RMS Speed — Oxygen (Worked Example)

Given

• T = 300 K

• M = 0.032 kg/mol

• R = 8.314 J/(mol·K)

Find

v_rms.

Principle

v_rms = √(3RT/M): heavier than nitrogen, so slightly slower.

Formula

Solution

1. Substitute: $v_{rms} = \sqrt{3 \times 8.314 \times 300/0.032}$

2. Inside: $233{,}831$

3. Square root: $v_{rms} = 484\ \mathrm{m/s}$

Unit check

√(J/(mol·K) × K ÷ kg/mol) = m/s. Dimensionally consistent — a speed.

What this means physically

484 m/s — a touch slower than nitrogen's 517 m/s, as expected for the heavier molecule. Both race around far faster than any wind you have ever felt.