# What is the rms speed of He atoms at 295 K?

Jun 19, 2016

The root-mean-square (RMS) speed of a gas is:

$\setminus m a t h b f \left({\upsilon}_{\text{RMS}} = \sqrt{\frac{3 {k}_{B} T}{m}} = \sqrt{\frac{3 R T}{{M}_{m}}}\right)$

where $\frac{{M}_{m}}{m} = \frac{R}{{k}_{B}}$.

If you feel the need, I do derive where it comes from, here.

• ${k}_{B} = 1.3806 \times {10}^{- 23} \text{J/K}$ is the Boltzmann constant.
• $R$ is the universal gas constant, which in this scenario is $\text{8.314472 J/mol"cdot"K}$.
• $T$ is the temperature in $\text{K}$.
• ${M}_{m}$ is the molar mass of the gas in $\setminus m a t h b f \left(\text{kg/mol}\right)$ (NOT $\text{g/mol}$ like it normally would be!!).
• $m$ would be the mass of one molecule (or atom) of the gas, in $\text{kg}$.

Use whichever version floats your boat, but I'm going to use the one on the right because I'm more used to working with relative atomic masses rather than the mass of one atom.

Remember that $\text{1 J}$ $=$ ${\text{1 kg"cdot"m"^2"/s}}^{2}$, which explains why ${M}_{m}$ is in $\text{kg/mol}$.

$\textcolor{b l u e}{{\upsilon}_{\text{RMS}}} = \sqrt{\frac{3 R T}{{M}_{m}}}$

= sqrt((3("8.314472 J/mol"cdot"K")("295 K"))/(4.0026xx10^(-3) "kg/mol"))

$= \sqrt{\left(3 \left(\text{8.314472" cancel("kg")cdot"m"^2"/s"^2cdotcancel("mol")cdotcancel("K"))("295" cancel("K")))/(4.0026xx10^(-3) cancel("kg")"/"cancel("mol}\right)\right)}$

$=$ $\text{1355.87 m/s}$

Or to three sig figs,

$=$ $\textcolor{b l u e}{\text{1360 m/s}}$