Problem — Temperature that doubles the pressure

Problem

An ideal gas enclosed in a rigid container is at $T = 300$ K, $p = 1{,}0$ atm. At what temperature does the pressure double?

Solution

The container is rigid, so the volume $V$ stays constant. For an ideal gas, at fixed $V$ and $n$, Gay-Lussac’s law relates pressure and absolute temperature in direct proportion: $\frac{p_1}{T_1}=\frac{p_2}{T_2}$ If the pressure doubles, $p_2 = 2\,p_1$, so the absolute temperature also doubles: $T_2 = T_1\,\frac{p_2}{p_1}=300\cdot 2$ $\ev{T_2 \approx 600\ \text{K}}$

Links

Topics: Kinetic theory of gases Concepts: Ideal gas law · Temperature · Pressure