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Maxwell Speed Distribution

by kashe
Tags: distribution, maxwell, speed
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kashe
#1
Feb8-11, 09:56 PM
P: 7
1. The problem statement, all variables and given/known data
Confirm that the mean speed of molecules of molar mass M at a temperature T is equal to (8RT/piM)^1/2. Hint: You will need an integral of the form ∫ (where a=0, and b=infinity) x^3*e^(-ax^2) dx = 1/2a^2.


2. Relevant equations
The Maxwell speed distribution formula we are using is f=F(s)delta s where F(s)=4pi*(M/2piRT)^1/2*s^2*e^(-Ms^2/2RT)


3. The attempt at a solution
I attempted to use the answer to the integral and say a= M/2RT, but that didn't work. I then thought a=M/2piRT, but I couldn't get it to be the correct answer. I don't think I need to do the integral just because it is already done for me, but I am stumped at how to relate the mean speed and this integral.
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vela
#2
Feb9-11, 04:26 AM
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PF Gold
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The normalization constant of the distribution function isn't correct. It should be

[tex]F(s) = \sqrt{\frac{2}{\pi}\left(\frac{M}{RT}\right)^3} s^2 e^{-Ms^2/2RT}[/tex]
dextercioby
#3
Feb9-11, 07:43 AM
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How do you define the mean speed for molecules ?

kashe
#4
Feb9-11, 09:26 AM
P: 7
Maxwell Speed Distribution

The mean speed is the sum off all the speeds of all the molecules divided by the number of molecules. I also know that the mean speed is related to the rms speed of the molecules by the equation mean speed= (8/3pi)^.5 times the rms speed.

Also I have been plugging in numbers and found the equation my book gave to me is equivalent to the normalization constant above.
vela
#5
Feb9-11, 12:42 PM
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PF Gold
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What do you mean by f=F(s)Δs? Perhaps that's where the confusion is arising.

My point about the normalization constant comes from the fact that

[tex]\int_0^\infty 4\pi\sqrt{\frac{M}{2\pi RT}}s^2e^{-\frac{Ms^2}{2RT}}\,ds = \frac{2\pi R T}{M} \ne 1[/tex]
kashe
#6
Feb9-11, 12:51 PM
P: 7
The f(s) delta s. Is the fraction of molecules between two speeds. The delta s is just the change in speed you have, I.e. if you want the fraction between 300 and 310 m/s, you do the normalization constant with 300 and then multiply it by 10 for the fraction
vela
#7
Feb9-11, 01:26 PM
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PF Gold
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OK, that's what I thought. Then your normalization constant is wrong for the reason I explained earlier. The integral of the probability density alone should be equal to 1.

http://en.wikipedia.org/wiki/Maxwell..._for_the_speed

Using the correct constant, I did indeed get the result you're trying to derive for the mean speed.
kashe
#8
Feb9-11, 02:53 PM
P: 7
Ok, I will attempt to integrate the constant you gave then, although I am not sure I know how to do it. I will work with it and post again and see if you can give some guiding help. Thanks.
kashe
#9
Feb9-11, 08:36 PM
P: 7
Ok I am stuck. I tried to integrate the distribution constant but got stuck, and also wondered why I need to do it if in the hint on the problem they gave it to me. Although in the hint I am not sure where "x" went, because when I integrate the constant the "x" term should still be there. Any help on where to start? I tried many things but haven't got anywhere.
vela
#10
Feb9-11, 10:53 PM
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PF Gold
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What do you mean by "integrate the distribution constant"? The point of my previous posts were to point out that the F(s) you gave in the original post is incorrect, specifically that the constant factor in front was wrong. I gave you F(s) with the correct constant in post #2.

It would help if you'd show us explicitly what you're doing.
kashe
#11
Feb9-11, 11:07 PM
P: 7
Sure. Since the integral was equal to 1/2a^2, it appeared the a value would be equal to M/RT, getting that from the exponent of e. Putting that value in for a however does not yield the answer I want. I then tried to see if integrating that integral in the hint would help me understand, I tried to use integration by parts with u=x^3 and v=e^-x^2. Using the tabular method I never got to 1/2a^2. That took a while but no help. I then thought the hint answer did not include the square root of 2/pi, but that didn't make much sense to me. I then tried to integrate again using u-sub, but did not get a result of 1/2a^2. I am now lost.
kashe
#12
Feb10-11, 12:42 AM
P: 7
I just tried this and got the correct answer....
Using the hint, i.e. 1/2a^2, I made a=M/2RT, and included the terms (square root 2/pi times (M/RT)^3) After plugging the a value in, I did all the algebra to get to (8RT/piM)^.5.


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