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10 Gases Solutions to Exercises (a) 1Pa = m² 1N = 1kg-m s² X 1 = m 1kg Change mass to kg and area to m². P = A a = 0.50 130 lb X 9.81 m X 0.454 1lb kg X 1m² = 1.798 10⁶ kg = 1.8 10⁶ Pa = 1.8 10³ kPa Check. [1.30 X 10 0.5 X ≈ (130 X 16,000) ≈ 2.0 X 10⁶ Pa ≈ 2.0 X 10³ kPa. The units are correct. (b) 1 atm = 101.325 kPa 1.798 X 10³ kPa X = 1.774 X 10¹ = 18 atm 1 atm 101.325 kPa (c) 1 atm = 14.70 17.74 atm X 14.70 = 260.8 = 2.6 10² 1 atm 10.16 P = m a/A; 1Pa = 1kg/m A = 3.0 cm X 4.1 cm X 4 = 49.2 = 49 49.2 262 kg X 9.81 s² m X = 5.224 X 10⁵ kg = 5.2 X 10⁵ Pa 10.17 Analyze. Given: 760 mm column of Hg, densities of Hg and Find: height of a column of H₂O at same pressure. Plan. We must develop a relationship between pressure, height of a column of liquid, and density of the liquid. Relationships that might prove useful: P = F/A; F = m a; m=dx V(density) (volume); V = A height Solve. P = F = = = dxAxhxa A A A A (a) = PH₂O; Using the relationship derived above: (d h = (d Since a, the acceleration due to gravity, is equal in both liquids, 1.00g/mL X = X 760 mm = 13.6 1.00 g/mL X 760 mm = 1.034 10⁴ = 1.03 10⁴ mm = 10.3 m (b) Pressure due to H₂O: 1 atm = 1.034 10⁴ mm H₂O (from part (a)) 39 ft X X 2.54 1in cm X 10 1cm mm X 1.034 1atm 10⁴ mm = 1.150 = 1.2 atm = +PH₂O = 0.97 atm +1.150 atm = 2.120 = 2.1 atm 10.18 Using the relationship derived in Solution 10.17 for two liquids under the influence of gravity, (d = At 749 torr, the height of an Hg barometer is 749 mm. 1.20g 1mL = 13.6g mL 760 mm; = 13.6 g/mL 1.20g/mL 749 mm = 8.49 10³ mm = 8.49 m 278