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

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 (A) greatest in container A (B) greatest in container B (C) greatest in container C (D) the same for all three containers Container (3.0 L) ABC GasMethaneEthaneButane FormulaCH 4 C2H6C2H6 C 4 H 10 Molar Mass (g/mol) 1630.58 Temp. (ºC)27 Pressure (atm) 2.04.02.0 How would you calculate the average KE per mole of gas? Avg. KE mol = 1.5 (8.31 J/mol K) 300.K Avg. KE mol = 3740 J/mol

4. The average velocity of the gas molecules is (A) greatest in container A (B) greatest in container B (C) greatest in container C (D) the same for all three containers Container (3.0 L) ABC GasMethaneEthaneButane FormulaCH 4 C2H6C2H6 C 4 H 10 Molar Mass (g/mol) 1630.58 Temp. (ºC)27 Pressure (atm) 2.04.02.0 How would you calculate the velocity of the methane gas? u rms = [3(8.31J/mol K)(300. K)/0.016 kg] ½ u rms = 680 m/s

5. The number of gas molecules is (A) greatest in container A (B) greatest in container B (C) greatest in container C (D) the same for all three containers Container (3.0 L) ABC GasMethaneEthaneButane FormulaCH 4 C2H6C2H6 C 4 H 10 Molar Mass (g/mol) 1630.58 Temp. (ºC)27 Pressure (atm) 2.04.02.0 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 (A) greatest in container A (B) greatest in container B (C) greatest in container C (D) the same for all three containers Container (3.0 L) ABC GasMethaneEthaneButane FormulaCH 4 C2H6C2H6 C 4 H 10 Molar Mass (g/mol) 1630.58 Temp. (ºC)27 Pressure (atm) 2.04.02.0 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 (A) greatest in container A (B) greatest in container B (C) greatest in container C (D) the same for all three containers Container (3.0 L) ABC GasMethaneEthaneButane FormulaCH 4 C2H6C2H6 C 4 H 10 Molar Mass (g/mol) 1630.58 Temp. (ºC)27 Pressure (atm) 2.04.02.0 Calculate how much faster the methane diffuses compared to the ethane. (rate meth /rate eth ) = (30. g/mol /16 g/mol ) ½ (rate meth /rate eth ) = 1.4 (times faster)

8. Consider the production of water at 450. K and 1.00 atm: 2 H 2(g) + O 2(g) --- > 2 H 2 O (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: n H2 = 0.00812 mol n O2 = 0.0149 mol You Need: Twice as many moles of H 2 than O 2 (from balanced reaction equation) Consequence: H 2 is the L.R. Since H 2 is the L.R. then 300.0 mL of it is used and only 150 mL of the O 2 is used. (2:1 mole ratio can be viewed as 2:1 volume ration as well due to Avogadro’s Hypothesis) Also, 300.0 mL of H 2 O will be produced. The reaction mixture remaining will be 600 mL in volume from the 300.0mL of water made and the 300.0 mL of O 2 unused.

9. UF 6 = 352 g U rms = [ 3(8.31Jmol -1 K -1 )(330.0K)/0.352kg ] ½ U rms =153 m/s

10. 1.03 mg O 2 = 3.22 x 10 -5 mol O 2 P O 2 = 0.00380 atm 0.41 mg He = 1.0 x 10 -4 mol HeP He = 0.012 atm P tot = 0.016 atm X O 2 (P tot ) = P O 2 X O 2 = P O 2 /(P tot ) = 0.24 X He = 0.76

11. P = dRT/MM MM air = 29 g/mol d air@25°C, 740 t = 1.2 g/L

12. U H 2 MM I 2 U I 2 MM H 2 = √ = Rate H 2 time I 2 Rate I 2 time H 2 = 52 s 253.8g time H 2 2.02 g = √ amount H 2 /time H 2 amount I 2 /time I 2 time H 2 = 4.6 s