1 mol at 0 °C and 1 atm
- Solve for
- Volume (V)
- Pressure
- 1
- Amount of gas
- 1
- Temperature
- 273.15
Result Result: 22.414
Unit of the result: L
Pressure in kPa: 101.325
Temperature in °C: 0
Molar volume (L/mol): 22.414
- R = 0.082057366 L·atm/(mol·K)
- V = nRT / P = 1 × 0.082057366 × 273.15 / 1
- Volume = 22.414 L
The classic molar volume: one mole of an ideal gas at 273.15 K and 1 atm fills about 22.414 L.