# 1.2.7 (1).pdf - 1 a At the same temperature and pressure...

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1. a. At the same temperature and pressure, the ratio of the effusion rates of the gases A and B are inversely proportional to the ratios of square roots of masses of A and B (rA/rB = sqr(MB)/sqr(MA), r = rate of effusion). Since, molar mass of N 2 = 28 g/mol, molar mass of O 2 = 32 g/mol, and molar mass of Ar = 40 g/mol, effusion rate of nitrogen/effusion rate of oxygen = sqr(32 g/mol)/sqr(28 g/mol) = 5.65/5.29. Therefore, the effusion rate of N 2 is 5.65 to that of oxygen 5.29 and argon is 6.32. Now we can conclude that argon has less effusion rate, oxygen has a high effusion rate, and nitrogen has the highest effusion rate. b. Percentage of N 2 will reduce the fastest, and the oxygen will be the next. Percentage of gases will change with this order of speed: the percentage of N 2 will decrease the fastest, that of O 2 will decrease faster than Argon, and the percentage of Argon will decrease the slowest. c. rA/rB = sqr(MB)/sqr(MA) (1.6 * 10 -3 M / 215 s)/rB = sqr(32 g/mol)/sqr(40 g/mol) (0.00000744 M/s)/rB = 0.8944 rB = 0.000008318 M/s rB in 215 s = 0.001788 M Therefore, 0.001788 moles of oxygen gas will effuse in 215 seconds. d. As the diffusion rate depends on molar masses of gases, if molar mass is less then the diffusion rate will be higher and vice versa. N 2 has the greatest molar mass among three gases, O 2 has a molar mass greater than Ar, and Ar has the lightest molar mass.