Where can I learn The first law of thermodynamics for Cambridge 9702?

This free lesson on MarkScheme covers syllabus point 16.2 (The first law of thermodynamics) for Cambridge A-Level Physics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9702/16-2-the-first-law-of-thermodynamics.

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16.2Physics25 min readPREMIUM

The first law of thermodynamics

Welcome to Thermodynamics, the fascinating study of heat and energy! Our first stop is understanding how a system's total energy content, called internal energy, can change. This is all captured by the foundational First Law of Thermodynamics, a universal principle of energy conservation. This pilot aligns with Senpai Corner A Level notes (2025–2027) for topic 16.2.

BY THE END, YOU CAN…

exam checklist — tick these off

1Define internal energy and explain how it relates to temperature.
2State and apply the First Law of Thermodynamics (ΔU = Q + W).
3Interpret the signs of Q, W, and ΔU correctly in thermodynamic processes.
4Apply the First Law to isovolumetric, isothermal, and adiabatic processes.
5Calculate the work done by a gas during expansion at constant pressure (W = pΔV).

STUDY MODE

READING

TYPEFACE

Literata — a reading serif made for e-books; set like a textbook, sits well beside maths.

SIZE

AIRY SPACING

Wider letter and line spacing. The one change with solid evidence for readers who lose their place — worth a try whoever you are.

Pick whatever reads fastest for you — it differs from person to person, by up to a third.

EXPLAIN SIMPLER

The same notes in plain English — no jargon. Key points and exam tips stay.

DIAGRAM NOTES

A small sketch beside each block of notes.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The first law of thermodynamics

Cambridge 9702 Paper 4 — The first law of thermodynamics (16.2). Senpai Corner diagram-backed pilot with premium structure and live visuals.

1
$\Delta U$ (Change in Internal Energy):
2
+ $\Delta U$ : System's internal energy increases (e.g., temperature rises).
3
- $\Delta U$ : System's internal energy decreases (e.g., temperature falls).
4
$Q$ (Energy Transferred via Heating):

What this topic covers

The official Cambridge syllabus points this lesson works through.

16.2.1
Recall and use $W = pΔV$ for the work done when the volume of a gas changes at constant pressure and understand the difference between the work done by the gas and the work done on the gas
16.2.2
Recall and use the first law of thermodynamics $ΔU = q + W$ expressed in terms of the increase in internal energy, the heating of the system (energy transferred to the system by heating) and the work done on the system

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

Figure Live interactive1 / 4

Loading simulation…

Fetching from 3JCN Physics Simulation by Thomas Nguyen…

Isobaric Process

Work done equals pΔV, the rectangle under the isobar.

If the embed does not load, use Open simulation.

3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0

Step-synced diagram — highlights what to look for in the simulation above.

Step 1

$\Delta U$ (Change in Internal Energy):

7 more simulations for this topic — run them in the Simulations section below

Simulations

Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.

