B.3 Gas Laws

IB DP PHYSICS Theme B: The Particulate Nature of Matter B.3 Gas Laws

Revision Notes · Standard and Higher Level Fahad H. Ahmad

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Original notes prepared for the IB Diploma Programme Physics course (first assessment 2025)

What the syllabus requires

B.3 is common to SL and HL. By the end of this topic you should be able to do everything in the table below. Use it as a final checklist before the exam.

Understanding You should be able to...

Pressure Use P = F / A; explain gas pressure in terms of molecular collisions with the container walls.

Amount of substance Convert between number of molecules N, number of moles n and mass, using n = N / NA and the molar mass.

Empirical gas laws State and apply Boyle's law, Charles's law and the pressure law; sketch and interpret the corresponding graphs.

Ideal gas equation Apply PV = nRT and PV = NkBT, and the combined form P1V1/T1 = P2V2/T2.

Kinetic theory List and justify the assumptions of the ideal gas model; outline how P = ⅓ρ⟨c2⟩ follows from molecular collisions.

Molecular energy Use Ek = (3/2)kBT for the average translational kinetic energy of a molecule.

Internal energy Use U = (3/2)NkBT = (3/2)nRT for an ideal monatomic gas.

Limits of the model State the conditions under which a real gas approximates an ideal gas, and when the model fails.

Exam note: Almost every lost mark in B.3 traces back to one of two slips: using a Celsius temperature inside a gas equation, or mixing units (kPa with m3, cm3 with Pa). Convert to kelvin and check units before substituting.

1. Pressure

Pressure is the normal (perpendicular) force exerted per unit area of a surface:

P=F/A

The SI unit is the pascal: 1 Pa = 1 N m-2. It is a small unit — atmospheric pressure at sea level is about 1.0 × 105 Pa (101 kPa, also written 1 atm). Pressure is a scalar: at a point in a gas it acts equally in all directions.

Where gas pressure comes from (microscopic picture)

A gas consists of an enormous number of molecules in rapid, random motion. Each time a molecule strikes a wall of the container it rebounds, so its momentum changes; by Newton's second law the wall exerts a force on the molecule, and by Newton's third law the molecule exerts an equal and opposite force on the wall. Billions of such collisions per second, averaged over the wall's area, produce a steady macroscopic pressure. This single idea explains all three empirical gas laws qualitatively — always answer 'explain' questions in terms of the rate of collisions and the momentum change per collision.

Worked example 1 — pressure from force and area

• A 55 kg person balances briefly on one stiletto heel of area 1.0 cm2. (b) A 5000 kg elephant stands on four feet, each of area 0.10 m2. Compare the pressures on the ground (g = 9.8 N kg-1).
• F = 55 × 9.8 = 539 N; A = 1.0 × 10-4 m2; P = 539 / 1.0 × 10-4 = 5.4 × 106 Pa.
• F = 5000 × 9.8 = 4.9 × 104 N; A = 0.40 m2; P = 4.9 × 104 / 0.40 = 1.2 × 105 Pa.

The heel exerts roughly 40 times the pressure of the elephant — pressure depends on area, not just force.

2. Counting molecules: the mole

Because a laboratory sample contains of order 1023 molecules, we count them in bundles. One mole is the amount of substance containing exactly 6.02 × 1023 elementary entities (atoms, molecules, ions). This number is the Avogadro constant:

NA = 6.02 × 1023 mol-1 n = N / NA where N is the number of molecules and n the number of moles. The molar mass M is the mass of one mole; numerically it equals the relative molecular mass in grams (careful: in SI calculations convert to kg mol-1). The three ways of measuring 'how much gas' are linked by:

mass m = nM N = nNA mass of one molecule = M / NA Worked example 2 — mass, moles and molecules

A cylinder contains 8.0 g of oxygen gas, O2 (molar mass 32 g mol-1). Find (a) the number of moles, (b) the number of molecules, (c) the mass of one molecule.

• n = m / M = 8.0 / 32 = 0.25 mol.
• N = nNA = 0.25 × 6.02 × 1023 = 1.5 × 1023 molecules.
• M / NA = 0.032 kg mol-1 / 6.02 × 1023 mol-1 = 5.3 × 10-26 kg.

Unit check: In PV = nRT everything is SI, so molar mass must be in kg mol-1 when you compute masses. 32 g mol-1 = 0.032 kg mol-1.

