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(A) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$\frac{P}{T} = 	ext{constant}$.Let the initial pressu

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$250\, K$

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(A) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$\frac{P}{T} = 	ext{constant}$.Let the initial pressure be $P_1$ and the initial temperature be $T_1$ (in Kelvin).When the temperature is increased by $1\,^{\circ}C$,the new temperature is $T_2 = T_1 + 1$.The pressure increases by $0.4\%$,so the new pressure is $P_2 = P_1 + 0.004 P_1 = 1.004 P_1$.Using the relation $\frac{P_1}{T_1} = \frac{P_2}{T_2}$,we get:$\frac{P_1}{T_1} = \frac{1.004 P_1}{T_1 + 1}$$T_1 + 1 = 1.004 T_1$$1 = 1.004 T_1 - T_1$$1 = 0.004 T_1$$T_1 = \frac{1}{0.004} = 250\, K$.

If the pressure of a gas contained in a closed vessel is increased by $0.4\%$ when heated by $1\,^{\circ}C$,the initial temperature must be

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