IB Chemistry SL
Topic 1 — The Behaviour of Ideal Gases
Paper 1 & 2
Core skill
~11 min read
The Ideal Gas Equation
One equation ties together everything about a gas — its pressure, volume, temperature and amount: PV = nRT. Get comfortable rearranging it and watching your units, and you can find any one of those quantities from the others.
📘 What you need to know
- The ideal gas equation is PV = nRT.
- P = pressure in pascals (Pa), V = volume in m3, n = moles, R = 8.31 J K-1 mol-1, T = temperature in Kelvin.
- Convert first: kPa → Pa (×1000), cm3/dm3 → m3, °C → K (+273).
- For a fixed amount of gas, P1V1/T1 = P2V2/T2.
- Combined with moles = mass ÷ M, PV = nRT can also give the molar mass of a gas.
The equation and its units
The ideal gas equation links all four gas quantities in one line:
Ideal gas equation
PV = nRT
The catch is units — the equation only works if every quantity is in its correct SI unit:
- P — pressure in pascals (Pa)
- V — volume in cubic metres (m3)
- n — amount in moles
- R — the gas constant, 8.31 J K-1 mol-1
- T — temperature in Kelvin
💡 Watch your units
- 1 m3 = 1,000,000 cm3 — to convert cm3 → m3, divide by 106; dm3 → m3, divide by 1000.
- kPa → Pa: multiply by 1000. °C → K: add 273.
- Unit slips are the single biggest source of lost marks in gas problems.
WORKED EXAMPLECalculate the volume, in dm3, occupied by 0.781 mol of oxygen at 220 kPa and 21 °C.
Rearrange: V = nRT ÷ P
P = 220 000 Pa, T = 294 K, R = 8.31.
V = (0.781 × 8.31 × 294) ÷ 220 000
= 0.00867 m³
= 8.67 dm³
Changing conditions of a fixed gas
If the amount of gas doesn’t change, n and R stay constant, and the equation simplifies to a handy before-and-after form:
Fixed amount of gas
P1V1 ÷ T1 = P2V2 ÷ T2
Here 1 = initial conditions, 2 = final conditions, and T is always in Kelvin. A neat feature: because you’re taking a ratio, volumes just need to be in the same unit — they don’t have to be converted to m3.
WORKED EXAMPLEAt 25 °C and 100 kPa a gas occupies 20 dm3. Find its new temperature, in °C, if the volume drops to 10 dm3 at constant pressure.
Pressure constant, so P cancels: T₂ = V₂T₁ ÷ V₁
V₁ = 20, V₂ = 10, T₁ = 25 + 273 = 298 K.
T₂ = (10 × 298) ÷ 20 = 149 K
= −124 °C
WORKED EXAMPLEA 2.00 dm3 container of oxygen at 80 kPa is heated from 20 °C to 70 °C, and the volume expands to 2.25 dm3. Find the final pressure.
Rearrange: P₂ = P₁V₁T₂ ÷ (V₂T₁)
T₁ = 293 K, T₂ = 343 K.
P₂ = (80 × 2.00 × 343) ÷ (2.25 × 293)
= 83 kPa
Finding molar mass
Because moles = mass ÷ molar mass, you can combine that with PV = nRT to find the molar mass (M) of an unknown gas: find n from the gas equation, then divide the measured mass by n.
WORKED EXAMPLEA 1000 cm3 flask holds 6.39 g of a gas at 300 kPa and 23 °C. Find the molar mass.
Step 1 — find moles: n = PV ÷ RT
P = 300 000 Pa, V = 1.0 × 10⁻³ m³, T = 296 K.
n = (300 000 × 1.0 × 10⁻³) ÷ (8.31 × 296) = 0.12 mol
Step 2 — molar mass = mass ÷ moles
6.39 ÷ 0.12
M = 53 g mol⁻¹
⚠️ Common mix-up
- For PV = nRT you must convert volume to m3 and pressure to Pa. But for the ratio form (P₁V₁/T₁ = P₂V₂/T₂) you only need matching units — so dm³ is fine there.
- Temperature is always in Kelvin, in both forms.
Up next: Real Gas Behaviour — why actual gases stray from PV = nRT, and the conditions where the ideal model breaks down.
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