Question

Assuming a homogeneous gas in the context of Kinetic Molecular Theory of gases, how does the root-mean-squared velocity relate to the average gas velocity in one dimension?

Answer 1

If you take the squared velocity in a certain direction `x` and then average it over `N` particles, you get the average squared velocity in the `x` direction:

`<< v_x^2 >> = 1/Nsum_(i=1)^(N) v_(ix)^2 = (v_(1x)^2 + v_(2x)^2 + . . . + v_(Nx)^2)/N`

For an homogeneous gas, its motion is isotropic, so that

`<< v_x^2 >> = << v_y^2 >> = << v_z^2 >>`,

and thus,

`<< v^2 >> = << v_x^2 >> + << v_y^2 >> + << v_z^2 >>`

`= 3<< v_x^2 >>`

If you then take the square root of `<< v^2 >>`, the average of the squared velocity in a radial direction, you get the root-mean-square (RMS) speed for gases.

`v_(RMS) = sqrt(<< v^2 >>)`

For gases that follow the Maxwell-Boltzmann Distribution, this is given by:

`v_(RMS) = sqrt((3RT)/M)`

where `R = "8.314472 J/mol"cdot"K"` and `T` is temperature in `"K"`. `M` is the molar mass in `"kg/mol"`.

The function of `v_(RMS)` is to express the average velocity of a gas while also factoring out the dependence of `v` on sign due to gases changing direction.