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Unformatted text preview: 10 Gases Visualizing Concepts 10.2 At constant temperature and volume, pressure depends on total number of particles (Charles’ Law). In order to reduce the pressure by a factor of 2, the number of particles must be reduced by a factor of 2. At the lower pressure, the container would have half as many particles as at the higher pressure. 10.3. (a) At constant pressure and temperature, the container volume is directly proportional to the number of particles present (Avogadro’s Law). As the reaction proceeds, 3 gas molecules are converted to 2 gas molecules, so the container volume decreases. If the reaction goes to completion, the final volume would be 2/3 of the initial volume. (b) At constant volume, pressure is directly proportional to the number of particles (Charles’ Law). Since the number of molecules decreases as the reaction proceeds, the pressure also decreases. At completion, the final pressure would be 2/3 the initial pressure. 10.5 (a) Partial pressure depends on the number of particles of each gas present. Red has the fewest particles, then yellow, then blue. P r ed < P y ellow < P b lue (b) P g as = χ g as P t . Calculate the mole fraction, χ g as = [mol gas / total moles] or [particles gas / total particles]. This is true because Avogadro’s number is a counting number, and mole ratios are also particle ratios. χ r ed = 2 red atoms / 10 total atoms = 0.2; P r ed = 0.2(0.90 atm) = 0.18 atm χ y ellow = 3 yellow atoms / 10 total atoms = 0.3; P y ellow = 0.3(0.90 atm) = 0.27 atm χ b lue = 5 blue atoms / 10 total atoms = 0.5; P b lue = 0.5(0.90 atm) = 0.45 atm 10.6 10.8 (a) Total pressure is directly related to total number of particles (or total mol particles). P(ii) 144 10 Gases Solutions to Exercises < P(i) = P(iii) (b) Partial pressure of He is directly related to number of He atoms (yellow) or mol He atoms. P He (iii) < P He (ii) < P H e (i) (c) Density is total mass of gas per unit volume. We can use the atomic or molar masses of He (4) and N 2 (28), as relative masses of the particles. mass(i) = 5(4) + 2(28) = 76 mass(ii) = 3(4) + 1(28) = 40 mass(iii) = 2(4) + 5(28) = 148 Since the container volumes are equal, d(ii) < d(i) < d(iii). (d) At the same temperature, all gases have the same “avg” kinetic energy. The average kinetic energies of the particles in the three containers are equal....
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- Fall '08
- mol, NH