Are the effusion rates of carbon dioxide and propane the same?

Apr 20, 2017

They are close... not quite the same.

From the root-mean-square speeds, one could derive the rate of effusion.

${v}_{R M S} = \sqrt{\frac{3 R T}{M}}$

The rate of effusion $z$ is directly proportional to the velocity. Thus, ${z}_{A} / {z}_{B} = {v}_{A} / \left({v}_{B}\right)$, and we have Graham's law of effusion:

$\boldsymbol{{z}_{A} / {z}_{B}} = \frac{\sqrt{\frac{3 R T}{M} _ A}}{\sqrt{\frac{3 R T}{M} _ B}}$

$= \boldsymbol{\sqrt{{M}_{B} / {M}_{A}}}$

Hence, if we choose ${\text{CO}}_{2}$ to be $A$, we have that

${z}_{C {O}_{2}} / {z}_{{C}_{3} {H}_{8}} = \sqrt{{M}_{C {O}_{2}} / {M}_{{C}_{3} {H}_{8}}}$

$= \sqrt{\text{44.01 g/mol"/"44.1 g/mol}}$

$= 0.9979$

So they're almost the same, but propane effuses slightly faster.

${z}_{C {O}_{2}} = 0.9979 {z}_{{C}_{3} {H}_{8}}$