An LPG cylinder contains gas at a pressure of 300kPa at 27 ° C . The cylinder can withstand the pressure of 1 . 2 × 10 ^ 6 Pa . The room in which the cylinder is kept catches fire. The minimum temperature at which the

Correct answer: 0 P _ 1 T _ 1 = P _ 2 T _ 2 ⇒ 300 × 10 ^ 3 300 = 1 . 2 × 10 ^ 6 T _ 2 ⇒ T _ 2 = 1200K T _ 2 = 927 ° C

JEE Main 202125 JulyEvening Shift

JEE Main 2021 · 25 July · Evening Shift · Chemistry · States of Matter

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

An LPG cylinder contains gas at a pressure of

300kPa

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 __________

C ∘

. (Nearest integer)

$\dfrac{{P}_{1}}{{T}_{1}} = \dfrac{{P}_{2}}{{T}_{2}} ⇒ \dfrac{300 \times {10}^{3}}{300} = \dfrac{1 . 2 \times {10}^{6}}{{T}_{2}}$

$⇒ {T}_{2} = 1200K$

${T}_{2} = 927 ° C$