The Ideal Gas Equation

The ideal gas equation combines all three gas laws into a single equation and introduces the important constants R (the molar gas constant) and k (the Boltzmann constant). This lesson also covers the concept of the mole, the Avogadro constant, and the assumptions that define an ideal gas.

Spec mapping (AQA 7408 §3.6.2.2 — The ideal gas equation): This lesson addresses the equation of state pV = nRT and its equivalent molecular form pV = NkT, the definitions of the mole, Avogadro constant N_A, molar gas constant R, and Boltzmann constant k, and the relation k = R/N_A. Candidates must convert between moles and molecules, calculate any one of p, V, N, n, or T given the others, and identify the assumptions that define an ideal gas (point particles, elastic collisions, no intermolecular forces, large N, random motion). (Refer to the official AQA specification document for exact wording.)

Synoptic links: The ideal gas equation is the synthesis of the empirical gas laws (§3.6.2.2) and the foundation for the molecular kinetic theory (§3.6.2.3), where equating pV = (1/3)Nm〈c²〉 = NkT yields the temperature-kinetic-energy relationship. It connects forward to kinetic energy and temperature (§3.6.2.3) via (3/2)kT, and to chemistry's stoichiometry through n = m/M_r (where M_r is the molar mass). Practically it underpins Boyle's-law verification in RP-style experiments and any question on gas heating or compression. The Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ also reappears in statistical-mechanics treatments of equipartition (each quadratic degree of freedom carries (1/2)kT of energy) — a synoptic link to undergraduate-level physics.

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The Equation of State for an Ideal Gas

The ideal gas equation can be written in two forms:

pV = nRT (using moles)

pV = NkT (using number of molecules)

where:

• p = pressure (Pa)
• V = volume (m³)
• n = number of moles (mol)
• R = molar gas constant = 8.31 J mol⁻¹ K⁻¹
• T = absolute temperature (K)
• N = number of molecules
• k = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹

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Moles, Molecules, and the Avogadro Constant

A mole is the SI unit for amount of substance. One mole of any substance contains exactly 6.022 × 10²³ particles (atoms, molecules, ions, etc.). This number is called the Avogadro constant (Nₐ).

Nₐ = 6.022 × 10²³ mol⁻¹

The relationship between the number of moles (n) and the number of molecules (N) is:

N = nNₐ

The molar mass (M) is the mass of one mole of a substance, measured in kg mol⁻¹ (or g mol⁻¹ in chemistry). For example:

• Hydrogen (H₂): M = 2.0 × 10⁻³ kg mol⁻¹
• Helium (He): M = 4.0 × 10⁻³ kg mol⁻¹
• Nitrogen (N₂): M = 28 × 10⁻³ kg mol⁻¹
• Oxygen (O₂): M = 32 × 10⁻³ kg mol⁻¹

The number of moles is:

n = m/M (mass divided by molar mass)

The mass of a single molecule is:

m_molecule = M/Nₐ

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Relationship Between R and k

The Boltzmann constant is the gas constant per molecule:

k = R/Nₐ = 8.31 / (6.022 × 10²³) = 1.38 × 10⁻²³ J K⁻¹

This can be verified:

pV = nRT = (N/Nₐ)RT = N(R/Nₐ)T = NkT ✓

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Assumptions of an Ideal Gas

An ideal gas is a theoretical model. The assumptions are:

1. The gas consists of a very large number of identical molecules in constant random motion.
2. The volume of the molecules themselves is negligible compared to the volume of the container.
3. There are no intermolecular forces between the molecules (except during collisions).
4. Collisions between molecules and with the container walls are perfectly elastic (kinetic energy is conserved).
5. The duration of a collision is negligible compared to the time between collisions.
6. The molecules obey Newton's laws of motion.

Exam Tip: You must be able to state the assumptions of an ideal gas. Examiners typically award one mark per distinct assumption, up to a maximum of 3–4 marks. The most commonly missed assumptions are the negligible volume of molecules and the negligible collision duration.

