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(D) For a gas in a closed vessel,the volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$,where $T$ is the absolute temperature in Kelv

Vedclass · Accepted Answer

$207$

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(D) For a gas in a closed vessel,the volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$,where $T$ is the absolute temperature in Kelvin.Let the initial pressure be $P$ and the initial temperature be $T$ (in Kelvin).When the temperature increases by $\Delta T = 2.4 \ K$,the pressure increases by $\Delta P = 0.005 P$.From $P/T = (P + \Delta P) / (T + \Delta T)$,we have:$P/T = (P + 0.005 P) / (T + 2.4)$$1/T = 1.005 / (T + 2.4)$$T + 2.4 = 1.005 T$$0.005 T = 2.4$$T = 2.4 / 0.005 = 480 \ K$.The initial temperature in Celsius is $t = T - 273 = 480 - 273 = 207^{\circ} C$.

When the temperature of a gas in a closed vessel is increased by $2.4^{\circ} C$,its pressure increases by $0.5 \%$. The initial temperature of the gas is (in $^{\circ} C$)

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