# Maxwell Boltzmann Distribution Question

Hello everyone

## Homework Statement

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

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

Not really sure

## 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

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G01
Homework Helper
Gold Member
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 ... ?

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?

G01
Homework Helper
Gold Member
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?
No. Evaluate the integral and then solve for C. That should be your answer.

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

G01
Homework Helper
Gold Member
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?

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

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

Now can you solve for C?

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

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

Now can you solve for C?
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.

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?

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