Avogadro's law and molar volume

Two balloons, one filled with hydrogen and one with carbon dioxide, are inflated to exactly the same size at the same temperature and pressure. Despite hydrogen molecules being far lighter than carbon dioxide molecules, both balloons hold the same number of gas molecules. That surprising equality is Avogadro's Law, and it hands chemists a shortcut worth knowing cold: at STP, one mole of any gas — no matter what it is made of — always occupies 22.4 dm³. This single number lets you jump between the volume of a gas, the number of moles, the number of molecules, and the mass, without ever needing a balance for the gas itself. This article builds up the law, its molar-volume consequence, and the calculations ICSE Class 10 expects you to do with it.

The core insight

Equal volumes of any gas at the same temperature and pressure contain equal numbers of molecules, so one mole of any gas occupies the same molar volume — 22.4 dm³ at STP.

The core idea

Avogadro's Law states: equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of molecules. Nothing about the gas's identity, molar mass, or molecular size enters the statement — only volume, temperature and pressure decide how many molecules are present.

Why this is remarkable. A litre of hydrogen (very light molecules) and a litre of carbon dioxide (much heavier molecules) under the same conditions hold exactly the same molecule count, even though the carbon dioxide sample is far heavier overall. Gas volume, unlike gas mass, does not depend on what the gas is made of — only on how many particles are bouncing around inside it.

The molar-volume consequence. Since equal volumes always hold equal numbers of molecules, it follows that one mole of any gas (6.023 × 10²³ molecules, by definition) occupies a fixed volume at fixed temperature and pressure. At STP — which in this syllabus means 0 °C and 1 atmosphere pressure — this volume is found experimentally to be:

Vm=22.4 dm3 mol−1 (=22.4 L/mol=22,400 cm3/mol)V_m = 22.4\ \text{dm}^3\,\text{mol}^{-1}\ \left(= 22.4\ \text{L/mol} = 22{,}400\ \text{cm}^3/\text{mol}\right)

where: VmV_m is the molar volume, the volume occupied by exactly one mole of any gas at STP. This gives the volume-based mole relation used throughout the Mole Concept chapter:

n=V22.4 dm3(gas at STP)n = \dfrac{V}{22.4\ \text{dm}^3}\quad(\text{gas at STP})

where: nn is the number of moles of gas and VV is its volume in dm³, measured at STP.

Where Avogadro's Law comes from historically. Gay-Lussac's experiments on combining volumes of reacting gases had shown simple whole-number volume ratios (for example, 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of steam), but this clashed with Dalton's picture of indivisible atoms — how can 2 volumes of hydrogen atoms give 2 volumes of product if atoms cannot split? Avogadro's resolution was that elemental gases like H₂, O₂, N₂ and Cl₂ exist as diatomic molecules, which can split apart and recombine during a reaction. This is also the tool used to deduce the molecular formula of a simple gaseous compound once its combining-volume ratio with known gases is known.

Figure
Two equal-volume boxes at the same temperature and pressure, one containing light hydrogen molecules and one containing heavier carbon dioxide molecules, both boxes holding the same number of molecule dots
Same volume, same T and P — same number of molecules Hydrogen, H2 light molecules, low mass Carbon dioxide, CO2 heavier molecules, higher mass Both boxes: 12 molecules, same volume, same T and P — molecule count does not depend on gas identity
Derivation AFrom Avogadro's Law to the 22.4 dm³ molar volume

The goal is to show, in fewer than 200 characters of reasoning made explicit, why Avogadro's Law forces the volume of one mole of any gas at STP to be a single fixed number rather than something that depends on the gas chosen — and to connect that fixed number to the mole–volume formula used in calculations.

Step 1. Start from the statement of the law: equal volumes of any two gases, at the same temperature and pressure, contain equal numbers of molecules. Turn this around — if two different gas samples contain the same number of molecules, and they are kept at the same temperature and pressure, then the law guarantees they occupy the same volume.

Step 2. Apply this to exactly one mole of each gas. One mole of hydrogen and one mole of carbon dioxide both contain exactly 6.023×10236.023 \times 10^{23} molecules, by the definition of the mole. Since both samples have the same molecule count and (at STP) the same temperature and pressure, Step 1 says they must occupy the same volume — regardless of the fact that a CO₂ molecule is roughly 22 times heavier than an H₂ molecule.

Step 3. Because this argument used no property specific to hydrogen or carbon dioxide, it holds for any gas. So there is one single volume, common to every gas, occupied by one mole at STP; this shared volume is defined as the molar volume VmV_m, and experiment fixes its value at Vm=22.4 dm3 mol−1V_m = 22.4\ \text{dm}^3\,\text{mol}^{-1}.

