Maxwell Boltzmann Distribution Question

[Crosshash]

Hello everyone

Homework Statement

The equivalent of the Maxwell-Boltzmann 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

Homework Equations

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

 

[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 ... ?

 

[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?

 

[G01]

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.

 

[Crosshash]

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]

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?

To evaluate your integral, evaluate the following instead:

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?

 

[Crosshash]

To evaluate your integral, evaluate the following instead:

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.

 

[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?