Hw11-G1K-Gas

# Hw11-G1K-Gas - HW Set 11 - KEY GAS LAWS; Kinetic Molecular...

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HW Set 11 - KEY GAS LAWS; Kinetic Molecular Theory CHM 2045 Hw11-G1K-Gas.wpd Kinetic-Molecular Theory 1. a) The number of impacts per unit time on a given container wall increases as: C the pressure (molecular concentration) of the gas increases. C the temperature increases. This causes the average molecular kinetic energy to increase, which means average molecular velocity (and collision frequency) increases. b) The average molecular kinetic energy varies directly with the absolute temperature, so decreasing the temperature will lower the average kinetic energy and decrease the average energy of molecular impact. c) Lowering the pressure of the gas (by increasing the volume of the gas and/or removing some of the gas) will increase the average distance between gas molecules. d) The average speed of molecules in the gas mixture is increased as the average molecular energy increases which increases as the absolute temperature increases. 2. a) Since both gases are at the same temperature (273 K), they both have identical average molecular kinetic energies. b) The average kinetic energy is determined from the molecular mass and the average molecular velocity (KE = ½mv 2 ). Since both gases have the same average kinetic energy, and N 2 molecules are much smaller in mass than SF 6 molecules, the N 2 molecules must have a greater average velocity than the SF 6 molecules. The frequency of molecular collisions with the container walls would be greater for the N 2 gas sample at STP. Ideal Gas Law 3. Ideal gas law: PV = nRT a) T = 273 + 38 = 311 K P = nRT = 3.32 mol x 0.0821 L C atm / mol C K x 311 K = 16.1 atm V 5.28 L b) T = 273 - 23 = 250 K V = nRT = 0.105 mol x 0.0821 L C atm / mol C K x 250 K = 0.431 L P 5.00 atm c) T = 273 + 27 = 300 K n = PV = 0.954 atm x 0.500 L = 0.0194 mol P = 725 mm x 1 atm = 0.954 atm RT 0.0821 L C atm / mol C K x 300 K 760 mm d) T = PV = 2.45 atm x 25.0 L = 1.11 x 10 3 K nR 0.670 mol x 0.0821 L C atm / mol C K 4. a) 1.53 mL x 10 -3 L = 1.53 x 10 -3 L P = nRT = 5.27 x 10 -2 mol x 0.0821 L C atm / mol C K x 250 K mL V 1.53 x 10 -3 L T = 273 - 23 = 250 K P = 707 atm b) T = 273 + 37 = 310 K V = nRT = 5.23 mol x 0.0821 L C atm / mol C K x 310 K = 150. L P 0.886 atm c) T = 273 - 15 = 258 K n = PV = 1.25 atm x 1.00 L = 0.0590 mol RT 0.0821 L C atm / mol C K x 258 K d) P = 745 mm x 1 atm = 0.980 atm T = PV = 0.980 atm x 5.25 x 10 -2 L = 160. K 760 mm nR 3.92 x 10 -3 mol x 0.0821 L C atm / mol C K V = 52.5 mL x 10 -3 L = 5.25 x 10 -2 L mL

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2 5. V = 0.250 L n = PV = 0.350 atm x 0.250 L = 3.58 x10 -3 mol Cl 2 T = 273 + 25 = 298 K RT 0.0821 L C atm / mol C K x 298 K Mass Cl 2 = 3.58 x 10 -3 mol Cl 2 x 70.9 g = 0.254 g Cl 2 mol 6. T = 273 + 250 = 523 K n = PV = 12.5 atm x 6.50 L N 2 = 1.89 mol N 2 RT 0.0821 L C atm / mol C K x 523 K Mass N 2 = 1.89 mol N 2 x 28.0 g N 2 = 52.9 g N 2 mol N 2 Gas Density 7. Recall that density equals the mass per unit volume, therefore we will find the number of grams of CH 4 in 1.00 L : T = 273 + 25 = 298 K n = PV
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## This note was uploaded on 11/16/2011 for the course CHM 2045 taught by Professor Mitchell during the Fall '07 term at University of Florida.

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Hw11-G1K-Gas - HW Set 11 - KEY GAS LAWS; Kinetic Molecular...

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