Explanation
$$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC_{P}\Delta T}=\frac{nR\Delta T}{nC_{P}\Delta T}=\frac{x/3}{x}$$ (standard result)
$$\displaystyle \Rightarrow C_{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$$
$$\textbf{Explanation}$$
$$\bullet$$Kinetic energy of particle can be given by the formula:
$$K = \dfrac{3}{2} K_b\ T$$, where $$K_b$$ is Boltzmann’s constant and T is temperature.
$$\bullet$$As temperature increases, it will results into increase of kinetic energy. Electrons or atoms in gaseous states are bound with either energy states or covalent bonds which results into increment of vibrational energy.
So, speed with which the molecules vibrate increases.
Hence, option A is the correct answer.
$$\displaystyle P=\dfrac {P_0}{1+\left (\dfrac {V}{V_0}\right )^3}=\dfrac {P_0}{2}\Rightarrow T=\dfrac {P_0V_0}{2R}$$
$$\therefore$$ Translational kinetic energy is equal to $$\dfrac {3}{2}RT=\dfrac {3R}{2}\dfrac {P_0V_0}{2R}=\dfrac {3P_0V_0}{4}$$
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