PhysicsKinetic Theory of GasesJEE · NEET · NSEP · INPhO

Kinetic Theory of Gases · Part 5 of 9

Distribution of Molecular Speeds

At a fixed temperature, gas molecules don't all move at the same speed: some crawl, some race. The Maxwell–Boltzmann distribution describes the spread, and three different averages summarise it.

Builds on: Part 2 · Gas Laws.

Video coming soon

Maxwell–Boltzmann

How speeds are spread

dN = f(v) dv is the number of molecules with speeds between v and v + dv; the area under the curve is N.

$f (v) = 4 π N v^{2} (\frac{M}{2 π R T})^{3/2} e^{- M v^{2} /2 R T}$

Heat the gas and the peak moves to higher speed and flattens, keeping the same area.

Two Maxwell–Boltzmann curves for nitrogen at 300 K and 1000 K. — N₂ at 300 K and 1000 K.

Speed distributions of oxygen, neon and helium at 300 K. — O₂, Ne and He at 300 K.

Same temperature

Heavier is slower

At the same temperature, heavier molecules have a lower, narrower spread of speeds: O₂ peaks below neon, and neon below helium.

Three speeds

v_mp < v_mean < v_rms

The average velocity is zero, since directions cancel. The three speeds are:

most probable $v_{mp} = \frac{2 R T}{M}$ , the peak
mean $v_{mean} = \frac{8 R T}{π M}$
rms $v_{rms} = \frac{3 R T}{M}$

Ratio $2 : 8/ π : 3$ ≈ 1 : 1.13 : 1.22. For N₂ at 300 K: 422, 476 and 517 m/s.

A speed distribution with v_mp, v_mean and v_rms marked. — N₂ at 300 K with the three speeds marked.

Summary

Key formulas

f (v) = 4 π N v^{2} (\frac{M}{2 π R T})^{3/2} e^{- M v^{2} /2 R T}

Maxwell–Boltzmann distribution

v_{rms} = \frac{3 R T}{M}

Average velocity, rms, mean and most probable speed

v_{mean} = \frac{8 R T}{π M}

Average velocity, rms, mean and most probable speed

v_{mp} = \frac{2 R T}{M}

Average velocity, rms, mean and most probable speed

Worked examples

From the video

1. Mean speed of air molecules at 25 °C (M = 29 g/mol).

$v_{mean} = \frac{8 \times 8.314 \times 298}{π \times 0.029}$ ≈ 466.5 m/s

2. 1300 mol of N₂ fill 8.0 m³ at 2.1 atm. Find v_rms.

T = PV/nR ≈ 157.5 K; $v_{rms} = \frac{3 \times 8.314 \times 157.5}{0.028}$ ≈ 374.6 m/s

JEE-style question

Your turn

O₂ has rms speed v at temperature T. At what temperature does H₂ have the same rms speed?

\frac{T}{16}

16 T

\frac{T}{4}

T

Show the answer and the traps

v_rms ∝ √(T/M): keep T/M equal. M drops from 32 to 2, so T drops to T/16. Option 1.

Option 2 inverts the ratio. Option 3 takes the square root of the mass ratio. Option 4 assumes rms speed depends on temperature alone.

Watch out

Common mistakes

Using M in g/mol.M = 0.029 kg/mol for air, not 29.

Confusing average velocity with mean speed.Average velocity is zero; mean speed is not.

Practice

Try these

1. Find v_rms of O₂ at 27 °C.

$\frac{3 \times 8.314 \times 300}{0.032}$ ≈ 484 m/s.

2. At what temperature does H₂ have the same v_rms as O₂ at 47 °C?

Same T/M: $T_{H_{2}} = 320 \times \frac{2}{32} = 20 K$ .

3. Arrange v_mp, v_mean and v_rms for any gas and give their ratio.

v_mp < v_mean < v_rms, in the ratio $2 : 8/ π : 3$ ≈ 1 : 1.13 : 1.22.

4. The temperature of N₂ is raised from 27 °C until its v_rms doubles. Final temperature?

$v_{rms} \propto T$ , so T × 4 = 1200 K (927 °C).

Distribution of Molecular Speeds

How speeds are spread

Heavier is slower

vmp < vmean < vrms

Key formulas

From the video

Your turn

Common mistakes

Try these

v_mp < v_mean < v_rms