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(C) Given:Heat at constant pressure: $Q_P = n C_P \Delta T_1 = 160 	ext{ cal}$,where $\Delta T_1 = 50^{\circ} C$.Heat at constant volume: $Q_V = n C_

Vedclass · Accepted Answer

$6$

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(C) Given:Heat at constant pressure: $Q_P = n C_P \Delta T_1 = 160 	ext{ cal}$,where $\Delta T_1 = 50^{\circ} C$.Heat at constant volume: $Q_V = n C_V \Delta T_2 = 240 	ext{ cal}$,where $\Delta T_2 = 100^{\circ} C$.From the first equation: $n C_P = \frac{160}{50} = 3.2$.From the second equation: $n C_V = \frac{240}{100} = 2.4$.Taking the ratio: $\frac{C_P}{C_V} = \gamma = \frac{3.2}{2.4} = \frac{4}{3}$.The relation between degrees of freedom $f$ and adiabatic index $\gamma$ is $\gamma = 1 + \frac{2}{f}$.Substituting $\gamma = \frac{4}{3}$: $\frac{4}{3} = 1 + \frac{2}{f} \Rightarrow \frac{1}{3} = \frac{2}{f} \Rightarrow f = 6$.

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