Question:
An LPG cylinder contains gas at a pressure of 300 $\mathrm{kPa}$ 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)
Solution:
$\frac{\mathrm{P}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2}}{\mathrm{~T}_{2}} \Rightarrow \frac{300 \times 10^{3}}{300}=\frac{1.2 \times 10^{6}}{\mathrm{~T}_{2}}$
$\Rightarrow \mathrm{T}_{2}=1200 \mathrm{~K}$
$\mathrm{T}_{2}=927^{\circ} \mathrm{C}$
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