The shape of the lesson
By the end of the lesson, learners can
- Work done by a gas: When a gas expands at constant pressure it pushes a piston back, doing work W = pΔV.
- The first law, and its signs: The first law of thermodynamics is ΔU = q + W, where q is the heat supplied to the gas and W is the work done on the gas.
- Internal energy: the total random kinetic and potential energy of the molecules; for an ideal gas it depends only on temperature.
- Work at constant pressure: W = pΔV is the work done by the gas, shown as the area under the line on a p–V diagram.
- First law: ΔU = q + W, with q the heat added to the gas and W the work done on the gas (W = −pΔV for an expansion).
One picture every idea returns to
A gas expanding at constant pressure does work W = pΔV. The first law, ΔU = q + W, says the increase in internal energy equals the heat added plus the work done on the gas.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Heat, work, and the first law.. The first law is just conservation of energy for a gas: whatever heat goes in either raises the internal energy or comes back out as work. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Work done by a gas | When a gas expands at constant pressure it pushes a piston back, doing work W = pΔV. Project tab 2 of the lab (The first law) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: The first law, and its signs | The first law of thermodynamics is ΔU = q + W, where q is the heat supplied to the gas and W is the work done on the gas. | Mix-Pair-Share |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (7 questions, 81 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The first law, and its signs, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
The first law of thermodynamics: topic worksheet
7 questions, 81 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A gas expands at a constant pressure of 1.5 × 10⁵ Pa, its volume increasing from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³. Fig. 2.1 shows this change on a p–V diagram.
(a) Show that the work done by a gas expanding at constant pressure is W = pΔV. [2]
(b) Calculate the work done by the gas in this expansion. [2]
(c) State what the work done is represented by on the p–V diagram. [1]
(d) During the expansion, 900 J of thermal energy is supplied to the gas. Use the first law to find the change in its internal energy. [3]
(e) State and explain whether the temperature of the gas rises, falls or stays the same. [2]
(f) State the sign of the work done on the gas if instead it were compressed at constant pressure. [1]
(g) The gas instead expands by the same amount with no thermal energy supplied. State and explain what happens to its temperature. [2]
A fixed mass of ideal gas is taken through several separate processes, each starting from the state S shown in Fig. 3.1.
(a) State the first law of thermodynamics in the form ΔU = q + W, defining each symbol and its sign convention. [3]
(b) The gas is heated at constant volume, gaining 600 J of thermal energy. State the work done on the gas and the change in internal energy. [2]
(c) In a separate isothermal expansion the gas does 400 J of work. State the change in internal energy and the heat transferred. [2]
(d) In a separate adiabatic compression, 250 J of work is done on the gas. State the change in internal energy and what happens to the temperature. [2]
(e) Using the first law, explain why compressing a gas rapidly raises its temperature. [2]
(f) Explain why ΔU = 0 for any process that returns the gas to its original state. [2]
A gas at a constant pressure of 1.5 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³, as shown in Fig. 6.1.
(a) Write the expression for the work done by a gas at constant pressure. [1]
(b) Calculate the work done by the gas during the expansion. [2]
(c) State the sign of the work done on the gas, and explain. [2]
(d) State what the work done corresponds to on the p-V diagram. [1]
(e) The gas is then compressed back to 2.0 × 10⁻³ m³ at a constant pressure of 2.5 × 10⁵ Pa. Calculate the work done on the gas. [2]
(f) State the net work done on the gas over the two stages. [2]
0.050 mol of an ideal monatomic gas is heated at constant volume; 250 J of thermal energy is supplied. Parts (a) to (c) use the first law; parts (d) and (e) use internal energy.
(a) State the first law of thermodynamics, defining each symbol. [2]
(b) State the work done on the gas at constant volume, and explain. [2]
(c) Hence state the increase in internal energy. [1]
(d) Explain, in terms of the molecules, what this increase in internal energy represents. [2]
(e) State and explain what happens to the temperature of the gas. [2]
(f) If instead the gas were allowed to expand as it was heated, state whether the temperature rise would be larger or smaller for the same heat supplied. [2]
An ideal gas expands isothermally. This question links internal energy with the first law.
(a) State what happens to the internal energy of the gas, and explain. [2]
(b) Apply the first law to this process. [2]
(c) State the direction of heat flow as the gas expands isothermally. [2]
(d) The gas does 600 J of work on the surroundings. State the work done on the gas and the heat supplied. [2]
(e) Describe the isothermal expansion on a p-V diagram and state what the area under it represents. [2]
(f) Explain why the temperature is unchanged even though the gas absorbs heat. [1]
A gas is compressed rapidly so that no heat enters or leaves it (an adiabatic change).
(a) State the value of q for an adiabatic process. [1]
(b) Apply the first law to an adiabatic change. [2]
(c) The gas is compressed, so work is done on it. State and explain what happens to its internal energy and temperature. [3]
(d) 300 J of work is done on the gas. State the change in internal energy. [1]
(e) Explain, in molecular terms, why an adiabatic compression raises the temperature. [2]
(f) State and explain what happens to the temperature during an adiabatic expansion. [2]
A fixed mass of ideal gas is taken through a cycle and returns to its original state, as suggested by Fig. 10.1. This question applies the first law to the steps and uses the fact that internal energy is a function of state.
