A Sealed Gas Cylinder Left in the Sun
A rigid steel cylinder of volume 0.020 m³ contains nitrogen gas at an initial pressure of 1.5 × 10⁵ Pa and temperature 290 K. It is left in direct sunlight and its temperature rises to 320 K. The cylinder does not expand measurably. A pressure-relief valve is set to open at 1.8 × 10⁵ Pa. Treat nitrogen as an ideal gas. Separately, a technician wonders whether the same amount of gas, if allowed to expand into a flexible bag at constant temperature, would reach the same pressure.
1. Predict the pressure at 320 K and state whether the relief valve opens.
Read the worked answer
Volume and amount are fixed, so this is an isochoric (constant-volume) process: $P/T = \text{const}$.
$P_2 = P_1 \dfrac{T_2}{T_1} = 1.5\times10^5 \times \dfrac{320}{290} = 1.5\times10^5 \times 1.103 \approx 1.66\times10^5$ Pa.
This is below the 1.8 × 10⁵ Pa set point, so the valve stays closed. Note temperatures must be in kelvin; using 47 °C and 17 °C directly would give a nonsense ratio.
2. The technician proposes letting the gas expand into a flexible bag at constant temperature to lower the pressure. Explain, using the ideal gas law, why constant-temperature expansion lowers pressure, and compute the volume needed to bring the pressure back to 1.5 × 10⁵ Pa at 320 K.
Read the worked answer
From $PV = nRT$, at fixed $n$ and $T$, pressure and volume are inversely proportional (Boyle's law): doubling volume halves pressure. Increasing volume gives molecules more wall area and longer mean free path between collisions, reducing collision rate per unit area.
To return to $1.5\times10^5$ Pa at 320 K, use $P_1V_1/T_1 = P_2V_2/T_2$ with the state (1.66 × 10⁵ Pa, 0.020 m³, 320 K) as the starting point:
$V_2 = V_1 \dfrac{P_1}{P_2} = 0.020 \times \dfrac{1.66\times10^5}{1.5\times10^5} \approx 0.0221$ m³.
So the bag must expand by about 2.1 litres beyond the cylinder volume. This is an isothermal expansion; the gas does work at the expense of heat absorbed from the surroundings.