[PChem] Maxwell-Boltzmann Molecule Mean Speed

  • Thread starter Coop
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Hi,

I am having trouble with this concept...

"A mean speed (c) is calculated by multiplying each speed by the fraction of molecules that have that speed, and then adding all the products together. When the speed varies over a continuous range, the sum is replaced by an integral. To employ this approach here, we note that the fraction of molecules with a speed in the range v to v + dv is f(v)dv, so the product of this fraction and the speed is vf(v)dv. The mean speed is obtained by evaluating the integral

[tex]c=\int vf(v)dv[/tex]."

This passage is dealing with the M-B Speed Distribution. So integrating the function would give you the probability of finding a molecule with the speed between the lower and upper bounds, correct? So if you just took the integral of the function with no bounds it should be equal to one. I am having trouble seeing why integrating the product of the function and the speed leads to an average...

...Could anyone try and clear this up for me?

Thanks
 

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  • #2
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The integral over the function "f(v)" would equal one. What you integrate here is the function "v*f(v)".
 
  • #3
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The integral over the function "f(v)" would equal one. What you integrate here is the function "v*f(v)".
Yes sorry, I see that my wording was ambiguous. I am having trouble seeing why integrating v*f(v) gives you an average.
 
  • #4
Dr Transport
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it is a weighted average
 
  • #5
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Yes sorry, I see that my wording was ambiguous. I am having trouble seeing why integrating v*f(v) gives you an average.
If f(x) is a probability distribution then the integral over x*f(x) pretty much is the definition of the average, mean, or expectancy value of the property x. Not sure what I can do to help you understanding it. Maybe an example that does not explicity use an integral: Rolling a dice has an equal probability 1/6 for each side of the dice. The expectancy value is <x> = 1*1/6 + 2*1/6 +3*1/6 + 4*1/6 +5*1/6 + 6*1/6 = [tex] \sum_{i=1}^6 i * p(i)[/tex], with p(i) being the probability for event/value i.

I am not sure that emphasizing "weighted average" really helps understanding (but this thread has a chance to prove me wrong). Technically, the integral over x*f(x) is a weighted average of x with f(x) being the weights. But in my experience the term "weighted average" is only used when a sensible "unweighted average" would exist, which is not the case here (there is no point in or sensible meaning of "the average real number").
 
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  • #6
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Thanks for following up :). Yes, I was confused at where the integral c=∫vf(v)dv was coming from (I don't like using random equations which I don't know where they came from) but I figured out it was pretty much just a definition.
 

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