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Kinetic Theory of Gases

Key idea: Derive pV = (1/3)Nm⟨c^2⟩ from the particle model, connect it to pV = NkT, and solve kinetic theory questions (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply the kinetic model to gas pressure and mean translational kinetic energy.
  • Derive pV = ⅓Nm⟨c²⟩ from the definition of pressure and a one-dimensional model of molecular collisions extended to three dimensions.

1. Definitions (Must Know)

  • Pressure, p (Pa): force per unit area, p = F/A.
  • Thermodynamic temperature, T (K): temperature on an absolute scale.
  • Boltzmann constant, k (J K⁻¹).
  • Number of particles, N (dimensionless): number of gas molecules/atoms.
  • Amount of substance, n (mol).
  • Avogadro constant, N_A = 6.02 × 10²³ mol⁻¹.
  • Molar gas constant, R (J mol⁻¹ K⁻¹): R = N_A k.
  • Mean square speed, ⟨c²⟩ (m² s⁻²): average of c² over all particles.
  • Root-mean-square speed, cᵣₘₛ (m s⁻¹): cᵣₘₛ = square root of (⟨c²⟩).

2. Key Ideas (What Earns Marks)

  • Equation of state (particle form):
    • pV = NkT
  • Moles vs particles:
    • N = nN_A, R = N_A k ⇒ Nk = nR ⇒ pV = nRT
  • Kinetic theory key result:
    • pV = (1/3)Nm⟨c²⟩
  • Mean translational kinetic energy:
    • (1/2)m⟨c²⟩ = (3/2)kT
    • so cᵣₘₛ = square root of (3kT/m) = square root of (3RT/M) where m is the mass of one molecule and M is molar mass (kg mol⁻¹).

Graph intuition: cᵣₘₛ vs T

Exam trap: doubling thermodynamic temperature does not double molecular speed. Since cᵣₘₛ ∝ square root of T, it only increases by a factor of square root of 2.

r.m.s. speed vs temperature (nitrogen, illustrative)

A square-root curve showing c_rms increasing with temperature; doubling T increases c_rms by √2.

Scroll across the graph to read all labels.

A square-root curve showing c_rms increasing with temperature; doubling T increases c_rms by √2.A square-root curve showing c_rms increasing with temperature; doubling T increases c_rms by √2.
The curve comes from c_rms = √(3RT/M). It rises quickly at low T but flattens because of the square root.
Open full-size graph
View figure data
Values for r.m.s. speed vs temperature (nitrogen, illustrative)
Temperature, T (K)Nitrogen (M = 0.028 kg mol⁻¹)
100298
200421
300516
400596
600730

3. Detailed Explanations

A. Assumptions of the kinetic theory (syllabus-level)

  • A gas contains a very large number of identical particles in random motion.
  • The particles occupy negligible volume compared to the container volume.
  • There are no intermolecular forces except during collisions.
  • Collisions between particles and with the container walls are perfectly elastic.
  • The duration of collisions is negligible compared to the time between collisions.

B. Derivation of pV = (1/3)Nm⟨c²⟩

Consider a cubical container of side length L (so volume V = L³).

Take one molecule of mass m with velocity component cₓ towards a wall perpendicular to the x-axis.

  1. Change in momentum in one elastic collision with the wall: Δ p = 2mcₓ

  2. Time between successive collisions with the same wall: Δ t = 2L/cₓ

  3. Average force on the wall from this molecule:

F = (Δ p)/(Δ t) = 2mcₓ/(2L/cₓ) = mcₓ²/L

For N molecules,

Fₜₒₜₐₗ = m/L∑ᵢ₌₁^N c_(x,i)² = Nm/L⟨cₓ²⟩

Pressure is p = F/A and the wall area is A = L², so

p = Fₜₒₜₐₗ/L² = Nm/L³⟨cₓ²⟩ = Nm/V⟨cₓ²⟩

So pV = Nm⟨cₓ²⟩

For random motion in 3D, the motion is symmetric so ⟨cₓ²⟩ = ⟨c_y²⟩ = ⟨c_z²⟩ = 1/3⟨c²⟩

