SL & HL Structure · Models of the particulate nature of matter BETA S1.5

Ideal gases

IB syllabus topic Structure 1.5 — Ideal gases

The ideal gas model treats a gas as countless tiny particles in constant random motion with negligible volume and no intermolecular forces; this revision note covers the ideal gas equation $PV = nRT$ , the combined gas law, the individual gas laws (Boyle, Charles, Gay-Lussac), molar volume at STP (22.7 dm³ mol^-1), and how and why real gases deviate from ideal behaviour.

Part of IB Chemistry (2025-2030 syllabus) — Standard and Higher Level.

Key points

An ideal gas is a model: its particles have negligible volume, exert no intermolecular forces, are in constant random motion, and undergo perfectly elastic collisions.
Temperature in every gas-law calculation must be in kelvin: $T(\text{K}) = \theta(^{\circ}\text{C}) + 273$ . Using Celsius is the most common cause of wrong answers.
The ideal gas equation $PV = nRT$ is given in the data booklet, with $R = 8.31$ J K $^{-1}$ mol $^{-1}$ ; use SI units (Pa, m $^3$ , mol, K) so the units cancel.
The combined gas law $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ (also in the booklet) relates two states of a fixed amount of gas; it contains Boyle's, Charles's and Gay-Lussac's laws as special cases.
Equal volumes of any ideal gases at the same temperature and pressure contain equal numbers of particles (Avogadro's law). At STP (0 C, 100 kPa) one mole of gas occupies 22.7 dm $^3$ .
Real gases deviate most from ideal behaviour at high pressure and low temperature, because particle volume and intermolecular attractions can no longer be neglected.
Gases that are most ideal are small, non-polar, low-mass particles (such as helium and hydrogen) with weak intermolecular forces; large or polar molecules deviate more.

Subtopic by subtopic

The ideal gas model

The ideal gas model explains gas behaviour using a few simple assumptions about the particles. These are the assumptions of the kinetic theory applied to a perfect gas:

Particles are in constant, random motion, travelling in straight lines until they collide.
The volume of the particles themselves is negligible compared with the volume of the container - the gas is mostly empty space.
There are no intermolecular forces between the particles (no attraction or repulsion except during collisions).
Collisions (with each other and the walls) are perfectly elastic, so no kinetic energy is lost overall.
The average kinetic energy of the particles is proportional to the absolute (kelvin) temperature.

Pressure arises because particles collide with the container walls; more frequent or harder collisions mean higher pressure.

Because the particles have no volume and no forces between them, an ideal gas would obey $PV = nRT$ exactly under all conditions. No real gas does this perfectly, but the model is an excellent approximation for everyday conditions and is the foundation for all the gas-law calculations in this topic. Raising the temperature increases the average particle speed (and kinetic energy); it does not change the mass of the particles.

Real versus ideal gases

A real gas is made of particles that do have a finite volume and do attract one another, so it follows $PV = nRT$ only approximately. Deviations become significant in two situations:

High pressure: the particles are squeezed close together, so their own volume is no longer negligible compared with the container, and the available space is less than the model assumes.
Low temperature: the particles move slowly, so intermolecular attractions have time to act and pull particles together, reducing the pressure below the ideal value.

So a gas behaves most ideally at low pressure and high temperature, when particles are far apart and moving fast.

Which gases are closest to ideal?

Small, low-mass, non-polar particles with weak intermolecular forces behave most ideally - for example helium and hydrogen.
Large molecules, or polar molecules with strong forces (hydrogen bonding, dipole-dipole), such as water vapour or ammonia, deviate more.

A practical reminder: if a gas is near its boiling point or under very high pressure, do not assume ideal behaviour. For typical exam conditions (around room temperature and atmospheric pressure) the ideal gas equation works well.

Molar volume of a gas

Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles. A direct consequence is that one mole of *any* ideal gas occupies the same volume under the same conditions - the molar volume $V_m$ .

The data booklet gives the molar volume at STP (0 C = 273 K and 100 kPa):

$V_m = 22.7\ \text{dm}^3\,\text{mol}^{-1}\ (= 2.27 \times 10^{-2}\ \text{m}^3\,\text{mol}^{-1})$

Use it to convert between the volume of a gas and the amount in moles at STP:

$n = \frac{V}{22.7}\quad (V \text{ in dm}^3)$

Key points:

The value 22.7 dm $^3$ mol $^{-1}$ is for STP. Be careful not to use the older RTP figure of 24.0 dm $^3$ mol $^{-1}$ - the current IB booklet uses 22.7.
It applies to gases only, not solids or liquids.
Because molar volume is the same for every gas, you can compare amounts of different gases just by comparing their volumes (at the same conditions).

