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Notes/Physics/Paper 4/Ideal Gases
CAIEA Level9702§15.1–15.3

Ideal Gases

The mole and the Avogadro constant, the equation of state pV = nRT and pV = NkT, the kinetic-theory assumptions and the derivation of pV = ⅓Nm⟨c²⟩, root-mean-square speed, and the deduction that a molecule's average translational kinetic energy is ³⁄₂kT.

150 min read 6 sub-topics
133
question parts
2021–2025 · 37 papers
8 marks
per paper
≈ 8% of the paper
2.0/3
avg difficulty
moderate
#6
most examined
of 16 topics by marks

A single party balloon contains of order 102310^{23} molecules, and every one of them is slamming into the balloon's inner surface billions of times each second. The steady push you feel as "pressure" is the average of that planetary-scale hail. The whole of this topic is the round trip between two languages for the same gas: the bulk language of gauges and thermometers (pp, VV, TT) and the molecular language of counts, masses and speeds (NN, mm, cc). We build the counting dictionary first (§01), compress all bulk behaviour into one equation (§02), then rebuild that same equation from Newton's laws applied to a single bouncing molecule (§03–§04) — and finally cash temperature itself in as molecular kinetic energy (§05–§06).

The bank rewards the investment heavily: 298 marks across 2021–2025, ranking this 6th of the 16 A2 topics, carried by 133 question parts in 37 paper sittings — about 60 marks a year. The yearly haul reads 57, 37, 68, 60, 76: a dip in 2022, then a climb to a record 76 in 2025. The mean difficulty of 2.05 sits mid-range because the marks have a split personality: roughly half are word-perfect recall — definitions, the Avogadro constant, the kinetic-theory assumptions — plus disciplined unit conversion, while the rest ride on two derivations and one sketch skill that barely change shape between papers.

The route through is: §01 the mole and the Avogadro constant — §02 the equation of state pV=nRTpV = nRT and pV=NkTpV = NkT, with k=R/NAk = R/N_A — §03 the kinetic-theory assumptions and how molecular movement causes pressure — §04 the syllabus-demanded derivation of pV=13Nm⟨c2⟩pV = \frac{1}{3}Nm\langle c^{2} \rangle — §05 root-mean-square speed and its proportionality behaviour — and §06 the capstone deduction EK=32kTE_K = \frac{3}{2}kT and the internal-energy family U=32NkTU = \frac{3}{2}NkT. You need AS tools throughout: kelvin conversion (AS Temperature note), p=F/Ap = F/A, momentum change and Newton's laws, and fluent standard-form arithmetic.

Before you start you should be able to

  • Convert temperatures fluently using T/K=θ/∘C+273.15T/\text{K} = \theta/^\circ\text{C} + 273.15, and know that 0 K0\ \text{K} is absolute zero (AS Temperature note) — every equation in this topic demands kelvin

  • Use pressure as force per unit area, p=F/Ap = F/A, in pascals, and volume in m3\text{m}^3

  • Handle momentum changes and Newton's laws: Δp=mv−mu\Delta p = mv - mu, force as rate of change of momentum, and the third-law force pair (AS Mechanics notes)

  • Recall kinetic energy Ek=12mv2E_k = \frac{1}{2}mv^2 and apply it to a single particle

  • Multiply and divide standard-form quantities with negative powers of ten without losing track: (1.38×10−23)×290(1.38 \times 10^{-23}) \times 290, 2.7×105\sqrt{2.7 \times 10^{5}}, and friends

By the end of this page you can

  • Recognise amount of substance as an SI base quantity with base unit the mole, and define the Avogadro constant NAN_A as the number of particles per unit amount of substance

  • Move fluently between mass, amount of substance and number of molecules using n=N/NAn = N/N_A and molar mass

  • Define an ideal gas as one obeying pV∝TpV \propto T at all values of pp, VV and TT, where TT is the thermodynamic temperature

  • Recall and use both forms of the equation of state, pV=nRTpV = nRT (n = number of moles) and pV=NkTpV = NkT (N = number of molecules), with k=R/NAk = R/N_A, converting units correctly before substituting

