Ideal Gas Law Calculator
Solve for
Absolute pressure. A tire or tank gauge reads above the atmosphere: add 1 atm (14.7 psi, 101.3 kPa).
Try an example
Volume
22.41 L
22.41 L · 0.02241 m³ · 0.7915 ft³
V = nRT / P = 1 mol × 8.3145 × 273.15 K ÷ 101,325 Pa = 0.02241 m³. R = 8.3145 J/(mol·K), with everything in SI units.
For air that is 28.97 g of gas, at a density of 1.292 g/L.
Solve the ideal gas law PV = nRT for pressure, volume, moles or temperature in the units you have, from atm, psi and kPa to liters, cubic feet, °C and °F, with the mass and density of common gases and the gauge reading for tires and tanks.
Formula
Worked Example
P = nRT ÷ V = 2 × 8.3145 × 298.15 ÷ 0.015 m³ = 330,500 Pa. A gauge reads pressure above the atmosphere, so it would show 47.94 − 14.70 = 33.24 psi.
Frequently Asked Questions
Which value of R should I use?
The one that matches your units: 8.314 J/(mol·K) with pascals and cubic meters, 0.08206 L·atm/(mol·K) with atmospheres and liters, 8.314 L·kPa/(mol·K) with kilopascals and liters, or 62.36 L·Torr/(mol·K) with torr. This calculator converts everything to SI first, so any input units work.
Why must the temperature be in kelvin?
The law needs absolute temperature, which starts at absolute zero. In Celsius, 20 °C is not twice as hot as 10 °C, and 0 °C would give zero volume. Add 273.15 to Celsius; for Fahrenheit, subtract 32, multiply by 5/9 and add 273.15.
What volume does one mole of gas take up?
22.414 L at 0 °C and 1 atm, 22.711 L at the IUPAC standard of 0 °C and 100 kPa, and 24.465 L at 25 °C and 1 atm. The figure is the same for any ideal gas, whatever its molar mass.