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18 Chemistry of the Environment Solutions to Exercises (b) Troposphere, 0-12 km; stratosphere, 12-50 km; mesosphere, 50-85 km; thermosphere, 85-110 km. 18.12 (a) Boundaries between regions of the atmosphere are at maxima and minima (peaks and valleys) in the atmospheric temperature profile. For example, in the troposphere, temperature decreases with altitude, while in the stratosphere, it increases with altitude. The temperature minimum is the tropopause boundary. (b) From Figure 18.1, atmospheric pressure in the troposphere ranges from 760 torr to 200 torr, while pressure in the stratosphere ranges from 200 torr to 20 torr. Gas density (g/L) is directly proportional to pressure. The much lower density of the stratosphere means it has the smaller mass, despite having a larger volume than the troposphere. 18.13 Analyze/Plan. Given O₃ concentration in ppm, calculate partial pressure. Use the definition of ppm to get mol fraction O₃. For gases mole fraction = pressure fraction. Use the ideal-gas law to find mol O₃/L air and Avogadro's number to get molecules/L. = 0.441 ppm O₃ = 0.441 10⁶ mol mol O₃ air = 4.41 10⁻⁷ = Solve. (a) = = 4.41 X 10⁻⁷ (0.67 atm) = 2.955 10⁻⁷ = 3.0 10⁻⁷ atm (b) n = PV RT = 2.955 298 atm mol K atm = 1.208 = 1.2 mol O₃ 1.208 mol O₃ 6.022 mol molecules = = 10¹⁵ O₃ molecules 18.14 PAr = XAr PAr = 0.00934 (1.05 = 0.009807 = 10⁻³ bar = = 0.000382 (1.05 = = 10⁻⁴ bar Pco₂ = 0.0004011 bar bar 101,325 760 torr torr 18.15 Analyze/Plan. Given CO concentration in ppm, calculate number of CO molecules in 1.0 L air at given conditions. ppm atm CO mol CO molecules CO. Use the ideal gas law to change atm CO to mol CO, then Avogadro's number to get molecules. Solve. 3.5 ppm CO = 1 3.5 10⁶ mol mol CO air 3.5 = Xco = Xco = 3.5 10⁻⁶ 759 torr 760 1 atm torr 3.495 10⁻⁶ = 3.5 10⁻⁶ atm = PcoV RT = 3.495 295 atm mol- K atm = 1.444 = 1.4 10⁻⁷ mol CO 1.444 10⁻⁷ mol CO 6.022 X 10²³ molecules = 8.695 10¹⁶ = 8.7 10¹⁶ molecules CO mol 553