A gas of 0.50\ mol occupies 0.012\ m^3 at 300\ K. Find its pressure. Take R = 8.31\ J mol^-1\ K^-1.

OCR OCR 2018-style practice: A sealed container of volume 2.0 × 10^-3\ m^3 holds an ideal gas at a pressure of 1.5 × 10^5\ Pa and a temperature of 290\ K. Calculate the number of molecules in the container. Take k = 1.38 × 10^-23\ J K^-1.

Use the molecular form of the ideal gas equation pV = NkT, rearranged for N. N = (pV)/(kT) = (1.5 × 10^5)(2.0 × 10^-3)(1.38 × 10^-23)(290). Numerator: (1.5 × 10^5)(2.0 × 10^-3) = 300. Denominator: (1.38 × 10^-23)(290) = 4.00 × 10^-21. So N = 3004.00 × 10^-21 = 7.5 × 10^22 molecules. Markers reward using pV = NkT, the substitution, and the value about 7.5 × 10^22.

OCR OCR 2021-style practice: Calculate the root mean square speed of helium atoms (mass 6.6 × 10^-27\ kg) at a temperature of 300\ K. Take k = 1.38 × 10^-23\ J K^-1.

Mean kinetic energy of a molecule: (1)/(2)m c^2 = (3)/(2)kT, so c^2 = (3kT)/(m). c^2 = 3(1.38 × 10^-23)(300)6.6 × 10^-27 = 1.242 × 10^-206.6 × 10^-27 = 1.88 × 10^6\ m^2\ s^-2. Root mean square speed: c_rms = 1.88 × 10^6 = 1.4 × 10^3\ m s^-1. Markers reward (1)/(2)m c^2 = (3)/(2)kT, finding the mean square speed, and the rms speed about 1.4 × 10^3\ m s^-1.

EnglandPhysicsSyllabus dot point

How do the random motions of gas molecules give rise to pressure and the gas laws?

Ideal gases and kinetic theory: the ideal gas equation, the Boltzmann constant, the assumptions of the kinetic model, the pressure equation, and the link between mean kinetic energy and absolute temperature.

A focused answer to the OCR H556 content on ideal gases and kinetic theory, covering the ideal gas equation in both molar and molecular forms, the Boltzmann constant, the assumptions of the kinetic model, the kinetic theory pressure equation, root mean square speed, and the link between mean molecular kinetic energy and absolute temperature.

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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Quick answer

An ideal gas obeys $pV = nRT$ (with $n$ in moles and $R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$ ), equivalently $pV = NkT$ (with $N$ molecules and the Boltzmann constant $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ ), where $k = \frac{R}{N_A}$ . The kinetic model assumes many small molecules in rapid random motion, negligible molecular volume, no intermolecular forces except during elastic collisions, and short collision times. This gives the pressure equation $pV = \frac{1}{3}Nm\langle c^2\rangle$ . Comparing the two expressions shows the mean translational kinetic energy of a molecule is $\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT$ , so absolute temperature is a direct measure of average molecular kinetic energy. The root mean square speed is $c_{\text{rms}} = \sqrt{\langle c^2\rangle}$ .

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What this dot point is asking

OCR wants you to use the ideal gas equation in both its molar and molecular forms, use the Boltzmann constant, state the assumptions of the kinetic model, use the kinetic theory pressure equation, define root mean square speed, and link the mean kinetic energy of a molecule to the absolute temperature.

The answer

The ideal gas equation

Temperature must always be in kelvin. The equation combines Boyle's law ( $p \propto \frac{1}{V}$ at constant $T$ ), the pressure law ( $p \propto T$ at constant $V$ ) and Charles's law ( $V \propto T$ at constant $p$ ).

