A gas at the temperature 250, K is contained | vedclass

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

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$0.4$

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(A) For a gas in a closed vessel, the volume $V$ remains constant.According to Gay-Lussac's Law, $P \propto T$ or $\frac{P}{T} = 	ext{constant}$.Given initial temperature $T_1 = 250\, K$ and change in temperature $\Delta T = 1\, K$.The final temperature is $T_2 = T_1 + \Delta T = 251\, K$.From the relation $\frac{P_1}{T_1} = \frac{P_2}{T_2}$, we have $\frac{P_2}{P_1} = \frac{T_2}{T_1}$.The fractional increase in pressure is $\frac{\Delta P}{P_1} = \frac{P_2 - P_1}{P_1} = \frac{T_2 - T_1}{T_1} = \frac{\Delta T}{T_1}$.Substituting the values: $\frac{\Delta P}{P_1} = \frac{1}{250}$.To find the percentage increase: $\left( \frac{\Delta P}{P_1} ight) 	imes 100 = \frac{1}{250} 	imes 100 = 0.4\% $.

$A$ gas at the temperature $250\, K$ is contained in a closed vessel. If the gas is heated through $1\, K$, then the percentage increase in its pressure will be .... $\%$

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