How to use this calculator
- Enter or choose solve for.
- Enter or choose pressure unit.
- Enter or choose volume unit.
- Enter or choose temperature unit.
- The result updates as you type. Use Copy result or Share to keep it.
Method
P V = n R T; R = 8.314462618 J/(mol·K); T(K) = T(°C) + 273.15
- P
- absolute pressure in Pa
- V
- volume in m³
- n
- amount of substance in mol
- R
- molar gas constant, 8.314462618… J/(mol·K), exact in the 2019 SI (N_A × k)
- T
- absolute temperature in kelvin
Worked example
- T = 0 + 273.15 = 273.15 K; P = 101,325 Pa.
- V = n R T ÷ P = 1 × 8.314462618 × 273.15 ÷ 101,325 = 0.0224140 m³.
- That is 22.414 L, matching the CODATA molar volume of an ideal gas at 0 °C and 1 atm (22.41397 L/mol).
One mole of an ideal gas at 0 °C and 1 atm occupies 22.414 litres.
What the result means
Real gases follow the ideal law closely at low pressure and well above their boiling point; air at room conditions deviates by well under 1%. Always use absolute pressure and absolute temperature.
Assumptions and limits
- Ideal gas: no intermolecular forces and negligible molecular volume.
- At high pressure or near condensation, use a real-gas equation (van der Waals, Peng–Robinson) or a compressibility factor Z.
- Gauge pressure readings must have atmospheric pressure added first.
Questions
Why must temperature be in kelvin?
The law is proportional to absolute temperature. Doubling from 10 °C to 20 °C is only a 3.5% rise in kelvin (283.15 to 293.15 K).
What value of R should I use?
R = 8.314462618 J/(mol·K), which equals 8.314462618 kPa·L/(mol·K) or 0.0820574 L·atm/(mol·K). This calculator converts units for you.