Molecular Kinetic Theory — AQA A-level Physics

SPEC 3.6.2.3 • THERMAL PHYSICS

Molecular Kinetic Theory Model

Brownian motion, kinetic theory derivation, and molecular energies

Learning Objectives

You Will Learn To

Explain Brownian motion as evidence for atoms

State the assumptions of the kinetic theory model

Derive and apply

p V = \frac{1}{3} N m \overline{c^{2}}

Calculate root mean square (rms) speed

And Also

Relate molecular KE to temperature:

\frac{1}{2} m \overline{c^{2}} = \frac{3}{2} k T

Explain why ideal gas internal energy is purely kinetic

Use kinetic theory to explain the gas laws

Key Idea

The kinetic theory model connects the microscopic behaviour of molecules (mass, speed) to the macroscopic properties of gases (pressure, temperature, volume).

Brownian Motion

Definition

Brownian motion is the random motion of larger particles in a fluid caused by collisions with surrounding particles.

Observing a weak solution of milk, and later pollen grains in suspension, with a high-powered microscope the Scottish physicist Robert Brown saw that the particles showed a violent and random motion.

Evidence for Atoms

Brownian motion contributed to the evidence for the existence of atoms and molecules.

Assumptions of the Kinetic Theory

The kinetic theory model makes the following assumptions about molecules in an ideal gas:

1

No intermolecular forces act on the molecules (except during collisions)

2

The duration of collisions is negligible compared to time between collisions

3

The motion of molecules is random , and they experience perfectly elastic collisions

4

The motion of the molecules follows Newton's laws

5

The molecules move in straight lines between collisions

6

The volume of molecules is infinitesimal compared with the volume of the gas

7

The average kinetic energy of molecules is proportional to the absolute temperature

Empirical vs Theoretical

The gas laws are empirical (from experiments). The kinetic theory model is theoretical and explains why the gas laws work.

Deriving the Kinetic Theory Equation (1)

Consider a cube with side lengths

l

, full of gas molecules. One molecule has mass

m

and travels towards the right wall with velocity

u

.

Step 1: Change in Momentum

Elastic collision with wall:

Δ p = m u - (- m u) = 2 m u

Step 2: Time Between Collisions

Distance travelled = 2l, so:

t = \frac{2 l}{u}

Deriving the Kinetic Theory Equation (2)

Step 3: Force and Pressure

F = \frac{2 m u}{2 l / u} = \frac{m u ^{2}}{l} P = \frac{m u ^{2} / l}{l ^{2}} = \frac{m u ^{2}}{V}

Step 4: Mean Square Speed

For N molecules with different speeds, use mean square speed

\overline{u^{2}}

:

P = \frac{N m u ^{2}}{V}

Step 5: Three Dimensions

\overline{c^{2}} = \overline{u^{2}} + \overline{v^{2}} + \overline{w^{2}}

and for random motion

\overline{u^{2}} = \overline{v^{2}} = \overline{w^{2}}

\overline{c^{2}} = 3 \overline{u^{2}} \Rightarrow \overline{u^{2}} = \frac{1}{3} \overline{c^{2}}

The Kinetic Theory Equation

p V = \frac{1}{3} N m (c_{r m s})^{2}

RMS Speed Equations

The kinetic theory equation can be rearranged to find the rms speed:

Using Temperature

c_{r m s} = \frac{3 RT}{M} = \frac{3 k T}{m}

M = molar mass (kg mol⁻¹), m = molecular mass (kg)

Using Density

c_{r m s} = \frac{3 p}{ρ}

ρ = density (kg m⁻³), p = pressure (Pa)

Deriving the Density Form

From

p V = \frac{1}{3} N m (c_{r m s})^{2}

, note that total mass = Nm, so:

p = \frac{1}{3} \times \frac{N m}{V} \times (c_{r m s})^{2} = \frac{1}{3} ρ (c_{r m s})^{2} \Rightarrow (c_{r m s})^{2} = \frac{3 p}{ρ}

1

Worked Example: RMS from Density

Question: The density of nitrogen at STP is 1.251 kg m⁻³. Calculate the rms speed of nitrogen molecules. [Standard pressure = 1.013 × 10⁵ Pa]

1

Use:

c^{2} = \frac{3 p}{ρ} = \frac{3 \times 1.013 \times 1 0 ^{5}}{1.251}

2

c^{2} = 2.430 \times 1 0^{5} m^{2} s^{- 2}

Answer

c_{r m s} = 2.430 \times 1 0^{5} = 493 m s^{- 1}

2

Worked Example: RMS from Temperature

Question: Calculate the rms speed of oxygen molecules (O₂) at 25°C. [M = 32 g mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹]

1

Convert: T = 25 + 273 = 298 K, M = 0.032 kg mol⁻¹

2

c_{r m s} = \frac{3 RT}{M} = \frac{3 \times 8.31 \times 298}{0.032}

3

c_{r m s} = 232000

Answer

c_{r m s} = 482 m s^{- 1}

3

Worked Example: Harder

Question: At what temperature would helium atoms (M = 4.0 g mol⁻¹) have an rms speed equal to 11.2 km s⁻¹ (Earth's escape velocity)?

