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
$$u=p^2$$
$$du=2pdp$$
we get
$$\int p^3 e^{-p^2} 2pdp$$
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

 

[paranormal]

.Dear student,

Thank you for your efforts in attempting to solve this problem. I can see that you have made some progress, but there are a few errors in your solution.

Firstly, the expression for the probability distribution p(E) that you have written is incorrect. It should be:

p(E)dE=2\sqrt{\frac{E}{\pi(kT)^3}}e^{-\frac{E}{kT}}dE

Note that the exponent should be -E/kT, not -E/.

Next, when you substitute E/kT=u, you need to also change the limits of integration accordingly. The new limits should be from 0 to \infty, not from 0 to \infty.

Finally, when you substitute u=E/kT into the integral, you need to also change the differential from dE to (kT)du.

With these changes, your integral should look like this:

Eav=\int_{0}^{\infty }\frac{E}{kT}\times2\sqrt{\frac{E}{\pi(kT)^3}}e^{-\frac{E}{kT}}(kT)du

Simplifying this expression will give you the desired result:

Eav=\frac{3}{2}kT

I hope this helps. Keep up the good work!