(a) In the first step (constant volume), 200 J of heat is supplied. State the work done on the gas and the change in internal energy. [2]
(b) In the second step (constant-pressure expansion), 500 J of heat is supplied and the gas does 180 J of work. Calculate the change in internal energy. [2]
(c) After the complete cycle the gas returns to its starting state. State the total change in internal energy over the cycle, and explain. [3]
(d) Over the whole cycle the net work done on the gas is +120 J. Calculate the net heat transferred, and state its direction. [3]
(e) Explain why the internal energy at the end of the second step is greater than at the start of the cycle. [2]
Answer key and marking notes
- (a)At constant pressure the force on the piston is F = pA ✓; if it moves a distance d, W = Fd = pA d = p(Ad) = pΔV ✓
- (b)W = pΔV = 1.5 × 10⁵ × (5.0 − 2.0) × 10⁻³ ✓ = 1.5 × 10⁵ × 3.0 × 10⁻³ = 450 J ✓
- (c)The area under the line (between the line and the V axis) ✓
- (d)ΔU = q + W, with W the work done on the gas = −pΔV = −450 J ✓; ΔU = 900 + (−450) ✓ = +450 J ✓
- (e)The temperature rises ✓; ΔU is positive, so the internal energy increases, and for an ideal gas a higher internal energy means a higher temperature ✓
- (f)Positive (work is done on the gas) ✓
- (g)q = 0, so ΔU = W = −450 J ✓; the internal energy falls, so the temperature falls ✓
- (a)ΔU is the increase in internal energy ✓; q is the thermal energy supplied to the gas (positive when heat is added) ✓; W is the work done on the gas (positive when the gas is compressed, negative when it expands) ✓
- (b)Constant volume means ΔV = 0, so W = 0 ✓; ΔU = q + 0 = +600 J ✓
- (c)Isothermal means T constant, so ΔU = 0 ✓; then 0 = q + W with W = −400 J, so q = +400 J (heat is absorbed) ✓
- (d)q = 0 and W = +250 J, so ΔU = +250 J ✓; the internal energy rises, so the temperature rises ✓
- (e)A rapid compression gives little time for heat to leave, so q ≈ 0 ✓; work is done on the gas (W positive), so ΔU = W is positive and the temperature rises ✓
- (f)Internal energy is a function of state ✓; returning to the same state means the same internal energy, so the net change ΔU is zero ✓
- (a)W = pΔV ✓
- (b)W = 1.5 × 10⁵ × (5.0 − 2.0) × 10⁻³ ✓ = 4.5 × 10² J ✓
- (c)Negative ✓: the gas expands and does work on the surroundings, so work done on the gas is −450 J ✓
- (d)The area under the line ✓
- (e)W on gas = pΔV = 2.5 × 10⁵ × 3.0 × 10⁻³ ✓ = +7.5 × 10² J ✓
- (f)Net = −450 + 750 = +3.0 × 10² J done on the gas ✓✓
- (a)ΔU = q + W ✓: q is the heat supplied to the gas, W is the work done on the gas ✓
- (b)W = 0 ✓: the volume is constant, so no work is done ✓
- (c)ΔU = q = +250 J ✓
- (d)An increase in the mean kinetic energy of the molecules (faster random motion) ✓✓
- (e)The temperature rises ✓, because the internal energy (and mean kinetic energy) of an ideal gas is proportional to T ✓
- (f)Smaller ✓: some energy leaves as work done by the gas, so less is left to raise the internal energy ✓
- (a)It is unchanged ✓: the internal energy of an ideal gas depends only on temperature, which is constant ✓
- (b)ΔU = 0, so q = −W ✓: the heat supplied equals the work done by the gas ✓
- (c)Heat flows into the gas ✓✓
- (d)Work done on the gas = −600 J ✓; q = +600 J ✓
- (e)A curve (pV = constant) ✓; the area under it is the work done by the gas ✓
- (f)All the absorbed heat leaves again as work, so the internal energy and hence the temperature stay the same ✓
- (a)q = 0 ✓
- (b)ΔU = W ✓: the change in internal energy equals the work done on the gas ✓
- (c)Work is done on the gas (W positive), so ΔU is positive ✓: the internal energy rises ✓ and the temperature rises (U ∝ T) ✓
- (d)ΔU = +300 J ✓
- (e)Molecules rebound from the inward-moving piston with greater speed, gaining kinetic energy ✓✓
- (f)The temperature falls ✓: with q = 0 the gas does work as it expands, so ΔU is negative ✓
- (a)W = 0 (constant volume) ✓; ΔU = q = +200 J ✓
- (b)ΔU = q + W = 500 + (−180) ✓ = +320 J ✓
- (c)ΔU over the cycle = 0 ✓: internal energy is a function of state, and the gas returns to the same temperature ✓✓
- (d)Over the cycle 0 = q_net + W_net, so q_net = −120 J ✓✓: 120 J of heat is released by the gas ✓
- (e)The temperature is higher there, and the internal energy of an ideal gas depends only on temperature ✓✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
When a gas expands at constant pressure p by a volume ΔV, the work done by the gas is:
- ApΔV
- BΔV / p
- C½pΔV
- Dp / ΔV
In the first law ΔU = q + W, the symbol W stands for:
- Athe work done on the gas
- Bthe work done by the gas
- Cthe weight of the gas
- Dthe heat lost
A gas absorbs 500 J of heat and does 200 J of work as it expands. Its internal energy changes by:
- A+300 J
- B+700 J
- C−300 J
- D−700 J
During an isothermal change of an ideal gas, the change in internal energy is:
- Azero, because the temperature does not change
- Bequal to the heat supplied
- Cequal to the work done
- Dalways negative
Quiz key
Work done by the gas at constant pressure is W = pΔV, the area under the line on a p–V diagram.
In this convention W is the work done on the gas; an expansion (gas doing work) makes W negative.
ΔU = q + W = 500 + (−200) = +300 J, since the gas doing 200 J of work means W = −200 J on the gas.
Isothermal means constant temperature, and for an ideal gas U depends only on T, so ΔU = 0.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.