The area under the Maxwell-Boltzmann distribution of speeds of molecules of mass m in a gas at temperature T above a speed v can be estimated as
1/2 (m/2πkT)^(1/2) v exp(-(mv^2)/kT)
Show that the maximum likely speed of a gas molecule in the air in a typical room (i.e. the speed above which the Maxwell-Boltzmann distribution gives a probability of 1/N, where N is the number of molecules in the room) is of order 4<v^2>^1/2 (where <v^2>^1/2 = (3kT/m)^1/2).
The Attempt at a Solution
I think the question means that the Maxwell-Boltzmann distribution curve is approx a 1/N curve at and above a certain value of N. I was going to equate the max-bolt curve equation to 1/N, except the question gives the equation for the area. Then I thought I should integrate the 1/N curve to get the area then 1/N dN gives ln N. Then I can equate the areas. Then use PV=NkT so that
ln(PV/kT) = 1/2 (m/2 pi kT) exp -(mv/kT)
I was then going to plug in values for T (293Kelvin) and m (mass of nitrogen molecule) but this is really hard to solve for v so I think it must be wrong.
Any help would be really great! Thanks