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10 Gases Solutions to Exercises V = πr²h = 3.14159 (1.25 X (5.5 10² cm) = 2.700 10³ cm³ = 2.7 L PV = MM g RT; g = MM RT PV ; P = 1.78 torr X = 2.342 X 10⁻³ = 2.34 X 10⁻³ atm 760 torr g = 20.18 Ne mol K X 2.342 10⁻³ atm 2.700 L = 5.049 10⁻³ = 5.0 10⁻³ g Ne 1 mol Ne 0.08206 L - atm 308 K 10.39 Analyze/Plan. Follow the strategy for calculations involving many variables. Solve. (a) V = 2.25 L; T = 273 + = = 735 torr 1atm = 0.96710 = 0.967 atm 760 torr PV = nRT, n = PV/RT, number of molecules (#) = n 6.022 10²³ # = 0.9671 atm X 2.25L X mol K X 6.022 X 10²³ molecules 310K 0.08206 atm mol = 5.15 X 10²² molecules (b) V = 5.0 10³ L; T = 273 + 0°C = 273 K; P = 1.00 atm; MM = 28.98 g/mol PV = MM g RT; RT g = 28.98 air mol K 1.00 atm 5.0 10³ L 1 mol air 0.08206 atm 273K = 6,468 g = 6.5 X 10³ g air = 6.5 kg air 10.40 (a) = 3.0 X 10⁻³ atm; = V = (exact) # of O₃ molecules = PV RT X 6.022 X 10²³ # 3.0 X 10⁻³ atm mol K X 6.022 10²³ molecules 250 K 0.08206 atm mol = 8.8 10¹⁹ O₃ molecules (b) # of CO₂ molecules = PV RT 6.022 10²³ 0.0004 #= 1.0 atm 2.0L X mol- K 6.022 X 10²³ mol molecules 0.0004 300 K 0.08206L atm = 1.957 X 10¹⁹ = 2 10¹⁹ molecules 10.41 Analyze/Plan. Follow the strategy for calculations involving many variables. Solve. (a) P = nRT V = 0.29 kg O₂ X 1kg 32.00 1 mol O₂ = 9.0625 = 9.1 mol; T = 273 + 9°C = K P = 9.0625 2.3L mol X 0.08206 atm 282 K = 91 atm mol- K (b) V = nRT P 9.0625 0.95 atm 0.08206 mol K atm 299 = 2.3 10² L 10.42 (a) 1 mol = 0.052154 = 0.0522 mol 284