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(D) The given equation is $\left( \frac{T^7}{P^2} ight)^{1/5} = 	ext{constant}$.This can be rewritten as $T^{7/5} P^{-2/5} = 	ext{constant}$.For a

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$2.5$

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(D) The given equation is $\left( \frac{T^7}{P^2} ight)^{1/5} = 	ext{constant}$.This can be rewritten as $T^{7/5} P^{-2/5} = 	ext{constant}$.For an adiabatic process,the equation relating temperature $T$ and pressure $P$ is $T^{\gamma} P^{1-\gamma} = 	ext{constant}$.Comparing the exponents of $T$ and $P$,we have $\gamma = 7/5$ and $1 - \gamma = -2/5$.Solving for $\gamma$,we get $\gamma = 1 + 2/5 = 7/5 = 1.4$.The molar specific heat at constant volume $C_v$ is given by the formula $C_v = \frac{R}{\gamma - 1}$.Substituting $\gamma = 7/5$,we get $C_v = \frac{R}{7/5 - 1} = \frac{R}{2/5} = 2.5 R$.

The equation of a certain gas can be written as: $\left( \frac{T^7}{P^2} \right)^{1/5} = \text{constant}$. The specific heat at constant volume of this gas is (in $\text{J/mol K}$): (in $R$)

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