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(C) For a gas in a closed vessel,the volume $V$ remains constant.According to Gay-Lussac's Law,for a constant volume,$P \propto T$,which implies $\fra

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$-73$

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(C) For a gas in a closed vessel,the volume $V$ remains constant.According to Gay-Lussac's Law,for a constant volume,$P \propto T$,which implies $\frac{P}{T} = 	ext{constant}$.Differentiating this,we get $\frac{dP}{P} = \frac{dT}{T}$.Given that the pressure increases by $0.5\%$,we have $\frac{dP}{P} = 0.005$.The change in temperature is given as $dT = 1\, K$ (since a change of $1\,^{\circ}C$ is equal to a change of $1\, K$).Substituting these values into the equation: $0.005 = \frac{1}{T}$.Solving for $T$,we get $T = \frac{1}{0.005} = 200\, K$.To convert the temperature from Kelvin to Celsius,we use the formula $t(^{\circ}C) = T(K) - 273.15$.Thus,$t = 200 - 273 = -73\, ^{\circ}C$.

$A$ gas is heated through $1\,^{\circ}C$ in a closed vessel. Its pressure is increased by $0.5\%$. The initial temperature of the gas is ...... $^{\circ}C$.

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