- #1

rtareen

- 162

- 32

##P(v) = 4\pi \left(\frac{M}{2\pi RT}\right)^{3/2}v^2e^{-Mv^2/2RT}##

If we want to find the fraction of molecules that range from speeds ##v_1## to ##v_2## we can find that by using:

##\int_{v_1}^{v_2}P(v)dv##

So what does P(v) itself represent? It has units of s/m.

Secondly, the book says that to find the average speed of the molecules, we

*weight*each value of v in the distribution. This means we multiply the integral by v and take the limits to go from zero to infinity. In other words:

##v_{av} = \int_0^{\infty}vP(v)dv##

Can somebody explain what is going on here because I am completely lost. How does this formula give the average speed? What is

*weighting*?

Similarly, we can find the rms speed with the following formula

##v_{rms} = \sqrt{\int_0^{\infty} v^2P(v)dv}##

Again, what's with the ##v^2## and how does it help finding the mean squared speed? When we take the square root of this it becomes the root mean square speed.

Finally, we have the most probable speed, in which to find we set dP/dv = 0. I don't know how to take that derivative, and from what I've seen from my calculator its a really messy derivative. But anyways we then solve the equation for v and we get

##v = \sqrt{\frac{2RT}{M}}##

and this is the most probable speed. Why do we set the derivative to zero to find the most probable speed. I have a feeling all of this has to do with statistics and probability, which I have no experience with. Do statistics and probability come up a a lot in physics? If so would it be a good idea to take an upper level probability course? I want to major in physics.