IB Chemistry HLTopic 1 — The Behaviour of Ideal GasesPaper 1 & 2Core skill~11 min read
The Ideal Gas Equation
One equation replaces all three gas laws, works at any conditions, and tells you how much gas you have. The chemistry is not hard. What catches almost everybody is the units — this equation is fussier about them than anything else in the course.
📘 What you need to know
PV = nRT links pressure, volume, amount and temperature for an ideal gas.
P in pascals (Pa), V in cubic metres (m3), n in mol, T in kelvin (K).
R = 8.31 J K−1 mol−1, the gas constant. It is in the data booklet, along with the equation.
Conversions you will need every time: kPa × 1000 = Pa, dm3 ÷ 1000 = m3, cm3 ÷ 1 000 000 = m3, °C + 273 = K.
Rearranged: V = nRT/P, P = nRT/V, n = PV/RT, T = PV/nR.
Combine it with n = m/M to find the molar mass of a gas from a measured mass, volume, pressure and temperature.
Where the equation comes from
You already have the three pieces. Boyle’s law says PV is constant, Charles’s law says V/T is constant, and the pressure law says P/T is constant. Put them together and the combination that stays fixed is PV/T.
Now ask what that constant depends on. Double the amount of gas in the same container at the same temperature and the pressure doubles, so the constant is proportional to n. Pull the amount out as a separate factor and whatever is left over is the same for every gas — that leftover is R.
The ideal gas equationPV = nRT
The fact that R is the same number for helium, chlorine and steam is the real headline here. It only works because the ideal gas model deliberately ignores everything that makes those gases different from one another.
The units it demands
This is where the marks go. The value 8.31 is quoted in joules, and a joule is built from pascals and cubic metres. Feed the equation kilopascals or cubic decimetres and the answer will be out by a factor of a thousand or a million.
If an answer is out by exactly 1000 or 1 000 000, you have a units problem, not a chemistry problem. Check the volume first.
Symbol
Quantity
Unit required
The trap
P
pressure
Pa
questions give kPa — multiply by 1000
V
volume
m3
questions give dm3 or cm3 — divide by 1000 or 1 000 000
n
amount of gas
mol
questions give a mass — convert with n = m/M first
R
gas constant
J K−1 mol−1
none — the value 8.31 is given in the data booklet
T
temperature
K
questions give °C — add 273
Two steps down the ladder, not one. Going straight from cm3 to m3 means dividing by a million, which is where the biggest errors come from.
Rearranging it
Only four rearrangements exist, and each one answers a different question.
The four formsV = nRT ÷ PP = nRT ÷ Vn = PV ÷ RTT = PV ÷ nR
🧩 The method that works every time
Write down what you are given in a short list, one line per quantity, with the unit you were given.
Convert every line to Pa, m3, mol and K. Write the converted value next to the original so you can check it.
Rearrange the equation for the unknown before you touch the calculator.
Substitute and evaluate, keeping the full calculator value.
Convert the answer back if the question asked for dm3 or kPa or °C, then round.
Finding the molar mass of a gas
Here is the trick that turns this into an identification tool. If you weigh a gas sample as well as measuring it, you have both the mass and the number of moles, and molar mass follows.
Molar mass from the ideal gas equationM = m ÷ n where n = PV ÷ RT
Do it in two steps, not one. You can write M = mRT/PV in a single line, but finding n first and then dividing is far easier to check and earns method marks even if the arithmetic slips.
Worked examples
WORKED EXAMPLE
Finding a volume
Calculate the volume, in dm3, occupied by 0.500 mol of nitrogen at 150 kPa and 30 °C.
Step 1: convert everythingP = 150 kPa = 150 000 Pan = 0.500 mol R = 8.31T = 30 + 273 = 303 KStep 2: rearrange for VV = nRT ÷ PStep 3: substituteV = (0.500 × 8.31 × 303) ÷ 150 000V = 1258.97 ÷ 150 000 = 0.0083931 m³Step 4: convert back to dm³V = 8.39 dm³an answer in m³ is nearly always a small decimal — a good sign you converted correctly
WORKED EXAMPLE
Finding a pressure
0.150 mol of argon is sealed in a rigid 2.50 dm3 vessel at 45 °C. Calculate the pressure inside, in kPa.
Step 1: convertV = 2.50 dm³ = 2.50 × 10⁻³ m³T = 45 + 273 = 318 KStep 2: rearrange for PP = nRT ÷ VStep 3: substituteP = (0.150 × 8.31 × 318) ÷ (2.50 × 10⁻³)P = 396.39 ÷ 0.00250 = 158 555 PaStep 4: answer in kPaP = 159 kPadivide by 1000 at the end — the question asked for kPa, not Pa
WORKED EXAMPLE
Finding a temperature
0.0800 mol of a gas occupies 2.00 dm3 at a pressure of 120 kPa. Calculate the temperature in °C.
Step 1: convertP = 120 000 Pa V = 2.00 × 10⁻³ m³Step 2: rearrange for TT = PV ÷ (nR)Step 3: substituteT = (120 000 × 2.00 × 10⁻³) ÷ (0.0800 × 8.31)T = 240 ÷ 0.6648 = 361.0 KStep 4: back to Celsius361.0 − 273 = 88.0T = 88.0 °Cthe equation always gives kelvin — subtract 273 only at the very end
WORKED EXAMPLE
Finding the molar mass of an unknown gas
A 500 cm3 flask is filled with an unknown gas and found to contain 1.15 g of it at 101 kPa and 25 °C. Calculate the molar mass of the gas.
Step 1: convert, watching the cm³P = 101 000 PaV = 500 cm³ = 5.00 × 10⁻⁴ m³T = 25 + 273 = 298 KStep 2: find the amount in molesn = PV ÷ (RT)n = (101 000 × 5.00 × 10⁻⁴) ÷ (8.31 × 298)n = 50.5 ÷ 2476.4 = 0.02039 molStep 3: molar mass = mass ÷ molesM = 1.15 ÷ 0.02039 = 56.39M = 56.4 g mol⁻¹but-1-ene, C₄H₈, has M = 56.12 — a good match
💡 Exam tip
Do the unit conversions as a labelled list before anything else. Almost every lost mark on this topic is a units mark.
Remember that cm3 to m3 is two steps, a division by a million in total.
The equation is in the data booklet and so is R. Spend your memory on the units instead.
If a question gives a mass rather than an amount, convert with n = m/M before you start.
Convert the answer back into whatever the question asked for. A correct value in the wrong unit still loses the final mark.
Sanity check: at ordinary conditions one mole of gas is roughly 0.02 to 0.03 m3. If your volume is in the hundreds, something is wrong.
⚠ Common mix-up
Leaving the pressure in kPa. This makes the answer 1000 times out and is the most frequent error of all.
Using dm3 for the volume. Same problem, opposite direction. The equation only accepts m3.
Forgetting to convert °C to K. At low temperatures this can make the answer negative, which is a very visible giveaway.
Subtracting 273 too early. Do the whole calculation in kelvin, then convert at the end if asked.
Using the molar volume 22.7 dm3 at non-STP conditions. That is exactly what this equation is for.
Confusing n with mass.PV = nRT gives moles, never grams.
Up next: Real Gas Behaviour — what happens when the two fragile assumptions finally give way, and how to predict which gases go wrong first.
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