A gas at 1.0 × 10^5 Pa and 2.0 × 10^-3 cubic metres is compressed at constant temperature to 0.50 × 10^-3 cubic metres. Find the new pressure. [2 marks]

Pearson Edexcel Edexcel 2021-style practice: A fixed mass of ideal gas has volume 2.0 × 10^-3 cubic metres at a pressure of 1.0 × 10^5 Pa and temperature 300 K. Determine the number of moles, then find the new pressure if the gas is heated to 450 K at constant volume. Take R = 8.31 J per mol per K.

Number of moles from pV = nRT: n = (pV)/(RT) = 1.0 × 10^5 × 2.0 × 10^-38.31 × 300 = (200)/(2493) = 0.080 mol. Constant volume: (p_1)/(T_1) = (p_2)/(T_2), so p_2 = p_1 × (T_2)/(T_1) = 1.0 × 10^5 × (450)/(300) = 1.5 × 10^5 Pa. Markers reward n = (pV)/(RT) = 0.080 mol, the pressure law at constant volume, and the value 1.5 × 10^5 Pa.

EnglandPhysicsSyllabus dot point

How is heat transferred and how do ideal gases behave?

Internal energy and temperature, specific heat capacity and specific latent heat, the ideal gas laws and equation of state, and the kinetic theory of gases.

A focused answer to the Edexcel 9PH0 thermal physics content, covering internal energy and temperature, specific heat capacity and latent heat, the ideal gas laws and equation of state, and the kinetic theory of gases.

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

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Edexcel wants you to relate internal energy to temperature, use specific heat capacity and specific latent heat, apply the ideal gas laws and the equation of state $pV = nRT$ , and explain gas behaviour using the kinetic theory.

The answer

Internal energy and temperature

Raising the temperature increases the average kinetic energy of the particles. Converting between Celsius and kelvin uses $T(\text{K}) = \theta(\text{deg C}) + 273$ . During a change of state the temperature (and so average kinetic energy) stays constant while the potential energy of the particles changes.

Specific heat capacity and latent heat

Specific heat capacity is the energy to raise $1$ kg by $1$ K. Latent heat is the energy to change the state of $1$ kg with no temperature change; this energy goes into separating the particles (breaking bonds) rather than speeding them up, which is why a heating curve has flat sections at the melting and boiling points.

The ideal gas laws

Temperatures in these laws must be in kelvin. An ideal gas is one that obeys $pV = nRT$ exactly; real gases approach this at low pressure and high temperature.

Kinetic theory

The kinetic theory models a gas as many tiny particles in random motion, colliding elastically with the walls and each other. Deriving the pressure from the rate of change of molecular momentum at the walls gives:

This shows directly that absolute temperature is proportional to the mean kinetic energy of the molecules, the microscopic meaning of temperature. The assumptions (point particles, no intermolecular forces except in collisions, elastic collisions, random motion) define the ideal gas.

Mean molecular speed

Find the root-mean-square speed of nitrogen molecules at $300$ K. Take the mass of a nitrogen molecule as $4.7 \times 10^{-26}$ kg and $k = 1.38 \times 10^{-23}$ J per K.

step 1: Relate kinetic energy to temperature

$\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$ , so $\overline{c^2} = \frac{3kT}{m}$ .

step 2: Substitute

$\overline{c^2} = \frac{3 \times 1.38 \times 10^{-23} \times 300}{4.7 \times 10^{-26}} = \frac{1.242 \times 10^{-20}}{4.7 \times 10^{-26}} = 2.64 \times 10^{5}$ .

step 3: Take the square root

$c_{\text{rms}} = \sqrt{2.64 \times 10^{5}} = 5.1 \times 10^{2}$ m per second, about $510$ m per second.

Examples in context

Water's high specific heat capacity makes it an excellent coolant and moderates coastal climates. Latent heat of vaporisation explains why sweating cools the body and why steam burns are worse than boiling-water burns. The gas laws govern engines, weather balloons, and the pressure changes in scuba diving. Kinetic theory underpins the design of pressure vessels and explains why gases exert more pressure when heated, the principle behind the internal combustion engine.

Try this

Q1. State what absolute temperature measures at the molecular level. [1 mark]

Cue. The average random kinetic energy per particle.

Q2. Find the energy to heat $2.0$ kg of water by $30$ K. Take $c = 4200$ J per kg per K. [2 marks]

Cue. $Q = mc\Delta\theta = 2.0 \times 4200 \times 30 = 2.5 \times 10^{5}$ J.

Q3. A gas at $1.0 \times 10^{5}$ Pa and $2.0 \times 10^{-3}$ cubic metres is compressed at constant temperature to $0.50 \times 10^{-3}$ cubic metres. Find the new pressure. [2 marks]

Cue. Boyle's law: $p_2 = \frac{p_1 V_1}{V_2} = \frac{1.0 \times 10^{5} \times 2.0 \times 10^{-3}}{0.50 \times 10^{-3}} = 4.0 \times 10^{5}$ Pa.

Exam-style practice questions

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

Edexcel 20184 marksCalculate the energy needed to raise the temperature of

0.50

kg of water from

20

degrees C to

100

degrees C, then boil

0.10

kg of it. Specific heat capacity of water

= 4200

J per kg per K; specific latent heat of vaporisation

= 2.3 \times 10^{6}

J per kg.

Show worked answer →

Heating: $Q_1 = mc\Delta\theta = 0.50 \times 4200 \times (100 - 20) = 0.50 \times 4200 \times 80 = 1.68 \times 10^{5}$ J.

Boiling $0.10$ kg: $Q_2 = mL = 0.10 \times 2.3 \times 10^{6} = 2.3 \times 10^{5}$ J.

Total: $Q = 1.68 \times 10^{5} + 2.3 \times 10^{5} = 4.0 \times 10^{5}$ J.

Markers reward $Q = mc\Delta\theta$ for the heating, $Q = mL$ for the boiling, and the total about $4.0 \times 10^{5}$ J.

Edexcel 20215 marksA fixed mass of ideal gas has volume

2.0 \times 10^{-3}

cubic metres at a pressure of

1.0 \times 10^{5}

Pa and temperature

300

K. Determine the number of moles, then find the new pressure if the gas is heated to

450

K at constant volume. Take

R = 8.31

J per mol per K.

Show worked answer →

Number of moles from $pV = nRT$ : $n = \frac{pV}{RT} = \frac{1.0 \times 10^{5} \times 2.0 \times 10^{-3}}{8.31 \times 300} = \frac{200}{2493} = 0.080$ mol.

Constant volume: $\frac{p_1}{T_1} = \frac{p_2}{T_2}$ , so $p_2 = p_1 \times \frac{T_2}{T_1} = 1.0 \times 10^{5} \times \frac{450}{300} = 1.5 \times 10^{5}$ Pa.

Markers reward $n = \frac{pV}{RT} = 0.080$ mol, the pressure law at constant volume, and the value $1.5 \times 10^{5}$ Pa.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Physics (9PH0) specification — Pearson Edexcel (2015)