Editing Graham's law (section)

==Examples==
'''First Example:''' Let gas 1 be H<sub>2</sub> and gas 2 be O<sub>2</sub>. (This example is solving for the ratio between the rates of the two gases)

:<math>{\mbox{Rate H}_2 \over \mbox{Rate O}_2} =\sqrt{M(O_2) \over M(H_2)} ={\sqrt{32} \over \sqrt{2}}= \sqrt{16} = 4</math>

Therefore, hydrogen molecules effuse four times faster than those of oxygen.<ref name=LM/>

Graham's law can also be used to find the approximate molecular weight of a gas if one gas is a known species, and if there is a specific ratio between the rates of two gases (such as in the previous example).  The equation can be solved for the unknown molecular weight.

:<math>{M_2}={M_1 \mbox{Rate}_1^2 \over \mbox{Rate}_2^2}</math>

Graham's law was the [[Gaseous diffusion|basis]] for separating [[uranium-235]] from [[uranium-238]] found in natural [[uraninite]] (uranium ore) during the [[Manhattan Project]] to build the first atomic bomb. The United States government built a gaseous diffusion plant at the [[Clinton Engineer Works]] in [[Oak Ridge, Tennessee]], at the cost of $479 million (equivalent to ${{format price|{{inflation|US-GDP|479,589,999|1945|r=2}} }} in {{Inflation/year|US-GDP}}). In this plant, [[uranium]] from uranium ore was first converted to [[uranium hexafluoride]] and then forced repeatedly to diffuse through porous barriers, each time becoming a little more enriched in the slightly lighter uranium-235 isotope.<ref name=Petrucci/>

'''Second Example:''' An unknown gas diffuses 0.25 times as fast as He. What is the molar mass of the unknown gas?

Using the formula of gaseous diffusion, we can set up this equation.

:<math>\frac{\mathrm{Rate}_\mathrm{unknown}}{\mathrm{Rate}_\mathrm{He}} = \frac{\sqrt{4}}{\sqrt{M_2}}</math>

Which is the same as the following because the problem states that the rate of diffusion of the unknown gas relative to the helium gas is 0.25.

:<math>0.25 = \frac{\sqrt{4}}{\sqrt{M_2}}</math>

Rearranging the equation results in

:<math>M = (\frac{\sqrt{4}}{0.25})^2 = \frac{\mathrm{64g}}{\mathrm{mol}}</math>