1

Convert: c = 11200 m s⁻¹, M = 0.004 kg mol⁻¹

2

Rearrange

c_{r m s} = \frac{3 RT}{M}

to get

T = \frac{M c _{r m s}^{2}}{3 R}

3

T = \frac{0.004 \times ( 11200 ) ^{2}}{3 \times 8.31} = \frac{0.004 \times 1.254 \times 1 0 ^{8}}{24.93}

This explains why helium escapes from Earth's atmosphere over geological time!

Answer

T = 20100 K \approx 2.0 \times 1 0^{4} K

Kinetic Energy of a Molecule

The average kinetic energy of randomly moving molecules is directly proportional to absolute temperature.

Mean Kinetic Energy

\frac{1}{2} m (c_{r m s})^{2} = \frac{3}{2} k T = \frac{3 RT}{2 N _{A}}

k = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹

Derivation

Equate

p V = N k T

with

p V = \frac{1}{3} N m (c_{r m s})^{2}

Cancel N:

k T = \frac{1}{3} m (c_{r m s})^{2}

Multiply by 3/2:

\frac{3}{2} k T = \frac{1}{2} m (c_{r m s})^{2}

Key Result

KE depends only on T

Same for all gases at same T

Independent of molecular mass

4

Worked Example: Kinetic Energy

Calculate:

(a) the KE of a gas molecule of mass 3.5 × 10⁻²⁶ kg moving at 600 m s⁻¹

(b) the average KE of gas molecules at 20°C [k = 1.38 × 10⁻²³ J K⁻¹]

(a)

K E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 3.5 \times 1 0^{- 26} \times 60 0^{2}

= 6.3 × 10⁻²¹ J

(b) T = 293 K

K E = \frac{3}{2} k T = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 293

= 6.07 × 10⁻²¹ J

5

Worked Example: Harder

Question: A mixture contains hydrogen (M = 2 g mol⁻¹) and oxygen (M = 32 g mol⁻¹) at 300 K. Calculate the ratio of their rms speeds.

1

At the same T, all molecules have the same average KE

2

Since

c_{r m s} = \frac{3 RT}{M}

, we have

c_{r m s} \propto \frac{1}{M}

3

\frac{c _{H_{2}}}{c _{O_{2}}} = \frac{M _{O_{2}}}{M _{H_{2}}} = \frac{32}{2} = 16

Lighter molecules move faster to have the same KE!

AnswerRatio = 4:1 (hydrogen molecules move 4× faster than oxygen)

Internal Energy of an Ideal Gas

An ideal gas has no intermolecular forces, so no potential energy. Internal energy = sum of kinetic energies.

Internal Energy

U = N \times \frac{3}{2} k T = \frac{3}{2} N k T = \frac{3}{2} n RT

Key Implications

Internal energy depends ONLY on temperature

Doubling T doubles the internal energy

At constant T, U is constant (even if p or V change)

For isothermal processes: ΔU = 0

6

Worked Example: Total Internal Energy

Question: A bottle contains 128 g of oxygen at 330 K. Find the total kinetic energy of all the molecules. [Molar mass O₂ = 32 g, N_A = 6.02 × 10²³, k = 1.38 × 10⁻²³ J K⁻¹]

1

Moles: n = 128/32 = 4 mol

2

Number of molecules: N = 4 × 6.02 × 10²³ = 2.408 × 10²⁴

3

KE per molecule:

\frac{3}{2} k T = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 330 = 6.83 \times 1 0^{- 21}

J

4

Total:

6.83 \times 1 0^{- 21} \times 2.408 \times 1 0^{24}

Or use: $U = \frac{3}{2} n RT = \frac{3}{2} \times 4 \times 8.31 \times 330 = 16450$ J

AnswerU = 16450 J ≈ 16.5 kJ

7

Worked Example: Harder

Question: 0.24 mol of an ideal gas at 27°C is heated at constant volume until its pressure doubles. Calculate: (a) the final temperature (b) the increase in internal energy [R = 8.31 J mol⁻¹ K⁻¹]

(a) At constant V, p ∝ T (Pressure Law)

1

If p doubles, T doubles: T₂ = 2 × (27 + 273) = 2 × 300 = 600 K

(b) Change in internal energy:

2

Δ U = \frac{3}{2} n R Δ T = \frac{3}{2} \times 0.24 \times 8.31 \times (600 - 300)

