Physics 9702 · Paper 4

Four Standard Processes

Nearly every first-law question is one of four situations in costume. Learn which term of $\Delta U = q + W$ dies in each, and the costume stops working.

Which term is zero?

Process	What is zero	The first law becomes
constant volume	W = 0 (no volume change, no work)	ΔU = q
constant pressure	nothing — all three terms live	ΔU = q − pΔV (expanding)
rapid compression / expansion	q = 0 (no time for heat transfer)	ΔU = W
isothermal (constant temperature, ideal gas)	ΔU = 0 (U depends only on T)	q = −W

The two zeros students confuse: rapid means $q = 0$ — the temperature does change; isothermal means $\Delta U = 0$ — heat does flow. They are opposite corners of the table, and exam tables test exactly this separation.

Worked example

The bicycle pump (a banked 3-marker)

Use the first law to explain why the air in a bicycle pump gets hot when the pump is pushed in quickly (9702/41/O/N/25 Q2(b)(i)) (9702/42/F/M/23 Q2(b)(ii)).

Quick stroke → no time for thermal energy to leave: $q = 0$ .
The piston does work on the gas: $W > 0$ .
So $\Delta U = q + W > 0$ : internal energy rises, and for a gas that means temperature rises.

Answer

q=0,\ W>0 \Rightarrow \Delta U > 0 \Rightarrow T\ \text{rises}

Why heating at constant pressure costs more

Heat a gas through the same temperature rise twice — once at constant volume, once at constant pressure. Same $\Delta T$ , so same $\Delta U$ (ideal gas: $U \propto T$ ).
At constant volume no work is done, so $q = \Delta U$ . At constant pressure the gas expands and pays $p\,\Delta V$ to the surroundings, so $q = \Delta U + p\,\Delta V$ — more thermal energy for the same temperature rise. The constant-volume specific heat capacity is therefore lower. This 3-mark comparison has appeared twice with near-identical mark schemes (9702/41/O/N/23 Q2(c)) (9702/42/F/M/24 Q2(c)).
This settles the point left open in the Temperature chapter: a gas has no single specific heat capacity — the value depends on what the gas is allowed to do while heated.

Constant V: $W=0$ . Rapid: $q=0$ . Isothermal: $\Delta U=0$ . Constant p: everything lives, and $q > \Delta U$ when expanding — which is why $c_p > c_V$ .

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