A gas cylinder can withstand a pressure of 15 | vedclass

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(B) According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,$P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Given: $P_1 = 12 \

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$102$

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(B) According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,$P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Given: $P_1 = 12 \ atm$,$T_1 = 27 \ ^oC = 27 + 273 = 300 \ K$.Maximum pressure the cylinder can withstand,$P_2 = 15 \ atm$.Substituting the values: $\frac{12}{300} = \frac{15}{T_2}$.$T_2 = \frac{15 	imes 300}{12} = 375 \ K$.Converting back to Celsius: $T_2( ^oC) = 375 - 273 = 102 \ ^oC$.Thus,the cylinder will not burst up to $102 \ ^oC$.

List-$I$	List-$II$
$(1)$ Boyle's Law	$(A)$ $p \propto T$
$(2)$ Charles's Law	$(B)$ $V \propto n$
$(3)$ Gay-Lussac's Law	$(C)$ $p \propto \frac{1}{V}$
$(4)$ Avogadro's Law	$(D)$ $pV = nRT$
$(5)$ Ideal Gas Equation	$(E)$ $V = k_2T$

$A$ gas cylinder can withstand a pressure of $15 \ atm$. At $27 \ ^oC$,the pressure in the cylinder is $12 \ atm$. Up to what temperature in $^oC$ will the cylinder not burst?

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