- #1
prace
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Homework Statement
Given Maxwell's probability distribution function,
[tex] P(v) = 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}}[/tex]
Where v = velocity, M = molar mass, R = Universal Gas Constant, n = # of mols, T = temperature, solve
[tex] \int P(v) dv =1[/tex] from 0 to [tex] \infty [/tex].
Homework Equations
Given above.
The Attempt at a Solution
So the idea here if it is not clear from the LaTeX above (sorry, still working on my LaTeX skills) is that I would like to prove that the integral of p(v) with respect to v is equal to one from 0 to infinity.
First, I figured that many of these variables are actually constants. Since we are only integrating with respect to v, they can be legally treated as such, and be taken out of the integral.
So, step 1:
[tex] \int 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}} dv = 1[/tex]
then becomes:
[tex] 4\pi\ (\frac{M}{2\pi RT})^{3/2} \int v^2 \cdot e^{\frac{-Mv^2}{nRT}} dv = 1[/tex]
This is where I get stuck. My thoughts are as follows: because everything but the v term in [tex] e^{\frac{-Mv^2}{nRT}} [/tex] is a constant, am I allowed to just give it a different variable, say x? This would leave me with [tex] e^{-xv^2} [/tex]. I don't see how this can really help me, but I just wanted to show some of my thought process.
With the integral as it is, I plugged it into my calculator and came up with infinity as an answer. This makes sense to me, but is not the correct answer since it does not equate to 1. I also tried solving the integral by integration by parts, but could not find a good way to do that either.
I apologize for the long and drawn out problem here, but any help would be great! Thanks for looking!