7 simulations · 5 to start with

Start herein this order — each one shows a different piece of the topic

3JCN PhysicsStart here · 19702 16.2 · 9702 16.1 · IB B.4
Isovolumetric Pro
Heat a gas at constant volume and watch pressure rise with no work done
Why this one: Heat at constant volume: W = 0, so ΔU equals the heat supplied and pressure rises.
Try this
1. Heat the gas at constant volume and watch the pressure rise.
2. Read the work done on the PV diagram.
Look for With no volume change no work is done, so all heat supplied raises internal energy.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 29702 16.2 · 9702 16.1 · IB B.4
Isothermal Process
Compress a gas at constant temperature and trace the isotherm on a PV diagram
Why this one: Compress at constant temperature: ΔU = 0, so the work done on the gas leaves as heat.
Try this
1. Compress the gas at constant temperature and trace the isotherm.
2. Halve the volume and read the pressure.
Look for At constant T, pV is constant, so the isotherm is a hyperbola and all heat supplied equals work done.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 39702 16.2 · 9702 16.1 · IB B.4
Adiabatic Process
Compress a gas with no heat exchange; see temperature rise on the adiabat
Why this one: Compress with q = 0 and watch the temperature climb — the work done goes entirely into ΔU.
Try this
1. Compress the gas with no heat exchange and watch the temperature.
2. Compare the adiabat's steepness with an isotherm through the same point.
Look for With q = 0 the work done on the gas raises its internal energy, so temperature rises and the adiabat is steeper than an isotherm.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 49702 16.2 · 9702 16.1 · IB B.4
A Cyclic Process
Run a gas around a closed PV cycle and read net work from the enclosed area
Why this one: Run a closed loop on the p–V diagram and read the net work from the enclosed area.
Try this
1. Run the gas around the closed PV cycle.
2. Read the net work from the enclosed area.
3. Note that the internal energy change over the full cycle is zero.
Look for Over a complete cycle ΔU = 0, so net heat in equals net work out, the area enclosed.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
PhETStart here · 59702 15.2 · 16.2
Gas Properties
Pump particles into a box; hold one variable fixed and change the others.
Why this one: Push the piston in fast and watch the temperature jump: work done on the gas raises its internal energy.
Try this
1. On “Ideal”, hold T fixed and push the wall in to halve V — read the pressure.
2. Hold V fixed and heat the gas — plot p against T in your head: straight line through 0 K.
3. Pump in twice the particles at fixed V and T — p doubles.
Look for pV = nRT: each pair of variables scales exactly as the equation predicts.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)

More simulations2 more on this topic — core ones first

3JCN PhysicsCore9702 16.2 · IB B.4
4-Stroke Engine
Watch intake, compression, power and exhaust strokes of a 4-stroke engine
Try this
1. Watch the intake, compression, power and exhaust strokes in turn.
2. Identify which stroke does work on the piston.
Look for Only the power stroke delivers work; the other three strokes prepare and clear the cylinder.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsCore9702 16.2 · 9702 16.1 · IB B.4
T in Adiabatic Proc.
Track temperature change during an adiabatic compression or expansion
Try this
1. Compress adiabatically and track the temperature.
2. Expand adiabatically and track it again.
Look for Adiabatic compression heats the gas and adiabatic expansion cools it, with no heat flowing either way.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0

Key formulas

Tap any symbol to reveal exactly what it means and its units.

KEY RELATION

$Work_{by} = p \Delta V$

Tap a symbol — great for exam definitions

KEY RELATION

$\Delta U = Q + W$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

What is Internal Energy (U)?

Every substance is made up of countless particles (atoms or molecules) that are constantly moving and interacting. The internal energy (U) of a body is the total sum of all the random kinetic energies (due to particle motion) and potential energies (due to forces between particles) of these particles. It’s a 'state function', meaning its value depends only on the current condition of the system, not how it arrived at that state.

Not clear?

Temperature and Internal Energy

The absolute temperature of a system is a direct measure of the average kinetic energy of its constituent particles. If you increase a system's temperature, you increase the average kinetic energy of its particles, and consequently, its overall internal energy rises. For an ideal gas, the internal energy is directly proportional to its absolute temperature.

Not clear?

Work Done on or by a Gas (W)

Besides heating, the other way to change a system's internal energy is by doing work. For a gas in a cylinder with a movable piston, work is done when the volume of the gas changes. If the gas expands, it pushes the piston outwards, doing work on its surroundings. If the gas is compressed, the surroundings do work on the gas.

For a process occurring at a constant pressure $p$ , the work done by the gas as it expands by a volume $\Delta V$ is given by:

Work_{by} = p $\Delta V$

It's crucial to relate this to the term $W$ in the First Law equation, $\Delta U = Q + W$ . In the Cambridge A-Level Physics syllabus, $W$ represents the work done on the system. Therefore:

When a gas expands ( $\Delta V > 0$ ), it does positive work on the surroundings. The work done on the gas is negative: $W = -p \Delta V$ .
When a gas is compressed ( $\Delta V < 0$ ), the surroundings do positive work on the gas. The work done on the gas is positive: $W = -p \Delta V$ (since $\Delta V$ is negative, $W$ becomes positive).

