Pressure of an ideal gas,contained in a close | vedclass

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(C) For an ideal gas in a closed vessel,the volume $V$ remains constant,which represents an isochoric process.According to Gay-Lussac's Law,for a cons

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$250 \ K$

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(C) For an ideal gas in a closed vessel,the volume $V$ remains constant,which represents an isochoric process.According to Gay-Lussac's Law,for a constant volume,the pressure $P$ is directly proportional to the absolute temperature $T$ $(P \propto T)$.This implies $\frac{P_2}{P_1} = \frac{T_2}{T_1}$,or in terms of small changes,$\frac{\Delta P}{P} = \frac{\Delta T}{T}$.Given that the pressure increases by $0.4 \%$,we have $\frac{\Delta P}{P} = \frac{0.4}{100}$.The change in temperature is given as $\Delta T = 1 \ K$ (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 $T$,we get $T = \frac{100}{0.4} = 250 \ K$.

Pressure of an ideal gas,contained in a closed vessel,is increased by $0.4 \%$ when heated by $1^{\circ} C$. Its initial temperature must be

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