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 N2= 28 g/mol, molar mass of O2= 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 N2is 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 N2will reduce the fastest, and the oxygen will be the next. Percentage of gases will change with this order of speed: the percentage of N2will decrease the fastest, that of O2will decrease faster than Argon, and the percentage of Argon will decrease the slowest. c.rA/rB = sqr(MB)/sqr(MA) (1.6 * 10-3M / 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. N2has the greatest molar mass among three gases, O2has a molar mass greater than Ar, and Ar has the lightest molar mass.