Not clear?

Introducing the First Law of Thermodynamics

The First Law of Thermodynamics is a powerful restatement of the principle of conservation of energy. It tells us that energy cannot be created or destroyed, only transferred or transformed. For a thermodynamic system, energy can move in or out through two primary mechanisms: heating (Q) or work done (W). The law quantifies how these transfers affect the system's internal energy.

$\Delta U = Q + W$

This formula links the change in a system's internal energy ( $\Delta U$ ) to the heat energy transferred ( $Q$ ) and the work done ( $W$ ). Understanding the signs of each term is absolutely critical for solving problems correctly.

Not clear?

Deciphering the Signs: Q, W, and $\Delta U$

$\Delta U$ (Change in Internal Energy):
+ $\Delta U$ : System's internal energy increases (e.g., temperature rises).
- $\Delta U$ : System's internal energy decreases (e.g., temperature falls).
$Q$ (Energy Transferred via Heating):
+ $Q$ : Heat energy is transferred into the system (system gets hotter).
- $Q$ : Heat energy is transferred away from the system (system cools down).
$W$ (Work Done):
+ $W$ : Work is done on the system (e.g., a gas is compressed, energy input).
- $W$ : Work is done by the system (e.g., a gas expands, energy output).

Not clear?

The First Law in Specific Processes

The First Law can be simplified for specific types of thermodynamic processes:

Isovolumetric (or Isochoric) Process: The volume of the system remains constant ( $\Delta V = 0$ ). Since no volume change occurs, no work is done ( $W = 0$ ). The First Law simplifies to $\Delta U = Q$ . All heat added goes directly into increasing the internal energy.
Isothermal Process: The temperature of the system remains constant ( $\Delta T = 0$ ). For an ideal gas, internal energy depends only on temperature, so the change in internal energy is zero ( $\Delta U = 0$ ). The First Law becomes $0 = Q + W$ , or $Q = -W$ . Any heat added to the system is immediately converted into work done by the system.
Adiabatic Process: No heat is transferred into or out of the system ( $Q = 0$ ). This can happen if the system is perfectly insulated or if the process occurs very rapidly. The First Law simplifies to $\Delta U = W$ . If the gas expands and does work, its internal energy decreases (and it cools down). If work is done on the gas (compression), its internal energy increases (and it heats up).
Isobaric Process: The pressure of the system remains constant ( $p = \text{constant}$ ). In this case, all terms in the First Law equation ( $\Delta U = Q + W$ ) can be non-zero. The work done is calculated as $W = -p \Delta V$ .

Not clear?

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

reveal slowly — mark each step

EXAMPLE 1

A gas in a cylinder absorbs 200 J of heat from its surroundings. Simultaneously, the gas expands, performing 70 J of work on the piston. Calculate the change in the internal energy of the gas.

1
Identify the system and processes: The system is the gas. It absorbs heat and expands.

EXAMPLE 2

A fixed mass of an ideal gas is held in a cylinder by a piston at a constant pressure of 2.5 x 10^5 Pa. The gas is heated, and its volume increases from 1.2 x 10^-3 m^3 to 1.9 x 10^-3 m^3. During this process, 450 J of thermal energy is supplied to the gas. Calculate the change in the internal energy of the gas.

