# Maxwell Boltzmann Distribution Question

1. Jan 3, 2009

### Crosshash

Hello everyone

1. The problem statement, all variables and given/known data
The equivalent of the Maxwell-Boltzman distribution for a two-dimensional
gas is
$P(v) = Cv e^-\frac {mv^2}{kt}$

Determine $C$ so that

$\int_0^\infty P(v)dv = N$

2. Relevant equations
Not really sure

3. The attempt at a solution
I wasn't really sure how to tackle this question so I figured i'd integrate $P(v)$ since the question says that'll equal N.

$\int_0^\infty P(v)dv$
$\int_0^\infty Cv e^-\frac {mv^2}{kt} dv$
$C\int_0^\infty v e^-\frac {mv^2}{kt} dv$

$u = \frac {mv^2}{kt}$

$\frac {du}{dv} = \frac {2mv}{kt}$

$dv = \frac {du kt}{2mv}$

$C\int_0^\infty v e^{-u} \frac {du kt}{2mv}$

$C\int_0^\infty e^{-u} \frac {du kt}{2m}$

$\frac {Ckt}{2m} \int_0^\infty e^{-u} du$

$= \frac {Ckt}{2m} \bigg[{-e^{-u}\bigg]_0^\infty$

$= \frac {Ckt}{2m} \bigg[{-e^{-\frac {mv^2}{kt}}\bigg]_0^\infty$

I'm not really sure where to go from here. How would I evaluate this between infinity and zero?

Thanks

2. Jan 3, 2009

### G01

HINT: You said the entire integral has to equal N, correct? Well, your last line is equal to the integral. So if A = B and B=C ... ?

3. Jan 3, 2009

### Crosshash

ok, so you're saying

$N = \frac {Ckt}{2m} \bigg[{-e^{-\frac {mv^2}{kt}}\bigg]_0^\infty$

which yeah, makes sense. But do you want me to rearrange it to make C the subject while not evaluating the integral?

4. Jan 3, 2009

### G01

No. Evaluate the integral and then solve for C. That should be your answer.

5. Jan 3, 2009

### Crosshash

I thought so. This might seem stupid, but I really don't know how to evaluate the integral when one of the limits is $\infty$. Could you shed some light on that please?

6. Jan 3, 2009

### G01

$$N = \frac {Ckt}{2m} \bigg[{-e^{-\frac {mv^2}{kt}}\bigg]_0^t$$

Now take that whole resulting expression and take the limit as t$\rightarrow\infty$.

Now can you solve for C?

7. Jan 3, 2009

### Crosshash

Ok, so $\bigg[{-e^{-\frac {mv^2}{kT}}\bigg]_0^t$

will give me

$\lim_{t \to \infty}({-e^{-\frac {mt^2}{kT}} + 1)$

yeah?

Sorry if this is annoying, I've never actually done something like this which seems a bit strange considering a question requires it. I appreciate the help.

8. Jan 12, 2009

### tswsl1989

I used a site that provides a nice integral and limit calculator:
http://www.numberempire.com/limitcalculator.php
From that, I get:

$$$\lim_{v\to0}\left(-{{k\,T\,e^ {- {{m\,v^2}\over{k\,T}} }}\over{2\,m}}\right) = -{{k\,T}\over{2\,m}}$$$
So:
$$$C \times \left[ \frac{kTe^{\frac{-mv^2}{kT}}}{2m}\right]^{\infty}_0=C\times \frac{kT}{2m}$$$

Have I missed something obvious?

Last edited: Jan 12, 2009