From weather balloons to car tyres, the ideal gas laws describe how pressure, volume and temperature are linked. These equations connect the macroscopic properties we measure to the microscopic world of atoms and molecules.
🔵State and apply Boyle's, Charles's and Gay-Lussac's (pressure) laws
🔢Use the ideal gas equation pV = nRT and pV = NkT
🌡️Explain the concept of absolute zero and the Kelvin temperature scale
⚗️Use the Avogadro constant N_A and Boltzmann constant k_B
📊Interpret pV, pT and VT graphs for ideal gas processes
🏗️Describe the assumptions of an ideal gas model
The Gas Laws
Three empirical (experimentally determined) gas laws describe how the macroscopic properties of a fixed mass of ideal gas are related. All temperatures must be in Kelvin.
Boyle's Law (1662) — Isothermal (constant T): At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.
pV = constant or p₁V₁ = p₂V₂
A p–V graph gives a hyperbola; a p–(1/V) graph gives a straight line through the origin.
Charles's Law (1787) — Isobaric (constant p): At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
V/T = constant or V₁/T₁ = V₂/T₂
A V–T graph (T in Kelvin) gives a straight line through the origin.
Gay-Lussac's (Pressure) Law (1809) — Isochoric (constant V): At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.
p/T = constant or p₁/T₁ = p₂/T₂
A p–T graph (T in Kelvin) gives a straight line through the origin.
⚠️ Always convert temperatures to Kelvin: T(K) = T(°C) + 273.15 ≈ T(°C) + 273. Using Celsius in gas law calculations is one of the most common exam errors.
The Ideal Gas Equation
Combining all three gas laws gives the combined gas law: pV/T = constant (for fixed mass). Introducing the amount of gas in moles (n), the proportionality constant is the molar gas constant R = 8.314 J mol⁻¹ K⁻¹:
pV = nRT
Alternatively, expressing in terms of the number of molecules N, using the Boltzmann constant k_B = R/N_A = 1.38 × 10⁻²³ J K⁻¹:
pV = NkT
The two forms are equivalent: n = N/N_A, so nRT = (N/N_A)(R)T = N(R/N_A)T = NkT ✓.
Key constants to remember:
Constant
Symbol
Value
Units
Molar gas constant
R
8.314
J mol⁻¹ K⁻¹
Avogadro constant
N_A
6.022 × 10²³
mol⁻¹
Boltzmann constant
k_B
1.38 × 10⁻²³
J K⁻¹
Standard pressure
p₀
1.013 × 10⁵
Pa
Absolute Zero and the Kelvin Scale
Absolute zero (0 K = −273.15°C) is the temperature at which an ideal gas would have zero volume and zero pressure — the minimum possible temperature in nature. At absolute zero, particles have minimum possible kinetic energy (only zero-point quantum energy remains).
The evidence for absolute zero comes from extrapolating the Charles's Law graph (V vs T) backward — all ideal gas lines converge on V = 0 at −273.15°C. Similarly, pressure-temperature graphs extrapolate to p = 0 at 0 K.
The Kelvin scale (also called the absolute or thermodynamic scale) starts at absolute zero and has degree intervals equal to Celsius degrees:
T(K) = θ(°C) + 273.15
The Kelvin scale is defined by two fixed points: absolute zero (0 K) and the triple point of water (273.16 K = 0.01°C). It is the fundamental scale used in all thermodynamic equations.
In practice we cannot reach absolute zero — the Third Law of Thermodynamics states it would require infinite steps/resources. The coldest temperatures achieved in labs are ~10⁻¹⁰ K above absolute zero using laser cooling and magnetic evaporation.
Assumptions of an Ideal Gas
The ideal gas model makes the following simplifying assumptions:
The gas consists of a large number of identical molecules (or atoms for monatomic gases) moving randomly.
The volume of the molecules themselves is negligible compared to the volume of the container.
There are no intermolecular forces between molecules (except during collisions).
All collisions are perfectly elastic (kinetic energy is conserved).
The duration of collisions is negligible compared to the time between collisions.
Newton's laws of motion apply to all molecular motion.
Real gases deviate from ideal behaviour at:
High pressures: molecules are close together — intermolecular repulsion matters and molecular volume is significant.
Low temperatures: molecules move slowly — intermolecular attractions become significant; the gas may condense.
Noble gases (He, Ne, Ar) behave most like ideal gases because their atoms are small, spherical and have very weak intermolecular forces. Polar molecules (H₂O, NH₃) deviate most due to strong dipole-dipole and hydrogen bonding interactions.
The Mole and Avogadro's Constant
The mole is the SI unit of amount of substance. One mole of any substance contains exactly N_A = 6.022 × 10²³ particles (atoms, molecules, ions, etc.) — the Avogadro constant.
The molar mass M (kg mol⁻¹) of a substance equals the mass of one mole. The number of moles n in a sample of mass m:
n = m/M
The number of molecules N in that sample: N = nN_A = mN_A/M.
The Boltzmann constant k_B = R/N_A = 8.314/6.022×10²³ = 1.38×10⁻²³ J K⁻¹ can be thought of as "R per molecule" — the gas constant scaled down to the level of individual molecules. It appears wherever thermal energy at the molecular scale is involved.
A sealed container holds gas at pressure 1.5 × 10⁵ Pa and temperature 27°C. It is heated to 127°C at constant volume. Find the new pressure.