1
Identify given values and signs:

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAThe first law of thermodynamics

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Key terms for this topic — skim now; the Check step will test them.

internal energy (U): The internal energy (U) of a body is the total sum of all the random kinetic energies (due to particle motion) and potential energies (due to forces between particles) of these particles. It’s a 'state function', meaning its value depends only on the current condition of the system, not how it arrived at that state.
conservation of energy: The First Law of Thermodynamics is a powerful restatement of the principle of conservation of energy. It tells us that energy cannot be created or destroyed, only transferred or transformed.
In the equation ( $\Delta$: $W = -p\Delta V$ , because $p\Delta V$ is work done by the gas, while $W$ in the equation is work done on the gas.
Isothermal process: A process that occurs at a constant temperature. For an ideal gas, this means $\Delta U = 0$ , so the First Law becomes $Q = -W$ .
Isovolumetric (isochoric) process: A process that occurs at a constant volume ( $\Delta V = 0$ ). No work is done ( $W = 0$ ), so the First Law becomes $\Delta U = Q$ .

10Apply

Quick check

Write your answer first, then compare it with the model one — the gap is what you would have lost.

0 of 6 answered

11Understand

Teach it back

If you can explain it simply, you own it — gaps here are marks you’d lose.

Teach it back

Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.

Your explanation — no notes

12Remember

Revision flashcards

Guess first, then flip — retrieval beats re-reading.

Your guess — write it before you look

1 / 15 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

close the tab only after these stick

$\Delta U$ (Change in Internal Energy):
+ $\Delta U$ : System's internal energy increases (e.g., temperature rises).
- $\Delta U$ : System's internal energy decreases (e.g., temperature falls).
$Q$ (Energy Transferred via Heating):
+ $Q$ : Heat energy is transferred into the system (system gets hotter).
- $Q$ : Heat energy is transferred away from the system (system cools down).
$W$ (Work Done):
+ $W$ : Work is done on the system (e.g., a gas is compressed, energy input).
- $W$ : Work is done by the system (e.g., a gas expands, energy output).

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Question 1 of 29702/42 · Q3(b)(iv)

PhysicsMay/June 2025

(b) (iv)

Explain whether the work done on the block is positive or negative.

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/M/J/25

Question 2 of 29702/42 · Q3(b)(i)

PhysicsOctober/November 2024

(b) (i)

Show that the magnitude of the work done on the substance when it vaporises is 1.7 kJ.

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/O/N/24

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

Why is internal energy called a state function?

What is the difference between heat and internal energy?

Does the First Law of Thermodynamics apply to all types of systems?

Can you summarize the First Law for different processes (isothermal, adiabatic, etc.)?

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do 9702/42 · Q3(b)(iv) on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

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Revise The first law of thermodynamics for Cambridge 9702 (16.2)

Looking for free Physics 9702 notes on The first law of thermodynamics? This A-Level lesson covers official syllabus point 16.2 on Paper 4: A Level Structured Questions. Use the visual steps, key terms, and quick checks below — then practise a real 9702 past-paper question with mark-scheme marking. Built for students searching "The first law of thermodynamics 9702 revision", "9702 16.2 notes", or a free alternative to paid note sites.

Full Physics 9702 course Mark a past paper →9702 subject hub

16.2Physics25 min readPREMIUM

The first law of thermodynamics

BY THE END, YOU CAN…

exam checklist — tick these off

1Define internal energy and explain how it relates to temperature.
2State and apply the First Law of Thermodynamics (ΔU = Q + W).
3Interpret the signs of Q, W, and ΔU correctly in thermodynamic processes.
4Apply the First Law to isovolumetric, isothermal, and adiabatic processes.
5Calculate the work done by a gas during expansion at constant pressure (W = pΔV).

STUDY MODE

READING

TYPEFACE

Literata — a reading serif made for e-books; set like a textbook, sits well beside maths.

SIZE

AIRY SPACING

Wider letter and line spacing. The one change with solid evidence for readers who lose their place — worth a try whoever you are.

Pick whatever reads fastest for you — it differs from person to person, by up to a third.

EXPLAIN SIMPLER

The same notes in plain English — no jargon. Key points and exam tips stay.

DIAGRAM NOTES

A small sketch beside each block of notes.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The first law of thermodynamics

Cambridge 9702 Paper 4 — The first law of thermodynamics (16.2). Senpai Corner diagram-backed pilot with premium structure and live visuals.

