Tutoring chemistry, you may mention kinetic energy of particles, diffusion, and effusion. The tutor gives a brief explanation.

In my post from Jan 18, 2017, I define effuse: it means to escape from a container through a porous boundary.

Graham’s Law of Effusion compares the rate at which two different gases will effuse, based on their comparative molecular masses. If e_A is the effusion rate of gas A, and e_B that of gas B, then

e_A/e_B = (MM_B/MM_A)^1/2

where MM_A is molecular mass of gas A, and MM_B, that of gas B.

The reasoning behind the formula is that effusion depends on molecular motion, which is quantified by kinetic energy, KE:

KE = 0.5MMv²

where, once again, MM means molecular mass, while v means velocity.

Two gases of the same temperature have the same kinetic energy:

0.5MM_Av_A²=0.5MM_Bv_B²

Dividing both sides by 0.5MM_Av_B² gives

v_A²/v_B² = MM_B/MM_A

square rooting both sides gives

v_A/v_B = (MM_B/MM_A)^1/2

Since effusion is motion through pores, the ratio of velocities is the ratio of effusion:

e_A/e_B=v_A/v_B = (MM_B/MM_A)^1/2

Example: Compare the effusion rate of methane, CH₄, with that of propane, C₃H₈.

Solution: the ratio of effusion should be

e_A/e_B=(MM_B/MM_A)^1/2

Note that methane, CH₄, has MM=16, while propane, C₃H₈, has MM=44. Therefore,

e_methane/e_propane=(44/16)^1/2=1.66

Methane should escape from a porous container 1.66 times the rate that propane escapes.

Note that, in a more general sense, this is a law of diffusion. Therefore, if propane and methane are released at one end of a room, methane should reach the other end 1.66 times as quickly as propane.

HTH:)

Source:

Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

White, J. Edmund. Physical Chemistry, College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.