A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature $50^{\ci

Question

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature $50^{\circ} \mathrm{C}$. The gas is now heated to double its temperature. The modified pressure is $\_\_\_\_$ Pa .

Professor PC Thomas Classes · Answer

For a closed cylinder of fixed volume, Gay-Lussac's law dictates that absolute pressure is directly proportional to absolute temperature ($P \propto T$), meaning $\frac{P_1}{T_1} = \frac{P_2}{T_2}$. The initial temperature is $T_1 = 50^\circ	ext{C} + 273.15 = 323.15 	ext{ K}$, and the initial pressure is $P_1 = 3.23 	ext{ kPa} = 3230 	ext{ Pa}$. The gas is heated to double its Celsius temperature, so $T_2 = 100^\circ	ext{C} = 373.15 	ext{ K}$. The new pressure is $P_2 = P_1 \frac{T_2}{T_1} = 3230 	imes \frac{373.15}{323.15} = 10 	imes 373.15 \approx 3730 	ext{ Pa}$.