Molar Gas Volume

TL;DR: One mole of any gas occupies 24 dm³ (24,000 cm³) at room temperature and pressure (r.t.p.: 20 °C, 1 atm). This allows chemists to convert between the volume of a gas and the amount in moles using the formula volume = moles × 24.

The molar gas volume is the volume occupied by one mole of any gas under specified conditions, and at room temperature and pressure (r.t.p., defined as 20 °C and 1 atmosphere) this volume is 24 dm³ (or 24,000 cm³). This remarkable constancy arises from Avogadro’s law, which states that equal volumes of all gases, at the same temperature and pressure, contain the same number of particles—meaning the identity or mass of the gas molecules is irrelevant to the volume they occupy. The practical formula linking these quantities is volume (dm³) = moles × 24, and its rearrangement moles = volume (dm³) ÷ 24 allows students to move fluidly between the macroscopic measurement of gas volume and the microscopic counting unit of the mole. At r.t.p., this relationship sits at the heart of quantitative chemistry, bridging laboratory measurements (gas collected in a syringe or over water) with stoichiometric calculations involving reacting masses, limiting reagents, and yields. Understanding molar gas volume unlocks a wide range of IGCSE problems, from calculating the volume of carbon dioxide produced when a carbonate reacts with acid, to determining the mass of metal that reacts to produce a given volume of hydrogen.

Definition

The molar gas volume is the volume occupied by one mole of a gas.

• At room temperature and pressure (r.t.p.) — 20 °C and 1 atmosphere (101 kPa) — the molar gas volume is 24 dm³ mol⁻¹.
• At standard temperature and pressure (s.t.p.) — 0 °C and 1 atmosphere — the molar gas volume is 22.4 dm³ mol⁻¹ (though IGCSE primarily uses r.t.p.).

The key insight is that this value is the same for all gases, regardless of the size or mass of the individual molecules.

The Key Formula

The relationship between gas volume and amount is:

Rearrangements

Converting Between cm³ and dm³

• 1 dm³ = 1,000 cm³
• 24 dm³ = 24,000 cm³

When working in cm³:

IGCSE tip: Always check whether a question gives or expects volumes in dm³ or cm³. If the question gives cm³, either convert to dm³ first (÷ 1,000) or use 24,000 directly.

Why It Is the Same for All Gases

Avogadro’s Law

Avogadro’s law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules.

This means the physical space a gas occupies depends on the number of particles, not on their size or mass. The reasoning is:

1. In a gas, the particles are widely separated — the distance between molecules is vastly larger than the size of the molecules themselves.
2. The volume of a gas is therefore determined primarily by the space between particles, not by the volume of the particles themselves.
3. At a given temperature and pressure, this inter-particle spacing is the same for all gases.
4. Consequently, the same number of particles (one mole, 6.02 × 10²³) occupies the same volume, regardless of the gas.

Comparison of Common Gases at r.t.p.

Despite masses differing by a factor of over 35 (from 2 g to 71 g), the volume is always 24 dm³. This is the power of Avogadro’s law.

Worked Examples

Example 1: Moles to Volume

Question: What volume does 0.50 mol of hydrogen gas occupy at r.t.p.?

Solution:

• volume = moles × 24
• volume = 0.50 × 24 = 12 dm³

Example 2: Volume to Moles

Question: A reaction produces 6.0 dm³ of carbon dioxide at r.t.p. How many moles of CO₂ were produced?

Solution:

• moles = volume ÷ 24
• moles = 6.0 ÷ 24 = 0.25 mol

Example 3: Using cm³

Question: 0.020 mol of oxygen is collected at r.t.p. What is its volume in cm³?

Solution:

• volume (cm³) = moles × 24,000
• volume = 0.020 × 24,000 = 480 cm³

Alternative: volume (dm³) = 0.020 × 24 = 0.48 dm³ → 0.48 × 1,000 = 480 cm³.

Example 4: Reacting Volumes

Question: Hydrogen reacts with oxygen to form water:

2H₂(g) + O₂(g) → 2H₂O(g)

What volume of oxygen is needed to react completely with 60 dm³ of hydrogen at r.t.p.?

Solution:

• From the equation, 2 mol H₂ reacts with 1 mol O₂.
• The mole ratio H₂ : O₂ = 2 : 1.
• By Avogadro’s law, the volume ratio is the same as the mole ratio for gases.
• So the volume ratio H₂ : O₂ = 2 : 1.
• Volume of O₂ = 60 ÷ 2 = 30 dm³

Key principle: For gases at the same temperature and pressure, reacting volumes are in the same ratio as the moles in the balanced equation.

Example 5: Mass to Gas Volume

Question: What volume of CO₂ (at r.t.p.) is produced when 10.0 g of calcium carbonate (CaCO₃) is heated?

CaCO₃(s) → CaO(s) + CO₂(g)

Solution:

1. Mᵣ of CaCO₃ = 40 + 12 + (3 × 16) = 100
2. Moles of CaCO₃ = mass ÷ Mᵣ = 10.0 ÷ 100 = 0.100 mol
3. From the equation, 1 mol CaCO₃ → 1 mol CO₂, so moles of CO₂ = 0.100 mol
4. Volume of CO₂ = moles × 24 = 0.100 × 24 = 2.4 dm³ (or 2,400 cm³)

Stoichiometric Calculations with Gases

When a reaction involves gases, the molar gas volume links the gas volume to the mole, which can then be connected to mass via Mᵣ or to solution concentration via c = n/V.

General Strategy

1. Write the balanced chemical equation.
2. Convert the given quantity (mass, gas volume, solution volume/concentration) into moles.
3. Use the mole ratio from the balanced equation to find moles of the target substance.
4. Convert moles of the target substance into the required quantity (mass, gas volume, concentration, etc.).

Common Conversions at a Glance

Relationship to Other Gas Laws

The Ideal Gas Equation

The molar gas volume at r.t.p. is a special case of the ideal gas equation:

Deriving the molar volume at r.t.p.:

For 1 mole of gas at r.t.p.:

• V = nRT / p
• V = (1 × 8.31 × 293) / 101,325
• V = 2,435 / 101,325 = 0.0240 m³ = 24.0 dm³

This confirms that the 24 dm³ value used at IGCSE is a convenient approximation derived from the ideal gas law. At s.t.p. (0 °C, 273 K), the same calculation yields 22.4 dm³.

Relationship to Avogadro’s Law

From pV = nRT, if p and T are constant (as at r.t.p.), then V ∝ n: volume is directly proportional to the number of moles. This is the mathematical expression of Avogadro’s law.

Summary of Key Relationships

Common Misconceptions

See Also

• Mole
• Avogadro Constant
• Concentration and Gas Volumes
• Gas Laws

Sources

• Cambridge IGCSE Chemistry Coursebook (2021), Chapters on Stoichiometry and Gases
• Cambridge IGCSE Chemistry Syllabus 0620, Topic: Stoichiometry — Molar Gas Volume
• Petrucci, R. H. et al. (2017). General Chemistry: Principles and Modern Applications (11th ed.)