This makes reacting-gas-volume problems quick: volume ratios of gases equal their mole ratios (Avogadro).

The ideal gas equation and gas laws

The ideal gas equation (in the data booklet) links all four variables:

$PV = nRT$

with $R = 8.31$ J K $^{-1}$ mol $^{-1}$ . To make the units cancel, use SI units: $P$ in pascals (Pa), $V$ in cubic metres (m $^3$ ), $T$ in kelvin (K), $n$ in moles. Useful conversions: $1\ \text{kPa} = 1000\ \text{Pa}$ ; $1\ \text{dm}^3 = 1 \times 10^{-3}\ \text{m}^3$ ; $1\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^3$ .

For a fixed amount of gas changing conditions, the combined gas law (also in the booklet) is more convenient:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Holding one variable constant gives the individual laws (you construct these by cancelling):

Boyle's law (constant $T$ ): $P_1 V_1 = P_2 V_2$ - $P$ and $V$ are inversely proportional.
Charles's law (constant $P$ ): $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ - $V \propto T$ .
Gay-Lussac's law (constant $V$ ): $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ - $P \propto T$ .

In the combined gas law, $T$ must be in kelvin, but $P$ and $V$ only need consistent units on both sides because they cancel.

Formulae

PV = nRT

Given in the data booklet. The ideal gas equation, relating pressure $P$ , volume $V$ , amount $n$ and absolute temperature $T$ . Use SI units: $P$ in Pa, $V$ in m $^3$ , $T$ in K, with $R = 8.31$ J K $^{-1}$ mol $^{-1}$ .

\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}

Given in the data booklet. The combined gas law, for a fixed amount of gas changing from state 1 to state 2. $T$ must be in kelvin; $P$ and $V$ units only need to match on both sides (they cancel).

P_1 V_1 = P_2 V_2

Boyle's law (constant $T$ and $n$ ): a special case of the combined gas law you CONSTRUCT by cancelling $T$ . At constant temperature, $P$ and $V$ are inversely proportional.

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}

Charles's law (constant $P$ and $n$ ): constructed by cancelling $P$ from the combined gas law. At constant pressure, $V \propto T$ (in kelvin).

\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}

Gay-Lussac's (pressure) law (constant $V$ and $n$ ): constructed by cancelling $V$ . At constant volume, $P \propto T$ (in kelvin).

n = \dfrac{V}{V_m} = \dfrac{V}{22.7}

Constructed by the student (not a separate booklet equation): amount of gas from its volume in dm $^3$ at STP, using the booklet molar volume $V_m = 22.7$ dm $^3$ mol $^{-1}$ .

T(\text{K}) = \theta(^{\circ}\text{C}) + 273

Constructed conversion (not in the booklet's equation list) from Celsius to absolute temperature; always apply it before any gas-law calculation.

Definitions

Ideal gas: A theoretical gas whose particles have zero volume and no intermolecular forces, are in constant random motion, and collide elastically; it obeys $PV = nRT$ exactly under all conditions.
Real gas: An actual gas whose particles have finite volume and experience intermolecular attractions, so it follows the ideal gas equation only approximately, deviating most at high pressure and low temperature.
Standard temperature and pressure (STP): The reference conditions of 0 C (273 K) and 100 kPa, at which one mole of an ideal gas occupies a molar volume of 22.7 dm $^3$ .
Molar volume of a gas, $V_m$: The volume occupied by one mole of a gas under stated conditions; for an ideal gas at STP it is 22.7 dm $^3$ mol $^{-1}$ , the same for every gas.
Absolute (kelvin) temperature: Temperature measured from absolute zero (0 K = -273 C), proportional to the average kinetic energy of the particles; it must be used in all gas-law equations.
Avogadro's law: Equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of particles (and therefore equal amounts in moles).
Gas constant, $R$: The universal proportionality constant in the ideal gas equation, equal to 8.31 J K $^{-1}$ mol $^{-1}$ and given in the data booklet.

Worked examples

(SL) A sample of gas occupies 250 cm³ at a pressure of 100 kPa and a temperature of 27 C. Calculate the amount of gas, in moles. ( $R = 8.31$ J K $^{-1}$ mol $^{-1}$ .)

