Ideal Gas Equation and Calculations — AQA A-level Chemistry

SPEC 3.1.2.3 • AMOUNT OF SUBSTANCE

The Ideal Gas Equation

Using pV = nRT to relate pressure, volume, moles, and temperature of gases

What is an Ideal Gas?

An ideal gas is a theoretical model that simplifies how we understand gas behaviour. While no real gas is truly ideal, most gases behave close to ideally under normal conditions.

Kinetic Theory Assumptions

Gas particles move rapidly and randomly — in straight lines until they collide

Particle volume is negligible — compared to the empty space between them

No intermolecular forces — no attractions or repulsions between particles

Collisions are elastic — no energy is lost when particles collide

Average kinetic energy depends only on temperature — higher T means faster particles

Key Point

A hypothetical gas obeying all these assumptions is called an ideal gas. Noble gases like helium and neon come closest to ideal behaviour because their intermolecular forces are minimal.

The Ideal Gas Equation

Under standard conditions, gases follow these relationships:

pressure is proportional to temperature (p ∝ T)

volume is proportional to temperature (V ∝ T)

pressure and volume are inversely proportional (p ∝ 1/V)

These combine to give the ideal gas equation:

The Equation

p V = n RT = \frac{m RT}{M _{r}}

SymbolQuantitySI Unit

p

PressurePascals (Pa)

V

VolumeCubic metres (m³)

n

Moles of gasmol

R

Gas constant8.31 J K⁻¹ mol⁻¹

T

TemperatureKelvin (K)

m

Massgrams (g)

M_{r}

Relative molecular massg mol⁻¹

Essential Unit Conversions

The ideal gas equation only works with SI units. You must convert before substituting!

QuantityGiven UnitConversionSI Unit

PRESSUREkPa× 1000Pa

atm× 101325Pa

VOLUMEdm³÷ 1000m³

cm³÷ 1,000,000m³

TEMPERATURE°C+ 273K

Memory Aid

dm³ → m³: 1 dm = 0.1 m, so 1 dm³ = (0.1)³ = 0.001 m³ = ÷ 1000

cm³ → m³: 1 cm = 0.01 m, so 1 cm³ = (0.01)³ = 0.000001 m³ = ÷ 10⁶

Common Mistake

Forgetting to convert units is the most common source of errors! Always check: Pa, m³, mol, K before substituting into pV = nRT.

01Worked Example

Calculate the volume, in dm³, occupied by 0.250 mol of an ideal gas at 100 kPa and 25°C.

R = 8.31 J K⁻¹ mol⁻¹

1

Convert units: p = 100 × 1000 = 100,000 Pa

2

Convert temperature: T = 25 + 273 = 298 K

3

Rearrange for V:

V = \frac{n RT}{p}

4

Substitute:

V = \frac{0.250 \times 8.31 \times 298}{100000}

5

Calculate: V = 6.19 × 10⁻³ m³

6

Convert to dm³: V = 6.19 × 10⁻³ × 1000 = 6.19 dm³

Answer: 6.19 dm³

02Worked Example

A sealed container of volume 500 cm³ contains 0.0400 mol of nitrogen gas at 27°C. Calculate the pressure in the container in kPa.

R = 8.31 J K⁻¹ mol⁻¹

1

Convert volume: V = 500 ÷ 1,000,000 = 5.00 × 10⁻⁴ m³

2

Convert temperature: T = 27 + 273 = 300 K

3

Rearrange for p:

p = \frac{n RT}{V}

4

Substitute:

p = \frac{0.0400 \times 8.31 \times 300}{5.00 \times 1 0 ^{- 4}}

5

Calculate: p = 199,536 Pa

6

Convert to kPa: p = 199,536 ÷ 1000 = 200 kPa (3 s.f.)

