The big idea: You can't weigh a gas easily, but you can measure its volume. Two tools turn a gas volume into an amount in moles.
- The ideal gas equation, PV = nRT, works at any pressure and temperature. - The molar volume, V_{m} = 22.7 dm³ mol⁻¹, is a shortcut that works only at STP (standard temperature and pressure).
What STP means: STP = standard temperature and pressure = 273 K (0 °C) and 100 kPa.
At STP, one mole of any ideal gas takes up the same volume: 22.7 dm³. So at STP you don't need PV = nRT — just divide the volume by 22.7.
At STP, the identity of the gas doesn't matter — equal volumes hold equal numbers of moles. So the amount of a gas is simply its volume divided by the molar volume, Vm = 22.7 dm³ mol⁻¹.
- amount of gas (mol)
- volume of gas at STP (dm³)
- molar volume at STP = 22.7 dm³ mol⁻¹
Watch the volume unit: Vm is in dm³ mol⁻¹, so the volume must be in dm³.
- 1 dm³ = 1000 cm³, so divide cm³ by 1000 to get dm³. - 1 dm³ = 1 litre, so 22.7 dm³ = 22.7 L.
Worked example — volume of gas at STP
Calculate the volume, at STP, of 0.150 mol of carbon dioxide, CO2.
Solution
- Formula first — rearrange n = V/Vm for the volume:
- Substitute (Vm = 22.7 dm³ mol⁻¹ at STP):
- Work it out — keep the unit:
Final answer
V = 3.41 dm³ (at STP).
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Access past paper questions with model answers. Learn exactly what earns marks and what doesn't.
When conditions are not STP, use the full ideal gas equation. It links pressure, volume, amount and temperature. Rearrange it for whatever you are asked to find.
- pressure (Pa)
- volume (m³)
- amount of gas (mol)
- gas constant = 8.31 J K⁻¹ mol⁻¹
- temperature (K)
SI units — the make-or-break step: PV = nRT only works with SI units, because R = 8.31 J K⁻¹ mol⁻¹ is in SI.
- P in pascals (Pa) — multiply kPa by 1000. - V in cubic metres (m³) — divide dm³ by 1000, or cm³ by 10⁶. - T in kelvin (K) — add 273 to °C.
Get a unit wrong and the answer is off by a factor of 1000 or more.
Worked example — moles from P, V and T
A flask of volume 2.00 dm³ holds nitrogen gas at a pressure of 120 kPa and a temperature of 25 °C. Calculate the amount, in moles, of nitrogen. (R = 8.31 J K⁻¹ mol⁻¹.)
Solution
- Convert to SI units first. Pressure: 120 kPa → Pa:
- Volume: 2.00 dm³ → m³ (÷1000):
- Temperature: 25 °C → K (+273):
- Formula first — rearrange PV = nRT for n:
- Substitute and solve:
- Work it out:
Final answer
n = 0.0969 mol (≈ 0.097 mol).
How this is tested: S1.5 gas calculations turn up as a Paper 1A one-step MCQ and as a marked Paper 2 calculation.
The two recurring asks are: 'find the volume of gas at STP' (use Vm = 22.7) and 'find the moles of a vaporized compound from P, V and T' (use PV = nRT). Often the answer then feeds into a molar-mass step.
The marks live in unit conversion — Pa, m³ and K for PV = nRT; dm³ for Vm.
Three easy marks: (1) Decide first: STP given → Vm; otherwise → PV = nRT. (2) Convert every value to the right unit before substituting. (3) Write the rearranged formula before the numbers, and carry the unit.
IB-style question — volume of hydrogen at STP
Zinc reacts with excess hydrochloric acid to produce 0.0800 mol of hydrogen gas, H2. Calculate the volume of hydrogen produced at STP. [1]
Solution
- STP is stated, so use the molar volume — formula first:
- Substitute (Vm = 22.7 dm³ mol⁻¹):
- Work it out:
Final answer
V = 1.82 dm³ at STP.
IB-style question — molar mass via the gas equation
When 0.512 g of a volatile liquid is fully vaporized, it occupies 195 cm³ at a pressure of 99.0 kPa and a temperature of 90.0 °C. Calculate the amount, in moles, of vapour, then its molar mass. (R = 8.31 J K⁻¹ mol⁻¹.) [2]
Solution
- Convert to SI units. P = 99.0 kPa, V = 195 cm³, T = 90.0 °C:
- Formula first — rearrange PV = nRT for n:
- Amount of vapour:
- Now use M = m/n to get the molar mass:
- Work it out:
Final answer
n = 6.40 × 10⁻³ mol; M = 80.0 g mol⁻¹.