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

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · States of Matter

An LPG cylinder contains gas at a pressure of 300 kPa at

2 7^{\circ} C

. The cylinder can withstand the pressure of

1.2 \times 1 0^{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)

Answer: 927

The cylinder is rigid and sealed, so the volume and the amount of gas are both fixed and the pressure is proportional to the absolute temperature:

\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}

With

P_{1} = 3 \times 1 0^{5}

Pa at

T_{1} = 300

K and the bursting pressure

P_{2} = 1.2 \times 1 0^{6}

Pa,

T_{2} = T_{1} \frac{P _{2}}{P _{1}} = 300 \times \frac{1.2 \times 1 0 ^{6}}{3 \times 1 0 ^{5}} = 1200 K

The cylinder bursts once the gas reaches

1200 - 273 = 927^{\circ} C

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer