Particles in an ideal monatomic gas

[Jason Gomez]

Prove that for particles in an ideal monatomic gas the average energy Eav can be given by:

Eav=\int_{0}^{\infty }Ep(E)dE=3/2kT

where the probablity distribution p(E) is given by:
p(E)dE=2/\sqrt{\pi}(kT)^{3/2}\times e^{-E/} dE

Homework Equations

Let E/kT

The Attempt at a Solution

after working this problem over and over, this is as far as I can get

2\pi^{-1/2}\int_{0}^{\infty}u^{3/2}e^{-u}de=2\pi^{-1/2}\left ( 2/5u^{5/2}e^{-u}-ue^{-u}u^{3/2} \right )

i have tried pulling variables out but get no where, I feel it is right up to this point but do not know where to go from here except factor out common variables, but once again I get know where

 

[vela]

Hint: look up the gamma function.

http://en.wikipedia.org/wiki/Gamma_function

 

[zzzoak]

Or making substitution
<br /> u=p^2<br />
<br /> du=2pdp<br />
we get
<br /> \int p^3 e^{-p^2} 2pdp<br />
and integrating by parts.

 

[Jason Gomez]

Thank you I think I see a similarity between that and the problem I am working but I am on my phone looking at it, I will let you know how it goes when I get home