Chemistry: Graham’s Law of Effusion

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 eA is the effusion rate of gas A, and eB that of gas B, then

eA/eB = (MMB/MMA)1/2

where MMA is molecular mass of gas A, and MMB, 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:

0.5MMAvA2=0.5MMBvB2

Dividing both sides by 0.5MMAvB2 gives

vA2/vB2 = MMB/MMA

square rooting both sides gives

vA/vB = (MMB/MMA)1/2

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

eA/eB=vA/vB = (MMB/MMA)1/2

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

Solution: the ratio of effusion should be

eA/eB=(MMB/MMA)1/2

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

emethane/epropane=(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.

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