The shape of the lesson
By the end of the lesson, learners can
- Adding up the molecules: Internal energy is not the heat in a body or its temperature; it is the grand total of the random kinetic and potential energies of every molecule.
- A function of state: Internal energy depends only on the present state of the system, not on how it got there, so any change ΔU between two states is the same by any route.
- 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
Internal energy is the sum of the random kinetic and potential energies of a system's molecules. It is a function of state, and for an ideal gas it depends only on the temperature.
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, The energy inside.. Internal energy is the bookkeeping total of all the molecular energies. 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: Adding up the molecules | Internal energy is not the heat in a body or its temperature; it is the grand total of the random kinetic and potential energies of every molecule. Project tab 1 of the lab (Internal energy) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: A function of state | Internal energy depends only on the present state of the system, not on how it got there, so any change ΔU between two states is the same by any route. | Think-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, 77 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 A function of state, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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.
Internal energy: topic worksheet
7 questions, 77 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.
Fig. 1.1 represents the molecules of a fixed mass of ideal gas in random motion.
(a) State what is meant by the internal energy of a system. [2]
(b) Explain why, for an ideal gas, the internal energy is entirely kinetic. [2]
(c) The temperature of the gas is raised. State and explain the effect on its internal energy. [2]
(d) State whether the internal energy of this gas depends on its pressure and volume separately or only on its temperature, and justify your answer briefly. [2]
(e) The gas is 0.50 mol of a monatomic ideal gas at 300 K. Using U = ₃⁄₂ nRT, calculate its internal energy. [3]
(f) The gas is now compressed to half its volume at constant temperature. State the change in its internal energy. [1]
This question is about the meaning of internal energy.
(a) Explain what is meant by the internal energy of a system. [2]
(b) State why the motion of the molecules is described as random. [1]
(c) For an ideal gas, state and explain which type of molecular energy contributes to the internal energy. [2]
(d) Hence explain why the internal energy of an ideal gas depends only on its temperature. [2]
(e) State what happens to the internal energy of an ideal gas if its temperature is held constant while p and V change. [1]
(f) State what happens to the internal energy of a fixed mass of ideal gas if its kelvin temperature is doubled. [2]
A real gas is compared with an ideal gas.
(a) State the two contributions to the internal energy of a substance. [2]
(b) Explain why the internal energy of an ideal gas has no potential-energy contribution. [2]
(c) A fixed mass of ideal gas is heated from 300 K to 600 K. State the ratio of its final to initial internal energy, and explain. [3]
(d) Explain why two samples of the same ideal gas at the same temperature but different volumes have the same internal energy per mole. [2]
(e) State whether the internal energy depends on the current state of the gas or on the path taken to reach it. [1]
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)Internal energy is the sum of the random distribution of kinetic and potential energies ✓ of all the molecules of the system ✓
- (b)An ideal gas has no intermolecular forces ✓, so there is no molecular potential energy; the internal energy is therefore all kinetic ✓
- (c)The internal energy increases ✓; a higher temperature means a greater mean molecular kinetic energy (₃⁄₂ kT per molecule) ✓
- (d)It depends only on the temperature ✓; since U is all kinetic and the mean KE depends only on T, the same T gives the same U whatever the p and V ✓
- (e)U = ₃⁄₂ nRT = 1.5 × 0.50 × 8.31 × 300 ✓✓ = 1.9 × 10³ J ✓
- (f)Zero (the temperature is unchanged, so U is unchanged) ✓
- (a)The sum of the random distribution of the kinetic and potential energies of the molecules of the system ✓✓
- (b)The molecules move in all directions with a wide range of speeds ✓
- (c)Only kinetic energy ✓: an ideal gas is assumed to have no intermolecular forces, so no molecular potential energy ✓
- (d)The internal energy is the total molecular kinetic energy, and the mean kinetic energy is proportional to temperature ✓✓
- (e)It is unchanged (ΔU = 0) ✓
- (f)U is proportional to T, so the internal energy also doubles ✓✓
- (a)The random kinetic energy and the potential energy of the molecules ✓✓
- (b)An ideal gas is assumed to have no intermolecular forces, so there is no molecular potential energy ✓✓
- (c)U ∝ T, so the ratio is 600/300 = 2 ✓✓; the internal energy depends only on temperature ✓
- (d)Internal energy of an ideal gas depends only on temperature, not on volume or pressure ✓✓
- (e)Only on the current state: internal energy is a function of state ✓
- (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.
The internal energy of a system is:
- Athe total of the random kinetic and potential energies of its molecules
- Bthe heat stored inside it
- Cthe work it can do
- Dthe same as its temperature
For an ideal gas the internal energy is entirely kinetic because:
- Athere are no intermolecular forces, so no molecular potential energy
- Bthe molecules are stationary
- Cthe gas has no temperature
- Dthe potential energy is converted to heat
Raising the temperature of a fixed mass of ideal gas:
- Aincreases its internal energy
- Bdecreases its internal energy
- Cleaves its internal energy unchanged
- Donly changes its pressure
Two equal masses of the same ideal gas are at the same temperature but different pressures. Their internal energies are:
- Aequal, because internal energy depends only on temperature
- Bdifferent, because the pressures differ
- Cdifferent, because the volumes differ
- Dimpossible to compare
Quiz key
Internal energy is the sum of the random kinetic and potential energies of all the molecules; it is not heat, work or temperature.
An ideal gas is assumed to have no forces between molecules, so there is no molecular potential energy and the internal energy is all kinetic.
Higher temperature means a greater mean molecular kinetic energy, so the internal energy increases.
For an ideal gas U depends only on temperature, so equal temperatures give equal internal energies regardless of pressure or volume.
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.