If the pressure of a gas in a closed vessel i | vedclass

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(A) According to Gay-Lussac's Law,for a fixed volume,$P \propto T$,which implies $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Let the initial pressure be $P_1

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) According to Gay-Lussac's Law,for a fixed volume,$P \propto T$,which implies $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Let the initial pressure be $P_1 = P$ and initial temperature be $T_1 = T$.The final pressure $P_2 = P + 0.004P = 1.004P$.The final temperature $T_2 = T + 1$.Substituting these values into the equation: $\frac{P}{T} = \frac{1.004P}{T + 1}$.$T + 1 = 1.004T$.$1 = 0.004T$.$T = \frac{1}{0.004} = \frac{1000}{4} = 250 \ K$.

If the pressure of a gas in a closed vessel increases by $0.4\%$ when heated by $1\,^oC$,its initial temperature is ....... $K$.

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