Gas Volume at STP

Also known as 22.4 L per mole · molar volume at STP

V=n VmV = n\,V_m

Worked example: 1 mol ideal gas at STP → 22.414 L — press Try an example to run it live, then adjust anything.

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Gas Volume at STP explained

Vn

Fix the temperature and pressure and a mole of any ideal gas occupies the same volume — so converting between moles and litres needs one constant and no information about the substance. This is Avogadro's principle, and it deserves a moment of surprise before it becomes routine. A mole of hydrogen weighs 2 g and a mole of sulfur hexafluoride weighs 146 g, seventy-three times more, yet at the same conditions they fill the same flask. The reason is that pressure comes from the number of impacts and their momentum, and at a common temperature the heavier molecules move proportionally slower. The mass cancels out of everything the container can feel.

This page uses classic STP, 0 °C and 1 atm, where Vm=22.414V_m = 22.414 L/mol. Burning one mole of methane, CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}, consumes 2 mol of oxygen and produces 1 mol of carbon dioxide — which is 44.83 L of O₂ in and 22.41 L of CO₂ out at STP. Note that the two moles of water do not appear in that volume tally, because at 0 °C the water is a liquid and the equation only counts gases.

The constant is not independent; it is PV=nRTPV = nRT evaluated once. Vm=RT/P=(8.314×273.15)/101 325=0.022414V_m = RT/P = (8.314 \times 273.15)/101\,325 = 0.022414 m³/mol, and every other molar volume in circulation is the same calculation at different conditions. Historically the logic ran the other way. Gay-Lussac reported in 1808 that gases combine in simple whole-number volume ratios — two volumes of hydrogen to one of oxygen — and Avogadro's 1811 hypothesis explained why: equal volumes hold equal counts, so the volume ratios are the mole ratios of the balanced equation. That insight was ignored for half a century before Cannizzaro revived it at Karlsruhe in 1860.

The error that dominates this page is using 22.4 L/mol for conditions that are not STP. Room temperature is not 0 °C. At 25 °C and 1 atm the molar volume is 24.47 L/mol, so applying 22.4 to a bench-top measurement understates the volume by 9%. And "standard conditions" is not one definition but several: classic STP gives 22.414 L/mol, IUPAC's post-1982 STP (0 °C, 100 kPa) gives 22.711, and SATP (25 °C, 100 kPa) gives 24.79. The spread between the smallest and largest is over 10%, which is far more than the precision most people think they are carrying. Check which convention a source assumes before you borrow its number, and when the conditions are anything other than a listed standard, abandon the shortcut and use the ideal gas law directly.

Two smaller cautions. This is an ideal-gas result, so it degrades for gases near their condensation point — ammonia and sulfur dioxide at STP already deviate by a percent or two, and water vapour at 0 °C is not a gas at all. And when you use volume ratios as mole ratios in a reaction, they apply only to the species that are actually gaseous at the stated conditions. A dissolved or condensed product contributes no volume, and counting it is the quiet way to get a stoichiometry problem wrong.

Gas Volume at STP formula

V=n VmV = n\,V_m
Where
  • VV= Gas volume at STP (L)
  • nn= Amount of gas (mol)

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