 L = 1 mole PV= nRT The way we choose to view it Non- STP P = dRT/MM Due to Avogadro’s Hypothesis.` Derivation KMT P = 2 nNA1/2mu2 3V urms= (3RT/MM)1/2 P = 2 nKEper mol Dimensionless Pts. 3V The Way It Really Is. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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)

  8. 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.

  9. 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

More Related