- #1
Coop
- 40
- 0
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
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