If 250 mL of methane,CH₄,effuses through a small hole in 20 s,the time required for the same volume of helium to pass through the hole under the same conditions will be A)40 s. B)5 s. C)1.3 s. D)10 s. E)80 s.

Question

Quizplus · Accepted Answer

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since methane (CH4, molar mass = 16 g/mol) is heavier than helium (He, molar mass = 4 g/mol), helium will effuse faster. The time it takes for gases to effuse is directly proportional to the square root of their molar masses. Therefore, the time for helium to effuse can be calculated as follows: $t_{He} = t_{CH4} \times \sqrt{\frac{M_{CH4}}{M_{He}}} = 20s \times \sqrt{\frac{16}{4}} = 20s \times 2 = 40s$. However, since the calculation was incorrect in the explanation, the correct calculation should be: $t_{He} = 20s \times \sqrt{\frac{4}{16}} = 20s \times \frac{1}{2} = 10s$, making the correct answer 10 seconds.

If 250 ML of Methane,CH4,effuses Through a Small Hole in 20

If 250 ML of Methane,CH₄,effuses Through a Small Hole in 20