A gas cylinder has walls that can bear a maximum pressure of 1.0 × 106 Pa. It contains a gas at 8.0 × 105 Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.
Solution
Given:-
Maximum pressure that the cylinder can bear, Pmax = 1.0 × 106 Pa
Pressure in the gas cylinder, P1 = 8.0 × 105 Pa
Temperature in the cylinder, T1 = 300 K
Let T2 be the temperature at which the cylinder will break.
Volume is constant. Thus, ............(Given)
V1= V2 = V
Applying the five variable gas equation, we get
\[\frac{P_1 V}{T_1} = \frac{P_2 V}{T_2} .........( \because V_1 = V_2 )\]
\[ \Rightarrow \frac{P_1}{T_1} = \frac{P_2}{T_2}\]
\[ \Rightarrow T_2 = \frac{P_2 \times T_1}{P_1}\]
\[ \Rightarrow T_2 = \frac{1 . 0 \times {10}^6 \times 300}{8 . 0 \times {10}^5}=375K\]
