RMS Speed — Nitrogen at Room Temperature (Worked Example)

Problem

Air is mostly nitrogen (M = 0.028 kg/mol). What is the root-mean-square speed of N₂ molecules at 300 K? (R = 8.314 J/(mol·K))

Given

  • T = 300 K

  • M = 0.028 kg/mol

  • R = 8.314 J/(mol·K)

Find

The rms speed v_rms.

Principle

Kinetic theory: temperature is molecular motion, v_rms = √(3RT/M).

Formula

Solution

  1. Substitute: vrms=3×8.314×300/0.028v_{rms} = \sqrt{3 \times 8.314 \times 300/0.028}

  2. Inside: 7482.6/0.028=267,2367482.6/0.028 = 267{,}236

  3. Square root: vrms=517 m/sv_{rms} = 517\ \mathrm{m/s}

517 m/s

Unit check

√(J/(mol·K) × K ÷ kg/mol) = √(m²/s²) = m/s — correct.

What this means physically

517 m/s — faster than a passenger jet, yet air feels still because molecules collide billions of times per second and go nowhere on average. Temperature is this motion.

Common mistake

Using M = 28 g/mol without converting. The formula needs kg/mol — 0.028, or the speed comes out ~31× too small.