Thermodynamics Revision Notes — 9702 Paper 4

Internal energy

Syllabus requirement · §16.1.1

“

understand that internal energy is determined by the state of the system and that it can be expressed as the sum of a random distribution of kinetic and potential energies associated with the molecules of a system

”

From one special case to a universal quantity

The Ideal Gases note closed with a tidy result: an ideal gas stores its energy entirely as molecular kinetic energy, $U = \tfrac{3}{2}NkT$ . That tidiness leaned on one assumption — no forces between molecules, hence nothing stored in the interactions. Solids and liquids are different: their molecules pull and push on each other constantly, so energy also sits in the separations. This section makes "internal energy" precise for every kind of system, and then turns it into the accounting variable the rest of the note runs on.

Learn the definition word-perfectly now, because it is asked verbatim and paid clause by clause (9702/41 M/J 2025 Q4(a)(i), with identical schemes across variants and years):

the sum of a random distribution of kinetic and potential energies associated with the molecules of a system.

The scheme splits that sentence into two payable halves: "sum of potential energy and kinetic energy" (B1), then "(total) energy of random motion of particles" (B1). Both clauses must appear — "sum of KE and PE" alone misses why these energies qualify as internal (their randomness), while "energy of random motion" alone drops the potential store entirely.

One sentence to carry

Internal energy is the sum of the random kinetic and potential energies of a system's molecules — fixed by the state of the system, never by the route taken to reach it.

Unpacking the two stores

Store 1 — molecular kinetic energy. The energy of the molecules' random motion. It scales with temperature — the Ideal Gases note made this exact ( $E_K = \tfrac{3}{2}kT$ per molecule). Whenever a question asks what "tracks temperature", this store is the answer.

Store 2 — molecular potential energy. Energy stored in the forces between molecules. Picture tiny springs joining neighbours: stretch them (pull molecules apart) and the store fills; let them slacken (molecules settle closer) and it empties. This store tracks separation and phase — stretching a wire, melting a solid and boiling a liquid all load it — and it has nothing to do with height.

Now the contrast examiners probe. A ball falling off a bench gains kinetic energy — but that is the ordered motion of the whole object, every molecule drifting the same way on top of their random dance. None of it is internal energy. Likewise a book lifted onto a shelf gains gravitational potential energy, but the shelf does not warm the room by storing books. Internal energy counts only the random, molecule-scale stores; whole-object kinetic and potential energy live outside the ledger. A falling ball is no hotter for having fallen.

One quantity, two zoom levels. Panel (a): the macroscopic view — a lifted or falling object carries whole-body kinetic and potential energy, ordered and directional, and NONE of it is internal energy. Panel (b): zoomed into the material — molecules race randomly (molecular KE, tracking temperature) and tug on neighbour springs (molecular PE, tracking separation). Internal energy = the sum of those random molecular stores.

A state function, not a travel diary

The syllabus prints a quiet half-clause that makes the whole subject workable: internal energy is determined by the state of the system. Specify the state — pressure, volume, temperature — and $U$ is fixed, just as a bank balance is fixed by the account today regardless of which shops you visited yesterday. Two identical gas samples in the same state hold the same internal energy however differently they arrived.

For an ideal gas the point is sharpest, because $U$ rides on temperature alone: any journey between two temperatures deposits exactly the same $\Delta U$ , whatever the order of operations along the way. The next demonstration banks that idea before the past papers test it.

Invented demo — two paths, one destination

A sample of an ideal gas contains $N = 4.0 \times 10^{23}$ molecules and starts at thermodynamic temperature $T_1 = 300\ \text{K}$ . It is taken to a final state at $T_2 = 340\ \text{K}$ by two different routes:
Path A: heat at constant volume from $300\ \text{K}$ to $340\ \text{K}$ , then compress at constant temperature $340\ \text{K}$ .
Path B: compress at constant temperature $300\ \text{K}$ , then heat at constant volume from $300\ \text{K}$ to $340\ \text{K}$ .
Show that both routes give the same overall change in internal energy.

