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30 ml of methane gas diffuses out through a small hole of a vessel in 15 minutes. What is the time taken by 20 ml of helium to diffuse out from the same hole?

Answer :

Using Graham's law of effusion, which states that the rates of effusion for two gases are inversely proportional to the square roots of their molar masses, it is determined that 20 ml of helium gas would take 7.5 minutes to effuse out of the same hole that methane took 15 minutes.

The question asks about the time taken by a specific volume of helium gas to diffuse out of a hole compared to methane gas, an application of Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, the lighter a gas is, the faster it will effuse. Methane (CH4) has a molar mass of approximately 16 g/mol, while helium (He) has a molar mass of about 4 g/mol. The time taken by different gases to effuse is inversely proportional to the square root of their molar masses.

Since 30 ml of methane gas takes 15 minutes to effuse, we can use the ratio of the square roots of their molar masses to find out how long it would take for 20 ml of helium to do the same:

Rate of effusion of methane (CH4) / Rate of effusion of helium (He) = (Square root of molar mass of He / Square root of molar mass of CH4)

Using Graham's law, we calculate:

(15 min / t)2 = (4/ 16)

t2 = 152 * (4 / 16)

t2 = 225 * 0.25

t2 = 56.25

t = √56.25

t = 7.5 minutes

Therefore, it would take 7.5 minutes for 20 ml of helium to diffuse out from the same hole.

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