  • State the basic assumptions of the kinetic theory of gases and explain how molecular movement causes the pressure exerted by a gas

  • Derive pV=13Nm⟨c2⟩pV = \frac{1}{3}Nm\langle c^{2} \rangle from a one-dimensional collision model extended to three dimensions, and use the density form p=13ρ⟨c2⟩p = \frac{1}{3}\rho\langle c^{2} \rangle

  • Understand that the root-mean-square speed is cr.m.s.=⟨c2⟩c_{\text{r.m.s.}} = \sqrt{\langle c^{2} \rangle}, use cr.m.s.=3kT/mc_{\text{r.m.s.}} = \sqrt{3kT/m}, compare gases at the same temperature, and sketch its variation with TT

  • Compare pV=13Nm⟨c2⟩pV = \frac{1}{3}Nm\langle c^{2} \rangle with pV=NkTpV = NkT to deduce EK=12m⟨c2⟩=32kTE_K = \frac{1}{2}m\langle c^{2} \rangle = \frac{3}{2}kT, recall and use it, and explain why the internal energy of an ideal gas is proportional to its thermodynamic temperature

01

The mole and the Avogadro constant

Syllabus requirement · §15.1

“

understand that amount of substance is an SI base quantity with the base unit mol; use molar quantities where one mole of any substance is the amount containing a number of particles of that substance equal to the Avogadro constant NA

”

Counting the uncountable

A laboratory flask of air holds roughly 102210^{22} molecules — a 1 followed by twenty-two zeros, far beyond any everyday counting. Yet nearly every gas calculation in this topic ends with a question about those molecules: how many are there, what does each one weigh? Physics needs a bridge between the laboratory scale (grams, litres) and the molecular scale (masses of 10−26 kg10^{-26}\ \text{kg}), and the SI built one deliberately: a seventh base quantity called amount of substance, with base unit the mole (mol). Say that parallel explicitly, because it is the whole idea: amount of substance stands beside mass exactly as kilograms do — an independent base quantity, not something you derive from mass or volume.

One mole is defined as the amount of substance containing a number of particles equal to the Avogadro constant NAN_A:

NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}

"Particles" means whatever the substance is actually made of — atoms for helium, molecules for oxygen and nitrogen, ions for a salt. The examiner's definition of the constant itself is worth memorising word-perfectly, because it is asked verbatim and paid as its own mark (9702/41 O/N 2024 Q3(a)(i), 9702/43 O/N 2024 Q3(a)(i)):

the number of particles per unit amount of substance (B1).

Notice the shape of that sentence: it defines NAN_A as a conversion factor between a count (particles) and an amount (moles) — which is precisely the job it does in every formula below.

One sentence to carry

Amount of substance measures how many particles a sample contains, in units of moles; one mole contains NA=6.02×1023N_A = 6.02 \times 10^{23} particles, so NAN_A converts counts into amounts and back.

Two bridges: mass → moles → molecules

Everything in §01 runs across two bridges. The first connects the molecular count NN to the amount nn:

n=NNA⟺N=nNAn = \frac{N}{N_A} \qquad\Longleftrightarrow\qquad N = nN_A

— one line read in both directions: divide molecules by 6.02×10236.02\times10^{23} to get moles, multiply moles by it to get molecules. The second bridge connects a measured mass to an amount, through the molar mass (the mass of one mole, quoted in g mol−1\text{g mol}^{-1}):

n=massmolar massn = \frac{\text{mass}}{\text{molar mass}}

Chain them and you can walk from "I weighed out 8.0 g" all the way to "my sample contains this many molecules". One unit discipline keeps the second bridge honest: molar masses are tabulated in grams per mole, but physics equations demand kilograms — so convert on sight. The mass of one mole in kg is the relative atomic/molecular mass ×10−3\times 10^{-3}: helium's 4.0 g mol−14.0\ \text{g mol}^{-1} becomes 4.0×10−3 kg mol−14.0 \times 10^{-3}\ \text{kg mol}^{-1}. Make that conversion its own line and it can never bite you.

