What Avogadro's law says
In 1811 the Italian scientist Amedeo Avogadro made a bold guess. Take two gases at the same temperature and the same pressure. If their volumes are equal, they hold the same number of molecules.
It does not matter which gas it is. Tiny H₂ molecules and big CO₂ molecules count the same. Why? In a gas, molecules are very far apart. The space between them is much bigger than the molecules. So the size of a molecule hardly matters; only the number does.
Short form: V ∝ n (at fixed T and P). Here V is volume and n is the amount in moles. So V₁ ÷ n₁ = V₂ ÷ n₂.
Why it works
Pressure comes from molecules hitting the walls. At the same temperature, molecules of every gas have the same average kinetic energy. So each molecule gives, on average, the same push. Same push from each, same total pressure, same volume → same number of molecules.
Molar volume: 22.4 L at STP
Because one mole of every gas has the same number of molecules (6.02 × 10²³, the Avogadro constant), one mole of every gas fills the same volume at the same T and P. This is the molar volume, Vm.
- At STP (0 °C = 273 K and 101.3 kPa = 1 atm): Vm = 22.4 L/mol (= 22.4 dm³/mol).
- At room conditions (about 25 °C, 1 atm): Vm ≈ 24.5 L/mol (many books round to 24 L).
- IUPAC now uses 0 °C and 100 kPa for STP, giving 22.7 L/mol. Use the value your question gives.
Three key links
- n = V ÷ Vm (moles from volume)
- N = n × NA (number of molecules)
- m = n × M (mass from moles)
These let you jump between litres, moles, grams and molecules.
Relative density of gases
Equal volumes have equal numbers of molecules, but the molecules have different masses. So 22.4 L of H₂ weighs 2 g, of O₂ 32 g, of CO₂ 44 g.
The relative density of gas 1 compared with gas 2 tells how many times heavier gas 1 is at the same volume, T and P:
D = M₁ ÷ M₂ (no unit).
- By hydrogen: D(H₂) = M ÷ 2, so M = 2 × D(H₂).
- By air: D(air) = M ÷ 29 (air has an average molar mass of about 29 g/mol). If D(air) < 1 the gas rises in air (like helium, methane); if D > 1 it sinks (like CO₂).
You can also find the molar mass from density: M = ρ × Vm. For example, a gas with density 1.25 g/L at STP has M = 1.25 × 22.4 = 28 g/mol (N₂ or CO).
Volumes in gas reactions
Before Avogadro, Gay-Lussac saw that gases react in simple whole-number volume ratios. Avogadro's law explains it: equal volumes = equal numbers of molecules, so volume ratio = mole ratio = ratio of coefficients.
2H₂(g) + O₂(g) → 2H₂O(g): 2 volumes + 1 volume → 2 volumes.
N₂(g) + 3H₂(g) → 2NH₃(g): 1 L + 3 L → 2 L.
Remember: this works only for gases at the same T and P. Liquids and solids do not follow it.
Try it
Put a spoon of baking soda in a small bottle, add vinegar and quickly stretch a balloon over the mouth. CO₂ gas forms and the balloon grows. Use two spoons with more vinegar: more moles of CO₂, bigger balloon. In the 3D, step 6 lets you change moles and watch the volume.
Limits and exam focus
Avogadro's law is exact only for an ideal gas. Real gases follow it well at low pressure and high temperature. At very high pressure or very low temperature the molecules' own size and their pull on each other matter, so small errors appear.
Common exam questions: state the law; find volume at STP from mass; find moles or molecules from volume; find molar mass from relative density; find volumes of gases in a reaction. Always write units (L, mol, g) and check you used the right Vm.
Key formulas and definitions
- V ∝ n (same T and P)
- V₁ ÷ n₁ = V₂ ÷ n₂
- n = V ÷ Vm; Vm = 22.4 L/mol at STP
- N = n × Nₐ; Nₐ = 6.02 × 10²³ mol⁻¹
- m = n × M
- Relative density D = M₁ ÷ M₂
- M = 2 × D(H₂); M = 29 × D(air)
- M = ρ × Vm
Worked examples
1. A balloon holds 2 mol of gas and has volume 48 L. At the same T and P, what is its volume with 3 mol?
V₁/n₁ = V₂/n₂ → 48 ÷ 2 = V₂ ÷ 3 → V₂ = 24 × 3 = 72 L.
2. What volume does 0.5 mol of oxygen fill at STP?
V = n × Vm = 0.5 × 22.4 = 11.2 L.
3. How many moles and molecules are in 5.6 L of CO₂ at STP?
n = 5.6 ÷ 22.4 = 0.25 mol. N = 0.25 × 6.02 × 10²³ = 1.505 × 10²³ molecules.
4. Find the volume at STP of 8 g of methane, CH₄.
M(CH₄) = 12 + 4 = 16 g/mol. n = 8 ÷ 16 = 0.5 mol. V = 0.5 × 22.4 = 11.2 L.
5. A gas has relative density 22 compared with hydrogen. Find its molar mass and say if it is heavier than air.
M = 2 × 22 = 44 g/mol. D(air) = 44 ÷ 29 ≈ 1.52 > 1, so it is heavier than air (it could be CO₂).
6. What volume of oxygen is needed to burn 10 L of methane completely, and what volume of CO₂ forms? (CH₄ + 2O₂ → CO₂ + 2H₂O; same T and P)
Volume ratio CH₄ : O₂ : CO₂ = 1 : 2 : 1. O₂ = 2 × 10 = 20 L. CO₂ = 10 L.
Common mistakes
- Thinking a bigger molecule takes more gas volume. In a gas, the space between molecules is huge, so equal volumes hold equal numbers, whatever the size.
- Using 22.4 L/mol at room temperature. 22.4 L is for 0 °C and 1 atm; at about 25 °C use about 24.5 L (or the value given).
- Using volume ratios for liquids or solids, for example for liquid water. Only gases follow Avogadro's law.
- Mixing up equal volume with equal mass. Equal volumes have equal numbers of molecules, but different masses.