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(B) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}.$Taking the natural logarith

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$f_2 > f_1$

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(B) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}.$Taking the natural logarithm on both sides, we get $\ln P + \gamma \ln V = \ln(	ext{constant}).$Rearranging this, we get $\ln P = -\gamma \ln V + \ln(	ext{constant}).$This is the equation of a straight line $y = mx + c,$ where the slope $m = -\gamma.$The magnitude of the slope is equal to the adiabatic index $\gamma.$From the given graph, the slope of line $X$ is greater than the slope of line $Y,$ so $\gamma_x > \gamma_y.$Since $\gamma = 1 + \frac{2}{f},$ a larger $\gamma$ corresponds to a smaller degree of freedom $f.$Therefore, $\gamma_x > \gamma_y \implies f_1  f_1.$

For two different gases $X$ and $Y,$ having degrees of freedom $f_1$ and $f_2$ and molar heat capacities at constant volume $C_{v_1}$ and $C_{v_2}$ respectively, for an adiabatic process, the $\ln P$ versus $\ln V$ graph is plotted as shown:

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