how-to · 5 min read
Molar volume and the ideal gas law, with a worked example
Molar volume is the volume one mole of gas occupies at a given temperature and pressure. For an ideal gas it depends only on those two conditions, not on which gas is present, so a single number can connect moles, mass and gas volume.
This guide explains where the familiar 22.4 L/mol at STP comes from, how it differs from IUPAC standard conditions and SATP, and how to use PV = nRT with a worked example. It also covers the unit conversions that trip up hand calculations and the conditions under which real gases stop following the ideal model.
What molar volume means
Molar volume is volume per mole: Vm = V / n. For an ideal gas, substituting into PV = nRT gives Vm = RT / P, so the molar volume depends only on temperature and pressure. At 273.15 K and 101.325 kPa the value is 22.414 L/mol; at 273.15 K and 100 kPa it is 22.71 L/mol; at 298.15 K and 100 kPa it is 24.79 L/mol.
Because the identity of the gas cancels out of the ideal gas law, equal volumes of two different gases at the same temperature and pressure contain equal numbers of particles. That is the basis of gas stoichiometry: volume ratios of gases at the same conditions follow the coefficients in a balanced equation.
STP, SATP and the 22.4 L value
The familiar 22.4 L/mol comes from the older definition of STP, 273.15 K and 1 atm (101.325 kPa). IUPAC later moved standard pressure to 100 kPa, which gives 22.71 L/mol at the same temperature, while SATP uses 298.15 K and 100 kPa and gives 24.79 L/mol. The temperature and pressure must both be stated, because the two definitions differ by about 1.3% in volume.
In course work the number 22.4 is often rounded from 22.414 and used as a shortcut for moles at STP. The shortcut is only valid at the stated conditions; at room temperature the molar volume is closer to 24.8 L/mol, so using 22.4 there overestimates the amount of gas by roughly a tenth.
Using PV = nRT: a worked example
How much volume does 0.5 mol of gas occupy at 298.15 K and 100 kPa? Convert pressure to pascals and use kelvin directly: V = nRT / P = (0.5 × 8.314462618 × 298.15) / 100000. The numerator is about 1239.5, so V = 0.012395 m³, which is 12.39 L.
The same answer follows from the molar volume: 0.5 mol × 24.79 L/mol = 12.39 L. The two routes agree because the molar volume is just RT/P for those conditions. When the gas amount is given as a mass, convert to moles first with the formula molar mass, then apply the same equation.
The equation works in reverse as well: a measured volume gives the amount of substance through n = PV/RT. The 12.39 L in the example converts back to 0.012395 m³, and 0.012395 × 100000 / (8.314462618 × 298.15) returns approximately 0.5 mol, which is a useful check that the unit conversions in both directions are consistent.
Converting units before you calculate
PV = nRT is only consistent in SI units: pascals, cubic meters and kelvin. A volume in liters must be divided by 1000, a pressure in kPa multiplied by 1000, and a Celsius temperature increased by 273.15. Doing the conversions first, and writing them down, avoids the most common arithmetic errors.
When reading a result, convert back to the unit that makes sense for the problem: gas volumes are usually reported in liters, while engineering calculations often use cubic meters. The calculator on this site performs these conversions for L, mL and m³ and for kPa, atm, bar, mmHg and Pa, so the working shows every step.
Where the ideal gas model stops working
The ideal gas law treats molecules as point particles with no forces between them. Real gases have finite molecular volume and attractive forces, so they occupy less or more volume than predicted depending on the conditions. The deviation becomes noticeable at high pressure, at low temperature and close to the boiling point where condensation begins.
The compressibility factor Z = PV/(nRT) measures the deviation: Z is close to 1 for an ideal gas and moves away from 1 as non-ideal behavior grows. For careful work near condensation, an equation of state or tabulated Z values should replace PV = nRT. For gases well above their boiling point at ordinary pressures, the ideal model is usually good enough.