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

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$250 \ K$

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(C) For a gas in a closed container,the volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$,where $P$ is pressure and $T$ is absolute temperature in Kelvin.Taking the differential form,we have $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.Given,$\frac{\Delta P}{P} = 0.4 \% = \frac{0.4}{100}$ and $\Delta T = 1 \ K$ (since a change of $1 ^\circ C$ is equivalent to a change of $1 \ K$).Substituting these values into the equation: $\frac{0.4}{100} = \frac{1}{T}$.Solving for $T$: $T = \frac{100}{0.4} = 250 \ K$.

When the temperature of a gas in a closed container is increased by $1 ^\circ C$,its pressure increases by $0.4 \%$. What is the initial temperature of the gas?

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