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Maxwellboltzmann distribution 
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#1
Aug109, 12:51 PM

P: 30

1. The problem statement, all variables and given/known data
The area under the MaxwellBoltzmann 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 MaxwellBoltzmann 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). 2. Relevant equations 3. The attempt at a solution I think the question means that the MaxwellBoltzmann distribution curve is approx a 1/N curve at and above a certain value of N. I was going to equate the maxbolt 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 


#2
Aug709, 02:07 AM

P: 292

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



#3
Aug709, 04:22 AM

P: 30

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.



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