When the pressure of a gas is increased by 0. | vedclass

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(A) Let the initial pressure be $P_1 = P$ and the initial temperature be $T_1 = T$ (in Kelvin).Given that the pressure increases by $0.4\%$,the new pr

Vedclass · Accepted Answer

$250\, 	ext{K}$

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(A) Let the initial pressure be $P_1 = P$ and the initial temperature be $T_1 = T$ (in Kelvin).Given that the pressure increases by $0.4\%$,the new pressure is $P_2 = P + 0.004P = 1.004P$.Given that the temperature increases by $1\,^{\circ}	ext{C}$,the new temperature is $T_2 = T + 1$.Assuming the volume remains constant,according to Gay-Lussac's Law,$\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Substituting the values: $\frac{P}{T} = \frac{1.004P}{T + 1}$.$T + 1 = 1.004T$.$1 = 0.004T$.$T = \frac{1}{0.004} = \frac{1000}{4} = 250\, 	ext{K}$.

When the pressure of a gas is increased by $0.4\%$,its temperature increases by $1\,^{\circ}\text{C}$. What is the initial temperature of the gas?

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