1Convert to Kelvin: T₁ = 27 + 273 = 300 K; T₂ = 127 + 273 = 400 K
A weather balloon contains 2.0 m³ of helium at ground level where p = 1.01 × 10⁵ Pa and T = 20°C. At altitude, p = 2.5 × 10⁴ Pa and T = −40°C. Find the volume of the balloon at altitude.
V = 22.4 L at STP — the molar volume of an ideal gas. This is a useful reference value.
Q1. A gas at 2.0 × 10⁵ Pa occupies 3.0 L. At the same temperature, its pressure is increased to 6.0 × 10⁵ Pa. What is the new volume?
Q2. The temperature of a gas is increased from 27°C to 327°C at constant pressure. What happens to its volume?
Q3. What is the value of the Boltzmann constant k_B?
Q4. Which of the following is NOT an assumption of the ideal gas model?
Q5. A gas has n = 2.0 mol at T = 400 K and V = 0.040 m³. What is its pressure? (R = 8.314 J mol⁻¹ K⁻¹)
Challenge 1. A diver's lungs contain 4.0 L of air at a depth of 20 m where the pressure is 3.0 × 10⁵ Pa. The diver ascends to the surface (pressure 1.0 × 10⁵ Pa) at constant temperature. (a) Find the new lung volume. (b) Explain why rapid ascent is dangerous for a scuba diver.
✓ (a) Boyle's Law (isothermal): p₁V₁ = p₂V₂ → V₂ = p₁V₁/p₂ = (3.0×10⁵ × 4.0)/(1.0×10⁵) = 12 L. (b) Lungs would need to expand to 12 L — far beyond their capacity (max ~6 L). If a diver holds their breath and ascends rapidly, the gas expands and can rupture the lungs (pulmonary barotrauma), forcing air bubbles into the bloodstream. Additionally, dissolved nitrogen in the blood (at higher pressure) forms bubbles as pressure drops — "the bends" or decompression sickness. This is why scuba divers must breathe continuously (never hold breath) and ascend slowly, making decompression stops.
Challenge 2. A car tyre contains air at gauge pressure 2.2 × 10⁵ Pa at 20°C. After a long drive the temperature rises to 60°C. (a) Calculate the new gauge pressure (assume volume is constant). Note: absolute pressure = gauge pressure + atmospheric pressure (1.013 × 10⁵ Pa). (b) Why is it important to check tyre pressures when the tyres are cold?
✓ (a) Absolute pressure: p₁ = 2.2×10⁵ + 1.013×10⁵ = 3.213×10⁵ Pa. T₁ = 293 K; T₂ = 333 K. At constant volume: p₂ = p₁ × T₂/T₁ = 3.213×10⁵ × 333/293 = 3.213×10⁵ × 1.1365 = 3.652×10⁵ Pa. New gauge pressure = 3.652×10⁵ − 1.013×10⁵ = 2.639×10⁵ Pa ≈ 2.6×10⁵ Pa — about 18% higher than when cold. (b) If you check tyre pressure when hot (after driving), the reading is higher than the actual cold-inflation pressure. You might think the tyre is over-inflated and release air. When the tyre cools, the pressure drops below the correct value — leaving the tyre under-inflated, which reduces fuel efficiency, increases tyre wear, and can be dangerous (affects handling). Always check cold.
Challenge 3. A sample of nitrogen gas (molar mass 28 g mol⁻¹) has mass 14 g at pressure 2.5 × 10⁵ Pa and temperature 350 K. (a) Calculate the volume occupied. (b) Calculate the number of molecules present. (c) At what temperature would the gas occupy twice this volume at the same pressure?
✓ (a) n = m/M = 14/28 = 0.50 mol. V = nRT/p = 0.50×8.314×350/(2.5×10⁵) = 1454.95/250 000 = 5.82×10⁻³ m³ ≈ 5.82 L. (b) N = nN_A = 0.50 × 6.022×10²³ = 3.01×10²³ molecules. (c) For isobaric (constant pressure) process: V/T = constant → T₂ = T₁ × V₂/V₁ = 350 × 2 = 700 K (426.85°C ≈ 427°C). Volume doubles when absolute temperature doubles at constant pressure — this is Charles's Law.
Challenge 4 (Synoptic). The pressure at the centre of the Sun is estimated at 2.5 × 10¹⁶ Pa and the temperature is about 1.5 × 10⁷ K. Assuming ideal gas behaviour and the gas is primarily protons (m = 1.67 × 10⁻²⁷ kg), estimate the number density n (particles per m³) and the mass density ρ (kg m⁻³). Compare with the density of water (1000 kg m⁻³).
✓ Using pV = NkT → N/V = p/(kT) = (2.5×10¹⁶)/(1.38×10⁻²³ × 1.5×10⁷) = (2.5×10¹⁶)/(2.07×10⁻¹⁶) = 1.21×10³² particles m⁻³. Mass density: ρ = (N/V) × m_proton = 1.21×10³² × 1.67×10⁻²⁷ = 2.02×10⁵ kg m⁻³ ≈ 2×10⁵ kg m⁻³. This is about 200 times denser than water! The solar core is extremely compressed. Note: real solar plasma does not behave ideally at these extremes (degeneracy pressure and radiation pressure are also significant), but this order-of-magnitude estimate is remarkably close to accepted values (~150 g cm⁻³ = 1.5×10⁵ kg m⁻³).