Invented demo — bookkeeping a helium balloon

A small balloon holds 4.0 g4.0\ \text{g} of helium. Helium-4 is monatomic — its particles are single atoms — and its molar mass is 4.0 g mol−14.0\ \text{g mol}^{-1}. Calculate (a) the amount of helium in mol, (b) the number of helium atoms, and (c) the number of atoms in one gram of helium.

Show full working
  1. 1

    (a) Convert the data to SI first, then cross the mass bridge:

    m=4.0 g=4.0×10−3 kg,M=4.0 g mol−1=4.0×10−3 kg mol−1m = 4.0\ \text{g} = 4.0 \times 10^{-3}\ \text{kg}, \qquad M = 4.0\ \text{g mol}^{-1} = 4.0 \times 10^{-3}\ \text{kg mol}^{-1} n=mM=4.0×10−34.0×10−3=1.0 moln = \frac{m}{M} = \frac{4.0 \times 10^{-3}}{4.0 \times 10^{-3}} = 1.0\ \text{mol}

    Both masses go to kilograms before dividing — matching units cancel cleanly and the answer reads 'one mole', which is the sanity anchor for everything downstream.

  2. 2

    (b) Cross the counting bridge:

    N=nNA=1.0×6.02×1023=6.02×1023 atomsN = nN_A = 1.0 \times 6.02 \times 10^{23} = 6.02 \times 10^{23}\ \text{atoms}

    One mole of atoms is, by definition, N_A atoms — the definition doing real work rather than sitting in a fact list.

  3. 3

    (c) Scale down from the whole sample:

    Nm=6.02×10234.0 g=1.5×1023 atoms per gram\frac{N}{m} = \frac{6.02 \times 10^{23}}{4.0\ \text{g}} = 1.5 \times 10^{23}\ \text{atoms per gram}

    A single gram carries a hundred billion billion atoms — this is why the mole exists at all: no other counting unit survives these powers of ten.

Answer

(a) n = 4.0×10⁻³ / 4.0×10⁻³ = 1.0 mol. (b) N = 6.02×10²³ atoms. (c) 1.5×10²³ atoms per gram.

Mass → moles → molecules is the standard chain. Write each bridge as its own line — schemes pay the two steps separately, and merged working hides slips.

Common mistakes

  • Answering "State what is meant by the Avogadro constant" with just "6.02×10236.02 \times 10^{23}".

    The number of particles per unit amount of substance (equal to 6.02×10²³ mol⁻¹).

    The B1 is paid for the defining phrase, not the value. A bare number states no relationship between counts and amounts.

  • Treating the mole as a unit of mass ("a mole of helium weighs 6.02×10²³ g").

    The mole counts amount of substance; the MASS of one mole depends on the substance (its molar mass).

    One mole of helium is 4 g, one mole of oxygen is 32 g — but both contain the same number of particles. That contrast is the entire point of the unit.

  • Dividing a mass in grams by a molar mass in kg mol⁻¹ (or vice versa).

    Match units first: grams pair with g mol⁻¹, kilograms with kg mol⁻¹ — convert one of the pair before dividing.

    A stray factor of 1000 rides silently through the rest of the question. Convert molar mass to kg mol⁻¹ on sight.

  • Feeding NN (a molecule count) into pV=nRTpV = nRT, or nn (moles) into pV=NkTpV = NkT.

    Check which letter the equation wants: lowercase n = moles, uppercase N = molecules. Preview of §02's biggest trap.

    The two forms of the gas law differ only in their counting basis — mixing them is off by a factor of 6×10²³, never a near miss.

Your turn

The verbatim definition, a straight counting drill with big powers of ten, and the full mass-to-molecules chain.

  1. 19702/41 O/N 2024 Q3(a)(i)1 mark

    State what is meant by the Avogadro constant.

    Stuck? Show hint

    Define it as a conversion: number of what, per unit of what?

    Show solution
    1. 1

      The number of particles per unit amount of substance (B1) — numerically 6.02×1023 mol−16.02 \times 10^{23}\ \text{mol}^{-1}.

      The phrase is the mark; the value is decoration. 'Number of particles in one mole' also scores — same content, different wording.

    Answer

    The number of particles per unit amount of substance (6.02×10²³ mol⁻¹).