Hence,

pV = Nm(1/3⟨c²⟩) = (1/3)Nm⟨c²⟩

C. Mean translational kinetic energy and temperature

From the two equations of state: pV = NkT and pV = (1/3)Nm⟨c²⟩

Equate them and cancel N: kT = (1/3)m⟨c²⟩

Multiply both sides by 3/2: (3/2)kT = (1/2)m⟨c²⟩

So the mean translational kinetic energy per particle is proportional to T: Eₖ bar = (3/2)kT

4. Common Mistakes

  • Using T in °C (must convert to K).
  • Using molar mass (kg mol⁻¹) as m (kg). If you use molar mass M, use cᵣₘₛ = square root of (3RT/M).
  • Mixing up mean speed and r.m.s. speed (exams usually want cᵣₘₛ).
  • Using V in cm³ or p in kPa without converting to SI.

5. Exam Tips

  • Decide early if you’re using:
    • pV = NkT (particle count), or
    • pV = nRT (moles).
  • Show your conversion step clearly: N = nN_A.
  • Unit check: pV has unit Pa·m³ = J.

6. Worked Examples

Modelled example 1

Find number of molecules using pV = NkT

Core

Problem

A gas has p = 1.00 × 10⁵ Pa, V = 2.0 × 10⁻³ m³, and T = 300 K. Find N. Take k = 1.38 × 10⁻²³ J K⁻¹.
Study the worked solution

  1. Choose the particle equation

    Method

    Use pV = NkT.

    Reason

    The unknown N is a number of molecules, so it pairs with Boltzmann’s constant.

    Working

    pV = NkT

  2. Rearrange and substitute

    Method

    Make N the subject and insert the SI quantities.

    Reason

    pV and kT both have energy units, leaving a dimensionless particle count.

    Working

    N = pV/kT = ((1.00 × 10⁵)(2.0 × 10⁻³))/((1.38 × 10⁻²³)(300))

  3. Evaluate

    Method

    The gas contains approximately 4.83 × 10²² molecules.

    Reason

    The result is a count, so it has no physical unit.

    Working

    N ≈ 4.83 × 10²²

Guided practice 2

Find cᵣₘₛ using cᵣₘₛ = square root of (3RT/M)

About 4 min

Problem

Estimate the r.m.s. speed of nitrogen with M = 0.028 kg mol⁻¹ at T = 300 K. Take R = 8.31 J mol⁻¹K⁻¹.

Try this before viewing the solution

Unit: m s^-1

Hints

Hint 1: match the mass scale
Because the mass is per mole, use cᵣₘₛ = square root of (3RT/M).
Hint 2: preserve the square root
Calculate 3RT/M first, then take its square root.
View solution step by step

  1. Select the molar form

    Method

    Use cᵣₘₛ = square root of (3RT/M).

    Reason

    The supplied mass is molar mass in kg mol⁻¹, not the mass of one molecule.

    Working

    cᵣₘₛ = square root of (3RT/M)

  2. Substitute

    Method

    Insert T in kelvin and M in SI units.

    Reason

    The expression under the square root then has units of speed squared.

    Working

    cᵣₘₛ = square root of (3(8.31)(300)/0.028)

  3. Evaluate

    Method

    The r.m.s. speed is about 5.2 × 10² m s⁻¹.

    Reason

    The square root converts mean-square-speed scale to speed.

    Working

    cᵣₘₛ ≈ 5.2 × 10² m s⁻¹

Common misconception 3

Use pV = (1/3)Nm⟨c²⟩

Find and correct the mistake

Learner claim

A sample has N = 2.0 × 10²², m = 4.7 × 10⁻²⁶ kg, V = 1.0 × 10⁻³ m³ and cᵣₘₛ = 500 m s⁻¹. A learner substitutes 500 for ⟨c²⟩. Diagnose the error and estimate p.

Try this before viewing the solution

Unit: Pa

View solution step by step

  1. Correct the speed quantity

    Method

    Use ⟨c²⟩ = cᵣₘₛ² = 500².

    Reason

    R.m.s. speed is defined as the square root of mean square speed.

    Working

    ⟨c²⟩ = (500 m s⁻¹)²

  2. Rearrange for pressure

    Method

    Divide the kinetic-theory result by volume.

