Calculate RMS speed of oxygen molecules at 300K

Question

Calculate the root mean square (RMS) speed of oxygen molecules at 300 K. (R = 8.314 J/mol·K, molar mass of O₂ = 32 g/mol)

Solution — Step by Step

The RMS speed of gas molecules is derived from the kinetic theory of gases:

v_{rms} = \sqrt{\frac{3RT}{M}}

where:

$R$ = universal gas constant = 8.314 J/mol·K
$T$ = absolute temperature in Kelvin
$M$ = molar mass in kg/mol (critical: must be in kg/mol, not g/mol)

$M$ of O₂ = 32 g/mol = $32 \times 10^{-3}$ kg/mol = $0.032$ kg/mol

This unit conversion is where most students lose marks.

v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}}

= \sqrt{\frac{7482.6}{0.032}}

= \sqrt{233831.25}

\approx 483.6 \text{ m/s}

The RMS speed of O₂ molecules at 300 K is approximately 484 m/s.

Why This Works

The kinetic theory of gases equates the average kinetic energy of molecules with temperature:

\frac{1}{2}mv^2 = \frac{3}{2}k_BT

Multiplying by Avogadro’s number ( $N_A$ ) and using $k_B N_A = R$ , $mN_A = M$ :

\frac{1}{2}Mv_{rms}^2 = \frac{3}{2}RT \Rightarrow v_{rms} = \sqrt{\frac{3RT}{M}}

Higher temperature → faster molecules (more kinetic energy). Lighter molecules ( $M$ smaller) → faster at the same temperature (explains why H₂ effuses faster than O₂).

Alternative Method — Using $v_{rms} \propto \sqrt{T/M}$ for Comparison

If you already know $v_{rms}$ for one gas at one temperature, you can find another using ratios:

\frac{v_{rms}(\text{O}_2)}{v_{rms}(\text{H}_2)} = \sqrt{\frac{M_{\text{H}_2}}{M_{\text{O}_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}

So H₂ molecules at the same temperature move 4 times faster than O₂ molecules.

Common Mistake

The biggest error: using $M = 32$ g/mol instead of $32 \times 10^{-3}$ kg/mol. Since $R$ is in J/mol·K = kg·m²/(s²·mol·K), the molar mass must be in kg/mol. If you forget the conversion, your speed comes out as $483.6 \times 1000 = 483{,}600$ m/s — far faster than light, which is the giveaway that something is wrong.

The three speed formulas from kinetic theory: $v_{mp} = \sqrt{2RT/M}$ (most probable), $v_{avg} = \sqrt{8RT/\pi M}$ (mean), $v_{rms} = \sqrt{3RT/M}$ (RMS). Their ratio is $v_{mp} : v_{avg} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1 : 1.128 : 1.225$ . RMS is always the largest.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using vrms∝T/Mv_{rms} \propto \sqrt{T/M}vrms​∝T/M​ for Comparison

Common Mistake

Try These Next

Alternative Method — Using $v_{rms} \propto \sqrt{T/M}$ for Comparison