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10 Gases Solutions to Exercises 1 mol He = 32.6 He 4.0026 He = 8.1447 = 8.14 mol He = 1.600 mol 0.08206 mol K atm X 292 K = 3.8338 = 3.84 atm = 8.1447 mol X 0.08206 mol K atm X 10.0 292 L = 19.5159 = 19.5 atm = 3.8338 + 19.5159 = 23.3497 = 23.3 atm 10.67 Analyze. Given 390 ppm CO₂ in the atmosphere; 390 L in 10⁶ total L air. Find: mole fraction in the atmosphere. Plan. Avogadro's law deals with the relationship between volume and moles of a gas. Solve. Avogadro's law states that volume of a gas at constant temperature and pressure is directly proportional to moles of the gas. Using volume fraction to express concentration assumes that the 390 L CO₂ and 10⁶ total L air are at the same temperature and pressure. That is, 390 L is the volume that the number of moles of CO₂ present in 10⁶ L air would occupy at atmospheric temperature and pressure. The mole fraction of CO₂ in the atmosphere is then just the volume fraction from the concentration by volume. 0.00039 10.68 = 4/100 = 0.04; = = (1 - 0.04)/2 = 0.48 Vₜ = 0.900 mm 0.300 mm 10.0 mm X 10³ 1 3 X 1000 = 2.70 10⁻⁶ L = 500 torr 760 torr = atm = PV RT = 0.657895 atm 0.08206 mol K atm X 298 10⁻⁶ L X 6.022 mol 10²³ atoms = 4.3743 10¹⁶ = 4.37 10¹⁶ total atoms Xe atoms = total atoms = 0.04(4.3743 X 10¹⁶) = 1.75 10¹⁵ = 2 10¹⁵ Xe atoms Ne atoms = He atoms = 0.48(4.3743 X 10¹⁶) = 2.10 10¹⁶ = 2.1 X 10¹⁶ Ne and He atoms Assumptions: In order to calculate total moles of gas and total atoms, we assumed a reasonable room temperature. Since '4% Xe' was not defined, we conveniently assumed mole percent. The 1:1 relationship of Ne to He is assumed to be by volume and not by mass. 10.69 Analyze. Given: mass at V, T; pressure of air at same V, T. Find: partial pressure of at these conditions, total pressure of gases at V,T. Plan. mol (via P = Pₜ = +Pair Solve. 5.50 g 44.01 1 mol = 0.12497 = 0.125 mol CO₂; T = 273 + 24° = 297 K g 290