Moles in Gases: Molar Volume

Gases behave very differently from solids and liquids. You cannot simply weigh a gas on a balance in the same way you weigh a solid. Instead, chemists use the relationship between volume and moles to quantify gases — and this relationship is remarkably simple under certain conditions.

Molar Volume at Room Temperature and Pressure (RTP)

At room temperature and pressure (RTP), defined as approximately 298 K (25 °C) and 101 kPa (1 atm), one mole of any gas occupies approximately 24.0 dm³ (24,000 cm³).

This value is called the molar volume and is the same for all gases at the same temperature and pressure (assuming ideal behaviour). It does not matter whether the gas is hydrogen, oxygen, carbon dioxide, or any other gas — one mole always occupies 24.0 dm³ at RTP.

At standard temperature and pressure (STP), defined as 273 K (0 °C) and 100 kPa, the molar volume is 22.7 dm³ mol⁻¹. Make sure you check which conditions the question specifies.

The key equation is:

n = V / Vₘ

Where:

n = number of moles (mol)
V = volume of gas (dm³)
Vₘ = molar volume (24.0 dm³ at RTP; 22.7 dm³ at STP)

Or equivalently: V = n × Vₘ

If the volume is given in cm³, convert first: V (dm³) = V (cm³) / 1000.

Worked Example 1

Calculate the volume of 0.150 mol of carbon dioxide at RTP.

V = n × 24.0 = 0.150 × 24.0 = 3.60 dm³

Worked Example 2

What is the number of moles in 480 cm³ of nitrogen gas at RTP?

V = 480 / 1000 = 0.480 dm³ n = 0.480 / 24.0 = 0.0200 mol

Worked Example 3

What mass of oxygen gas occupies 1.20 dm³ at RTP?

Step 1: n = 1.20 / 24.0 = 0.0500 mol

Step 2: M(O₂) = 2 × 16.0 = 32.0 g mol⁻¹

Step 3: m = 0.0500 × 32.0 = 1.60 g

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 is why the molar volume is the same for all gases under the same conditions.

This law has an important consequence for reactions involving gases: the ratio of gas volumes in a reaction is the same as the mole ratio, provided all gases are measured at the same temperature and pressure.

Example

In the reaction: N₂(g) + 3H₂(g) → 2NH₃(g)

If 100 cm³ of N₂ reacts, it requires 300 cm³ of H₂ and produces 200 cm³ of NH₃ (all at the same temperature and pressure). The volume ratio is 1 : 3 : 2, exactly matching the mole ratio.

Worked Example 4 — Gas Volume Ratios

20.0 cm³ of propane (C₃H₈) is burned in excess oxygen. What volume of CO₂ is produced? (all gases at the same T and P)

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

Mole ratio C₃H₈ : CO₂ = 1 : 3

Volume of CO₂ = 20.0 × 3 = 60.0 cm³

What volume of oxygen is consumed? Mole ratio C₃H₈ : O₂ = 1 : 5 Volume of O₂ = 20.0 × 5 = 100 cm³

The Ideal Gas Equation

For conditions other than RTP, we use the ideal gas equation:

pV = nRT

Where:

p = pressure (Pa)
V = volume (m³)
n = number of moles (mol)
R = gas constant = 8.314 J K⁻¹ mol⁻¹
T = temperature (K)

Unit Conversions — Critical Reference Table

You must use SI units in this equation:

Quantity	Given in	Convert to	Method
Pressure	kPa	Pa	× 1000
Pressure	atm	Pa	× 101325
Volume	dm³	m³	÷ 1000
Volume	cm³	m³	÷ 1,000,000
Temperature	°C	K	+ 273

The most common error is mixing units — for instance, using kPa for pressure but forgetting to convert dm³ to m³. If your answer seems absurdly large or small, check your units first.

Worked Example 5

Calculate the volume occupied by 0.100 mol of gas at 300 K and 200 kPa.

Step 1: Convert units. p = 200 × 1000 = 200,000 Pa T = 300 K (already in K)

Step 2: Use pV = nRT rearranged to V = nRT / p. V = (0.100 × 8.314 × 300) / 200,000 V = 249.42 / 200,000 V = 0.001247 m³ = 1.25 dm³

Worked Example 6

A gas syringe contains 60.0 cm³ of gas at 25 °C and 101 kPa. How many moles of gas are present?

Step 1: Convert units. p = 101 × 1000 = 101,000 Pa V = 60.0 / 1,000,000 = 6.00 × 10⁻⁵ m³ T = 25 + 273 = 298 K

Step 2: n = pV / RT n = (101,000 × 6.00 × 10⁻⁵) / (8.314 × 298) n = 6.06 / 2477.6 n = 2.45 × 10⁻³ mol

Worked Example 7 — Finding Molar Mass from Gas Data

0.350 g of a volatile liquid was vaporised. At 100 °C and 101 kPa it occupied 150 cm³. Calculate the molar mass of the liquid.

Step 1: Convert units. p = 101,000 Pa; V = 1.50 × 10⁻⁴ m³; T = 373 K

Step 2: n = pV / RT = (101,000 × 1.50 × 10⁻⁴) / (8.314 × 373) = 15.15 / 3101.1 = 4.885 × 10⁻³ mol

Step 3: M = m / n = 0.350 / 4.885 × 10⁻³ = 71.6 g mol⁻¹

This method is used experimentally to determine the molar mass of volatile liquids.

When Does the Ideal Gas Equation Fail?

The ideal gas equation assumes:

Gas particles have negligible volume compared to the container.
There are no intermolecular forces between gas particles.

