Kinetic Theory of Gases

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, NA = 6.02 × 10²³ mol⁻¹.
• Molar gas constant, R (J mol⁻¹ K⁻¹): R = NA k.
• Mean square speed, langle c² rangle (m² s⁻²): average of c² over all particles.
• Root-mean-square speed, cᵣₘₛ (m s⁻¹): cᵣₘₛ = sqrtlangle c² rangle.

2. Key Ideas (What Earns Marks)

• Equation of state (particle form):
  • pV = NkT
• Moles vs particles:
  • N = nNA, R = NA k Rightarrow Nk = nR Rightarrow pV = nRT
• Kinetic theory key result:
  • pV = 1/3Nmlangle c² rangle
• Mean translational kinetic energy:
  • 1/2mlangle c² rangle = 3/2kT
  • so cᵣₘₛ = sqrt3kT/m = sqrt3RT/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 propto sqrtT, it only increases by a factor of sqrt2.

Data table

Nitrogen (M = 0.028 kg mol⁻¹)
Temperature, T (K)	r.m.s. speed, c_rms (m s⁻¹)
100	298
200	421
300	516
400	596
600	730

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 = 1/3Nmlangle c² rangle

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:

For N molecules,

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

So pV = Nmlangle cₓ² rangle

For random motion in 3D, the motion is symmetric so langle cₓ² rangle = langle cy² rangle = langle cz² rangle = 1/3langle c² rangle

Hence,

C. Mean translational kinetic energy and temperature

From the two equations of state: pV = NkT quadand pV = 1/3Nmlangle c² rangle

Equate them and cancel N: kT = 1/3mlangle c² rangle

Multiply both sides by 3/2: 3/2kT = 1/2mlangle c² rangle

So the mean translational kinetic energy per particle is proportional to T: overlineEₖ = 3/2kT

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 = sqrt3RT/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 = nNA.
• Unit check: pV has unit Pa·m³ = J.

6. Worked Examples

Example 1: Find number of molecules using pV = NkTCore

A gas has p = 1.00 × 10⁵ Pa, V = 2.0 × 10⁻³ m³, and T = 300 K. Find N.

Show Answer

N = pV/kT; = (1.00 × 10⁵)(2.0 × 10⁻³)/(1.38 × 10⁻²³)(300); ≈ 4.83 × 10²²

Example 2: Find cᵣₘₛ using cᵣₘₛ = sqrt3RT/MCore

Estimate the r.m.s. speed of nitrogen (M = 0.028 kg mol⁻¹) at T = 300 K.

Show Answer

cᵣₘₛ = sqrt3RT/M; = sqrt3(8.31)(300)/0.028; ≈ 5.2 × 10² m s⁻¹

Example 3: Use pV = 1/3Nmlangle c²rangleCore

A sample contains N = 2.0 × 10²² molecules of mass m = 4.7 × 10⁻²⁶ kg in volume V = 1.0 × 10⁻³ m³. If cᵣₘₛ = 500 m s⁻¹, estimate p.

Show Answer

Since langle c²rangle = cᵣₘₛ², p = 1/3Nmlangle c²rangle/V; = 1/3(2.0 × 10²²)(4.7 × 10⁻²⁶)(500²)/1.0 × 10⁻³; ≈ 7.8 × 10⁴ Pa

Example 4: Mean kinetic energy → temperatureCore

The mean translational kinetic energy of gas molecules is langle Eₖrangle = 6.21 × 10⁻²¹ J.

Find the temperature. Take k = 1.38 × 10⁻²³ J K⁻¹.

Show Answer

Use langle Eₖrangle = 3/2kT: T = 2langle Eₖrangle/3k = 2(6.21 × 10⁻²¹)/3(1.38 × 10⁻²³) T = 12.42 × 10⁻²¹/4.14 × 10⁻²³ = 3.00 × 10² K

Example 5: Density form: p = tfrac13ρ cᵣₘₛ²Core

A gas has density ρ = 1.2 kg m⁻³ and r.m.s. speed cᵣₘₛ = 500 m s⁻¹.

Estimate the pressure.

Show Answer

Using p = tfrac13ρ cᵣₘₛ²: p = 1/3(1.2)(500²) = 1/3(1.2)(2.50 × 10⁵) = 1.00 × 10⁵ Pa

7. Mind Stretchers

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

Show Answer

Use N = nNA: pV = NkT = (nNA)kT = n(NAk)T

Since R = NAk, this gives pV = nRT.

Mind stretcher 2: Link pressure to mean kinetic energyExtension

Show that: pV = 2/3Nlangle Eₖrangle where langle Eₖrangle is the mean translational kinetic energy per particle.

Show Answer

From kinetic theory: pV = 1/3Nmlangle c²rangle But langle Eₖrangle = 1/2mlangle c²rangle, so mlangle c²rangle = 2langle Eₖrangle.

Substitute: pV = 1/3N(2langle Eₖrangle) = 2/3Nlangle Eₖrangle

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.

8. Practice, Quiz and Next Step

Close your notes and use Kinetic Theory of Gases in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: An unfamiliar data set or physical system requires you to apply Kinetic Theory of Gases while stating the model, regime and assumptions.

1. Retrieve: define kinetic theory of gases in your own words, including units, sign or conditions where relevant.
2. Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
3. Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.

Check the response before looking back

• The model, regime, coordinates and assumptions are explicit.
• The derivation or multi-step reasoning is visible rather than implied.
• The conclusion is tested against units, data quality and a limiting case.
• A practical control, uncertainty or model limitation is evaluated where applicable.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.