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(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = 	ext{constant}$.Using the ideal gas equation $PV = \m

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$2.5$

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(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = 	ext{constant}$.Using the ideal gas equation $PV = \mu RT$,we have $V = \frac{\mu RT}{P}$.Substituting this into the adiabatic equation: $P \left( \frac{\mu RT}{P} ight)^{\gamma} = 	ext{constant}$.This simplifies to $P^{1-\gamma} T^{\gamma} = 	ext{constant}$.Rearranging for $P$: $P^{1-\gamma} = \frac{	ext{constant}}{T^{\gamma}} = 	ext{constant} \cdot T^{-\gamma}$.$P = 	ext{constant} \cdot T^{-\gamma / (1-\gamma)} = 	ext{constant} \cdot T^{\gamma / (\gamma - 1)}$.Comparing with $P \propto T^{c}$,we get $c = \frac{\gamma}{\gamma - 1}$.For a monoatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.Substituting the value of $\gamma$: $c = \frac{5/3}{(5/3) - 1} = \frac{5/3}{2/3} = \frac{5}{2} = 2.5$.

For an adiabatic process,the pressure and temperature of a monoatomic gas are related as $P \propto T^{c}$. Then,$c = $.......

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