Q30 / 90 · JEE Main 2021 (25 Jul Shift 2) · Chemistry

An LPG cylinder contains gas at a pressure of $300 kPa$ at $27 ° C$ . The cylinder can withstand the pressure of $1.2 \times 10^{6} Pa$ . The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is __________ ${}^{\circ}C$ . (Nearest integer)

Correct answer: 927

Solution

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow \frac{300 \times 10^{3}}{300} = \frac{1.2 \times 10^{6}}{T_{2}}$

$\Rightarrow T_{2} = 1200 K$

$T_{2} = 927 ° C$

Hence, the correct answer is 927.