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

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Answer 1 · 2017-04-20T19:09:49

They are close... not quite the same.

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

#v_(RMS) = sqrt((3RT)/M)#

The rate of effusion #z# is directly proportional to the velocity. Thus, #z_A/z_B = v_(A)/(v_(B))#, and we have Graham's law of effusion:

#bb(z_A/z_B) = sqrt((3RT)/M_A)/sqrt((3RT)/M_B)#

#= bb(sqrt(M_B/M_A))#

Hence, if we choose #"CO"_2# to be #A#, we have that

#z_(CO_2)/z_(C_3H_8) = sqrt(M_(CO_2)/M_(C_3H_8))#

#= sqrt("44.01 g/mol"/"44.1 g/mol")#

#= 0.9979#

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

#z_(CO_2) = 0.9979z_(C_3H_8)#