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(B) For an ideal gas in a closed vessel, the volume $V$ is constant. According to Gay-Lussac's Law, $P \propto T$ or $\frac{P_1}{P_2} = \frac{T_1}{T_2

Vedclass · Accepted Answer

$127$

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(B) For an ideal gas in a closed vessel, the volume $V$ is constant. According to Gay-Lussac's Law, $P \propto T$ or $\frac{P_1}{P_2} = \frac{T_1}{T_2}$.Let the initial pressure be $P$ and initial temperature be $T$ (in Kelvin).Given that the pressure increases by $0.5\%$, the new pressure $P_2 = P + 0.005P = 1.005P$.The new temperature $T_2 = T + 2$.Substituting these values into the gas law equation:$\frac{P}{1.005P} = \frac{T}{T + 2}$$\frac{1}{1.005} = \frac{T}{T + 2}$$T + 2 = 1.005T$$0.005T = 2$$T = \frac{2}{0.005} = 400 \, K$.To convert Kelvin to Celsius: $T(^\circ C) = T(K) - 273 = 400 - 273 = 127^\circ C$.

If the pressure of an ideal gas contained in a closed vessel is increased by $0.5\%,$ the increase in temperature is $2 \, K.$ The initial temperature of the gas is ...... $^\circ C$

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