One equation covers every gas. Pressure times volume equals moles times temperature times a constant that is the same for oxygen, helium and steam alike — which is remarkable, and is what makes gas calculations so much simpler than they have any right to be.
The symbols, and the units that go with them
| Symbol | Meaning | Unit used here |
|---|---|---|
| P | Pressure | kPa |
| V | Volume | L, the same as dm³ |
| n | Amount of gas | mol |
| R | The gas constant, 8.314 | J K⁻¹ mol⁻¹ |
| T | Absolute temperature | K |
These four units work together with R = 8.314 and need no conversion, because 1 kPa × 1 L is exactly 1 J. That is not a coincidence worth memorising so much as a convenience worth using.
Temperature must be in kelvin. Not Celsius, ever. Add 273.15 to a Celsius reading first. The reason is that the law says volume is proportional to temperature, and a proportionality only makes sense from a true zero — 0 °C is not “no thermal energy”, it is just the freezing point of water.
What ‘ideal’ is assuming
The law treats gas particles as having no volume of their own and no attraction to one another. Neither is true, but both are close enough to true under ordinary conditions that the equation works well.
It starts to drift when:
The pressure is high. Squeeze a gas hard and the particles’ own volume stops being negligible compared with the container.
The temperature is low. Chill a gas towards its boiling point and the attractions between particles start to matter — which is exactly why gases can be liquefied at all, something an ideal gas could never do.
At room temperature and around atmospheric pressure the error is typically well under one per cent, which is why you can use it freely on exam questions without a second thought.
A worked example
What volume does 2.50 mol of nitrogen occupy at 25 °C and 101.325 kPa?
Convert the temperature first, before anything else:
T = 25 + 273.15 = 298.15 K
Rearrange for volume:
V = nRT ⁄ P = (2.50 × 8.314 × 298.15) ⁄ 101.325
V = 6196.0 ⁄ 101.325 = 61.2 L
Check it for sense. One mole of any gas occupies about 24 L at room temperature and pressure, so 2.50 mol should be near 60 L. 61.2 L is right where it should be, and that estimate would have caught a Celsius-for-kelvin error immediately — using 25 K instead of 298.15 K would have given 5.1 L.
Notice that nitrogen never entered the calculation. The law does not care which gas it is, which is the whole point of it.
Where marks get lost
Celsius instead of kelvin. The single most common error on this topic, and it produces an answer that is wrong by a factor of ten or more without looking obviously absurd. Convert first, every time.
Mismatched pressure and volume units. If P is in pascals, V must be in cubic metres. If P is in kPa, V is in litres. Mixing atmospheres with litres needs R = 0.08206 instead of 8.314, and using the wrong R is a silent error.
Using mass instead of moles. n is an amount, not a mass. Convert with n = m ⁄ M first — which is why that formula turns up inside so many gas questions.
Forgetting that a diatomic gas has a doubled molar mass. Nitrogen gas is N₂ at 28.02 g/mol, not 14.01.
Applying it to a liquid or a solid. It is a gas law. The moment a substance condenses it stops obeying it entirely.
Questions students actually ask
Why must temperature be in kelvin?
Because the law makes volume proportional to temperature, and a proportion needs a genuine zero. The kelvin scale starts at absolute zero; the Celsius scale starts at the freezing point of water, which is an arbitrary place to begin.
Which value of R should I use?
It depends on your units. Use 8.314 J K⁻¹ mol⁻¹ with pressure in kPa and volume in litres, or with pascals and cubic metres. Use 0.08206 L atm K⁻¹ mol⁻¹ if the pressure is in atmospheres.
What is molar volume, and where does 22.4 L come from?
It is the volume one mole of gas occupies. At standard temperature and pressure — 0 °C and 100 kPa — that comes to about 22.7 L; the older 22.4 L figure uses 101.325 kPa. At room temperature it is nearer 24 L.
When does the ideal gas law stop working?
At high pressures and low temperatures, where the particles’ own volume and the attractions between them stop being negligible. The van der Waals equation adds correction terms for both.
How do I use it when a gas changes conditions?
If the amount of gas is fixed, PV⁄T stays constant, so P₁V₁⁄T₁ = P₂V₂⁄T₂. That combined form saves you working out n at all when a gas is compressed or heated.
Related formulas
Moles, mass and molar mass · The dilution formula · The equations of motion
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