1
$\Delta U$ (Change in Internal Energy):
2
+ $\Delta U$ : System's internal energy increases (e.g., temperature rises).
3
- $\Delta U$ : System's internal energy decreases (e.g., temperature falls).
4
$Q$ (Energy Transferred via Heating):

What this topic covers

The official Cambridge syllabus points this lesson works through.

16.2.1
Recall and use $W = pΔV$ for the work done when the volume of a gas changes at constant pressure and understand the difference between the work done by the gas and the work done on the gas
16.2.2
Recall and use the first law of thermodynamics $ΔU = q + W$ expressed in terms of the increase in internal energy, the heating of the system (energy transferred to the system by heating) and the work done on the system

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

Figure Live interactive1 / 4

Loading simulation…

Fetching from 3JCN Physics Simulation by Thomas Nguyen…

Isobaric Process

Work done equals pΔV, the rectangle under the isobar.

If the embed does not load, use Open simulation.

3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0

Step-synced diagram — highlights what to look for in the simulation above.

Step 1

$\Delta U$ (Change in Internal Energy):

7 more simulations for this topic — run them in the Simulations section below

Simulations

Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.

7 simulations · 5 to start with

Start herein this order — each one shows a different piece of the topic

3JCN PhysicsStart here · 19702 16.2 · 9702 16.1 · IB B.4
Isovolumetric Pro
Heat a gas at constant volume and watch pressure rise with no work done
Why this one: Heat at constant volume: W = 0, so ΔU equals the heat supplied and pressure rises.
Try this
1. Heat the gas at constant volume and watch the pressure rise.
2. Read the work done on the PV diagram.
Look for With no volume change no work is done, so all heat supplied raises internal energy.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 29702 16.2 · 9702 16.1 · IB B.4
Isothermal Process
Compress a gas at constant temperature and trace the isotherm on a PV diagram
Why this one: Compress at constant temperature: ΔU = 0, so the work done on the gas leaves as heat.
Try this
1. Compress the gas at constant temperature and trace the isotherm.
2. Halve the volume and read the pressure.
Look for At constant T, pV is constant, so the isotherm is a hyperbola and all heat supplied equals work done.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 39702 16.2 · 9702 16.1 · IB B.4
Adiabatic Process
Compress a gas with no heat exchange; see temperature rise on the adiabat
Why this one: Compress with q = 0 and watch the temperature climb — the work done goes entirely into ΔU.
Try this
1. Compress the gas with no heat exchange and watch the temperature.
2. Compare the adiabat's steepness with an isotherm through the same point.
Look for With q = 0 the work done on the gas raises its internal energy, so temperature rises and the adiabat is steeper than an isotherm.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 49702 16.2 · 9702 16.1 · IB B.4
A Cyclic Process
Run a gas around a closed PV cycle and read net work from the enclosed area
Why this one: Run a closed loop on the p–V diagram and read the net work from the enclosed area.
Try this
1. Run the gas around the closed PV cycle.
2. Read the net work from the enclosed area.
3. Note that the internal energy change over the full cycle is zero.
Look for Over a complete cycle ΔU = 0, so net heat in equals net work out, the area enclosed.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
PhETStart here · 59702 15.2 · 16.2
Gas Properties
Pump particles into a box; hold one variable fixed and change the others.
Why this one: Push the piston in fast and watch the temperature jump: work done on the gas raises its internal energy.
Try this
1. On “Ideal”, hold T fixed and push the wall in to halve V — read the pressure.
2. Hold V fixed and heat the gas — plot p against T in your head: straight line through 0 K.
3. Pump in twice the particles at fixed V and T — p doubles.
Look for pV = nRT: each pair of variables scales exactly as the equation predicts.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)

