The bicycle pump

Problem

When quickly inflating a bicycle tyre, the pump heats up noticeably.

• (a) Why does it heat up? Where does the energy come from?
• (b) If you pumped the same air very slowly, would the pump heat up just as much?
• (c) Is the fast process adiabatic or isothermal? What about the slow one? Justify by comparing the characteristic times of compression and heat exchange.

Solution

(a) Why it heats up. Fast compression is approximately adiabatic: there is no time for heat to flow out of the gas. By the first law with $Q = 0$: $\Delta U = -L = L_\text{sul gas}$ The work the muscles do on the gas is converted entirely into internal energy, i.e. into a temperature rise. The energy comes from the muscular work of whoever is pumping.

(b) Slow pumping. No, the pump does not heat up noticeably. Compressing slowly, the gas has plenty of time to release heat to the surroundings as it is compressed: the temperature stays practically constant (an isothermal process) and the energy of the work leaves as heat instead of accumulating.

(c) Comparing the times. Let $\tau_\text{compr}$ be the compression time and $\tau_\text{scambio}$ the characteristic time for heat exchange with the surroundings:

• Fast: $\tau_\text{compr} \ll \tau_\text{scambio}$ → heat has no time to leave → adiabatic.
• Slow: $\tau_\text{compr} \gg \tau_\text{scambio}$ → the gas stays in thermal equilibrium with the surroundings → isothermal.

$\ev{\text{fast} = \text{adiabatic},\quad \text{slow} = \text{isothermal}}$

It is the same process (the same compression) giving opposite thermal outcomes depending on its speed (Povey 2015).

Links

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