Problem — Expansion at constant pressure

Problem

A balloon contains $V_1 = 0{,}50$ L of air at $T_1 = 20\,^\circ$C. It goes outside at $T_2 = -10\,^\circ$C, with $p$ unchanged. Find the new volume.

Solution

At constant pressure Charles’s law holds: $V \propto T$, with $T$ in kelvin. First I convert the temperatures: $T_1 = 20 + 273{,}15 = 293{,}15\ \text{K}, \qquad T_2 = -10 + 273{,}15 = 263{,}15\ \text{K}$ From the proportionality $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ I find: $V_2 = V_1 \frac{T_2}{T_1}$ $V_2 = 0{,}50 \cdot \frac{263{,}15}{293{,}15}$ $\ev{V_2 \approx 0{,}449\ \text{L}}$ As the gas cools the volume decreases. Careful: using temperatures in kelvin is essential — with degrees Celsius one would get an absurd result.

Links

Topics: Kinetic theory of gases Concepts: Ideal gas law · Temperature Skills: Reading graphs