3. The empirical gas laws

Historically, three separate laws were found by experiment, each holding one variable fixed for a fixed mass of gas. All temperatures must be in kelvin: T(K) = t(°C) + 273.

Law Held constant

Statement Graph shape

Boyle's law T, n P ∝ 1/V, i.e. PV = constant P vs V: falling curve (hyperbola); P vs 1/V: straight line through origin

Law Held constant

Statement Graph shape

Charles's law P, n V ∝ T, i.e. V/T = constant V vs T(K): straight line through origin; vs t(°C): line hitting axis at -273 °C

Pressure law (Gay-Lussac)

V, n P ∝ T, i.e. P/T = constant P vs T(K): straight line through origin

Boyle's law: isotherms

On a P–V diagram, each constant-temperature curve PV = constant is called an isotherm. Curves further from the origin correspond to higher temperatures (at any given V, a hotter gas has larger P). Microscopically: halving the volume doubles the number density of molecules, so collisions with the walls occur twice as often and the pressure doubles.

Figure 1. Boyle's law isotherms for 1 mol of ideal gas. Each curve is P ∝ 1/V, so the product P·V = nRT is constant along it

(2.49, 3.33 and 4.16 kJ at 300, 400 and 500 K). A hotter gas lies on a higher isotherm.

Charles's law and absolute zero

At constant pressure, the volume of a gas plotted against Celsius temperature is a straight line that, extrapolated backwards, cuts the temperature axis at -273 °C — the same intercept for every gas. This universal intercept defines absolute zero, the natural zero of the kelvin scale, the temperature at which an ideal gas would occupy zero volume (and molecular translational kinetic energy would reach its minimum). The pressure law gives the same intercept when P is extrapolated to zero at constant volume.

Figure 2. Pressure law (Gay-Lussac) for a fixed volume of gas: P ∝ T. Extrapolating the straight line back to P = 0 meets the temperature axis at absolute zero, t = −273.15 °C — the same intercept for every gas.

Typical experimental arrangements

• Boyle: a fixed mass of dry air trapped in a sealed graduated syringe or tube; pressure changed slowly (to keep T constant) and read from a pressure sensor; plot P against 1/V and look for a straight line through the origin.
• Charles: a bead of air trapped below a mercury or oil thread in a capillary tube open at the top (constant pressure = atmospheric + thread); tube warmed in a water bath, length of air column recorded against temperature.
• Pressure law: a rigid flask of air (constant volume) in a water bath, connected to a pressure gauge; record P as the bath temperature is varied, then extrapolate the P–t line to P = 0.

Worked example 3 — Boyle's law (trapped air column)

Air is trapped in a smooth syringe at 100 kPa; the column of gas is 30.0 cm long. The plunger is pushed in slowly until the column is 20.0 cm long, at constant temperature. Find the new pressure.

The cross-sectional area is constant, so V ∝ length. Boyle: P1V1 = P2V2.

P2 = 100 × (30.0 / 20.0) = 150 kPa.

'Slowly' matters: a rapid compression would warm the gas, so T would not be constant and Boyle's law would not apply.

4. The ideal gas equation

The three empirical laws combine into a single equation of state. For n moles, or equivalently N molecules:

PV = nRT PV = NkBT

Constant Value Meaning

Gas constant R 8.31 J K-1 mol-1 energy per kelvin per mole

Boltzmann constant kB 1.38 × 10-23 J K-1 energy per kelvin per molecule

Link R = NAkB 6.02 × 1023 × 1.38 × 10-23 ≈ 8.31

An ideal gas is defined as one that obeys PV = nRT exactly at all pressures and temperatures. For a fixed mass of gas (n constant) taken between two states, the equation collapses to the combined gas law:

P1V1 / T1 = P2V2 / T2

In this ratio form, P and V may stay in any consistent units (kPa, cm3, litres) on both sides — but T must always be in kelvin.

Worked example 4 — tyre pressure on a hot day

A car tyre contains air at an absolute pressure of 320 kPa on a cold morning at 7 °C. After a long drive the air warms to 35 °C. Assuming the tyre volume is unchanged, find the new pressure.

Constant V and n, so P/T = constant. Convert: T1 = 280 K, T2 = 308 K.

P2 = 320 × (308 / 280) = 352 kPa ≈ 350 kPa.