Assumptions of the kinetic model

The kinetic theory pressure equation

Temperature and molecular kinetic energy

Pressure change at constant volume

A fixed mass of ideal gas in a rigid container is at a pressure of $1.0 \times 10^{5}\ \text{Pa}$ and a temperature of $300\ \text{K}$ . It is heated to $450\ \text{K}$ . Find the new pressure.

step 1: Identify what is constant

The container is rigid, so the volume and amount of gas are constant. Then $\frac{p}{T}$ is constant (the pressure law).

step 2: Apply the relation

$\frac{p_1}{T_1} = \frac{p_2}{T_2}$ , so $p_2 = p_1 \times \frac{T_2}{T_1} = 1.0 \times 10^{5} \times \frac{450}{300}$ .

step 3: Evaluate

$p_2 = 1.0 \times 10^{5} \times 1.5 = 1.5 \times 10^{5}\ \text{Pa}$ .

Examples in context

The kinetic model explains why a car tyre's pressure rises on a hot day or after fast driving, and why an aerosol can warns against heating. Diffusion of gases and the operation of the kinetic theory underpin processes from breathing to chemical engineering. The fact that lighter molecules have higher mean speeds at a given temperature is why hydrogen and helium have escaped from the Earth's atmosphere over geological time, while heavier gases remain.

Try this

Q1. State two assumptions of the kinetic theory of an ideal gas. [2 marks]

Cue. Any two of: negligible molecular volume, no intermolecular forces except in collisions, perfectly elastic collisions, negligible collision time, random motion obeying Newton's laws.

Q2. A gas of $0.50\ \text{mol}$ occupies $0.012\ \text{m}^3$ at $300\ \text{K}$ . Find its pressure. Take $R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$ . [2 marks]

Cue. $p = \frac{nRT}{V} = \frac{0.50 \times 8.31 \times 300}{0.012} = 1.04 \times 10^{5}\ \text{Pa}$ .

Q3. State how the mean kinetic energy of a gas molecule depends on temperature. [1 mark]

Cue. It is directly proportional to the absolute temperature, $\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT$ .

Exam-style practice questions

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

OCR 20184 marksA sealed container of volume

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

holds an ideal gas at a pressure of

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

and a temperature of

290\ \text{K}

. Calculate the number of molecules in the container. Take

k = 1.38 \times 10^{-23}\ \text{J K}^{-1}

Show worked answer →

Use the molecular form of the ideal gas equation $pV = NkT$ , rearranged for $N$ .

$N = \frac{pV}{kT} = \frac{(1.5 \times 10^{5})(2.0 \times 10^{-3})}{(1.38 \times 10^{-23})(290)}$ .

Numerator: $(1.5 \times 10^{5})(2.0 \times 10^{-3}) = 300$ . Denominator: $(1.38 \times 10^{-23})(290) = 4.00 \times 10^{-21}$ .

So $N = \frac{300}{4.00 \times 10^{-21}} = 7.5 \times 10^{22}$ molecules.

Markers reward using $pV = NkT$ , the substitution, and the value about $7.5 \times 10^{22}$ .

OCR 20213 marksCalculate the root mean square speed of helium atoms (mass

6.6 \times 10^{-27}\ \text{kg}

) at a temperature of

300\ \text{K}

. Take

k = 1.38 \times 10^{-23}\ \text{J K}^{-1}

Show worked answer →

Mean kinetic energy of a molecule: $\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT$ , so $\langle c^2\rangle = \frac{3kT}{m}$ .

$\langle c^2\rangle = \frac{3(1.38 \times 10^{-23})(300)}{6.6 \times 10^{-27}} = \frac{1.242 \times 10^{-20}}{6.6 \times 10^{-27}} = 1.88 \times 10^{6}\ \text{m}^2\ \text{s}^{-2}$ .

Root mean square speed: $c_{\text{rms}} = \sqrt{1.88 \times 10^{6}} = 1.4 \times 10^{3}\ \text{m s}^{-1}$ .

Markers reward $\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT$ , finding the mean square speed, and the rms speed about $1.4 \times 10^{3}\ \text{m s}^{-1}$ .

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Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)