    Reason

    pV = (1/3)Nm⟨c²⟩.

    Working

    p = (Nm cᵣₘₛ²)/3V

  3. Calculate

    Method

    The pressure is approximately 7.8 × 10⁴ Pa.

    Reason

    Squaring the speed supplies the correct dimensions and magnitude.

    Working

    p = ((2.0 × 10²²)(4.7 × 10⁻²⁶)(500²))/(3(1.0 × 10⁻³)) ≈ 7.8 × 10⁴ Pa

Examiner practice 4

Mean kinetic energy → temperature

3 marks

Examination question

The mean translational kinetic energy of gas molecules is ⟨Eₖ⟩ = 6.21 × 10⁻²¹ J. Find the temperature using k = 1.38 × 10⁻²³ J K⁻¹. [3 marks]

Try this before viewing the solution

Unit: K

View solution step by step

  1. State the energy relation

    1 mark

    Method

    Use ⟨Eₖ⟩ = (3/2)kT.

    Reason

    Thermodynamic temperature is proportional to mean translational kinetic energy per particle.

    Working

    ⟨Eₖ⟩ = (3/2)kT

  2. Rearrange and substitute

    1 mark

    Method

    Make T the subject and insert the energy and k.

    Reason

    The energy is per molecule, matching Boltzmann’s constant.

    Working

    T = (2⟨Eₖ⟩)/3k = (2(6.21 × 10⁻²¹))/(3(1.38 × 10⁻²³))

  3. Evaluate

    1 mark

    Method

    The temperature is 3.00 × 10² K.

    Reason

    The absolute-temperature unit follows from joules divided by joules per kelvin.

    Working

    T = 300 K

Challenge 5

Density form: p = (1/3)ρ cᵣₘₛ²

Minimal support

Independent transfer

A gas has density ρ = 1.2 kg m⁻³ and r.m.s. speed cᵣₘₛ = 500 m s⁻¹. Estimate the pressure using p = (1/3)ρ cᵣₘₛ².

Try this before viewing the solution

Unit: Pa

Hints

Hint 1: recognise the compressed representation
The density form has already combined total molecular mass and volume into ρ.
View solution step by step

  1. Interpret the density form

    Method

    Use the given bulk density directly with the molecular r.m.s. speed.

    Reason

    ρ = Nm/V compresses the particle-count, molecular-mass and volume factors in the original relation.

    Working

    p = (1/3)ρ cᵣₘₛ²

  2. Substitute and calculate

    Method

    The pressure is 1.00 × 10⁵ Pa.

    Reason

    Density times speed squared has pressure units.

    Working

    p = (1/3)(1.2)(500²) = 1.00 × 10⁵ Pa

7. Mind Stretchers

Mind stretcher 1: Show pV = nRT starting from pV = NkTExtension

Show Answer

Use N = nN_A: pV = NkT = (nN_A)kT = n(N_Ak)T

Since R = N_Ak, this gives pV = nRT.

Mind stretcher 2: Link pressure to mean kinetic energyExtension

Show that: pV = (2/3)N⟨Eₖ⟩ where ⟨Eₖ⟩ is the mean translational kinetic energy per particle.

Show Answer

From kinetic theory: pV = (1/3)Nm⟨c²⟩ But ⟨Eₖ⟩ = (1/2)m⟨c²⟩, so m⟨c²⟩ = 2⟨Eₖ⟩.

Substitute: pV = (1/3)N(2⟨Eₖ⟩) = (2/3)N⟨Eₖ⟩

Mind stretcher 3: Optional (Enrichment)Extension

A. Pressure units you may see outside SI

  • 1 atm = 1.013 × 10⁵ Pa (often rounded to 1.0 × 10⁵ Pa)
  • 1 bar = 1.0 × 10⁵ Pa, and 1 mbar = 100 Pa (same as 1 hPa)
  • 760 mmHg ≈ 1 atm, so 1 mmHg ≈ 133 Pa
  • 1 inHg ≈ 3.39 × 10³ Pa

For barometer details (vacuum vs trapped gas), see Barometer.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027