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Worked Examples

Worked Example 1 — Using pV = nRT

Question: Calculate the volume occupied by 2.0 mol of an ideal gas at a temperature of 300 K and a pressure of 1.0 × 10⁵ Pa.

Solution:

V = nRT/p = (2.0 × 8.31 × 300) / (1.0 × 10⁵)

V = 4986 / 100 000 = 0.0499 m³ ≈ 0.050 m³ (50 litres)

Worked Example 2 — Using pV = NkT

Question: A container holds 5.0 × 10²⁴ molecules of nitrogen gas at 400 K and a pressure of 2.0 × 10⁵ Pa. Calculate the volume of the container.

Solution:

V = NkT/p = (5.0 × 10²⁴ × 1.38 × 10⁻²³ × 400) / (2.0 × 10⁵)

V = (5.0 × 1.38 × 400 × 10) / (2.0 × 10⁵)

V = 27 600 / (2.0 × 10⁵) = 0.138 m³

Worked Example 3 — Finding the Number of Molecules

Question: A room has dimensions 5.0 m × 4.0 m × 3.0 m. The air is at a temperature of 293 K and a pressure of 1.01 × 10⁵ Pa. Calculate the number of air molecules in the room.

Solution:

V = 5.0 × 4.0 × 3.0 = 60 m³

N = pV/(kT) = (1.01 × 10⁵ × 60) / (1.38 × 10⁻²³ × 293)

N = 6.06 × 10⁶ / (4.043 × 10⁻²¹)

N = 1.50 × 10²⁷ molecules

To find the number of moles: n = N/Nₐ = 1.50 × 10²⁷ / 6.022 × 10²³ = 2490 mol

Worked Example 4 — Mass of a Single Molecule

Question: Calculate the mass of a single nitrogen molecule. (Molar mass of N₂ = 28 × 10⁻³ kg mol⁻¹)

Solution:

m = M/Nₐ = (28 × 10⁻³) / (6.022 × 10²³) = 4.65 × 10⁻²⁶ kg

Worked Example 5 — Changing Conditions

Question: A sealed flask contains 0.050 mol of helium at 20 °C and 1.0 × 10⁵ Pa. The flask is heated to 200 °C. Calculate the new pressure.

Solution:

Since the flask is sealed, n and V are constant. Therefore p/T = nR/V = constant.

p₁/T₁ = p₂/T₂

T₁ = 20 + 273 = 293 K T₂ = 200 + 273 = 473 K

p₂ = p₁ × T₂/T₁ = 1.0 × 10⁵ × 473/293 = 1.61 × 10⁵ Pa

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Real Gases vs Ideal Gases

Real gases deviate from ideal behaviour under two conditions:

At high pressures:

• Molecules are forced close together, so their volume is no longer negligible.
• Intermolecular forces (van der Waals forces) become significant.
• The measured volume is larger than predicted (molecules take up space).
• The measured pressure is slightly lower than predicted (attractive forces reduce the impact on walls).

At low temperatures:

• Molecules move slowly enough for intermolecular attractions to influence their paths.
• The gas may liquefy, which cannot happen for an ideal gas.

Real gases behave most like ideal gases at high temperatures and low pressures, when molecules are far apart and moving fast.

Condition	Ideal Gas Prediction	Real Gas Behaviour
High T, low p	Accurate	Close to ideal
High p	pV = nRT	pV > nRT (at very high p)
Low T	Gas always remains a gas	Gas may liquefy
Near condensation	No prediction of liquefaction	Significant deviation

Exam Tip: If asked when a real gas behaves most like an ideal gas, always answer "at high temperature and low pressure." This is because at high T, molecules have enough kinetic energy to overcome intermolecular attractions, and at low p, molecules are far apart so their volume is negligible.

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Common Misconceptions

1. "n in pV = nRT is the number of molecules." No — n is the number of moles. The number of molecules is N, used in pV = NkT.

