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(C) For an adiabatic process, the relation between pressure and volume is given as $PV^{\gamma} = 	ext{constant}$. Here, $\gamma = 3/2$.Using the ide

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$\sqrt{2}T$

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(C) For an adiabatic process, the relation between pressure and volume is given as $PV^{\gamma} = 	ext{constant}$. Here, $\gamma = 3/2$.Using the ideal gas law $PV = nRT$, we can write $P = nRT/V$.Substituting this into the adiabatic equation: $(nRT/V)V^{\gamma} = 	ext{constant}$, which simplifies to $TV^{\gamma-1} = 	ext{constant}$.Given $\gamma = 3/2$, the relation becomes $TV^{(3/2 - 1)} = TV^{1/2} = 	ext{constant}$.Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.Given $T_1 = T$ and $V_2 = V_1/2$.Using $T_1 V_1^{1/2} = T_2 V_2^{1/2}$:$T_2 = T_1 (V_1/V_2)^{1/2} = T (V_1 / (V_1/2))^{1/2} = T (2)^{1/2} = \sqrt{2}T$.

The relation obeyed by a perfect gas during an adiabatic process is $PV^{3/2} = \text{constant}$. The initial temperature of the gas is $T$. When the gas is compressed to half of its initial volume, the final temperature of the gas is:

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