Explain the relationship between the rate of effusion of a gas and its molar mass

Question

Explain the relationship between the rate of effusion of a gas and its molar mass.

Methane gas (CH₄) effuses 3.4 times faster than an unknown gas. Determine the molar mass of the unknown gas. Show your work or explain your answer, giving specific values used to determine the answer. 

Answer 1

Graham's Law of Effusion

rate gas 1 / rate gas 2 = sqrt MW gas 2 / sqrt MW gas 1

Graham's law of effusion essential states that the rate of effusion (diffusion) of a gas is inversely proportional to the square root of its molar mass. So, the heavier a gas is, the slower it will effuse, and vice versa.

In the current problem, we have methane (CH₄) that has a molar mass of 16 g / mole effusing 3.4 times FASTER than an unknown gas. We are to find the molar mass of the unknown. Since CH4 effuses FASTER, that means the unknown must have a molar mass GREATER than 16 g / mole. We use Graham's law to find the molar mass of the unknown.

rate methane / rate unknown = sqrt MW unknown / sqrt MW methane

Let rate of methane be 3.4, and then the rate of the unknown (X) will be 1

3.4 /1 = sqrt X / sqrt 16

3.4/1 = sqrt X / 4

sqrt X = 13.6 and now square both sides to solve for X...

x = 185 g / mole = molar mass of unknown