Using 7 and 35 instead of 280 and 308 would give the absurd answer 1600 kPa — the classic Celsius error.

Worked example 5 — how much air is in a room?

A room measures 5.0 m × 4.0 m × 2.5 m. The air is at 101 kPa and 20 °C (molar mass of air ≈ 29 g mol-1). Find (a) the number of moles, (b) the number of molecules, (c) the mass of air.

• V = 50 m3, T = 293 K. n = PV / RT = (1.01 × 105 × 50) / (8.31 × 293) = 2.1 × 103 mol.
• N = nNA = 2074 × 6.02 × 1023 ≈ 1.2 × 1027 molecules.
• m = nM = 2074 × 0.029 ≈ 60 kg — the air in a room weighs as much as a person.

Worked example 6 — two connected vessels

Vessel A (volume 2.0 L, pressure 300 kPa) is connected through a closed tap to vessel B (3.0 L, 100 kPa) at the same temperature. The tap is opened and the temperature returns to its initial value. Find the final pressure.

Molecules are conserved: nfinal = nA + nB. With T common and using n = PV/RT:

P(VA + VB) = PAVA + PBVB

P = (300 × 2.0 + 100 × 3.0) / 5.0 = 900 / 5.0 = 180 kPa.

Litres are acceptable here because volume units cancel throughout.

5. Kinetic theory: the microscopic model

Kinetic theory derives the gas laws from Newtonian mechanics applied to molecules. The assumptions of an ideal gas — learn this list — are:

Assumption Why it is reasonable / what it buys us

A gas consists of a very large number of identical molecules in random motion.

Large numbers make statistical averages (pressure, temperature) steady and meaningful.

The volume of the molecules themselves is negligible compared with the volume of the container.

True when the gas is dilute; lets us treat molecules as point particles and use the full container volume V.

There are no intermolecular forces except during collisions.

Molecules are far apart on average, so attractions are tiny; this removes potential energy from the model.

All collisions (molecule-molecule and molecule-wall) are perfectly elastic.

Kinetic energy is conserved, so a gas left alone does not spontaneously cool down.

The duration of a collision is negligible compared with the time between collisions.

Molecules spend almost all their time in free flight, so pressure is a smooth average.

Deriving the pressure equation (logic sketch)

Consider one molecule of mass m moving with velocity component cx perpendicular to a wall of a cubical box of side L:

• Momentum change per collision: the molecule rebounds elastically, so its momentum changes from +mcx to -mcx: magnitude Δp = 2mcx.
• Collision rate: it next hits the same wall after travelling to the far wall and back, a distance 2L, so the time between hits is 2L/cx.
• Average force of one molecule: F = Δp/Δt = 2mcx ÷ (2L/cx) = mcx 2/L.
• Add up all N molecules: total force = Nm⟨cx 2⟩/L, where ⟨ ⟩ denotes the average over all molecules.
• Random directions: on average ⟨cx 2⟩ = ⟨cy 2⟩ = ⟨cz 2⟩, and they sum to ⟨c2⟩, so ⟨cx 2⟩ = ⟨c2⟩/3.
• Divide by wall area L2: P = Nm⟨c2⟩ / 3L3 = Nm⟨c2⟩ / 3V. Since the density is ρ = Nm/V:

P = (1/3) ρ ⟨c2⟩ where ⟨c2⟩ is the mean square speed of the molecules. The syllabus asks for the reasoning above qualitatively, not a fully rigorous derivation — but steps 1 (momentum change) and 5 (averaging over three directions gives the factor 1/3) are the two ideas examiners probe.

Where temperature enters: Comparing PV = (1/3)Nm⟨c2⟩ with PV = NkBT forces (1/2)m⟨c2⟩ = (3/2)kBT. Temperature is a direct measure of average molecular kinetic energy.

6. Molecular kinetic energy and rms speed

The average translational kinetic energy of one molecule of an ideal gas depends only on the absolute temperature — not on the type of gas:

Ek = (1/2)m⟨c2⟩ = (3/2)kBT

The square root of the mean square speed is the root-mean-square (rms) speed, a representative molecular speed. Rearranging (and using m = M/NA, R = NAkB):

crms = √(3kBT / m) = √(3RT / M)

• At fixed T: lighter molecules move faster, crms ∝ 1/√M.
• For a given gas: crms ∝ √T — quadrupling the kelvin temperature doubles the rms speed.
• In a mixture at one temperature, every species has the same average kinetic energy but different rms speeds.

