A steel cylinder contains helium gas at 250 , | vedclass

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(D) Given: $P_1 = 250 \, kPa$,$T_1 = 300 \, K$,$T_2 = 1800 \, K$.Since the volume of the cylinder is constant,we use Gay-Lussac's Law: $\frac{P_1}{T_1

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$1500 \, kPa$,it will burst

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(D) Given: $P_1 = 250 \, kPa$,$T_1 = 300 \, K$,$T_2 = 1800 \, K$.Since the volume of the cylinder is constant,we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Substituting the values: $\frac{250 \, kPa}{300 \, K} = \frac{P_2}{1800 \, K}$.$P_2 = \frac{250}{300} 	imes 1800 = 1500 \, kPa$.The maximum pressure the cylinder can withstand is $1 	imes 10^6 \, Pa = 1000 \, kPa$.Since the pressure at the melting point $(1500 \, kPa)$ is greater than the maximum pressure the cylinder can withstand $(1000 \, kPa)$,the cylinder will burst before it melts.

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