Physics 9702 · Paper 4

The First Law: ΔU = q + W

Only two ways exist to put energy into a system from outside: heat it, or do work on it. The first law says the books always balance.

The statement (two marks)

\Delta U = q + W

Symbols

$\Delta U$ = increase in internal energy of the system (J)
$q$ = thermal energy transferred TO the system (by heating) (J)
$W$ = work done ON the system (J)

The first law as an energy count: ΔU = q + W. Both q and W are counted positive when energy flows INTO the system — heat it or squeeze it, and U rises by exactly what went in.

In words: the increase in internal energy equals the energy transferred to the system by heating plus the work done on the system. The relation earns the first mark; the directions — to the system, on the system — earn the second (9702/42/M/J/23 Q3(a)) (9702/42/O/N/25 Q4(a)).
It is the principle of conservation of energy written for thermal physics, and it applies to anything — gases, liquids, wires, springs — not only to gases.
Energy leaving is not a third case: cooling means $q$ is negative, expansion (the system doing work) means $W$ is negative. One equation, signed entries.

Worked example

Smallest cases: both signs

(a) A gas receives 250 kJ by heating while 500 kJ of work is done on it (compression). (b) The same gas receives 250 kJ by heating but expands, doing 200 kJ of work on the surroundings. Find $\Delta U$ in each case.

(a) Both entries are into the system: $\Delta U = 250 + 500 = +750\ \text{kJ}$ .
(b) This time the gas gives out 200 kJ as work, so $W = -200$ kJ: $\Delta U = 250 + (-200) = +50\ \text{kJ}$ .

Answer

(a)\ +750\ \text{kJ};\quad (b)\ +50\ \text{kJ}

Common mistake

Older textbooks write the first law with

W

as work done by the gas. In 9702,

W

is always work done on the system, and mark schemes withhold the second mark for a statement with the directions missing or reversed. Fix the convention now and every sign question in this chapter becomes mechanical.

Using the law to find q

A standard chain gives you $\Delta U$ and $W$ and asks for the thermal energy (9702/41/O/N/23 Q2(b)(iii)): rearrange to $q = \Delta U - W$ and respect the signs.
Notice what the law reveals: heating a gas at constant pressure, some of $q$ leaks straight back out as expansion work — so $q$ is bigger than $\Delta U$ . A 2024 paper made the point with a vaporising liquid: $\Delta U = 17.6$ kJ but $q = 17.6 + 1.7 = 19.3$ kJ, because 1.7 kJ went into pushing back the atmosphere (9702/42/O/N/24 Q3(b)).

PreviousInternal Energy NextWork = pΔV