Compressed air heats up

Problem

When you quickly compress the air in a bicycle pump, the pump heats up. Explain in five lines in terms of work, internal energy and heat-exchange timescales.

Solution

Rapid compression ≈ adiabatic. The compression happens in a fraction of a second: the gas has no time to exchange heat with the surroundings, so $Q\approx 0$.

First law. With $Q = 0$ the first law $\Delta U = Q - L$ becomes $\Delta U = -L$. Work is done ON the gas ($L < 0$), so $\Delta U > 0$: internal energy increases, temperature rises and the pump heats up.

Comparison with the slow case. If you compressed it slowly, the gas would gradually release heat to the surroundings (an almost isothermal process) and would stay at room temperature: no noticeable heating.

This is the same principle behind ignition in Diesel engines (Povey 2015).

$\ev{\text{rapid compression}\approx\text{adiabatic}\to \Delta U > 0\to T\uparrow}$

Links

Topics: Thermodynamics Concepts: Internal energy · Thermodynamic processes Objects: Ideal gas