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?

1 Answer
Nov 5, 2017

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.