Question

Two gases are put in two (identical) separate containers under the same conditions, and let effuse completely out of their container. If for Ar takes 120.0 s to effuse out, while for the second gas takes 60.0 s, what is most likely the second gas?

Answer 1

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Where Rate1 is the rate of effusion of gas 1, which is argon.

Rate 2 is the  rate of effusion of gas 2.

M1 is molar mass of gas 1

M2 is molar mass of gas 2.

Now rate of effusion= volume of gas /time taken

Let V be the volume of the both the gases [ as both the gases are placed in identical containers under similar conditions.]

t1 be the time taken by gas1(Ar)

t2 be the time taken by gas2

Therefore,

Rate1 = V/t1= V/120s

Rate2=V/t2= V/60s

Taking square of both the sides,

M2 / M1= 1/4

M2 /40 gmol^-1 = 1/4 [Molar mass of gas1 Argon= 40 gmol^-1]

M2= (1/4) x 40 gmol^-1 =10 gmol^-1

Therefore the second gas will be a gas with molar mass 10 gmol^-1