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(C) For a gas in a closed vessel,volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$ (where $T$ is in Kelvin).Let the initial pressure

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$227$

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(C) For a gas in a closed vessel,volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$ (where $T$ is in Kelvin).Let the initial pressure be $P_1$ and initial temperature be $T_1$ (in Kelvin).When the temperature increases by $5^{\circ}C$,the new temperature $T_2 = T_1 + 5$.The pressure increases by $1\%$,so the new pressure $P_2 = P_1 + 0.01 P_1 = 1.01 P_1$.Using the relation $\frac{P_1}{P_2} = \frac{T_1}{T_2}$,we get:$\frac{P_1}{1.01 P_1} = \frac{T_1}{T_1 + 5}$$T_1 + 5 = 1.01 T_1$$0.01 T_1 = 5$$T_1 = \frac{5}{0.01} = 500 \ K$.To convert Kelvin to Celsius: $t(^{\circ}C) = T(K) - 273$.$t = 500 - 273 = 227^{\circ}C$.

When a gas in a closed vessel was heated so as to increase its temperature by $5^{\circ}C$,there occurred an increase of $1\%$ in its pressure. The original temperature of the gas was ...... $^{\circ}C$

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