- #1

Aleoa

- 128

- 5

In the first volume of his lectures (cap 6-4) Feynman presents Maxwell's distribution of velocities of the molecules in a gas.

And, referring to the PDF graph he says:

"If we consider the molecules in a typical container (with a volume of, say, one

liter), then there are a very large number N of molecules present (N ≈ [itex]10^{22}[/itex]).

Since p(v) ∆v is the probability that one molecule will have its velocity in ∆v,

by our definition of probability we mean that the expected number <∆N> to be

found with a velocity in the interval ∆v is given by: [itex]<\Delta N>=Np(v)\Delta v[/itex].

[...]

I have not understand this part, how it's possible to derive this 1/√N deviation ( or a value the gives the idea of what Feynman is saying) ?

Thanks for your support

And, referring to the PDF graph he says:

"If we consider the molecules in a typical container (with a volume of, say, one

liter), then there are a very large number N of molecules present (N ≈ [itex]10^{22}[/itex]).

Since p(v) ∆v is the probability that one molecule will have its velocity in ∆v,

by our definition of probability we mean that the expected number <∆N> to be

found with a velocity in the interval ∆v is given by: [itex]<\Delta N>=Np(v)\Delta v[/itex].

[...]

**Since with a gas we are usually dealing with large numbers of molecules,**

we expect the deviations from the expected numbers to be small (like 1/√N), so

we often neglect to say the “expected” number"we expect the deviations from the expected numbers to be small (like 1/√N), so

we often neglect to say the “expected” number

I have not understand this part, how it's possible to derive this 1/√N deviation ( or a value the gives the idea of what Feynman is saying) ?

Thanks for your support