A gas of 0.50\ mol occupies 0.012\ m^3 at 310\ K. Find its pressure (R = 8.31\ J mol^-1\ K^-1). [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A sealed container of volume 2.0 × 10^-3\ m^3 holds gas at a pressure of 1.5 × 10^5\ Pa and a temperature of 290\ K. Calculate the number of moles of gas. Take R = 8.31\ J mol^-1\ K^-1.

Rearrange the equation of state pV = nRT for n: n = pVRT. n = (1.5 × 10^5)(2.0 × 10^-3)(8.31)(290) = 3002410 = 0.124\ mol, about 0.12\ mol. Markers reward n = (pV)/(RT), correct substitution, and about 0.12\ mol.

WJEC Eduqas Eduqas 2021-style practice: A fixed mass of ideal gas at 300\ K occupies 0.025\ m^3 at a pressure of 1.0 × 10^5\ Pa. It is heated at constant pressure until its volume is 0.030\ m^3. Calculate the new temperature.

At constant pressure (and fixed amount of gas), VT is constant, so V_1T_1 = V_2T_2. Rearrange for T_2: T_2 = T_1 V_2V_1 = 300 × 0.0300.025 = 300 × 1.2 = 360\ K. Markers reward using (V)/(T) constant at constant pressure, the rearrangement, and the new temperature 360\ K.

EnglandPhysicsSyllabus dot point

How are the pressure, volume, temperature and amount of a gas related, and what is an ideal gas?

Kinetic theory: the gas laws, the absolute temperature scale, the equation of state pV = nRT (and pV = NkT), and the conditions under which a real gas behaves ideally.

A focused answer to the Eduqas A-Level Physics Component 1 ideal gas content, covering the experimental gas laws, the absolute temperature scale, the equation of state pV = nRT and pV = NkT, and the conditions under which a real gas behaves ideally.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Eduqas wants you to state the experimental gas laws, explain the absolute temperature scale, use the ideal gas equation of state $pV = nRT$ (equivalently $pV = NkT$ ), and state the conditions under which a real gas behaves ideally.

The answer

The gas laws

The absolute temperature scale

The equation of state

When a real gas behaves ideally

Boyle's law for a compressed gas

A gas occupies $0.040\ \text{m}^3$ at a pressure of $1.0 \times 10^{5}\ \text{Pa}$ . It is compressed at constant temperature to $0.010\ \text{m}^3$ . Find the new pressure.

step 1: Apply Boyle's law

At constant temperature (and fixed amount), $pV$ is constant, so $p_1 V_1 = p_2 V_2$ .

step 2: Rearrange for the new pressure

$p_2 = p_1 \dfrac{V_1}{V_2} = (1.0 \times 10^{5}) \times \dfrac{0.040}{0.010} = (1.0 \times 10^{5}) \times 4.0$ .

step 3: Evaluate

$p_2 = 4.0 \times 10^{5}\ \text{Pa}$ . Reducing the volume to a quarter raises the pressure four times.

Examples in context

The ideal gas equation underpins the engineering of engines, compressors, refrigerators and pneumatic systems, and the behaviour of weather balloons and scuba tanks. Charles's law explains why a balloon expands when warmed; Boyle's law explains why a diver's lungs and bubbles expand as they rise. Meteorology and the study of planetary atmospheres rely on the gas laws to relate pressure, temperature and density.

Try this

Q1. State Boyle's law. [1 mark]

Cue. At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.

Q2. A gas of $0.50\ \text{mol}$ occupies $0.012\ \text{m}^3$ at $310\ \text{K}$ . Find its pressure ( $R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$ ). [2 marks]

Cue. $p = \frac{nRT}{V} = \frac{0.50 \times 8.31 \times 310}{0.012} = 1.1 \times 10^{5}\ \text{Pa}$ .

Q3. State the conditions under which a real gas behaves most like an ideal gas. [2 marks]

Cue. Low pressure and high temperature (low density), where the molecules are far apart and intermolecular forces are negligible.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20194 marksA sealed container of volume

2.0 \times 10^{-3}\ \text{m}^3

holds gas at a pressure of

1.5 \times 10^{5}\ \text{Pa}

and a temperature of

290\ \text{K}

. Calculate the number of moles of gas. Take

R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}

Show worked answer →

Rearrange the equation of state $pV = nRT$ for $n$ : $n = \dfrac{pV}{RT}$ .

$n = \dfrac{(1.5 \times 10^{5})(2.0 \times 10^{-3})}{(8.31)(290)} = \dfrac{300}{2410} = 0.124\ \text{mol}$ , about $0.12\ \text{mol}$ .

Markers reward $n = \frac{pV}{RT}$ , correct substitution, and about $0.12\ \text{mol}$ .

Eduqas 20214 marksA fixed mass of ideal gas at

300\ \text{K}

occupies

0.025\ \text{m}^3

at a pressure of

1.0 \times 10^{5}\ \text{Pa}

. It is heated at constant pressure until its volume is

0.030\ \text{m}^3

. Calculate the new temperature.

Show worked answer →

At constant pressure (and fixed amount of gas), $\dfrac{V}{T}$ is constant, so $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ .

Rearrange for $T_2$ : $T_2 = T_1 \dfrac{V_2}{V_1} = 300 \times \dfrac{0.030}{0.025} = 300 \times 1.2 = 360\ \text{K}$ .

Markers reward using $\frac{V}{T}$ constant at constant pressure, the rearrangement, and the new temperature $360\ \text{K}$ .

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)