Ideal Gas Law Calculator

R = 8.314462618 J·mol⁻¹·K⁻¹. Since the 2019 SI redefinition that figure is exact rather than measured, because R is the product of two constants that are now fixed by definition: the Avogadro constant, 6.02214076×10²³ mol⁻¹, and the Boltzmann constant, 1.380649×10⁻²³ J·K⁻¹. The litre-atmosphere form you see in chemistry classes, 0.082057366 L·atm·mol⁻¹·K⁻¹, is that same number divided by 101.325 — one litre-atmosphere is exactly 101.325 joules, because an atmosphere is defined as 101325 pascals and a litre as a thousandth of a cubic metre.

That is the molar volume of an ideal gas at STP as most textbooks quote it. Putting P = 1 atm and T = 273.15 K into V = RT/P gives 22.413969545 L per mole, which rounds to 22.414. The defaults here are that whole state — 1 atm, 22.414 L, 1 mol, 273.15 K — so whichever quantity you solve for, the answer comes back as the value you started from, to within the 1.4 parts per million the rounding costs.

The arithmetic does; your typing does not. PV = nRT needs an absolute temperature, so 20 °C has to become 293.15 K before it is used — halving the kelvin temperature halves the pressure, while halving the Celsius reading means nothing at all. Pick °C or °F in the selector and the conversion happens for you, including on the value already in the box. Anything at or below absolute zero is refused rather than answered.

Absolute, measured from a vacuum. A tyre gauge, a manometer and most cylinder regulators read the difference between the gas and the surrounding air, so a gauge showing 32 psi is really about 46.7 psi absolute. Add roughly 14.7 psi, 101.3 kPa, 1.013 bar or 1 atm to a gauge reading before entering it, or the answer will be badly wrong at low pressures and slightly wrong at high ones.

It assumes molecules take up no space and do not attract each other, which is close enough for most gases near room temperature and around one atmosphere — typically within a percent or so. It degrades as you compress a gas, cool it towards its boiling point, or work with strongly polar molecules such as water vapour and ammonia. The usual next steps are the van der Waals equation, which adds terms for molecular volume and attraction, or a tabulated compressibility factor Z in PV = ZnRT.

Divide the mass by the molar mass of the substance: 8.00 g of oxygen gas, whose molar mass is 32.00 g/mol, is 0.250 mol. Molar masses come from the periodic table, summed over the formula — O₂ is two oxygens at 16.00, CO₂ is 12.01 + 2 × 16.00 = 44.01 g/mol. Going the other way, an amount solved here becomes a mass by multiplying by the molar mass, which is how a gas law problem turns into a weighable quantity.