# Some possibilities for modeling the shape of the

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Some possibilities for modeling the shape of the balloon include a sphere (not a very good model), a cylinder (better, but probably over-estimates the volume), and two base-to-base cones (probably under-estimates the volume). Approximating the saucer as a cylinder that is 5 feet high and 20 feet wide gives the volume as ( d /2) 2 ( h ) = (20 ft /2) 2 (5 ft) = 1600 ft 3 Converting ft 3 to L gives 1600 ft 3 [(12 in / 1 ft) (2.54 cm / 1 in)] 3 (1 L / 1000 cm 3 ) ≈ 5 10 4 L Assuming 1.0 atm and 25°C, and rearranging PV = nRT gives 4 (1 atm)(5 10 L) 2000 mol gas (0.0821 L atm/K mol)(25 273 K) PV n RT 195 CHAPTER 5: GASES Given that the average molar mass of air is 29 g/mol (see solution to Problem 5.171), and the molar mass of helium gas is 4 g/mol, the mass displaced by the balloon will be 2000 mol (29 4) g/mol = 50,000 g = 50 kg 50 kg (1000 g / 1 kg) (1 lb / 453.6 g) ≈ 110 lb The average weight for a healthy six-year-old boy is around 50-70 lbs, so yes, it looks like the balloon could have lifted the boy, but remember that our model probably overestimates the volume of the balloon, and we have ignored the mass of the balloon. ANSWERS TO REVIEW OF CONCEPTS Section 5.2 (p. 177) 1) (b) < (c) < (a) < (d) . 2) It would be easier to drink water with a straw at the foot of Mt. Everest because the atmospheric pressure is greater there, which helps to push the water up the straw. Section 5.3 (p. 183) (a) Volume doubles . (b) Volume increases 1.4 times. Section 5.4 (p. 186) Greatest volume, (b) . Greatest density, (c) . Section 5.5 (p. 195) Only for the combustion of methane, CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O( g ). Section 5.6 (p. 202) Blue sphere: 0.43 atm . Green sphere: 1.3 atm . Red sphere: 0.87 atm . Section 5.7 (p. 210) (c) and (d) . Section 5.8 (p. 213) High pressure and low temperature . 196 • • • 