What Is the Maximum Likely Speed of a Gas Molecule in a Typical Room?

[CatWoman]

Homework Statement

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).

Homework Equations

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

 

[Geezer]

If you recall from calc, when you want to find a max or min, you need to take a derivative...

 

[CatWoman]

Oh yeah! So if iI have the formula for the area then I need to differentiate twice then set equal to zero to get the maxima? I did this and got V^2=(18 pi k T/m)^1/2. This doesn't match the answer so I don't know if my working is wrong or something else.