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(B) Given: Initial temperature $T_1 = 100^{\circ} C = 373 \ K$. The volume is constant in a closed vessel. According to Gay-Lussac's Law,for a fixed m

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$14.92$

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(B) Given: Initial temperature $T_1 = 100^{\circ} C = 373 \ K$. The volume is constant in a closed vessel. According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,$P \propto T$. Therefore,$\frac{P_2}{P_1} = \frac{T_2}{T_1}$. Given that the pressure increases by $4 \%$,$P_2 = P_1 + 0.04 P_1 = 1.04 P_1$. Substituting this into the ratio: $\frac{T_2}{T_1} = 1.04$. $T_2 = 1.04 	imes 373 \ K = 387.92 \ K$. The increase in temperature $\Delta T = T_2 - T_1 = 387.92 \ K - 373 \ K = 14.92 \ K$. Since a change in temperature in Kelvin is equal to a change in temperature in Celsius,$\Delta T = 14.92^{\circ} C$.

$A$ gas is contained in a closed vessel. The initial temperature of the gas is $100^{\circ} C$. If the pressure of the gas is increased by $4 \%$,what is the increase in the temperature of the gas (in $^{\circ} C$)?

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