1
Convert every quantity to SI units before substituting into $PV = nRT$ .
2
Pressure: $P = 100\ \text{kPa} = 100\,000 = 1.00 \times 10^5$ Pa.
3
Volume: $V = 250\ \text{cm}^3 = 250 \times 10^{-6} = 2.50 \times 10^{-4}$ m³.
4
Temperature: $T = 27 + 273 = 300$ K.
5
Rearrange: $n = \dfrac{PV}{RT}$ .
6
$n = \dfrac{(1.00 \times 10^5)(2.50 \times 10^{-4})}{(8.31)(300)} = \dfrac{25.0}{2493}$ .
7
$n = 0.01003 \approx 1.00 \times 10^{-2}$ mol.

Answer: $n = \dfrac{PV}{RT} = \dfrac{(1.00\times10^5)(2.50\times10^{-4})}{(8.31)(300)} = 1.00 \times 10^{-2}\ \text{mol}$ (0.0100 mol).

(SL) A gas occupies 1.20 dm³ at 300 K and 101 kPa. It is heated to 400 K while the pressure rises to 202 kPa. Calculate the new volume.

1
Fixed amount of gas changing state, so use the combined gas law $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ .
2
Temperatures are already in kelvin: $T_1 = 300$ K, $T_2 = 400$ K. $P$ and $V$ units cancel, so kPa and dm³ are fine.
3
State 1: $P_1 = 101$ kPa, $V_1 = 1.20$ dm³. State 2: $P_2 = 202$ kPa, $V_2 = ?$
4
Rearrange for $V_2$ : $V_2 = \dfrac{P_1 V_1 T_2}{T_1 P_2}$ .
5
$V_2 = \dfrac{(101)(1.20)(400)}{(300)(202)} = \dfrac{48\,480}{60\,600}$ .
6
$V_2 = 0.800$ dm³.

Answer: $V_2 = \dfrac{P_1 V_1 T_2}{T_1 P_2} = \dfrac{(101)(1.20)(400)}{(300)(202)} = 0.800\ \text{dm}^3$ . (Doubling the pressure halves the volume, but the temperature rising by a factor 4/3 partly offsets it.)

(SL) Calculate the volume occupied by 3.2 g of methane, CH₄, at STP. ( $M(\text{CH}_4) = 16.05$ g mol $^{-1}$ ; molar volume at STP = 22.7 dm $^3$ mol $^{-1}$ .)

1
First find the amount in moles: $n = \dfrac{m}{M} = \dfrac{3.2}{16.05}$ .
2
$n = 0.1994$ mol ( $\approx 0.20$ mol to 2 s.f.).
3
At STP, volume = amount x molar volume: $V = n \times V_m = n \times 22.7$ .
4
Carry the unrounded amount through to avoid rounding error: $V = 0.1994 \times 22.7 = 4.53$ dm³.
5
Round to 2 s.f. (the data is given to 2 s.f.): $V = 4.5$ dm³.

Answer: $V = n \times 22.7 = 0.1994 \times 22.7 = 4.53\ \text{dm}^3$ , which rounds to $4.5\ \text{dm}^3$ (2 s.f.).

Common mistakes

×Leaving the temperature in degrees Celsius. Every gas law needs absolute temperature, so always convert with $T(\text{K}) = \theta(^{\circ}\text{C}) + 273$ first.
×Forgetting to convert volume and pressure to SI units in $PV = nRT$ : using dm $^3$ instead of m $^3$ , or kPa instead of Pa, throws the answer off by powers of ten.
×Using the old RTP molar volume of 24.0 dm $^3$ mol $^{-1}$ . The current IB data booklet quotes 22.7 dm $^3$ mol $^{-1}$ at STP - use 22.7.
×Claiming gases are most ideal at high pressure. It is the opposite: real gases behave most ideally at LOW pressure and HIGH temperature, when particles are far apart and fast-moving.

Exam tips

✓For $PV = nRT$ questions, write out the SI conversions for $P$ , $V$ and $T$ as a first step - most marks are lost on units, not on the algebra.
✓Choose the right equation: use the combined gas law $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$ when a fixed sample changes conditions, and $PV=nRT$ when you need (or are given) the amount in moles.
✓When explaining real-gas deviations, name BOTH causes - finite particle volume AND intermolecular attractions - and link each to high pressure or low temperature.
✓In the combined gas law, only $T$ must be in kelvin; $P$ and $V$ can stay in any units (kPa, dm $^3$ ) as long as they are the same on both sides, since they cancel.

All IB Chemistry topics