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?

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:

$\left\langle{v}_{x}^{2}\right\rangle = \frac{1}{N} {\sum}_{i = 1}^{N} {v}_{i x}^{2} = \frac{{v}_{1 x}^{2} + {v}_{2 x}^{2} + . . . + {v}_{N x}^{2}}{N}$

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

$\left\langle{v}_{x}^{2}\right\rangle = \left\langle{v}_{y}^{2}\right\rangle = \left\langle{v}_{z}^{2}\right\rangle$,

and thus,

$\left\langle{v}^{2}\right\rangle = \left\langle{v}_{x}^{2}\right\rangle + \left\langle{v}_{y}^{2}\right\rangle + \left\langle{v}_{z}^{2}\right\rangle$

$= 3 \left\langle{v}_{x}^{2}\right\rangle$

If you then take the square root of $\left\langle{v}^{2}\right\rangle$, the average of the squared velocity in a radial direction, you get the root-mean-square (RMS) speed for gases.

${v}_{R M S} = \sqrt{\left\langle{v}^{2}\right\rangle}$

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

${v}_{R M S} = \sqrt{\frac{3 R T}{M}}$

where $R = \text{8.314472 J/mol"cdot"K}$ and $T$ is temperature in $\text{K}$. $M$ is the molar mass in $\text{kg/mol}$.

The function of ${v}_{R M S}$ is to express the average velocity of a gas while also factoring out the dependence of $v$ on sign due to gases changing direction.