  2. 21 mark

    A sample of an ideal gas contains N=3.4×1022N = 3.4 \times 10^{22} molecules. Calculate the amount of substance nn of the sample, in mol.

    Stuck? Show hint

    n=N/NAn = N/N_A — watch the power of ten in the division.

    Show solution
    1. 1

      Cross the counting bridge:

      n=NNA=3.4×10226.02×1023n = \frac{N}{N_A} = \frac{3.4 \times 10^{22}}{6.02 \times 10^{23}}

      Divide the front numbers, then subtract the powers: 3.4/6.02 = 0.56, and 10²²⁻²³ = 10⁻¹.

    2. 2
      n=0.56×10−1=0.056 moln = 0.56 \times 10^{-1} = 0.056\ \text{mol}

      Fewer than one mole — correct, since the sample holds about a twentieth of Avogadro's number of molecules. This is the exact arithmetic the O/N 2025 paper pays an A1 for.

    Answer

    n = 3.4×10²² / 6.02×10²³ = 0.056 mol.

  3. 32 marks

    Oxygen gas consists of diatomic molecules, O2\text{O}_2, of molar mass 32 g mol−132\ \text{g mol}^{-1}. A vessel contains 8.0 g8.0\ \text{g} of oxygen gas. Calculate the number of oxygen molecules in the vessel.

    Stuck? Show hint

    Two bridges in sequence: mass → moles, then moles → molecules.

    Show solution
    1. 1

      Mass bridge first:

      n=8.0 g32 g mol−1=0.25 moln = \frac{8.0\ \text{g}}{32\ \text{g mol}^{-1}} = 0.25\ \text{mol}

      Grams against g mol⁻¹ pair consistently here, so no conversion is needed — but say WHY they pair rather than hoping.

    2. 2

      Then the counting bridge:

      N=nNA=0.25×6.02×1023=1.5×1023 moleculesN = nN_A = 0.25 \times 6.02 \times 10^{23} = 1.5 \times 10^{23}\ \text{molecules}

      A quarter of a mole times Avogadro's number. Note the particles here are MOLECULES — the question said diatomic, and 'atoms' would double the answer wrongly.

    Answer

    n = 8.0/32 = 0.25 mol, so N = 0.25 × 6.02×10²³ = 1.5×10²³ molecules.

Practise mole and Avogadro questionsReal past-paper questions · The mole and the Avogadro constant

The rest of this note

Checking your access…

Can you do all of these?

  • Quote the Avogadro constant definition verbatim: number of particles per unit amount of substance

  • Decide n versus N BEFORE choosing the equation: n moles → pV = nRT; N molecules → pV = NkT — never mix R with k or vice versa

  • Convert BEFORE substituting: add 273 only to absolute temperatures (never to ΔT), and cm³ → m³ by ×10⁻⁶

  • Define an ideal gas with BOTH clauses: obeys pV ∝ T at all values of p, V and T, AND T is the thermodynamic temperature

  • Reproduce the kinetic-theory assumptions word-perfectly: constant random motion · perfectly elastic collisions · no forces except during collisions · negligible molecular volume · instantaneous collisions

  • Run the pressure chain in order: collide with wall → momentum changes → wall exerts force on molecule → molecule exerts force on wall → many molecules over the area → pressure

  • Rebuild the derivation on demand: Δp = 2mc → Δt = 2L/c → F = mc²/L → pV = Nmc² → replace c² by ⟨cx²⟩ = ⅓⟨c²⟩

  • Read ⟨c²⟩ as the mean of the SQUARES, and c_r.m.s. as its square root — r.m.s. speed is not the mean speed

  • On a pV-versus-kT graph: straight line through the origin ⇒ ideal gas; gradient = N

  • State E_K = ³⁄₂kT as depending on T alone — hydrogen and oxygen at the same T have equal molecular kinetic energies but different speeds

  • Use U = ³⁄₂pV as the fast internal-energy route when p and V are given

  • Explain U ∝ T in two clauses: no intermolecular forces ⇒ zero potential energy ⇒ internal energy is all kinetic, and KE ∝ T

Now do the questions
133 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes
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