Answer: 200 kPa

Rearranging pV = nRT

You need to be able to rearrange the ideal gas equation to solve for any variable:

Find Pressure

p = \frac{n RT}{V}

Find Volume

V = \frac{n RT}{p}

Find Moles

n = \frac{p V}{RT}

Find Temperature

T = \frac{p V}{n R}

Method

To rearrange: identify what you need to find, then isolate that variable by dividing both sides by everything else. For example, to find n: divide both sides by RT to get n = pV/RT.

03Worked Example

A gas cylinder contains 12.0 dm³ of gas at a pressure of 150 kPa and a temperature of 30°C. Calculate the amount of gas, in moles.

R = 8.31 J K⁻¹ mol⁻¹

1

Convert pressure: p = 150 × 1000 = 150,000 Pa

2

Convert volume: V = 12.0 ÷ 1000 = 0.0120 m³

3

Convert temperature: T = 30 + 273 = 303 K

4

Rearrange for n:

n = \frac{p V}{RT}

5

Substitute:

n = \frac{150000 \times 0.0120}{8.31 \times 303}

6

Calculate: n = 0.714 mol

Answer: 0.714 mol

04Worked Example

0.500 mol of an ideal gas occupies a volume of 15.0 dm³ at a pressure of 120 kPa. Calculate the temperature of the gas in °C.

R = 8.31 J K⁻¹ mol⁻¹

1

Convert pressure: p = 120 × 1000 = 120,000 Pa

2

Convert volume: V = 15.0 ÷ 1000 = 0.0150 m³

3

Rearrange for T:

T = \frac{p V}{n R}

4

Substitute:

T = \frac{120000 \times 0.0150}{0.500 \times 8.31}

5

Calculate: T = 433 K

6

Convert to °C: T = 433 − 273 = 160°C

Answer: 160°C

Finding Molar Mass

A powerful application of the ideal gas equation is finding the molar mass (Mᵣ) of an unknown gas or volatile liquid.

Derivation

We know that

n = \frac{m}{M}

where m = mass and M = molar mass.

Substituting into pV = nRT:

p V = \frac{m}{M} RT

Rearranging for M:

M = \frac{m RT}{p V}

Practical Application

This is used in the "volatile liquid" experiment — you vaporise a known mass of liquid, measure the volume of gas produced at known temperature and pressure, then calculate the molar mass.

05Worked Example

A sample of 0.185 g of a volatile liquid is vaporised. The vapour occupies 72.0 cm³ at 100 kPa and 95°C. Calculate the molar mass of the liquid.

R = 8.31 J K⁻¹ mol⁻¹

1

Convert pressure: p = 100 × 1000 = 100,000 Pa

2

Convert volume: V = 72.0 ÷ 1,000,000 = 7.20 × 10⁻⁵ m³

3

Convert temperature: T = 95 + 273 = 368 K

4

Use:

M = \frac{m RT}{p V}

5

Substitute:

M = \frac{0.185 \times 8.31 \times 368}{100000 \times 7.20 \times 1 0 ^{- 5}}

6

Calculate: M = 78.6 g mol⁻¹

Answer: 78.6 g mol⁻¹ (This is consistent with benzene, C₆H₆)

06Worked Example

Calculate the mass of oxygen gas (O₂) in a 5.00 dm³ cylinder at 200 kPa and 20°C.

R = 8.31 J K⁻¹ mol⁻¹, Mᵣ(O₂) = 32.0

1

Convert: p = 200,000 Pa, V = 0.00500 m³, T = 293 K

2

Find moles using:

n = \frac{p V}{RT}

3

Substitute:

n = \frac{200000 \times 0.00500}{8.31 \times 293}

4

Calculate: n = 0.410 mol

5

Calculate mass: m = n × M = 0.410 × 32.0

6

m = 13.1 g

Answer: 13.1 g

07Worked Example (Challenging)

When 0.500 g of calcium carbonate is reacted with excess hydrochloric acid, carbon dioxide gas is produced. Calculate the volume of CO₂ produced at 25°C and 100 kPa.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

