A gas at 27 ^C is heated to 327 ^C at constant volume. By what factor does the pressure change? [2 marks]

AQA AQA 2018-style practice: A sealed container holds 0.40 mol of an ideal gas at a temperature of 27 ^C in a volume of 0.010 m^3. Calculate the pressure of the gas. Take R = 8.31 J K^-1 mol^-1.

Convert the temperature to kelvin: T = 27 + 273 = 300 K. Rearrange the ideal gas equation pV = nRT to p = nRTV. p = (0.40)(8.31)(300)0.010 = 997.20.010 = 9.97 × 10^4 Pa ≈ 1.0 × 10^5 Pa. Markers reward converting to kelvin, the correct rearrangement, and substitution. A common error is leaving the temperature in degrees Celsius.

EnglandPhysicsSyllabus dot point

How are the pressure, volume and temperature of a fixed amount of gas related?

The gas laws, the ideal gas equation in molar and molecular forms, absolute zero and the experimental basis of the gas laws.

A focused answer to AQA A-Level Physics 3.6.2.2, covering Boyle's, Charles's and the pressure law, absolute temperature, and the ideal gas equation in both molar and molecular forms.

Generated by Claude Opus 4.89 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
The gas laws
Absolute temperature
The ideal gas equation
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What this dot point is asking

AQA specification point 3.6.2.2 wants you to state and use the gas laws, understand absolute (Kelvin) temperature and absolute zero, and use the ideal gas equation in both its molar form $pV = nRT$ and its molecular form $pV = NkT$ , along with the Avogadro constant, molar gas constant and Boltzmann constant.

The gas laws

For a fixed mass of gas, three experimental laws relate the pressure, volume and temperature:

Boyle's law: at constant temperature, $pV = \text{constant}$ , so pressure is inversely proportional to volume. Squeezing a gas into half the volume doubles its pressure.
Charles's law: at constant pressure, $\dfrac{V}{T} = \text{constant}$ , so volume is proportional to absolute temperature. A balloon expands when warmed.
The pressure law (Gay-Lussac's law): at constant volume, $\dfrac{p}{T} = \text{constant}$ , so pressure is proportional to absolute temperature. A sealed can heated over a flame can burst.

Each law was established experimentally before the kinetic theory explained them. They are special cases of the combined gas law $\dfrac{pV}{T} = \text{constant}$ . Boyle established his law in 1662 by trapping air with mercury and measuring volume against pressure; Charles and Gay-Lussac later studied the temperature dependence. The kinetic theory of gases (covered in the related dot point) explains all three from the random motion of molecules: pressure arises from molecular collisions with the walls, and raising the temperature raises the mean molecular speed.

Absolute temperature

Plotting the pressure of a fixed-volume gas against temperature in degrees Celsius gives a straight line that extrapolates back to zero pressure at $-273 \text{ }^{\circ}\text{C}$ , which is one experimental route to absolute zero and the Kelvin scale.

The ideal gas equation

The Boltzmann constant links the two: $k = \dfrac{R}{N_A}$ , where $N_A = 6.02 \times 10^{23} \text{ mol}^{-1}$ is the Avogadro constant (the number of particles in one mole). An ideal gas is a model with point molecules, no intermolecular forces (except during collisions) and perfectly elastic collisions; real gases approach this behaviour at low pressure and high temperature.

Try this

Q1. State Boyle's law and the condition under which it applies. [2 marks]

Cue. $pV = \text{constant}$ for a fixed mass of gas at constant temperature.

Q2. A gas at $27 \text{ }^{\circ}\text{C}$ is heated to $327 \text{ }^{\circ}\text{C}$ at constant volume. By what factor does the pressure change? [2 marks]

Cue. $300 \text{ K}$ to $600 \text{ K}$ , so the pressure doubles.

Q3. State the value of absolute zero in degrees Celsius. [1 mark]

Cue. About $-273 \text{ }^{\circ}\text{C}$ .

Exam-style practice questions

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

AQA 20184 marksA sealed container holds

0.40 \text{ mol}

of an ideal gas at a temperature of

27 \text{ }^{\circ}\text{C}

in a volume of

0.010 \text{ m}^3

. Calculate the pressure of the gas. Take

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

Show worked answer →

Convert the temperature to kelvin: $T = 27 + 273 = 300 \text{ K}$ .

Rearrange the ideal gas equation $pV = nRT$ to $p = \dfrac{nRT}{V}$ .

$p = \dfrac{(0.40)(8.31)(300)}{0.010} = \dfrac{997.2}{0.010} = 9.97 \times 10^4 \text{ Pa} \approx 1.0 \times 10^5 \text{ Pa}$ .

Markers reward converting to kelvin, the correct rearrangement, and substitution. A common error is leaving the temperature in degrees Celsius.

AQA 20224 marksA fixed mass of ideal gas at a pressure of

1.0 \times 10^5 \text{ Pa}

and temperature

300 \text{ K}

is heated at constant volume to

450 \text{ K}

. Calculate the new pressure and explain, using the gas laws, why the pressure changes.

Show worked answer →

At constant volume the pressure law applies: $\dfrac{p}{T} = \text{constant}$ , so $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$ .

$p_2 = p_1 \dfrac{T_2}{T_1} = (1.0 \times 10^5)\dfrac{450}{300} = 1.5 \times 10^5 \text{ Pa}$ .

The pressure rises because raising the absolute temperature increases the mean kinetic energy and hence the speed of the molecules, so they strike the walls harder and more often, increasing the pressure for the fixed volume.

Markers reward using the pressure law with kelvin temperatures, the correct new pressure, and a molecular explanation of why pressure increases.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)