The big idea: A gas's pressure P, volume V and temperature T are all tied together by one equation — the ideal gas law.
It links them to how much gas there is, measured two ways: the amount in moles (n) or the raw number of molecules (N).
A mole is just a fixed-size 'pack' of particles — one mole = 6.02 × 10²³ of them (the Avogadro constant).
Spot it: Same P, V and T → same number of molecules N — no matter what the gas is.
So two equal boxes at the same conditions hold the same N, even if one gas is heavier; the heavier gas just has the bigger mass and density.
The ideal gas law is given in the data booklet in two equal forms — use moles n with R, or molecules N with the Boltzmann constant kB. They describe the same gas:
- pressure (Pa)
- volume (m³)
- amount of gas (mol)
- gas constant = 8.31 J K⁻¹ mol⁻¹ (given)
- absolute temperature (K — always kelvin)
- number of molecules (no unit — just a count)
- Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹ (given)
Two things to get right: 1. Temperature T is always in kelvin — add 273 to a Celsius value first.
2. Use n with R (8.31), or N with k_B (1.38 × 10⁻²³) — never mix them.
To swap between the amount in moles n and the raw number of molecules N, use the Avogadro constant NA. It's a simple quotient, so a formula triangle helps:
- amount of gas (mol)
- number of molecules (just a count)
- Avogadro constant = 6.02 × 10²³ mol⁻¹ (given)
[Diagram: phys-formula-triangle] - Available in full study mode
Worked example — number of molecules in a gas
A sealed flask holds gas at a pressure of 1.2 × 10⁵ Pa in a volume of 8.0 × 10⁻⁴ m³ at a temperature of 300 K. Find the number of gas molecules N. (kB = 1.38 × 10⁻²³ J K⁻¹.)
Solution
- Start with the given formula in its N-form:
- Rearrange for N:
- Put in the numbers (P = 1.2 × 10⁵, V = 8.0 × 10⁻⁴, T = 300):
- Work it out — N is just a count, no unit:
Final answer
N ≈ 2.3 × 10²² molecules. (That is about 0.039 mol, since n = N ÷ NA.)
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How this is tested: The ideal gas law is one of the most-tested ideas in Theme B.
- Paper 1A: quick MCQs comparing two samples — equal N → compare density/pressure; or which sample of equal P, V, T has the smallest mass. - Paper 1B / Paper 2: calculate the number of molecules or moles in a sample, or a ratio of amounts between two containers.
Classic trap: leaving the temperature in degrees Celsius — it must be in kelvin (add 273). Another: mixing n with k_B or N with R.
Comparing two samples: Write PV = NkT for each sample.
Whatever is the same (P, V or T) cancels when you divide one equation by the other, leaving a simple ratio of the rest. This is how every 'compare the two containers' question is solved.
IB-style question — ratio of amount of substance in two containers
Container X holds gas at pressure P, volume V and temperature 300 K. Container Y holds the same type of gas at the same pressure P, but volume 2V and temperature 400 K. Find the ratio of the amount of gas in Y to that in X (nY ÷ nX).
Solution
- Start with the given formula for each container (rearranged for n):
- Divide Y by X — the pressure P and R cancel (same in both):
- Put in the numbers (VY = 2V, VX = V, TX = 300, TY = 400):
- Work it out:
Final answer
nY ÷ nX = 1.5 — container Y holds 1.5 times as much gas as X.