DSE Physics | First Law of Thermodynamics: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary Physics

Students can compute delta U equals Q minus W and still freeze when asked whether work is positive when a gas is compressed. This lesson fixes the direction of three quantities once, and every later question becomes substitution.

Objectives — Students can: (1) apply delta U equals Q minus W to find a rise of 500 joules when 800 joules is absorbed and 300 joules of work is done by the gas; (2) compute work as p times delta V, getting 300 joules at 2.0 times 10 to the 5 pascals with a volume increase of 1.5 times 10 to the minus 3 cubic metres; (3) state that an isothermal process has delta U equal to zero and an adiabatic process has Q equal to zero; (4) use delta U equals n Cv delta T to get 250 joules for 0.5 moles with Cv of 12.5 and a 40 kelvin rise.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Heat and temperature	Roughly how much energy raises 1 kg of water by 1 degree Celsius?	About 4,200 joules
Gas laws	What happens to volume when temperature rises at constant pressure?	Volume increases, since volume is proportional to temperature
Conservation of energy	Can energy disappear?	No, it only changes form
Work done	How much work does 10 newtons over 3 metres do?	30 joules

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Opening conflict	Ask why a bicycle pump barrel heats up	Answer with compression and work done
10-22 min	Three quantities	Define Q, W and delta U with their directions	Write delta U equals Q minus W
22-34 min	Work as p delta V	Use 2.0 times 10 to the 5 pascals and 1.5 times 10 to the minus 3 cubic metres	Get 300 joules
34-46 min	Special processes	Tabulate isothermal, adiabatic and isobaric cases	Name which quantity is zero
46-56 min	Microscopic link	Internal energy depends only on temperature, so use n Cv delta T	Get 250 joules
56-60 min	Close	Ask students to state the signs during compression	Write the first exit question

03 Teaching script (read it out)

Scene 1: Fix the direction before touching the formula

In delta U equals Q minus W, Q is heat entering the system and W is work done by the system. The errors are rarely arithmetic: during compression the surroundings do work on the gas, so W is negative and the internal energy rises. Do not present three rearranged versions at once. Write one, then ask of each quantity whether it goes in or out.

Board line: ask which way the energy flows before asking how much.

Scene 2: Work as p delta V only holds at constant pressure

At constant pressure the work is pressure times change in volume, so 2.0 times 10 to the 5 pascals times 1.5 times 10 to the minus 3 cubic metres gives 300 joules. If volume is fixed, W is zero and all the heat becomes internal energy. If pressure varies, the formula fails and the area under the p-V graph is needed. Students who use it on isothermal compression find their numbers no longer balance.

Board line: the moment pressure varies, p delta V is out.

Scene 3: Internal energy depends only on temperature

For an ideal gas the internal energy is set by temperature alone. Isothermal means delta U is zero, so all heat absorbed becomes work done; adiabatic means Q is zero, so doing 120 joules of work drops the internal energy by 120 joules and the temperature falls. Microscopically, delta U equals n Cv delta T gives 250 joules for 0.5 moles, Cv of 12.5 and a 40 kelvin rise, which must match the macroscopic Q minus W.

Board line: unchanged temperature means unchanged internal energy.

04 Question types and worked steps

Type	Example	Steps	Answer
Internal energy change	A gas absorbs 800 joules and does 300 joules of work	Write delta U equals Q minus W, treat heat in and work out as positive, then subtract	Plus 500 joules
Work done	Constant pressure 2.0 times 10 to the 5 pascals, volume from 1.0 to 2.5 times 10 to the minus 3 cubic metres	Find delta V, multiply by pressure, keep pascal cubic metres as joules	300 joules
Special process	Isothermal compression with 250 joules of work done on the gas	Delta U is zero, so Q equals W, and the system releases 250 joules	250 joules released
Microscopic calculation	0.5 moles heated by 40 kelvin with Cv of 12.5	Use n Cv delta T, multiply, and state that it is an increase	Plus 250 joules

Sanity check (teach this): Sign check: write the in-positive and out-positive statements before any numbers, and label the final answer as absorbed or released. Order check: 0.5 moles rising 40 kelvin changes internal energy by a few hundred joules, so an answer in tens of thousands points to a mole or temperature slip. Energy check: the largest of the three numbers is not always Q, since W can be negative while Q is small. Isothermal questions must state delta U equal to zero, which is the step most often omitted.

05 Independent practice (3 questions)

Question	Answer	What to watch
A gas releases 200 joules while 350 joules of work is done on it. Find delta U.	Plus 150 joules	Two negative signs to track
At constant pressure 1.5 times 10 to the 5 pascals, volume falls by 2.0 times 10 to the minus 3 cubic metres. Find W.	Minus 300 joules, work done on the gas	The sign carries the direction
Adiabatic expansion does 120 joules of work. What happens to internal energy and temperature?	Delta U is minus 120 joules and temperature falls 19.2 kelvin	Did you use Q equals zero

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating work as positive during compression	The surroundings do 300 joules, so W is plus 300	W is work done by the system. Work done on the system is minus 300, otherwise the internal energy comes out falling.
Claiming internal energy rises in an isothermal process	Isothermal compression raises internal energy because work is done	For an ideal gas internal energy depends only on temperature, so delta U is zero and the work leaves as heat.
Using p delta V when pressure changes	Pressure rises from 1 to 3 atmospheres, so I multiply 3 by delta V	That formula requires constant pressure. When pressure varies, use the area under the p-V graph or the data the question supplies.
Mixing units	I multiply p in atmospheres by V in cubic centimetres	Convert first: 1 atmosphere is about 1.0 times 10 to the 5 pascals and 1 cubic centimetre is 10 to the minus 6 cubic metres.

07 Homework (paper, section and question numbers)

Physics chapter index — Revise the gas laws, then do questions 1 and 2
Gas laws — Do question 4 on isobaric expansion, then questions 3 to 5
Electric field and potential — Optional: compare the meaning of work in mechanics and in thermodynamics

08 Exit check (last 5 minutes)

Question	Answer	Decision
Absorbing 800 joules and doing 300 joules of work, what is delta U?	Plus 500 joules	Error: wrote 1,100 joules
Work at constant pressure 2.0 times 10 to the 5 pascals with delta V of 1.5 times 10 to the minus 3 cubic metres	300 joules	Error: forgot to convert cubic centimetres
Delta U in an isothermal process	Zero	Error: said it equals Q
Q in an adiabatic process	Zero	Error: said it equals minus W

09 Tutor FAQ

Why do textbooks show delta U equals Q minus W and also Q plus W?

Both are correct. The difference is whether W means work done by the system or work done on it. Pick one convention per question and stay with it.

Why does a compressed gas heat up?

The surroundings do work on the gas, so energy enters the system. If heat cannot escape quickly, internal energy and temperature rise, which is exactly what happens in a bicycle pump.

When can I use delta U equals n Cv delta T?

For an ideal gas in any process, because internal energy depends only on temperature. Cv depends on the gas, and a monatomic gas is about 12.5 joules per mole per kelvin.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.