Thermodynamics question bank (Physics)

In the following $p-V$ diagram the equation of state along the curved path is given by $(V-2)^{2}=4 a p$ where $a$ is a constant. The total work done in the closed path is

A poly-atomic molecule ( $C_V=3 R, C_P=4 R$, where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.$ $\left.\mathrm{Pa}, \mathrm{V}_{\mathrm{C}}=8 \times 10^{-6} \mathrm{~m}^3\right)$. A to B is an adiabatic path and $B$ to $C$ is an isothermal path. The net heat absorbed per unit mole by the system is :