More simulations2 more on this topic — core ones first

3JCN PhysicsCore9702 16.2 · IB B.4
4-Stroke Engine
Watch intake, compression, power and exhaust strokes of a 4-stroke engine
Try this
1. Watch the intake, compression, power and exhaust strokes in turn.
2. Identify which stroke does work on the piston.
Look for Only the power stroke delivers work; the other three strokes prepare and clear the cylinder.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsCore9702 16.2 · 9702 16.1 · IB B.4
T in Adiabatic Proc.
Track temperature change during an adiabatic compression or expansion
Try this
1. Compress adiabatically and track the temperature.
2. Expand adiabatically and track it again.
Look for Adiabatic compression heats the gas and adiabatic expansion cools it, with no heat flowing either way.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0

Key formulas

Tap any symbol to reveal exactly what it means and its units.

KEY RELATION

$Work_{by} = p \Delta V$

Tap a symbol — great for exam definitions

KEY RELATION

$\Delta U = Q + W$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

What is Internal Energy (U)?

Not clear?

Temperature and Internal Energy

Not clear?

Work Done on or by a Gas (W)

For a process occurring at a constant pressure $p$ , the work done by the gas as it expands by a volume $\Delta V$ is given by:

Work_{by} = p $\Delta V$

It's crucial to relate this to the term $W$ in the First Law equation, $\Delta U = Q + W$ . In the Cambridge A-Level Physics syllabus, $W$ represents the work done on the system. Therefore:

When a gas expands ( $\Delta V > 0$ ), it does positive work on the surroundings. The work done on the gas is negative: $W = -p \Delta V$ .
When a gas is compressed ( $\Delta V < 0$ ), the surroundings do positive work on the gas. The work done on the gas is positive: $W = -p \Delta V$ (since $\Delta V$ is negative, $W$ becomes positive).

Not clear?

Introducing the First Law of Thermodynamics

$\Delta U = Q + W$

Not clear?

Deciphering the Signs: Q, W, and $\Delta U$

$\Delta U$ (Change in Internal Energy):
+ $\Delta U$ : System's internal energy increases (e.g., temperature rises).
- $\Delta U$ : System's internal energy decreases (e.g., temperature falls).
$Q$ (Energy Transferred via Heating):
+ $Q$ : Heat energy is transferred into the system (system gets hotter).
- $Q$ : Heat energy is transferred away from the system (system cools down).
$W$ (Work Done):
+ $W$ : Work is done on the system (e.g., a gas is compressed, energy input).
- $W$ : Work is done by the system (e.g., a gas expands, energy output).

Not clear?

The First Law in Specific Processes

The First Law can be simplified for specific types of thermodynamic processes:

Isovolumetric (or Isochoric) Process: The volume of the system remains constant ( $\Delta V = 0$ ). Since no volume change occurs, no work is done ( $W = 0$ ). The First Law simplifies to $\Delta U = Q$ . All heat added goes directly into increasing the internal energy.
Isothermal Process: The temperature of the system remains constant ( $\Delta T = 0$ ). For an ideal gas, internal energy depends only on temperature, so the change in internal energy is zero ( $\Delta U = 0$ ). The First Law becomes $0 = Q + W$ , or $Q = -W$ . Any heat added to the system is immediately converted into work done by the system.
Adiabatic Process: No heat is transferred into or out of the system ( $Q = 0$ ). This can happen if the system is perfectly insulated or if the process occurs very rapidly. The First Law simplifies to $\Delta U = W$ . If the gas expands and does work, its internal energy decreases (and it cools down). If work is done on the gas (compression), its internal energy increases (and it heats up).
Isobaric Process: The pressure of the system remains constant ( $p = \text{constant}$ ). In this case, all terms in the First Law equation ( $\Delta U = Q + W$ ) can be non-zero. The work done is calculated as $W = -p \Delta V$ .

Not clear?

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

reveal slowly — mark each step

EXAMPLE 1

A gas in a cylinder absorbs 200 J of heat from its surroundings. Simultaneously, the gas expands, performing 70 J of work on the piston. Calculate the change in the internal energy of the gas.

