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(C) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus, $T_1 V_1^

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Less than $1/2$

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(C) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus, $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given that the temperature is doubled, $T_2 = 2 T_1$, so $\frac{T_2}{T_1} = 2$.Substituting this into the relation: $\frac{T_2}{T_1} = \left( \frac{V_1}{V_2} ight)^{\gamma - 1} = 2$.Taking the reciprocal: $\left( \frac{V_2}{V_1} ight)^{\gamma - 1} = \frac{1}{2}$.Therefore, the ratio of final volume to initial volume is $\frac{V_2}{V_1} = \left( \frac{1}{2} ight)^{\frac{1}{\gamma - 1}}$.Since for any gas $\gamma > 1$, the exponent $\frac{1}{\gamma - 1} > 1$.Raising $1/2$ to a power greater than $1$ results in a value less than $1/2$.Hence, $\frac{V_2}{V_1} < \frac{1}{2}$.

$A$ gas is compressed adiabatically until its temperature is doubled. The ratio of its final volume to initial volume will be

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