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(C) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$P \propto T$,which means $\frac{P_1}{T_1} = \frac{P_2

Vedclass · Accepted Answer

$160$

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(C) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$P \propto T$,which means $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Let the initial pressure be $P$ and the initial temperature be $T$. Then $P_1 = P$ and $T_1 = T$.The pressure is increased by $2.5 \%$,so $P_2 = P + 0.025P = 1.025P$.The temperature increases by $4 \ K$,so $T_2 = T + 4$.Substituting these into the relation $\frac{P}{T} = \frac{1.025P}{T + 4}$:$T + 4 = 1.025T$$4 = 1.025T - T$$4 = 0.025T$$T = \frac{4}{0.025} = \frac{4000}{25} = 160 \ K$.Thus,the initial temperature of the gas is $160 \ K$.

When the pressure of the gas contained in a closed vessel is increased by $2.5 \%$,the temperature of the gas increases by $4 \ K$. The initial temperature of the gas is (in $K$)

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