← All notesPhysics · Ideal gases⬇ Download PDF
A Level Physics 9702 · A Level · Topic 15 · ~4 min read

Ideal gases

The mole and the Avogadro constant, the ideal gas equation pV = nRT = NkT, the kinetic theory assumptions, deriving pV = ⅓Nm⟨c²⟩, and linking the mean kinetic energy of a molecule to temperature.

Amount of substanceAvogadro constantIdeal gas equationBoltzmann constantKinetic theory assumptionspV = ⅓Nm⟨c²⟩Mean kinetic energyr.m.s. speed
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • The mole: 1 mol contains N_A = 6.02 × 10²³ particles.
  • An ideal gas obeys pV = nRT (n moles) = NkT (N molecules), with T in kelvin. R = 8.31 J K⁻¹ mol⁻¹; k = R/N_A = 1.38 × 10⁻²³ J K⁻¹.
  • Kinetic theory: molecules bouncing off the walls change momentum, giving pressure: pV = ⅓Nm⟨c²⟩.
  • Temperature measures the mean kinetic energy of the molecules: ½m⟨c²⟩ = (3/2)kT.

By the end you can

  • Understand amount of substance in moles and the Avogadro constant.
  • Recall and use pV = nRT and pV = NkT.
  • State the basic assumptions of the kinetic theory of gases.
  • Derive and use pV = ⅓Nm⟨c²⟩, and link mean kinetic energy to temperature.

1The ideal gas equation

pV = nRT

Ideal gas — a gas that obeys pV ∝ T at all pressures, volumes and temperatures (T in kelvin).

pV = nRT = NkT ; k = R/N_A
p in Pa, V in m³, T in K
VPisobaric (P const)isovolumetricisothermaladiabatic (steeper)work done by gas= area under curveW = PΔV (isobaric)
At constant temperature, p ∝ 1/V; the hotter isotherm sits further out.
Worked example

Find the number of moles of gas in 0.024 m³ at 1.0 × 10⁵ Pa and 290 K.

  1. n = pV/RT = 1.0 × 10⁵ × 0.024/(8.31 × 290) C1
  2. n = 1.0 mol A1
Worked example

A sealed rigid container of gas is heated from 300 K to 375 K. By what factor does the pressure change?

  1. constant V: p/T is constant C1
  2. p₂ = p₁ × 375/300 = 1.25 p₁ A1
Common mistakeUsing litres or cm³ and °C. Convert: 1 dm³ = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³, and T in kelvin.

2Kinetic theory

Assumptions

AssumptionWhy it matters
A large number of molecules in random motionpressure is the same in all directions, and statistics work
Molecules have negligible volume compared with the containerV is the volume available for motion
No intermolecular forces except during collisionsno potential energy: internal energy is all kinetic
Collisions are elasticno kinetic energy is lost
The time of a collision is negligible compared with the time between collisionscollisions are instantaneous
Each collision with a wallreverses the momentumcomponent: Δp = 2mv.Many collisions per secondgive a steady force, soa pressure P = F/A.P = ⅓ρ⟨v²⟩
Each wall collision reverses a molecule's velocity component, changing its momentum by 2mu.

Deriving pV = ⅓Nm⟨c²⟩

  • One molecule in a cube of side L: it hits a wall with velocity u, changing momentum by 2mu. Time between hits on that wall: 2L/u. Average force = 2mu ÷ (2L/u) = mu²/L.
  • N molecules: total force = Nm⟨u²⟩/L, so pressure = Nm⟨u²⟩/L³ = Nm⟨u²⟩/V.
  • Three dimensions: ⟨c²⟩ = ⟨u²⟩ + ⟨v²⟩ + ⟨w²⟩ = 3⟨u²⟩ (random motion), so pV = ⅓Nm⟨c²⟩.
pV = ⅓Nm⟨c²⟩ ; ½m⟨c²⟩ = (3/2)kT DATA & FORMULAE
combine with pV = NkT
Worked example

Find the mean kinetic energy of a molecule at 300 K, and the r.m.s. speed of nitrogen molecules (mass 28 u).

  1. E = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.2 × 10⁻²¹ J C1 A1
  2. crms = √(3kT/m) = √(3 × 1.38 × 10⁻²³ × 300/(28 × 1.66 × 10⁻²⁷)) = 517 m s⁻¹ A1
Common mistakeConfusing ⟨c²⟩ with ⟨c⟩². The r.m.s. speed is √⟨c²⟩: square the speeds, find the mean, then square-root.
Linking questionInternal energy and the first law are in Thermodynamics; the kelvin scale is in Temperature; momentum change is from Dynamics.

Test yourself — cover the answers first

Quick check

  1. State the Avogadro constant.
  2. What happens to the mean KE of molecules if T (in K) doubles?
  3. Name one assumption of kinetic theory.
  4. Find the r.m.s. speed of speeds 2, 4, 6 m s⁻¹.
  5. How many molecules are in 2 mol?

Exam-style

  1. A gas at 1.2 × 10⁵ Pa occupies 3.0 × 10⁻³ m³ at 27 °C. Find the number of molecules.
  2. By what factor does c_rms change if T rises from 300 K to 1200 K?
  3. Show that p = ⅓ρ⟨c²⟩.
Show answers
  1. 6.02 × 10²³ mol⁻¹
  2. it doubles
  3. e.g. collisions are elastic
  4. √(56/3) = 4.32 m s⁻¹
  5. 1.2 × 10²⁴
  6. N = pV/kT = 360/(1.38 × 10⁻²³ × 300) = 8.7 × 10²²
  7. c_rms ∝ √T, so it doubles
  8. pV = ⅓Nm⟨c²⟩ and ρ = Nm/V, so p = ⅓ρ⟨c²⟩
Practise this chapter
A Level practice papers
Next chapterThermodynamics →
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.