Real gases deviate from ideal behaviour at high pressure (particles are forced close together, so their volume matters) and low temperature (particles move slowly, so intermolecular forces become significant). Gases with strong intermolecular forces (e.g. NH₃, HCl) deviate more than noble gases.

Combining Gas Volumes with Mole Calculations

Many exam questions require you to convert between gas volumes and masses within the same problem.

Worked Example 8

What volume of hydrogen gas, at RTP, is produced when 2.43 g of magnesium reacts with excess hydrochloric acid?

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Step 1: Moles of Mg = 2.43 / 24.3 = 0.100 mol

Step 2: From the equation, 1 mol Mg → 1 mol H₂. So moles of H₂ = 0.100 mol.

Step 3: Volume of H₂ = 0.100 × 24.0 = 2.40 dm³ (or 2400 cm³)

Worked Example 9

In an experiment, 120 cm³ of CO₂ gas at RTP was collected from a reaction. What mass of CaCO₃ decomposed to produce this gas?

CaCO₃(s) → CaO(s) + CO₂(g)

Step 1: n(CO₂) = 0.120 / 24.0 = 5.00 × 10⁻³ mol

Step 2: 1 mol CaCO₃ → 1 mol CO₂, so n(CaCO₃) = 5.00 × 10⁻³ mol

Step 3: m(CaCO₃) = 5.00 × 10⁻³ × 100.0 = 0.500 g

Common Mistakes with Gas Calculations

Mistake	Effect	Prevention
Using 24.0 dm³ at non-RTP conditions	Wrong answer	Use pV = nRT for non-RTP
Forgetting volume unit conversions	Off by factors of 1000 or 10⁶	Convert to m³ for pV = nRT; dm³ for n = V/24.0
Applying n = V/24.0 to liquids or solids	Nonsensical answer	Only gases have a molar volume
Confusing dm³ and cm³ in volume ratios	Factor of 1000 error	Both volumes must be in the same units
Forgetting °C to K conversion	Off by ~10% at room temp	Always add 273

Choosing the Right Gas Equation

flowchart TD
    A["Gas calculation needed"] --> B{"Conditions specified?"}
    B -- "RTP (25°C, 101 kPa)" --> C["Use n = V / 24.0 (V in dm³)"]
    B -- "STP (0°C, 100 kPa)" --> D["Use n = V / 22.7 (V in dm³)"]
    B -- "Other T and P" --> E["Use pV = nRT (SI units!)"]
    B -- "Same T and P, comparing gases" --> F["Use volume ratios = mole ratios"]

Key Data for Gas Calculations

Condition	Temperature	Pressure	Molar volume
RTP	298 K (25 °C)	101 kPa	24.0 dm³ mol⁻¹
STP	273 K (0 °C)	100 kPa	22.7 dm³ mol⁻¹
R value	—	—	8.314 J K⁻¹ mol⁻¹

Worked Example 10 — Using STP

Calculate the volume occupied by 0.250 mol of gas at STP.

V = 0.250 × 22.7 = 5.68 dm³

If this question had said RTP instead of STP, the answer would be 0.250 × 24.0 = 6.00 dm³. Always check which conditions are specified.

Real-World Application: Airbags

Car airbags use the rapid decomposition of sodium azide (NaN₃) to produce nitrogen gas:

2NaN₃(s) → 2Na(s) + 3N₂(g)

An airbag needs about 67 dm³ of gas. At RTP: n(N₂) = 67 / 24.0 = 2.79 mol From the equation, 2 mol NaN₃ → 3 mol N₂, so moles NaN₃ = 2.79 × 2/3 = 1.86 mol Mass NaN₃ = 1.86 × 65.0 = 121 g

However, an airbag inflates at much higher temperature (and the gases are hot), so the actual mass needed is different — but this illustrates how gas volume calculations have real engineering applications.

Gas calculations are tested extensively at A-Level, both in standalone questions and as part of multi-step problems involving titrations, moles, and redox chemistry.

A-Level Deep Dive: Moles in Gases: Molar Volume

Spec mapping

Edexcel 9CH0 specification Topic 5 — Formulae, Equations and Amounts of Substance, sub-strand on gases, requires students to use the molar volume of a gas (24.0 dm $^3$ mol $^{-1}$ at room temperature and pressure, RTP) and apply the ideal gas equation $pV = nRT$ for non-RTP conditions, with $R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}$ (refer to the official specification document for exact wording). Examined across Papers 1, 2 and 3 — Paper 1 in physical chemistry contexts (kinetics, equilibrium Kp), Paper 2 in synthesis (gas-collection volumes), Paper 3 in practical CP9 alternatives. The data sheet provides $R$ , the molar volume at RTP (24.0 dm $^3$ mol $^{-1}$ ) and standard temperature/pressure values.

Worked example with full mark scheme

Question (8 marks):

(a) A 0.196 g sample of magnesium reacts completely with excess dilute hydrochloric acid at RTP. Calculate the volume of hydrogen gas produced. ( $A_r(\text{Mg}) = 24.3$ , molar volume at RTP = 24.0 dm $^3$ mol $^{-1}$ .) (3)

(b) A 0.250 g sample of an unknown gas occupies 95.0 cm $^3$ at 298 K and $1.013 \times 10^5\,\text{Pa}$ . Calculate $M_r$ of the gas, and identify it from the list: NO (30.0), CO $_2$ (44.0), N $_2$ O (44.0), SO $_2$ (64.1). (5)

Solution with mark scheme:

(a) Step 1 — balanced equation.

$\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$ .

Step 2 — moles of Mg.

$n(\text{Mg}) = 0.196/24.3 = 8.07 \times 10^{-3}\,\text{mol}$ .

M1.