RMS Speed of Gas Molecules

The pressure derivation of Section 9.3 involves $\overline{v^2}$, the MEAN of the squared molecular speeds, rather than the simple average speed $\bar{v}$ -- and this is not an arbitrary choice: since a molecule's kinetic energy depends on $v^2$ (not $v$), it is the mean square speed that connects most directly and most simply to the gas's energy content. The square root of this mean square speed is called the root-mean-square (RMS) speed,

To find $v_{rms}$ in terms of measurable quantities, rearrange the pressure relation of Section 9.3, $PV = \tfrac{1}{3}Nm\overline{v^2}$, and combine it with the ideal gas equation $PV = nRT$ (where $n$ is the number of moles). Equating the two expressions for $PV$,

For $n=1$ mole, $N = N_A$ (Avogadro's number), and $N_A m = M$, the molar mass of the gas. Substituting and solving for $\overline{v^2}$,

Equivalently, in terms of Boltzmann's constant $k_B = R/N_A$ and the mass $m$ of a single molecule ($M = N_Am$), the same result can be written per-molecule as $v_{rms} = \sqrt{3k_BT/m}$. Both forms show the same two physical facts: $v_{rms}$ INCREASES with the square root of the absolute temperature $T$ (hotter gas $\Rightarrow$ faster molecules), and $v_{rms}$ DECREASES with the square root of the molar mass $M$ (heavier molecules move more sluggishly at the same temperature) -- which is why, at any given temperature, light gases such as hydrogen and helium have far higher RMS speeds than heavy gases such as carbon dioxide or oxygen. …