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

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Question

(B) According to Gay-Lussac's Law for a gas in a closed vessel (constant volume),$P \propto T$,which implies $\frac{\Delta P}{P} = \frac{\Delta T}{T}$

Vedclass · Accepted Answer

$250 \ K$

Vedclass · Answer

(B) According to Gay-Lussac's Law for a gas in a closed vessel (constant volume),$P \propto T$,which implies $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.Given: $\frac{\Delta P}{P} = 0.4 \ \% = 0.004$ and $\Delta T = 1 \ K$ (a change of $1 \ ^\circ C$ is equivalent to a change of $1 \ K$ in absolute temperature).Substituting the values: $0.004 = \frac{1}{T}$.Therefore,$T = \frac{1}{0.004} = 250 \ K$.Converting to Celsius: $T( ^\circ C) = 250 - 273 = -23 \ ^\circ C$.However,based on the standard interpretation of such problems where the result is expected in Kelvin or the temperature difference is treated as absolute,the initial temperature is $250 \ K$.

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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