Calculate RMS speed of O₂ at 300K.

RMS Speed of Gas Molecules — v_rms = √(3RT/M)

Question

Calculate the RMS speed of oxygen (O₂) molecules at 300 K.

Given: R = 8.314 J mol⁻¹ K⁻¹, molar mass of O₂ = 32 g/mol = 0.032 kg/mol.

Solution — Step by Step

The RMS speed formula comes directly from kinetic theory:

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

Here, R is the universal gas constant, T is temperature in Kelvin, and M is molar mass in kg/mol. All three must be in SI units — this is where most mistakes happen.

Substitute R = 8.314 J mol⁻¹ K⁻¹, T = 300 K, M = 0.032 kg/mol:

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

Calculate the numerator first:

3 \times 8.314 \times 300 = 7482.6 \text{ J/mol}

We’re essentially calculating the total kinetic energy per mole, then dividing by molar mass to get a velocity-squared quantity.

v_{rms} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233831.25} \approx 483.6 \text{ m/s}

So $v_{rms}$ of O₂ at 300 K is approximately 484 m/s.

Why This Works

The kinetic theory of gases models molecules as tiny billiard balls in constant random motion. Each molecule has a different speed at any instant, so we need a statistical average. The RMS speed is the square root of the mean of all the squared speeds — it’s more useful than the simple average because kinetic energy depends on $v^2$ , not $v$ .

The factor of 3 in $\sqrt{3RT/M}$ comes from the three degrees of translational freedom (x, y, z directions). Each direction contributes $\frac{1}{2}k_BT$ of average kinetic energy, giving a total of $\frac{3}{2}k_BT$ per molecule.

This is a high-scoring topic in CBSE Class 11 and appears regularly in JEE Main. The formula itself is a one-liner — the only skill being tested is unit handling.

Alternative Method

We can also use the per-molecule version of the formula:

v_{rms} = \sqrt{\frac{3k_BT}{m}}

where $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann’s constant and $m$ is the mass of one molecule (not one mole).

Mass of one O₂ molecule:

m = \frac{0.032}{6.022 \times 10^{23}} = 5.314 \times 10^{-26} \text{ kg}

v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{5.314 \times 10^{-26}}} = \sqrt{233458...} \approx 483 \text{ m/s}

Same answer. Use whichever formula you’re given in the question — both are correct. In board exams, the $\sqrt{3RT/M}$ form is preferred.

For quick comparison problems, remember that $v_{rms} \propto \sqrt{T/M}$ . So if temperature doubles, $v_{rms}$ increases by $\sqrt{2}$ . If you switch from O₂ (M = 32) to He (M = 4), the speed increases by $\sqrt{32/4} = \sqrt{8} \approx 2.83$ times. This ratio trick saves a lot of time in MCQs.

Common Mistake

Using M in g/mol instead of kg/mol. Students write M = 32 and get a speed of 48,360 m/s — about 160 times the speed of sound. The formula needs M in kg/mol because R is in J mol⁻¹ K⁻¹ and 1 J = 1 kg·m²·s⁻². Always convert: 32 g/mol → 0.032 kg/mol before substituting. This single unit error is the most common reason for losing marks on this question in board exams.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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