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1.2.7 Practice: Intermolecular Forces and the Properties of Gases Rodolfo Marquez Valencia Mrs. Hepburn AP Chemistry January 22, 2021
1. A spaceship is found to have a tiny leak that allows the effusion of air into space. Air in the spaceship is a mixture of nitrogen (N 2 ), oxygen (O 2 ), argon (Ar), and trace amounts of other gases. a. Predict which of the three main gases will have the lowest rate of effusion and which will have the highest rate. Explain your answer. Answer: Well, the gas with lowest rate of effusion will be the one with the highest molar mass, so that is argon. The gas with the highest rate of effusion will be the one with the lowest molar mass, so that is nitrogen. b. Predict how the percentages of the gases in the air will change over time. Answer: Oxygen: 20% Nitrogen: 79% Argon: 1% I predict that the first one to drop will be nitrogen and the last one to drop will be argon because of their behavior of ideal or non-ideal gases. c. Argon is found to effuse through the hole at a rate of 1.6 × 10 –3 mol in 215 s. How much O 2 would effuse through the hole in the same amount of time? Answer: VAr = 1.6 * 10 -3 / 215 = 7.4 * 10 -6
VAr /VO 2 = sqrt (MO 2 / MAr) = (32/40) 1/2 VAr /VO 2 = 0.89 VO 2 = VAr / 0.89 VO 2 = (7.4 * 10 -6 ) / 0.89 = 8.31 * 10 -6 nO 2 = 8.31 * 10 -6 * 215 = 1.79 * 10 -3 mol d. Suppose the three gases were placed at 1.00 atm and 273 K in closed containers, as shown below. A tiny hole is opened in each container, allowing each gas to diffuse into another container filled with neon gas. What is the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen? Is this greater than, less than, or the same as the ratio of the rates of effusion for the two gases? Answer:
VNe/VO 2 = (MO 2 / MNe) 1/2 = (32/20.2) 1/2 VNe/VO 2 = 1.26 VNe = 1.26 * VO 2 1.26 * VO 2 = 1.18 * VN 2 VO 2 = 1.18 * VN 2 / 1.26 VO 2 = 0.94 * VN 2 This is equal to the ratio of rates of effusion for the two gases.

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