An LPG cylinder contains gas at a pressure of 300 kPa at 27^°. The cylinder can withstand the pressure of 1.2 x 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 __________^° C. (Nearest integer)

P_1 / T_1 = P_2 / T_2 => 300 x 10^3 / 300 = 1.2 x 10^6 / T_2 => T_2 = 1200 K => T_2 = 927^° C

An LPG cylinder contains gas at a pressure of 300 kPa a... | (Chemistry) | DivineJEE

Exam

Chemistry

An LPG cylinder contains gas at a pressure of 300 kPa at 27

^\circ

. The cylinder can withstand the pressure of 1.2

\times

10⁶ 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)

${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}} \Rightarrow {{300 \times {{10}^3}} \over {300}} = {{1.2 \times {{10}^6}} \over {{T_2}}}$

$\Rightarrow {T_2} = 1200$ K

$\Rightarrow$ ${T_2} = 927^\circ$ C

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