# 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 = 0.5MMv^{2}

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

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

0.5MM_{A}v_{A}^{2}=0.5MM_{B}v_{B}^{2}

Dividing both sides by 0.5MM_{A}v_{B}^{2} gives

v_{A}^{2}/v_{B}^{2} = 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, CH_{4}, with that of propane, C_{3}H_{8}.

Solution: the ratio of effusion should be

e_{A}/e_{B}=(MM_{B}/MM_{A})^{1/2}

Note that methane, CH_{4}, has MM=16, while propane, C_{3}H_{8}, 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.

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