If oxygen, #O_2#, effuses (spreads out) at a rate of #622.0# #ms^-1# under certain conditions of temperature and pressure, how fast will methane, #CH_4#, effuse under the same conditions?

1 Answer
Feb 17, 2016

Answer:

We know the mass of an #O_2# molecule is #32# #gmol^-1# and the mass of a #CH_4# molecule is #16# #gmol^-1#, so Graham's Law of Effusion shows that the rate of effusion will be #879.6# #ms^-1#.

Explanation:

This question requires Graham's Law of Effusion, which states:

#"Rate"_1/"Rate"_2 = sqrt(M_2/M_1)# where #M_1# and #M_2# are the molar masses of the effusing gases.

To make the math easier, let's call the #CH_4# rate #"Rate"_1#. (Which we call which doesn't matter, as long as we're consistent: the math will sort itself out!)

#"Rate"_1=?#
#M_1=16# #gmol^-1#
#"Rate"_2=622.0# #ms^-1#
#M_2=32# #gmol^-1#

Rearranging:

#"Rate"_1 = sqrt(M_2/M_1)*"Rate"_2#
#=sqrt(32/16)*622.0=879.6# #ms^-1#