Physics 9702 · Paper 4

First-Law Tables & Cycles

The highest-mark thermodynamics questions hand you a gas going around a loop and a half-empty table. Two counting rules fill in everything.

The two rules

Row rule: every stage must satisfy $\Delta U = q + W$ . Given any two entries in a row, the third is forced — and mark schemes award a specific mark for rows that balance (9702/41/M/J/25 Q4(b)).
Cycle rule: after a complete loop the gas is back at its starting state, so its internal energy is back too: the $\Delta U$ column sums to zero. That line is its own mark (9702/41/M/J/23 Q4(b)(iii)).
Fill the easy zeros first: constant-volume stages have $W = 0$ , isothermal stages have $\Delta U = 0$ , and for an ideal gas $U \propto T$ converts temperature statements into $\Delta U$ entries.

Worked example

A 2025 table, reconstructed

An ideal gas with internal energy $U$ at temperature $T$ is compressed rapidly to temperature $3T$ , the work done on it being $W$ . It is then cooled at constant volume to $2T$ . Complete the table of (work done on gas / thermal energy supplied / increase in internal energy) (9702/41/M/J/25 Q4(b)).

$U \propto T$ : at $3T$ internal energy is $3U$ , at $2T$ it is $2U$ .
Compression row: work on gas $= W$ ; $\Delta U = 3U - U = +2U$ ; row rule → thermal energy $q = 2U - W$ .
Cooling row: constant volume → work $= 0$ ; $\Delta U = 2U - 3U = -U$ ; row rule → $q = -U$ (thermal energy leaves).

Answer

\text{compression: } W,\ 2U-W,\ +2U;\quad \text{cooling: } 0,\ -U,\ -U

Reasoning about a whole cycle

On a p–V diagram, work for each constant-pressure stage is the area under that line (previous lesson) — compression counts into the gas, expansion counts out. A 2025 cycle gave $W = +8XY$ on compression and $-20XY$ on expansion, with the two constant-volume sides contributing nothing (9702/42/O/N/25 Q4(c)).
Then the cycle rule finishes it: total $\Delta U = 0$ , so the net thermal energy supplied must exactly cover the net work done by the gas: $q = 20XY - 8XY = +12XY$ (9702/42/O/N/25 Q4(d)).
The same three-mark skeleton — net work done by the gas, cycle $\Delta U = 0$ , therefore net thermal input — scored a 2023 describe-and-explain question word for word (9702/42/M/J/23 Q3(b)(ii)).

Common mistake

In a cycle question, do not hunt for new physics: every mark is one of the two counting rules or a

p\,\Delta V

area. If a filled table row cannot satisfy

\Delta U = q + W

, an entry is wrong — the rules are also your checking machine.

Where this leads

A gas that absorbs heat and delivers net work every loop is an engine; run the loop backwards and work pumps heat from cold to hot — a refrigerator. 9702 stops at the energy counting, but the second law of thermodynamics (why no engine converts heat to work completely) starts exactly here.

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