Maxwell-Boltzmann Distribution and average speed

  • Thread starter Math Jeans
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  • #1
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Homework Statement


Given the integral
integral(0,infinity)(v^3*e^(-a*v^2)dv)=a^-2/2

Calculate the average speed v(average) of molecules in a gas using the Maxwell-Boltzmann distribution function.


Homework Equations


v^2(average)=integral(0,infinity)(v^2*v(t)dv)


The Attempt at a Solution



I took the general approach.

integral(0,infinity)(f(v)dv)=(4/pi)*(m/(2*k*T))^(3/2)*integral(0,infinity)(v^2*e^(-mv^2/(2*k*t))dv)

However, I could not find out how to make a substitution for a in order to get the v^3 into the equation.
 

Answers and Replies

  • #2
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Maxwell-Boltzman distribution is as you wrote. So if you are looking for average speed you just multiply it with v, and than calculate the integral. It's exactly the integral you have given and of course a = m/(2kt).
 
  • #3
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Maxwell-Boltzman distribution is as you wrote. So if you are looking for average speed you just multiply it with v, and than calculate the integral. It's exactly the integral you have given and of course a = m/(2kt).

So I should just multiply by v and do integration by parts?
 
  • #4
49
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noup. You have already given the solution of that integral in your problem statement. You should only put in a = m/(2kt) and then multiply the result with all that constants in front (4/pi ...).

Maxwell-Boltzman distribution is: [tex]f(v)\propto v^2 e^{-av^2} [/tex] where a is as mentioned.
the average speed is [tex] \int_{0}^{\infty} v\cdot f(v)[/tex]

the solution of this integral is in your problem statement.
 

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