Select the law, choose which variable to solve for, and enter the three known values.
Using R = 8.314 J/mol/K. Enter values in SI units: P in Pa, V in m³, T in K (or use °C — the calculator converts automatically).
P₁V₁ = P₂V₂ at constant temperature and fixed moles. Units for P and V must be consistent.
V₁/T₁ = V₂/T₂ at constant pressure and fixed moles. Temperature must be in Kelvin.
P₁V₁/T₁ = P₂V₂/T₂ — Combined gas law. Enter 5 known values to find the 6th. Temperature in K.
The ideal gas law combines Boyle's Law, Charles' Law, and Avogadro's Law into one equation that describes the state of any ideal gas:
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Pressure | Pa (Pascals) |
| V | Volume | m³ |
| n | Amount of substance | mol |
| R | Gas constant | 8.314 J/mol/K |
| T | Temperature | K (Kelvin) |
n = 2 mol, T = 25 + 273 = 298 K, P = 100,000 Pa
V = nRT/P = (2 × 8.314 × 298) / 100,000
V = 4955.2 / 100,000 = 0.04955 m³
V = 49.55 dm³ (litres)
P = 101,325 Pa, V = 0.0224 m³ (22.4 L), T = 273 K
n = PV/RT = (101325 × 0.0224)/(8.314 × 273)
n = 2269.7 / 2269.7 = 1.000 mol
At constant temperature and constant amount of gas, pressure and volume are inversely proportional.
A gas at 200 kPa occupies 3.0 dm³. What is the volume at 500 kPa?
V₂ = P₁V₁/P₂ = (200 × 3.0)/500 = 1.20 dm³
At constant pressure and constant amount of gas, volume is directly proportional to absolute temperature (in Kelvin).
A gas occupies 5.0 dm³ at 27°C (300 K). What volume at 127°C (400 K)?
V₂ = V₁ × T₂/T₁ = 5.0 × 400/300 = 6.67 dm³
Temperature in Charles' Law must be in Kelvin. Using Celsius will give completely wrong answers. Always add 273.15 (or 273) before calculating.
| Condition | Temperature | Pressure | Molar Volume |
|---|---|---|---|
| STP (IUPAC 1982) | 0°C (273 K) | 1 atm (101,325 Pa) | 22.414 L/mol |
| STP (IUPAC 2013) | 0°C (273 K) | 1 bar (100,000 Pa) | 22.711 L/mol |
| RTP (A-Level UK) | 25°C (298 K) | 1 atm (101,325 Pa) | 24.465 L/mol |
| RTP (exam standard) | 25°C (298 K) | ~101 kPa | 24.0 dm³/mol |
For A-Level examinations in the UK, you will most commonly use 24.0 dm³/mol at RTP unless otherwise stated in the question.
An ideal gas has no intermolecular forces and molecules with negligible volume. Real gases deviate from this at:
where a accounts for intermolecular forces and b accounts for molecular volume. Each gas has unique constants a and b.
The ideal gas law is PV = nRT, where P is pressure in Pa, V is volume in m³, n is the amount in moles, R is the universal gas constant (8.314 J/mol/K), and T is temperature in Kelvin. It describes the behaviour of an ideal gas where molecules have no intermolecular forces and negligible volume compared to the container.
For PV = nRT using R = 8.314 J/mol/K, pressure must be in Pa (Pascals), volume must be in m³, and temperature must be in Kelvin (K = °C + 273.15). Remember: 1 dm³ = 1 L = 0.001 m³; 1 atm = 101,325 Pa; 1 bar = 100,000 Pa. Always check units before calculating.
Boyle's Law states that for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional: P₁V₁ = P₂V₂. If the pressure doubles, the volume halves. The relationship holds because compressing a gas forces molecules into a smaller space, increasing the frequency of collisions with the container walls.
Charles' Law states that for a fixed amount of gas at constant pressure, volume is directly proportional to absolute temperature: V₁/T₁ = V₂/T₂. Temperature must be in Kelvin — never Celsius. If the absolute temperature doubles, the volume doubles. This occurs because higher temperature gives molecules more kinetic energy, requiring a larger volume to maintain the same pressure.
The molar volume is the volume occupied by one mole of any ideal gas at a specified temperature and pressure. At STP (0°C, 101,325 Pa), it is 22.4 L/mol. At RTP (25°C, ~101 kPa), it is approximately 24.5 L/mol. UK A-Level examinations typically use 24.0 dm³/mol at RTP for calculations unless stated otherwise.
Real gases deviate from ideal behaviour at high pressures (molecular volume becomes significant — real gas volume exceeds ideal prediction) and low temperatures (intermolecular forces dominate — the gas may condense). Noble gases such as helium and neon behave most like ideal gases due to their very weak intermolecular forces. Gases with strong intermolecular forces (e.g. NH₃, CO₂) show the greatest deviation.
The Van der Waals equation modifies PV = nRT to account for real gas behaviour: (P + a/Vm²)(Vm − b) = RT. The constant a corrects for intermolecular attractions (reducing effective pressure) and b corrects for the finite volume of molecules. Larger values of a indicate stronger intermolecular forces; larger values of b indicate larger molecules.