Figure 3. Maxwell-Boltzmann speed distribution for N2 at 300 K and 1000 K. The most-probable speed vp = √(2kBT/m) shifts from 422 to 771 m s−1, and the peak lowers and broadens as T rises (the area under each curve stays 1).

Worked example 7 — hydrogen vs oxygen at 300 K

Compare the average kinetic energy and the rms speeds of H2 (M = 2.0 g mol-1) and O2 (M = 32 g mol-1) at 300 K.

Average KE (both gases): Ek = (3/2) × 1.38 × 10-23 × 300 = 6.2 × 10-21 J — identical, because T is the same.

O2: crms = √(3 × 8.31 × 300 / 0.032) = 480 m s-1.

H2: crms = √(3 × 8.31 × 300 / 0.0020) = 1.9 × 103 m s-1 — exactly 4 times faster, since the mass ratio is 16 and speed scales as 1/√M.

This is why hydrogen escapes Earth's atmosphere far more readily than oxygen.

7. Internal energy of an ideal monatomic gas

The internal energy U of a system is the total of the random kinetic energies plus the intermolecular potential energies of its particles. For an ideal gas there are no intermolecular forces, so there is no potential energy term: the internal energy is purely the sum of the molecular kinetic energies. For a monatomic gas (He, Ne, Ar) each atom carries only translational kinetic energy (3/2)kBT, so multiplying by the number of atoms:

U = (3/2)NkBT = (3/2)nRT = (3/2)PV

Consequences worth quoting in explanations:

• U depends only on T (and the amount of gas) — an isothermal change of an ideal gas leaves U unchanged.
• Doubling the kelvin temperature doubles the internal energy.
• The last form, (3/2)PV, lets you read internal energy straight off a P–V diagram point (used heavily in HL thermodynamics, B.4).

Worked example 8 — internal energy of helium

A cylinder holds 2.0 mol of helium at 300 K. Find (a) the internal energy of the gas, (b) the average kinetic energy per atom, (c) the new internal energy if the gas is heated to 600 K.

• U = (3/2)nRT = 1.5 × 2.0 × 8.31 × 300 = 7.5 × 103 J.
• Ek = (3/2)kBT = 1.5 × 1.38 × 10-23 × 300 = 6.2 × 10-21 J.
• U ∝ T, so doubling T doubles U: 1.5 × 104 J.

8. Real gases vs the ideal model

No real gas is ideal, but most gases are excellently approximated by the ideal model under everyday conditions. The model relies on molecules being far apart and interacting only in brief collisions, so it fails when either assumption breaks:

Condition Which assumption fails Effect

High pressure / high density Molecular volume is no longer negligible compared with V

Less free space than the model assumes; measured PV deviates from nRT

Low temperature (near condensation / boiling point)

Intermolecular attractions are no longer negligible — slow molecules linger near each other

Attractions reduce the momentum delivered to the walls; pressure falls below the ideal prediction; eventually the gas liquefies, which the ideal model cannot describe at all

A real gas therefore behaves most ideally at low pressure (molecules far apart) and high temperature (kinetic energy overwhelms any attraction) — conditions far from liquefaction. This is also why an ideal gas has no intermolecular potential energy: with no intermolecular forces there is nothing to store potential energy, so U is purely kinetic. In a real gas, part of the internal energy is potential, which is why real gases cool when they expand rapidly (work is done against attractions) — an effect absent in the ideal model.

Exam phrasing: 'State two conditions under which a real gas approximates ideal behaviour' — answer: low pressure (or low density) and a temperature well above the boiling point. Then justify each by naming the assumption it protects.

9. Common pitfalls

• Celsius in a gas equation — the number one error in B.3. Convert every temperature to kelvin before touching PV = nRT, ratios included.
• Mixing unit systems: with R = 8.31, pressure must be in Pa and volume in m3 (1 m3 = 103 L = 106 cm3). Ratio problems tolerate any consistent units except temperature.
• Using gauge pressure where absolute pressure is needed — gas laws require absolute pressure (gauge + atmospheric).
• Molar mass in g mol-1 substituted into √(3RT/M) — it must be kg mol-1, otherwise speeds come out about 32 times too large.
• Saying temperature measures 'the energy of the gas' — be precise: it measures the average translational kinetic energy per molecule.
• Claiming all molecules move at crms — speeds are distributed; the rms value is a statistical average.
• Applying U = (3/2)nRT to diatomic gases in contexts where rotation matters — the IB formula is stated for monatomic ideal gases.
• Forgetting that in a two-vessel or leaking-container problem it is the number of moles, not PV/T of one vessel, that is conserved.

