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Maxwell Boltzmann Distribution Question

  • Thread starter Crosshash
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  • #1
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Hello everyone

Homework Statement


The equivalent of the Maxwell-Boltzman distribution for a two-dimensional
gas is
[itex]P(v) = Cv e^-\frac {mv^2}{kt}[/itex]

Determine [itex]C[/itex] so that

[itex]\int_0^\infty P(v)dv = N[/itex]


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 [itex]P(v)[/itex] since the question says that'll equal N.

[itex]\int_0^\infty P(v)dv[/itex]
[itex]\int_0^\infty Cv e^-\frac {mv^2}{kt} dv[/itex]
[itex]C\int_0^\infty v e^-\frac {mv^2}{kt} dv[/itex]

[itex] u = \frac {mv^2}{kt}[/itex]

[itex]\frac {du}{dv} = \frac {2mv}{kt}[/itex]

[itex]dv = \frac {du kt}{2mv}[/itex]

[itex]C\int_0^\infty v e^{-u} \frac {du kt}{2mv}[/itex]

[itex]C\int_0^\infty e^{-u} \frac {du kt}{2m}[/itex]

[itex]\frac {Ckt}{2m} \int_0^\infty e^{-u} du[/itex]

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

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

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

Thanks
 

Answers and Replies

  • #2
G01
Homework Helper
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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
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ok, so you're saying

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

which yeah, makes sense. But do you want me to rearrange it to make C the subject while not evaluating the integral?
 
  • #4
G01
Homework Helper
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ok, so you're saying

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

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.
 
  • #5
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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 [itex]\infty[/itex]. Could you shed some light on that please?
 
  • #6
G01
Homework Helper
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I thought so. This might seem stupid, but I really don't know how to evaluate the integral when one of the limits is [itex]\infty[/itex]. Could you shed some light on that please?
To evaluate your integral, evaluate the following instead:

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

Now take that whole resulting expression and take the limit as t[itex]\rightarrow\infty[/itex].

Now can you solve for C?
 
  • #7
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To evaluate your integral, evaluate the following instead:

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

Now take that whole resulting expression and take the limit as t[itex]\rightarrow\infty[/itex].

Now can you solve for C?
Ok, so [itex]\bigg[{-e^{-\frac {mv^2}{kT}}\bigg]_0^t[/itex]

will give me

[itex]\lim_{t \to \infty}({-e^{-\frac {mt^2}{kT}} + 1)[/itex]

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

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

Have I missed something obvious?
 
Last edited:

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