Step 4. Turn the fixed molar volume into a working formula. If 1 mole occupies 22.4 dm³, then nn moles occupy V=n×22.4 dm3V = n \times 22.4\ \text{dm}^3, so rearranging for nn gives the relation used in every mole–volume calculation: n=V22.4 dm3n = \dfrac{V}{22.4\ \text{dm}^3}, valid only for a gas at STP.

Worked example

Problem: A flask contains 5.6 dm³ of oxygen gas at STP. Calculate (a) the number of moles of O₂, (b) the number of oxygen molecules present, and (c) the mass of oxygen present. (RAM: O = 16)

Step 1 — Moles of O₂ from volume. Using the mole–volume relation from the derivation above, n=V22.4 dm3=5.622.4=0.25 moln = \dfrac{V}{22.4\ \text{dm}^3} = \dfrac{5.6}{22.4} = 0.25\ \text{mol}.

Step 2 — Number of molecules. Each mole contains Avogadro's number of molecules, so Number of molecules=n×NA=0.25×6.023×1023=1.50575×1023 molecules\text{Number of molecules} = n \times N_A = 0.25 \times 6.023 \times 10^{23} = 1.50575 \times 10^{23}\ \text{molecules}.

Step 3 — Mass of oxygen. The molar mass of O₂ is M(O2)=2(16)=32 g/molM(\text{O}_2) = 2(16) = 32\ \text{g/mol}, so mass=n×M=0.25×32=8 g\text{mass} = n \times M = 0.25 \times 32 = 8\ \text{g}.

Answer: 5.6 dm³ of O₂ at STP contains 0.25 mol, equal to 1.50575×10231.50575 \times 10^{23} molecules, with a mass of 8 g.

Competitive stretch

Extend the same flask: what volume would 8.8 g of carbon dioxide occupy under the same STP conditions? (RAM: C = 12, O = 16)

Step 1. Molar mass of CO₂: M=12+2(16)=44 g/molM = 12 + 2(16) = 44\ \text{g/mol}.

Step 2. Moles of CO₂: n=8.844=0.2 moln = \dfrac{8.8}{44} = 0.2\ \text{mol}.

Step 3. Volume at STP: V=n×22.4 dm3=0.2×22.4=4.48 dm3V = n \times 22.4\ \text{dm}^3 = 0.2 \times 22.4 = 4.48\ \text{dm}^3.

Answer: 8.8 g of CO₂ occupies 4.48 dm³ at STP — note this is exactly double the 2.24 dm³ that 4.4 g of the same gas would occupy, since the mass has doubled.

Common traps

  • Applying the 22.4 dm³ molar volume relation to a liquid or a solid — it applies only to gases, and only when that gas is measured at STP.
  • Forgetting that "STP" in this syllabus is 0 °C and 1 atm; using room temperature values, or the modern IUPAC STP of 25 °C and 1 bar, gives a wrong molar volume.
  • Assuming molar volume depends on the identity, density or molar mass of the gas — Avogadro's Law guarantees it is the same for every gas under identical conditions.
  • Mixing up volume-based mole calculations (start from n=V/22.4n = V/22.4) with mass-based ones (start from n=mass/molar massn = \text{mass}/\text{molar mass}); always check which quantity the question actually gives you before choosing the formula.

Ask the tutor: Want a walkthrough converting between gas volume, moles, molecules and mass, or a check on why molar volume is the same for every gas? Ask the AI tutor.

Why do equal volumes of gases contain equal numbers of molecules, even though molecules have different sizes

At the pressures and temperatures used in this syllabus, the actual volume taken up by the molecules themselves is a tiny fraction of the total gas volume — most of a gas sample is empty space between fast-moving, widely-spaced molecules. So the total volume a gas occupies is set almost entirely by how much empty space the molecules need to move around in, not by the molecules' own size. Since this spacing behaviour is essentially the same for every gas at a given temperature and pressure, equal volumes end up holding equal numbers of molecules regardless of what the gas is.

Does the molar volume of 22.4 dm3 change if I use a different gas or a different amount of gas

The value 22.4 dm³ never changes with the identity of the gas — that is exactly what Avogadro's Law guarantees, and it is why the number is so useful. What does change is the total volume if you have more or less than one mole: 2 moles of any gas at STP occupy 2 times 22.4 = 44.8 dm³, and 0.1 mole occupies 0.1 times 22.4 = 2.24 dm³. The 22.4 dm³ figure is specifically the volume of one mole; always scale it by the number of moles you actually have.

Can I use the 22.4 dm3 rule for a gas that is not at STP, like a gas collected at room temperature

Not directly. The value 22.4 dm³ per mole is tied specifically to STP conditions of 0 °C and 1 atmosphere pressure; at any other temperature or pressure, one mole of gas occupies a different volume. This ICSE Class 10 syllabus does not require you to correct volumes for other temperatures or pressures using the ideal gas equation — that is beyond this chapter's scope — so every molar-volume question you are asked here will specify or assume STP conditions.

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