Ideal Gases

Chemistry SL· 12 min read

1. Core Assumptions of the Ideal Gas Model★★☆☆☆⏱ 8 min

The ideal gas model is a simplified framework derived from kinetic molecular theory that describes gas particle behaviour, eliminating complex real-world interactions to enable predictable, consistent calculations.

📘 Definition

Ideal Gas

A hypothetical gas that perfectly follows all 5 kinetic molecular theory assumptions for gas particle behaviour, with no intermolecular forces and negligible total particle volume relative to the container.

Gas particles have negligible total volume compared to the total container volume
There are no attractive or repulsive intermolecular forces between gas particles
Gas particles move in constant, random, straight-line motion
All collisions between gas particles and container walls are perfectly elastic, with no net kinetic energy loss
The average kinetic energy of gas particles is directly proportional to absolute temperature in Kelvin

✓ Quick check

Test your understanding of core ideal gas rules

Which of the following is NOT a valid assumption of the ideal gas model?
- Particles have no intermolecular forces
- Particles have zero mass
- Collisions are perfectly elastic
- Particle volume is negligible
Reveal answer
Particles have zero mass —
Ideal gas particles have mass, just negligible total volume relative to the container.

2. The Ideal Gas Law and Unit Consistency★★★☆☆⏱ 10 min

Combining Boyle’s Law, Charles’s Law, Avogadro’s Law and Gay-Lussac’s Law gives the unified ideal gas equation, which relates all four state variables for a fixed amount of gas.

PV = nRT

P = pressure, measured in Pascals (Pa)
V = volume, measured in cubic metres (m³)
n = amount of substance, measured in moles
R = universal gas constant = 8.314 J mol⁻¹ K⁻¹
T = absolute temperature, measured in Kelvin (K)

📐 Worked Example

Calculate the pressure exerted by 0.250 mol of helium gas in a 0.010 m³ container at 25 °C.

1
First convert temperature from Celsius to Kelvin:
$T = 25 + 273.15 = 298.15 \text{ K}$
2
Rearrange the ideal gas law to solve for pressure P:
$P = \frac{nRT}{V}$
3
Substitute all known values into the equation:
$P = \frac{0.250 \times 8.314 \times 298.15}{0.010}$
4
Calculate the final pressure, rounded to 3 significant figures:
$P = 61900 \text{ Pa} = 61.9 \text{ kPa}$

3. Molar Volume of an Ideal Gas at STP★★☆☆☆⏱ 7 min

IB defines standard temperature and pressure (STP) as 273 K and 100 kPa, not the older 1 atm definition. Under these conditions, one mole of any ideal gas occupies exactly 22.7 dm³, a value you can use directly for stoichiometry calculations.

📘 Definition

Molar Volume

Vₘ

The volume occupied by one mole of an ideal gas at specified temperature and pressure, equal to 22.7 dm³ mol⁻¹ at IB STP.

📐 Worked Example

Calculate the volume of carbon dioxide produced when 5.0 g of calcium carbonate fully decomposes at STP.

1
Write the balanced decomposition reaction:
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
2
Calculate moles of CaCO₃, molar mass = 100.09 g mol⁻¹:
$n(CaCO_3) = \frac{5.0}{100.09} = 0.050 \text{ mol}$
3
1:1 mole ratio gives n(CO₂) = 0.050 mol
4
Multiply moles by molar volume at STP:
$V(CO_2) = 0.050 \times 22.7 = 1.1 \text{ dm}^3$

4. Deviations of Real Gases from Ideal Behaviour★★★☆☆⏱ 6 min

Real gases only approximate ideal behaviour at low pressure and high temperature. Two extreme conditions break the core ideal gas assumptions, leading to measurable deviations from PV = nRT predictions.

At high pressure, gas particles are forced very close together, so their total volume is no longer negligible relative to the container volume
At low temperature, particle kinetic energy drops, so weak intermolecular forces become significant, slowing particles near collision points

5. Common Pitfalls

Wrong move:

Using Celsius instead of Kelvin for temperature in PV=nRT

Why:

Celsius values give negative or incorrectly low pressure/volume outputs, losing 1-2 calculation marks

Correct move:

Always convert temperature to Kelvin first before substituting into the ideal gas law

Wrong move:

Using 22.4 dm³ mol⁻¹ as molar volume at STP

Why:

IB uses the updated 100 kPa STP definition, not the old 1 atm standard, so 22.4 is incorrect

Correct move:

Memorise 22.7 dm³ mol⁻¹ as the IB specified molar volume at STP

Wrong move:

Using units of dm³ for volume directly in PV=nRT

Why:

The gas constant R = 8.314 uses m³, so dm³ values will give answers 1000x too small

Correct move:

Convert all volume values to m³ by dividing dm³ by 1000 before calculation

Wrong move:

Stating that real gases deviate because particles have mass

Why:

Ideal gas assumptions do not state particles have zero mass, only that their volume is negligible

Correct move:

Link deviations explicitly to non-negligible particle volume and existing intermolecular forces

Wrong move:

Forgetting to give final answers to 3 significant figures

Why:

IB mark schemes penalise answers that do not match the precision of given data

Correct move:

Round all final calculation outputs to 3 significant figures unless specified otherwise

6. Quick Reference Cheatsheet

Quantity	SI Unit	Conversion Factor	Ideal Gas Value at STP
Pressure	Pascal (Pa)	1 atm = 101325 Pa, 1 kPa = 1000 Pa	100000 Pa
Volume	Cubic metre (m³)	1 dm³ = 0.001 m³	0.0227 m³ mol⁻¹
Temperature	Kelvin (K)	°C + 273.15 = K	273 K
Gas Constant R	J mol⁻¹ K⁻¹	Fixed value	8.314

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Paper 2
Calculate gas volume at STP
2023 · Paper 1
Identify valid ideal gas assumption
2022 · Paper 2
Explain real gas deviation

What's Next

Mastering ideal gas calculations is a critical foundational skill for higher difficulty stoichiometry questions that combine gas volume data with titration and percentage yield calculations, which make up 15-20% of the IB SL Paper 2 marks. You will next apply these rules to solve reacting gas volume problems using Avogadro’s law, then move on to explore the properties of liquids and intermolecular forces in the periodicity unit. Regular practice of unit conversion and full working for calculation steps will eliminate avoidable mark losses in your exam.