# Maxwell's Speed Distribution

1. Sep 19, 2007

### prace

1. The problem statement, all variables and given/known data

Given Maxwell's probability distribution function,

$$P(v) = 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}}$$

Where v = velocity, M = molar mass, R = Universal Gas Constant, n = # of mols, T = temperature, solve

$$\int P(v) dv =1$$ from 0 to $$\infty$$.

2. Relevant equations

Given above.

3. 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:

$$\int 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}} dv = 1$$

then becomes:

$$4\pi\ (\frac{M}{2\pi RT})^{3/2} \int v^2 \cdot e^{\frac{-Mv^2}{nRT}} dv = 1$$

This is where I get stuck. My thoughts are as follows: because everything but the v term in $$e^{\frac{-Mv^2}{nRT}}$$ is a constant, am I allowed to just give it a different variable, say x? This would leave me with $$e^{-xv^2}$$. 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!

2. Sep 19, 2007

### fikus

Ok you should set a new variable $$x=Cv^2~ where ~C=\frac{M}{2RT}$$
then $$dx=2Cvdv ~ ~and ~ of ~ course ~~ v =\sqrt{1/C}\sqrt{x}$$

Put that in, and you'll get:$$\frac{1}{2}C^{-3/2}\int_{0}^{\infty}\sqrt{x} e^{-x}dx$$

If you look in any mathematical handbook you'll find that that integral is well known gamma function and its value is (you can find it in tabels)

$$\Gamma(3/2) =\frac{\sqrt{\pi}{2}$$

3. Sep 19, 2007

### fikus

sorry I had some problems.

Ok you should set a new variable $$x=Cv^2~ where ~C=\frac{M}{2RT}$$
then $$dx=2Cvdv ~ ~and ~ of ~ course ~~ v =\sqrt{1/C}\sqrt{x}$$

Put that in, and you'll get:$$\frac{1}{2}C^{-3/2}\int_{0}^{\infty}\sqrt{x} e^{-x}dx$$

If you look in any mathematical handbook you'll find that that integral is well known gamma function and its value is (you can find it in tabels)

$$\Gamma(3/2) =\frac{\sqrt{\pi}}{2}$$

You could also do this integral with complex integration but it's too "complex" :)
So the whole integral with all the constants give you exactly 1.
Actualy all this constants in front are there just to ensure that this integral will be 1(normalization of distribution).

4. Sep 19, 2007

### prace

Thanks for the reply, but I still seem to be a little lost. If I let $$C = \frac{M}{2RT}$$, that leaves me with $$4 \pi \cdot ( \frac{C}{ \pi})^{3/2} \int \frac{x}{C} ...$$. I don't see how making the $$C = \frac{M}{2RT}$$ helps me in the latter half of the integral because I have a n term in there.

Thanks again!

5. Sep 19, 2007

### fikus

I think that n - # of mols is wrong in your equation. There should be 2 instead of n (of course the distribution of speeds can not depend on number of molecules).
Then when you write the integral with that new variable you put in
$$vdv = dx/(2C)$$ and for the v which stays you write $$v =\sqrt{1/C}\sqrt{x}$$
So you get $$1/2 \cdot C^{-3/2} \cdot \Gamma(3/2)$$
now put in the result I wrote for gamma function. And multiply with all constants in front $$4\pi (\frac{C}{\pi})^{-3/2}$$.

The result is 1.
Hope I was clear enough.

6. Sep 19, 2007

### prace

Oh ok, so you are saying that the original equation should be:

$$\int 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{2RT}} dv = 1$$

$$\int 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}} dv = 1$$

?

7. Sep 19, 2007

### fikus

you are rigth. Basicly all physics is in distribution $$v^2 e^{...}$$. Once you get this distribution from physics behind the problem you should normalize it. You integrate it from zero to infinity and then you divide the distribution with that result. That's how you ensure that the probability os finding a molekule with speed between 0 and intinity is 1.

8. Sep 19, 2007

### prace

Sorry to keep badgering you here, but I do not see how you get

$$\frac{1}{2}C^{-3/2}\int_{0}^{\infty}\sqrt{x} e^{-x}dx$$

The $$\sqrt{x}$$ should be an x/C no? We let x = Cv², so that makes v² = x/C. The term you are substituting for there is v² not v. That would make the integral:

$$\int_{0}^{\infty} \frac{x}{C} e^{-x}dx$$

Am I way out in left field here? However, if I solve this integral, and plug in all the constants, I get 0 = 1, haha... which is definitely not the case.

9. Sep 19, 2007

### fikus

ok I see the problem now, you are not familiar with integration with new variable.
In first integral you are integrating over v so you have dv in the end of integral. When you introduce new variable that include v, as in this case x, you then integrate over x. So you have to find out what dx is. Since you have the equation for x, you just derivate it and get dx. In our case $$dx=2Cvdv$$. Instead of vdv in first integral you write now dx/(2C). There still remains one v in integral since you have v^2 dv. So you express this v with x, v=sqrt(x/C). That's it. I hope you understand.
When you put new variable that contain the old one in integral you should also write what dx is. Cos it's not the same as dv.