Show full working

1
Name the working rule and its pieces. For an ideal gas,
$\Delta U = \tfrac{3}{2}\,Nk\,\Delta T$
with $N = 4.0\times10^{23}$ molecules, $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ and $\Delta T = 340 - 300 = 40\ \text{K}$ — positive because the temperature rises.
Identify all three factors before multiplying — the habit that makes the sign and the power of ten automatic rather than hopeful.
2
Evaluate once:
$\tfrac{3}{2} \times 1.38\times10^{-23} \times 40 = 8.28\times10^{-22}\ \text{J per molecule}$ $\Delta U = 8.28\times10^{-22} \times 4.0\times10^{23} = +330\ \text{J}$
Per-molecule first, then scale up: the powers of ten cancel (−22 + 23), leaving a modest few hundred joules.
3
Walk Path A. Heating leg: $\Delta U_{\text{heat}} = +330\ \text{J}$ . Compression at constant temperature $340\ \text{K}$ : $U$ depends on $T$ alone and $T$ does not change, so $\Delta U_{\text{compress}} = 0$ . Total: $+330\ \text{J}$ .
The isothermal leg contributes nothing to the ledger — squeezing the gas rearranges energy but cannot change a quantity pinned to temperature.
4
Walk Path B. Compression at constant $300\ \text{K}$ : $\Delta U = 0$ . Heating leg: $\Delta U = +330\ \text{J}$ . Total: $+330\ \text{J}$ — identical to Path A, as the state-function promise demanded.
Different journeys, one destination, one ledger entry. This is exactly the reasoning the §06 cycle questions pay for.

Answer

ΔU = ³⁄₂NkΔT = ³⁄₂ × 4.0×10²³ × 1.38×10⁻²³ × 40 ≈ +330 J on either path — the constant-temperature legs contribute zero because U tracks T alone.

When a question changes the route but keeps the start and end states, write the sentence: internal energy is determined by the state, so the change depends only on start and end. It is a bankable mark and it saves recomputing.

The ideal-gas special case, upgraded

Fold the Ideal Gases conclusion into the new vocabulary. No intermolecular forces (except during instant collisions) means the spring store stands empty: molecular potential energy $= 0$ . What remains of internal energy is pure random kinetic energy, and since that scales with temperature,

U = \tfrac{3}{2}NkT \qquad\text{(PE zero, so all of } U \text{ is kinetic)}

The examiner asks for precisely this two-clause upgrade (9702/44 O/N 2025 Q3(a)), paying "total kinetic energy associated with random motion of molecules" (B1) and "potential energy is zero" (B1). Mind the logic direction when the question says "with reference to molecular potential energy": the PE is zero because ideal-gas molecules exert no forces on each other — tie the zero to its cause.

The definition, marked to scheme

9702/41 M/J 2025 Q4(a)(i)2 marks

State what is meant by the internal energy of a system. [2]

Show full working

1
Clause one — both stores, summed: the internal energy of a system is the sum of the potential energy and the kinetic energy (of its particles) (B1).
'Sum' is doing real work: neither store may be dropped, even though for some systems one of them happens to be empty.
2
Clause two — randomness: it is the (total) energy of the random motion of the particles (B1).
'Random' is what makes these energies internal rather than whole-object — and it is the clause the second B1 lives on.

Answer

The sum of the potential and kinetic energies associated with the random motion of the molecules of the system.

Two clauses, two marks, near-identical wording every sitting. Write them as two separate sentences so neither blurs into the other.

The ideal-gas variant, with the cause attached

9702/44 O/N 2025 Q3(a)2 marks

With reference to molecular kinetic energy and molecular potential energy, explain what is meant by the internal energy of an ideal gas. [2]

Show full working

1
Kinetic clause: the internal energy of an ideal gas is the total kinetic energy associated with the random motion of its molecules (B1).
Same first clause as the general definition — the random-motion KE survives untouched.
2
Potential clause, with its cause: there are no intermolecular forces between ideal-gas molecules (except during instantaneous collisions), so the molecular potential energy is zero (B1).
The scheme pays for the explicit zero AND its justification — 'small' or 'negligible' will not do, and an unexplained zero risks missing the reference the question demanded.

Answer

Total kinetic energy of the randomly moving molecules; molecular potential energy is zero because there are no intermolecular forces.

This pairing — random KE plus PE-is-zero-because-no-forces — is the exact shape the scheme reprints year after year.

Common mistakes

Defining internal energy as "the kinetic energy of the molecules".
The sum of BOTH stores: random kinetic AND potential energies of the molecules.
The potential store earns its own B1 — even in the ideal-gas variant, where the expected answer states explicitly that it is zero.
Counting whole-object energy as internal ("a book on a high shelf has more internal energy").
Only RANDOM molecular energies count; ordered whole-object motion and height sit outside the internal ledger.
Gravitational potential energy belongs to the book–Earth system, not to the molecules' separations. The zoom level is the entire distinction.
Treating internal energy as path-dependent ("more heating on the way means more internal energy at the end").
$U$ is determined by the STATE: equal states carry equal $U$ , whatever the history.
The state-function clause is printed in the syllabus outcome and becomes the engine of every §06 cycle argument.
Saying ideal-gas molecular potential energy is "small" or omitting it altogether.
It is ZERO — because ideal-gas molecules exert no intermolecular forces (apart from instant collisions).
The B1 demands the explicit zero tied to the force-free assumption; hedged wording leaves the marker nothing to credit.