3

Δ U = \frac{3}{2} \times 0.24 \times 8.31 \times 300

AnswerΔU = 897 J

Kinetic Theory and the Gas Laws

Boyle's Law

p \propto \frac{1}{V}

at constant T

Increase V → molecules further apart → less frequent collisions → pressure decreases

Charles's Law

V \propto T

at constant p

Increase T → molecules faster → to keep p constant, they must spread out → V increases

Pressure Law

p \propto T

at constant V

Increase T → molecules faster → more frequent & harder collisions → pressure increases

Mathematical Derivation of Charles's Law

At constant p:

p V_{1} = \frac{1}{3} M (c_{1 r m s})^{2}

and

p V_{2} = \frac{1}{3} M (c_{2 r m s})^{2}

\frac{V _{1}}{V _{2}} = \frac{( c _{1 r m s} ) ^{2}}{( c _{2 r m s} ) ^{2}} = \frac{K E _{1}}{K E _{2}} = \frac{T _{1}}{T _{2}}

8

Worked Example: Challenging

Question: A container of volume 0.050 m³ holds nitrogen gas at a pressure of 2.0 × 10⁵ Pa and temperature 300 K. Calculate: (a) the rms speed (b) the total number of molecules (c) the total mass of gas [M = 28 g mol⁻¹, R = 8.31, N_A = 6.02 × 10²³]

(a)

c_{r m s} = \frac{3 RT}{M} = \frac{3 \times 8.31 \times 300}{0.028} = 267107

= 517 m s⁻¹

(b) From pV = NkT:

N = \frac{p V}{k T} = \frac{2.0 \times 1 0 ^{5} \times 0.050}{1.38 \times 1 0 ^{- 23} \times 300} = \frac{10000}{4.14 \times 1 0 ^{- 21}}

= 2.42 × 10²⁴ molecules

(c) n = N/N_A = 2.42 × 10²⁴ / 6.02 × 10²³ = 4.02 mol

Mass = n × M = 4.02 × 28 = 113 g

Practice Problem

Question: The rms speed of hydrogen molecules at 300 K is 1920 m s⁻¹.

(a) What is the rms speed at 1200 K?

(b) At what temperature would oxygen (M = 32 g mol⁻¹) have the same rms speed as hydrogen (M = 2 g mol⁻¹) at 300 K?

(a)

c_{r m s} \propto T

, so

\frac{c _{2}}{c _{1}} = \frac{1200}{300} = 2

→ c₂ = 3840 m s⁻¹

(b) Same c means same T/M:

\frac{300}{2} = \frac{T _{O_{2}}}{32}

→ T = 4800 K

Practice Problem

Question: An ideal gas has a density of 1.80 kg m⁻³ at a pressure of 1.50 × 10⁵ Pa.

(a) Calculate the rms speed of the molecules.

(b) If the gas is nitrogen (M = 28 g mol⁻¹), calculate the temperature.

(a)

c_{r m s} = \frac{3 p}{ρ} = \frac{3 \times 1.50 \times 1 0 ^{5}}{1.80} = 2.5 \times 1 0^{5}

= 500 m s⁻¹

(b) From

c_{r m s} = \frac{3 RT}{M}

:

T = \frac{M c _{r m s}^{2}}{3 R} = \frac{0.028 \times ( 500 ) ^{2}}{3 \times 8.31}

= 280 K

Key Equations Summary

EquationDescription

p V = \frac{1}{3} N m (c_{r m s})^{2}

Kinetic theory equation

\frac{1}{2} m (c_{r m s})^{2} = \frac{3}{2} k T = \frac{3 RT}{2 N _{A}}

Mean KE per molecule

c_{r m s} = \frac{3 RT}{M} = \frac{3 k T}{m}

RMS speed (temperature)

c_{r m s} = \frac{3 p}{ρ}

RMS speed (density)

U = \frac{3}{2} N k T = \frac{3}{2} n RT

Internal energy

Common Mistakes

Convert molar mass to kg mol⁻¹ (divide g by 1000)

Always use Kelvin for temperature

rms speed ≠ mean speed (they're different!)

Summary

Key Concepts

ConceptDescription

Brownian MotionEvidence for atoms; random motion from molecular collisions

AssumptionsNo forces, negligible volume, random motion, elastic collisions, straight lines

Kinetic Equation

p V = \frac{1}{3} N m (c_{r m s})^{2}

KE & Temperature

\frac{1}{2} m (c_{r m s})^{2} = \frac{3}{2} k T

— same for all gases at same T

Internal Energy

U = \frac{3}{2} n RT

— depends only on T

Historical Note

Gas laws arose from experiments (empirical). Kinetic theory arose from theoretical reasoning and explains why they work. Scientific understanding evolves with evidence.