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MaxwellBoltzmann distribution 
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#1
Sep905, 11:13 AM

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Hello, I have a question regarding the MaxwellBoltzmann distribution.
As you know, the distribution basically looks like [tex]n(v) = constant \cdot v^2e^{\frac{mv^2}{2kT}}[/tex] , where v is the speed, m is the particle mass, is k the boltzmann constant and T is the absolute temperature. Now, one can calculate the mean value of the squared velocity [tex]<v^2>[/tex] by evaluating the integral [tex]\frac{\int_0^\infty v^2 n(v) dv}{\int_0^\infty n(v) dv} = \frac{3kT}{m}[/tex] From here, we can calculate the average (translational)kinetic energy of particles, [tex]<\frac{mv^2}{2}> = \frac{m}{2} <v^2>=\frac{3kT}{2}[/tex] Here comes the question, say we have air of temperature T. How can one easily show that the fraction of (for example) oxygen molecules with kinetic energy greater than [tex]\frac{3kT}{2}[/tex] is less than 50%? My idea is basically to create a new distribution, say [tex]n_1(v^2)[/tex] by substituting [tex]v[/tex] with [tex]v^2[/tex] in the original distribution, integrating from 0 to [tex]\frac{3kT}{m}[/tex] and then from [tex]\frac{3kT}{m}[/tex] to infinity. One can then compare these integrals and conclude whether the fraction of molecules with kinetic energy greater than [tex]\frac{3kT}{2}[/tex] is less than 50%. However, this approach seems extremely superflous and I am not even sure the integrals converge (even though I think so). There has to be an easier argument!! Can anyone help please? 


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