R = 8.31 J K⁻¹ mol⁻¹, Mᵣ(CaCO₃) = 100.1

1

Calculate moles of CaCO₃: n = m ÷ M = 0.500 ÷ 100.1 = 0.00500 mol

2

From equation: 1 mol CaCO₃ produces 1 mol CO₂, so n(CO₂) = 0.00500 mol

3

Convert units: p = 100,000 Pa, T = 298 K

4

Use:

V = \frac{n RT}{p} = \frac{0.00500 \times 8.31 \times 298}{100000}

5

Calculate: V = 1.24 × 10⁻⁴ m³ = 124 cm³

Answer: 124 cm³

08Worked Example (Challenging)

A sealed gas syringe contains 80.0 cm³ of gas at 20°C and 100 kPa. The syringe is heated to 150°C while the pressure is increased to 250 kPa. Calculate the new volume of gas.

Hint: Use $\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}$ (derived from pV = nRT when n is constant)

1

Identify initial conditions: p₁ = 100 kPa, V₁ = 80.0 cm³, T₁ = 20 + 273 = 293 K

2

Identify final conditions: p₂ = 250 kPa, V₂ = ?, T₂ = 150 + 273 = 423 K

3

Since n is constant:

\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}

4

Rearrange:

V_{2} = \frac{p _{1} V _{1} T _{2}}{T _{1} p _{2}}

5

Substitute:

V_{2} = \frac{100 \times 80.0 \times 423}{293 \times 250}

6

Calculate: V₂ = 46.2 cm³

Answer: 46.2 cm³

When Gases Deviate from Ideal

Real gases don't behave ideally under all conditions. The ideal gas assumptions break down when:

Summary

High pressure: Particles are forced close together, so their own volume is no longer negligible

Low temperature: Particles move slowly, so intermolecular forces have more effect — gases may condense

Ideal Behaviour Conditions

Gases behave most ideally under conditions where the assumptions are most valid:

Ideal Conditions

LOW

PressureHIGH

Temperature

Why?

Low pressure: Particles are far apart, so their volume is negligible compared to the container

High temperature: Particles move fast, so they spend very little time near each other — intermolecular forces don't have time to act

Exam Context

You may be asked to explain why real gases deviate from ideal behaviour, or to state conditions under which gases behave most ideally.

Practice

Question 1:

R = 8.31 J K⁻¹ mol⁻¹

1

Convert: p = 98,000 Pa, T = 293 K

2

V = \frac{n RT}{p} = \frac{0.0500 \times 8.31 \times 293}{98000}

3

V = 1.24 × 10⁻³ m³ = 1240 cm³

Answer: 1240 cm³

Practice

Question 2:

R = 8.31 J K⁻¹ mol⁻¹, Mᵣ(N₂) = 28.0

1

Convert: p = 105,000 Pa, V = 0.00250 m³, T = 298 K

2

(a)

n = \frac{p V}{RT} = \frac{105000 \times 0.00250}{8.31 \times 298}

3

n = 0.106 mol

4

(b) m = n × M = 0.106 × 28.0 = 2.97 g

Answers: (a) 0.106 mol, (b) 2.97 g

Practice

Question 3:

R = 8.31 J K⁻¹ mol⁻¹

1

Convert: p = 100,000 Pa, V = 9.50 × 10⁻⁵ m³, T = 363 K

2

M = \frac{m RT}{p V} = \frac{0.250 \times 8.31 \times 363}{100000 \times 9.50 \times 1 0 ^{- 5}}

3

M = 79.4 g mol⁻¹

4

This is close to 78 — likely benzene (C₆H₆)

Answer: 79.4 g mol⁻¹, likely benzene

Summary

Key Formulas

PurposeFormula

Ideal gas equation

p V = n RT

Find moles

n = \frac{p V}{RT}

Find molar mass

M = \frac{m RT}{p V}

Unit Conversions

FromTo SIMethod

kPaPa× 1000

dm³m³÷ 1000

cm³m³÷ 1,000,000

°CK+ 273

Ideal Gas Behaviour

Gases behave most ideally at low pressure and high temperature. Deviations occur when particle volume or intermolecular forces become significant.