Verified Solution • 2025–26 Session moderate

QJEEM21Ch520d: JEE Main 2021 Chemistry - Class 11 JEEM21Ch520d Solution: An LPG cylinder contains gas at a pressure of 300 kPa at 27 .…

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)

Step-by-Step Verified Solution

P1T1=P2T2⇒300×103300=1.2×106T2{{{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}}}

⇒T2=1200\Rightarrow {T_2} = 1200 K

⇒\Rightarrow T2=927∘{T_2} = 927^\circ C

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