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(D) For an adiabatic process,the relationship between temperature $T$ and pressure $P$ is given by $T^\gamma P^{1-\gamma} = 	ext{constant}$.This can

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$-142^\circ C$

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(D) For an adiabatic process,the relationship between temperature $T$ and pressure $P$ is given by $T^\gamma P^{1-\gamma} = 	ext{constant}$.This can be rewritten as $T \propto P^{\frac{\gamma - 1}{\gamma}}$.Given: Initial temperature $T_1 = 27^\circ C = 300K$,final pressure $P_2 = \frac{1}{8}P_1$,and $\gamma = 5/3$.Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} ight)^{\frac{\gamma - 1}{\gamma}}$.Substituting the values: $\frac{T_2}{300} = \left( \frac{1}{8} ight)^{\frac{5/3 - 1}{5/3}} = \left( \frac{1}{8} ight)^{\frac{2/3}{5/3}} = \left( \frac{1}{8} ight)^{2/5}$.$T_2 = 300 	imes (8^{-1})^{2/5} = 300 	imes (2^3)^{-2/5} = 300 	imes 2^{-6/5} \approx 300 	imes 0.435 = 131K$.Converting to Celsius: $T_2 = 131 - 273 = -142^\circ C$.

At $27^\circ C$,a gas is suddenly compressed such that its pressure becomes $1/8$th of the original pressure. The temperature of the gas will be $(\gamma = 5/3)$.

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