Estimate: work done by a bubble of hot air

Problem

Estimate the work done by the atmosphere when, on a sunny day, a bubble of hot air with initial volume $V_A \approx 1\ \text{m}^3$ expands at roughly constant (atmospheric) pressure, increasing its volume by about $10\%$. Is the work done by the gas on the surroundings or the other way round?

Solution

Change in volume. A $10\%$ increase on $1\ \text{m}^3$: $\Delta V \approx 0{,}10\ \text{m}^3$

Pressure. Roughly atmospheric: $p \approx 10^5\ \text{Pa}$.

Work at constant pressure. $L = p\,\Delta V = 10^5\cdot 0{,}10 = 10^4\ \text{J}$

$\ev{L \sim 10^4\ \text{J}}$

The bubble expands, pushing back the atmosphere: the work is done BY the gas ON the surroundings (the atmosphere is on the receiving end).

Links

Topics: Thermodynamics Concepts: Thermodynamic transformations Skills: Fermi estimates Objects: Ideal gas