10. Quick reference

Result Statement

Pressure P = F / A (Pa = N m-2; 1 atm ≈ 1.0 × 105 Pa)

Amount of substance n = N / NA = m / M; NA = 6.02 × 1023 mol-1

Ideal gas equation PV = nRT = NkBT; R = NAkB = 8.31 J K-1 mol-1

Combined gas law P1V1/T1 = P2V2/T2 (fixed mass; T in kelvin)

Kinetic theory pressure P = (1/3)ρ⟨c2⟩

Average molecular KE Ek = (1/2)m⟨c2⟩ = (3/2)kBT; crms = √(3RT/M)

Internal energy (monatomic) U = (3/2)NkBT = (3/2)nRT = (3/2)PV

Ideal behaviour best at low pressure and high temperature, far from liquefaction

11. Test yourself

Attempt these without notes; full answers follow. Take R = 8.31 J K-1 mol-1, kB = 1.38 × 10-23 J K-1, NA = 6.02 × 1023 mol-1.

• A laboratory is at 25 °C. State this temperature in kelvin, and explain why Celsius temperatures can never be used directly in PV = nRT.
• Find the number of moles and the number of molecules in 0.044 kg of carbon dioxide (molar mass 44 g mol-1).
• A gas occupies 250 cm3 at 100 kPa. It is compressed slowly at constant temperature to 100 cm3. Find the new pressure.
• At constant pressure, 1.20 L of gas at 27 °C is heated to 127 °C. Find the new volume.
• A sealed rigid aerosol can holds gas at 105 kPa at 290 K. It is thrown into a fire and reaches 580 K. Find the pressure, and explain microscopically why it rises.
• A fixed mass of gas occupies 3.0 L at 100 kPa and 300 K. It is compressed to 2.0 L while cooling to 250 K. Find the final pressure.
• A balloon of volume 0.030 m3 contains gas at 100 kPa and 300 K. Find the number of molecules and the number of moles.
• Calculate the average translational kinetic energy of a gas molecule at 400 K. By what factor does the rms speed of a gas change when its kelvin temperature rises from 100 K to 400 K?
• Calculate the rms speed of nitrogen molecules (molar mass 28 g mol-1) at 293 K.
• Explain (a) why the internal energy of an ideal gas contains no potential energy term, and (b) under what two conditions a real gas deviates most from ideal behaviour.

Answers

• T = 25 + 273 = 298 K. Gas-law relations are proportionalities through the origin of the absolute scale (e.g. P ∝ T requires P = 0 at T = 0); the Celsius zero is arbitrary, so ratios of Celsius temperatures have no physical meaning.
• n = 44 g / 44 g mol-1 = 1.0 mol; N = 1.0 × 6.02 × 1023 = 6.0 × 1023 molecules.
• Boyle: P2 = 100 × 250/100 = 250 kPa.
• Charles: T1 = 300 K, T2 = 400 K; V2 = 1.20 × 400/300 = 1.60 L.
• P2 = 105 × 580/290 = 210 kPa. At higher T the molecules move faster: each wall collision transfers more momentum, and collisions are more frequent, so the average force per unit area — the pressure — increases (volume unchanged).
• P2 = P1V1T2 / (T1V2) = 100 × 3.0 × 250 / (300 × 2.0) = 125 kPa.
• N = PV/kBT = (1.0 × 105 × 0.030) / (1.38 × 10-23 × 300) = 7.2 × 1023 molecules; n = N/NA = 1.2 mol.
• Ek = (3/2) × 1.38 × 10-23 × 400 = 8.3 × 10-21 J. Since crms ∝ √T, the factor is √(400/100) = 2.
• crms = √(3 × 8.31 × 293 / 0.028) = √(2.61 × 105) ≈ 510 m s-1.
• (a) The ideal model assumes no intermolecular forces (except instantaneous collisions); with no forces there is no way to store intermolecular potential energy, so U is entirely molecular kinetic energy. (b) At high pressure/density (molecular volume no longer negligible) and at low temperature near the condensation point (intermolecular attractions become significant).