2. "R and k are the same thing." They are related (k = R/Nₐ) but have different values and are used with different quantities. R is used with moles; k is used with individual molecules.

3. "Volume in the ideal gas equation can be in litres." No — volume must be in m³. Remember: 1 litre = 1 × 10⁻³ m³ and 1 cm³ = 1 × 10⁻⁶ m³.

4. "Ideal gases can be liquefied." An ideal gas cannot be liquefied because there are no intermolecular forces to hold the molecules together in a liquid state.

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Additional Worked Example — Density of an Ideal Gas

The density ρ of an ideal gas can be derived directly from the ideal gas equation. Starting from pV = nRT and recognising that n = m/M (where m is mass of gas and M is molar mass): pV = (m/M)RT p = (m/V) × (RT/M) p = ρ(RT/M) ρ = pM/(RT).

This rearrangement is exam-frequent. Use it to calculate:

(a) The density of dry air (molar mass M_air = 29.0 × 10⁻³ kg mol⁻¹) at 1.01 × 10⁵ Pa and 20 °C. (b) The density of hydrogen gas (M_H₂ = 2.02 × 10⁻³ kg mol⁻¹) at the same conditions. (c) Comment on why hydrogen-filled balloons are buoyant in air.

(a) Density of air: T = 20 + 273.15 = 293.15 K. ρ_air = pM/(RT) = (1.01 × 10⁵ × 29.0 × 10⁻³) / (8.31 × 293.15) ρ_air = 2929 / 2436 = 1.20 kg m⁻³.

This is the well-known room-temperature density of air, used routinely in fluid mechanics and aerodynamics.

(b) Density of hydrogen: ρ_H₂ = (1.01 × 10⁵ × 2.02 × 10⁻³) / (8.31 × 293.15) ρ_H₂ = 204 / 2436 = 0.0838 kg m⁻³.

Hydrogen has about 1/14 the density of air at the same temperature and pressure — a direct consequence of its much smaller molar mass.

(c) Buoyancy. Archimedes' principle: an object immersed in a fluid experiences an upthrust equal to the weight of fluid displaced. A hydrogen balloon of volume V displaces ρ_air × V of air; the balloon itself plus contents weighs ρ_H₂ × V × g (ignoring balloon-membrane mass for now). The net upward force per unit volume is (ρ_air − ρ_H₂)g = (1.20 − 0.084) × 9.81 = 10.9 N m⁻³.

For a 1 m³ hydrogen balloon, the lifting capacity is 10.9 N — enough to lift about 1.1 kg of payload (after subtracting membrane mass). A 10 m³ hot-air-balloon-sized hydrogen envelope can lift 11 kg. This is why hydrogen was used in early airships (e.g., Hindenburg), before helium replaced it for safety reasons. Helium (M = 4.00 × 10⁻³ kg mol⁻¹) has density 0.166 kg m⁻³ — slightly denser than hydrogen, giving slightly less lift, but non-flammable.

Density variation with conditions:

• Doubling pressure at constant T doubles density (mass crammed into the same volume).
• Doubling kelvin temperature at constant p halves density (gas expands to maintain pressure).
• Lighter gas (smaller M) means lower density at same p, T — basis of buoyancy and meteorology.

Exam Tip: The formula ρ = pM/(RT) is not in the AQA formula booklet but is derivable from pV = nRT in one line. Practise the derivation and have it ready — questions on gas density and buoyancy are popular Paper 1 sub-parts.

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Additional Worked Example — Multi-stage Ideal-Gas Calculation

A scuba diver's cylinder of internal volume 12.0 litres is filled to a pressure of 230 bar at 20.0 °C. (1 bar = 1.00 × 10⁵ Pa; 1 litre = 1.00 × 10⁻³ m³.) Calculate: (a) the number of moles of air in the cylinder; (b) the number of molecules; (c) the volume the same air would occupy at sea-level atmospheric pressure (1.01 × 10⁵ Pa) and 15 °C.