Your turn

The recall variant, the two-store explanations the examiner pays three marks apiece for, and the doorway into §04.

19702/42 O/N 2020 Q2(a)2 marks
State what is meant by the internal energy of a system.
Stuck? Show hint
Two clauses: what is summed, and what kind of motion qualifies it.
Show solution
1. 1
  The sum of the potential energy and kinetic energy (of the particles) (B1);
  Both stores named and added — the first payable half.
2. 2
  (total) energy of the random motion of the particles (B1).
  The randomness clause again. Nearly every sitting pays this exact pair.
Answer
The sum of the potential and kinetic energies of the particles — the total energy of their random motion.
29702/42 M/J 2024 Q3(b)(i)(ii)6 marks
With reference to molecular kinetic and potential energies, describe and explain how the internal energy of the system changes when:
(i) a gas is heated at constant volume so that its temperature increases;
(ii) a wire is stretched within its elastic limit at constant temperature.
Stuck? Show hint
Each answer is three clauses: what happens to separation (PE), what happens to temperature (KE), then the verdict on U.
Show solution
1. 1
  (i) Constant volume means the molecular separations do not change: no change in separation, so no change in molecular potential energy (B1).
  Rigid walls lock the separations — the potential store cannot move, however much energy arrives.
2. 2
  The temperature increases, so the molecular kinetic energy increases (B1);
  Temperature rise IS kinetic-energy rise — the §01 Store-1 statement doing its job.
3. 3
  KE increases while PE is unchanged, so the internal energy increases (B1).
  The verdict clause adds the stores explicitly — schemes pay separately for the conclusion, so always close with it.
4. 4
  (ii) Constant temperature means no change in molecular kinetic energy (B1);
  Whatever stretching does, it does not touch the random-motion store here — temperature pins it.
5. 5
  Stretching pulls the molecules further apart: separation increases, so molecular potential energy increases (B1);
  The spring picture again — a stretched wire has loaded its intermolecular springs, within the elastic limit.
6. 6
  PE increases while KE is unchanged, so the internal energy increases (B1).
  Mirror image of part (i): there the kinetic store moved, here the potential store does. Same three-clause skeleton both times.
Answer
(i) PE unchanged (no separation change), KE up (temperature up), so U increases. (ii) KE unchanged (constant T), PE up (separation up), so U increases.
39702/41 O/N 2025 Q2(a)2 marks
State two ways in which the first law of thermodynamics describes that the internal energy of a system may be changed.
1 ________
2 ________
Stuck? Show hint
The first law names exactly two delivery routes for energy. Name them both.
Show solution
1. 1
  Work done on / by the system (B1);
  
  Route one is mechanical — force acting through a displacement at the boundary. §03 quantifies it.
2. 2
  Thermal energy supplied to / removed from the system (B1).
  
  Route two is heating. Together they ARE the first law — this question is the doorway to §04, where the law gets its symbols and signs.
Answer
By work done on (or by) the system, and by thermal energy supplied to (or removed from) the system.

Practise internal-energy questionsReal past-paper questions · Internal energy as random molecular kinetic and potential energy

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Can you do all of these?

Quote internal energy word-perfectly: the sum of a random distribution of kinetic and potential energies associated with the molecules of a system
Keep the two molecular stores straight: kinetic energy tracks random motion (and temperature); potential energy tracks separation and phase — never height or whole-object speed
Separate macroscopic KE/PE of the whole object from internal energy — a ball gains no internal energy by falling off a bench
Explain constant-volume heating in three clauses: no separation change ⇒ PE fixed; temperature rises ⇒ KE rises; so U rises
Explain boiling in three clauses: separation increases ⇒ PE rises; KE unchanged ⇒ temperature stays at 100 °C
Decide the sign of W from volume behaviour alone: expansion ⇒ work done BY the gas ⇒ W (on gas) negative; compression ⇒ positive; constant volume ⇒ zero
Rebuild W = pΔV from force = pA acting through displacement Δx, and read work as the AREA under a p–V graph
State the first law in words AND symbols: increase in internal energy = work done on the system + energy transferred to the system by heating; ΔU = q + W
Interpret negative signs physically: q < 0 means the system loses thermal energy; W < 0 means the system does work on its surroundings
Fill q/w/ΔU tables column by column: w from volume behaviour, ΔU from temperature or state, then close the last unknown with ΔU = q + W
Use ΔU = ³⁄₂NkΔT for an ideal gas, keeping the sign of ΔT
Close every cycle with ΔU(total) = 0 ⇒ net thermal energy supplied = net work done BY the gas
In mcΔθ hybrids, correct the AS working with the first law: q = ΔU − W, watching the sign of W