1
Identify the system and processes: The system is the gas. It absorbs heat and expands.

EXAMPLE 2

1
Identify given values and signs:

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAThe first law of thermodynamics

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Key terms for this topic — skim now; the Check step will test them.

internal energy (U): The internal energy (U) of a body is the total sum of all the random kinetic energies (due to particle motion) and potential energies (due to forces between particles) of these particles. It’s a 'state function', meaning its value depends only on the current condition of the system, not how it arrived at that state.
conservation of energy: The First Law of Thermodynamics is a powerful restatement of the principle of conservation of energy. It tells us that energy cannot be created or destroyed, only transferred or transformed.
In the equation ( $\Delta$: $W = -p\Delta V$ , because $p\Delta V$ is work done by the gas, while $W$ in the equation is work done on the gas.
Isothermal process: A process that occurs at a constant temperature. For an ideal gas, this means $\Delta U = 0$ , so the First Law becomes $Q = -W$ .
Isovolumetric (isochoric) process: A process that occurs at a constant volume ( $\Delta V = 0$ ). No work is done ( $W = 0$ ), so the First Law becomes $\Delta U = Q$ .

10Apply

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11Understand

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12Remember

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Key takeaways

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$\Delta U$ (Change in Internal Energy):
+ $\Delta U$ : System's internal energy increases (e.g., temperature rises).
- $\Delta U$ : System's internal energy decreases (e.g., temperature falls).
$Q$ (Energy Transferred via Heating):
+ $Q$ : Heat energy is transferred into the system (system gets hotter).
- $Q$ : Heat energy is transferred away from the system (system cools down).
$W$ (Work Done):
+ $W$ : Work is done on the system (e.g., a gas is compressed, energy input).
- $W$ : Work is done by the system (e.g., a gas expands, energy output).

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Question 1 of 29702/42 · Q3(b)(iv)

PhysicsMay/June 2025

(b) (iv)

Explain whether the work done on the block is positive or negative.

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/M/J/25

Question 2 of 29702/42 · Q3(b)(i)

PhysicsOctober/November 2024

(b) (i)

Show that the magnitude of the work done on the substance when it vaporises is 1.7 kJ.

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/O/N/24

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Frequently asked

Why is internal energy called a state function?

What is the difference between heat and internal energy?

Does the First Law of Thermodynamics apply to all types of systems?

Can you summarize the First Law for different processes (isothermal, adiabatic, etc.)?

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Revise The first law of thermodynamics for Cambridge 9702 (16.2)

In simple terms

What this topic covers

Explore the concept

Simulations

Isovolumetric Pro

Isothermal Process

Adiabatic Process

A Cyclic Process

Gas Properties

4-Stroke Engine

T in Adiabatic Proc.

Key formulas

Full topic notes

What is Internal Energy (U)?

Temperature and Internal Energy

Work Done on or by a Gas (W)

Introducing the First Law of Thermodynamics

Deciphering the Signs: Q, W, and \(\Delta U\)

The First Law in Specific Processes

Worked examples

How it all connects

Glossary

Quick check

Teach it back

Revision flashcards

Key takeaways

Practice — then mark it

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss The first law of thermodynamics

Revise The first law of thermodynamics for Cambridge 9702 (16.2)

In simple terms

What this topic covers

Explore the concept

Simulations

Isovolumetric Pro

Isothermal Process

Adiabatic Process

A Cyclic Process

Gas Properties

4-Stroke Engine

T in Adiabatic Proc.

Key formulas

Full topic notes

What is Internal Energy (U)?

Temperature and Internal Energy

Work Done on or by a Gas (W)

Introducing the First Law of Thermodynamics

Deciphering the Signs: Q, W, and \(\Delta U\)

The First Law in Specific Processes

Worked examples

How it all connects

Glossary

Quick check

Teach it back

Revision flashcards

Key takeaways

Practice — then mark it

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss The first law of thermodynamics