When the temperature of a gas in a closed ves | vedclass

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(C) For a gas in a closed vessel,the volume $V$ is constant. According to Gay-Lussac's Law,$\frac{P}{T} = 	ext{constant}$,which means $\frac{P_1}{T_1

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$250 \ K$

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(C) For a gas in a closed vessel,the volume $V$ is constant. According to Gay-Lussac's Law,$\frac{P}{T} = 	ext{constant}$,which means $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Let the initial pressure be $P_1 = P$ and the initial temperature be $T_1 = T$ (in Kelvin).The final pressure is $P_2 = P + 0.4\% 	ext{ of } P = P + 0.004P = 1.004P$.The final temperature is $T_2 = T + 1$.Substituting these values into the equation: $\frac{P}{T} = \frac{1.004P}{T+1}$.Dividing both sides by $P$: $\frac{1}{T} = \frac{1.004}{T+1}$.Cross-multiplying: $T + 1 = 1.004T$.$1 = 0.004T$.$T = \frac{1}{0.004} = 250 \ K$.

When the temperature of a gas in a closed vessel is increased by $1^{\circ}C$,its pressure increases by $0.4\%$. The initial temperature of the gas is ...

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