Gas Law Problems - Practice

Gas Law Problems - Practice 1. A sample of gas occupies volume of 95.2 mL at pressure of 710 torr. What will its volume be in liters at 1.00 atm, with n and T constant? 710 torr/(760 torr/atm) = 0.93 atm 0.0952 L x 0.93 atm = V2 x 1.00 atm & so 95.2 mL (1 L/1000 mL) = 0.0952 V2 = 0.089 L 2. You have 0.50 mol of O2 gas with a volume of 12.2 liters at 1 atm and at room temperature. All the gas in the flask is converted to O3 (ozone) at the same P & T. What is the volume of O 3 produced? Equation: 3 O2 (g) ------ > 2 O3 (g) 0.50 mol O2 x (3 mol O2/2 mol O3) = 0.33 mol O 3 12.2 L = V2 0.50 mol 0.33 mol V2 = 8.12 L 3. An unknown liquid is vaporized in a 245.0 ml bulb at a pressure of 656 mm Hg and a temperature of 90.0 oC. The mass of the unknown liquid before it is vaporized is 0.520 grams. Determine the molecular mass of the unknown liquid. MM = (0.520 g x 0.8206 L atm/K mol x 363 K) / 0.860 atm x 0.245 L) = 73.6 g/mol 4. Methane gas, CH4 and O2 are combined and ignited to form CO2 and H2O. Determine the volume of CO2 gas formed given the following data. CH4 O2 CO2 P T V 1.65 atm 25.0 oC 2.80 L 1.25 atm 31.0 oC 3.10 L 2.50 atm 125.0 oC to find Need a balanced reaction: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g) Find the moles of each gas then determine the theoretical yield CH4: (1.65 atm x 2.80 L) / (R x 298 K) = 0.189 mol CH4 0.189 mol CH4 x (1 mol CO2/1 mol CH4) = 0.189 mol CO2 O2: (1.25 atm x 3.10 L) / (R x 304 K) = 0.155 mol O 2 0.155 mol O 2 x (1 mol CO2/2 mol O2) = 0.0775 mol CO2 Volume of CO2 that would be produced: V = ( 0.0775 mol CO2 x R x 398 K) / 2.50 atm V = 1.10 L CO2 5. A simple of gas contains 0.35 mol of Argon at 13.0 oC, 566 torr. The gas is heated to 56.0 oC and a pressure of 897 torr. Determine the change in volume of the argon under these conditions. V1 = (0.350 mol x R x 298 K) / 0.747 atm = 11.0 L V2 = (0.350 mol x R x 329 K) / 1.18 atm = 8.0 L ΔV = 11.0 L – 8.0 L = 3.0 L 6. The density of a gas at a pressure of 1.50 atm and a temperature of 27.0 oC is 1.95 g/L. Determine the molecular mass of the gas. MM = ( 1.95 g/L x R x 700 K) / 1.50 atm = 32.0 g/mol 7. Mixture of Helium and oxygen is used in scuba diving to help prevent the bends in deep sea diving. For a particular dive 4.6 x 104 ml of oxygen at 25.0 oC and at 1.00 atm along with 1.200 x 104 mL of helium at same temperature and pressure are pumped into a 5.00 L tank. Determine the partial pressure of each gas and the total pressure of the gases in the scuba tank. n O2 = (1.00 atm x 46 L) / (R x 298 K) = 1.92 mol O2 P O2 = (1.92 mol O2 x R x 298 K) / 5.00 L = 9.3 atm O2 n P P He He = (1.00 atm x 12 L) / (R x 298 K) = 0.49 mol He = (0.49 mol He x R x 298 K) / 5.00 L = 2.40 atm He total: (2.40 atm + 9.3 atm) = 11.7 atm 8. In an experiment H2 gas was collected under water at a temperature of 22.0 oC. If the atmospheric pressure was 698 mm Hg and the vapor press of water at 22.0 oC is 21.0 torr, determine the pressure of H2 gas (dry) in atm. 698 torr – P O2 + 21.0 torr P H2 = 677 torr or 0.891 atm 9. In an experiment oxygen is produced by the decomposition of potassium chlorate. Because of the flammability of oxygen it is collected under water. The volume of gas collected in this experiment is 650.0 ml at room pressure of 754 torr and a temperature of 22.0 oC. Determine the amount of KClO3 decomposed and the amount of oxygen collected in this reaction. The vapor pressure at 22.0 oC is 21.0 torr. P O2: (754 torr – 21.0 torr) = 733 torr or 0.964 atm nO2: (0.964 atm x 0.650 L) / ( R x 295 K) = 2.59 x 10-2 mol O2 2.59 x 10-2 mol O2 x 32 g/mol O 2 = 0.829 g O2 would be used 2.59 x 10-2 mol O 2 x (2 mol KClO3 / 3 mol O 2) x (122.6 g/mol KClO3) = 2.12 g KClO3 would be decomposed 10. The partial pressure of oxygen in air is 156 torr when the atmospheric pressure is 743 torr. Determine the mole fraction of oxygen present in air under these conditions. Χ O2 = P Χ O2 = 156 torr/ 743 torr = 0.210 O2 / P total 11. The mole fraction of nitrogen in air is 0.7808. Determine its partial pressure in atm, when the atmospheric pressure is 760 torr. Assuming only oxygen and nitrogen are present, determine the mole fraction and partial pressure of oxygen. PN2 = Χ N2 x Ptotal 0.7808 x 760 torr = 593 torr or 0.781 atm is PN2 Χ N2 + Χ O2 = 1 0.7808 + Χ O2 = 1 where Χ O2 = 0.219