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10 Gases Solutions to Exercises 10.19 (a) The tube can have any cross-sectional area. (The height of the Hg column in a barometer is independent of the cross-sectional area. See the expression for pressure derived in Solution 10.17.) (b) At equilibrium, the force of gravity per unit area acting on the mercury column is not equal to the force of gravity per unit area acting on the atmosphere. (F = ma; the acceleration due to gravity is equal for the two substances, but the mass of Hg for a given cross-sectional area is different than the mass of air for this same area.) (c) The column of mercury is held up by the pressure of the atmosphere applied to the exterior pool of mercury. (d) If you took the mercury barometer with you on a trip from the beach to high mountains, the height of the mercury column would decrease with elevation. (Atmospheric pressure decreases as elevation increases.) 10.20 The mercury would fill the tube completely; there would be no vacuum at the closed end. This is because atmospheric pressure will support a mercury column higher than 70 cm, while our tube is only 50 cm. No mercury flows from the tube into the dish and no vacuum forms at the top of the tube. 10.21 Analyze/Plan. Follow the logic in Sample Exercise 10.1. Solve. (a) 265 torr 760 torr = 0.349 atm (b) 265 torr X mm 1 torr = 265 mm Hg (c) 265 torr 1.01325 10⁵ Pa = 3.53 10⁴ Pa 760 torr (d) 265 torr 1.01325 760 torr 10⁵ Pa X 1 1 10⁵ bar Pa = 0.353 bar (e) 265 torr 760 torr 14.70 1 atm psi = 5.13 psi 10.22 (a) 0.912 atm X 760 1 torr = 693 torr atm (b) 0.685 bar 1 1 bar Pa 1 1kPa Pa = 68.5 kPa 1 atm (c) 655 mm = 0.862 atm 760 mm (d) 10⁵ 1.323 Pa = 1.3057 = 1.306 atm 1.01325 X 10⁵ Pa (e) 2.50 atm 14.70 1atm psi = = 36.8 psi 10.23 Analyze/Plan. Follow the logic in Sample Exercise 10.1. Solve. (a) 30.45 in Hg 25.4 1in mm X 1 mm 1 torr = 773.4 torr [The result has 4 sig figs because 25.4 mm/in is considered to be an exact number.] 279