Question

[Key Concept: For one mole V = M/ρ, so a ρ–P diagram converts directly to a P–V cycle; P ∝ ρ means PV constant (isothermal).]

$$\begin{matrix} & A \rightarrow B \\ & P \propto \rho \\ & \ \therefore PV = \text{~Constant~} \smallsetminus \text{~Isothermal pressure~} \\ & W_{AB} = nRTln\frac{v_{2}}{v_{1}} = P_{0}\frac{M}{\rho_{0}}\mathcal{l}n\frac{\rho_{0}}{2\rho_{0}} = - \frac{P_{0}M}{\rho_{0}}\mathcal{l}n2 \\ & W_{AB} = - \frac{\rho_{0}M}{\rho_{0}}\mathcal{l}n2\ \text{~Option (a)~} \\ & B \rightarrow C\text{~Isobaric pressure~} \\ & W = P\Delta V = 2P_{0}\left( \frac{M}{\rho_{0}} - \frac{M}{2\rho_{0}} \right) = \frac{P_{0}M}{\rho_{0}} \\ & Q_{AB} = - \frac{P_{0}M}{\rho_{0}}\mathcal{l}n2 \end{matrix}$$

$$\begin{matrix} & Q_{BC} = nC_{P}\Delta T = \frac{5}{2}\left( P_{2}V_{2} - P_{1}V_{1} \right) = \frac{5}{2}\left\lbrack 2P_{0}\frac{M}{\rho_{0}} - 2P_{0}\frac{M}{2\rho_{0}} \right\rbrack \\ & \ = \frac{5}{2}P_{0}\frac{M}{\rho_{0}} \\ & Q_{CA} = nC_{V}\Delta T = \frac{3}{2}\left( P_{2}V_{2} - P_{1}V_{1} \right) = \frac{3}{2}\left( P_{0}\frac{M}{\rho_{0}} - 2P_{0}\frac{M}{\rho_{0}} \right) \\ & \ = - \frac{3}{2}\frac{P_{0}M}{\rho_{0}} \\ & Q_{a} = Q_{BC} = \frac{5}{2}\frac{P_{0}M}{\rho_{0}} \\ & W = \Sigma W = \Sigma Q = \frac{P_{0}M}{\rho_{0}}(1 - \mathcal{l}n2) \\ & \eta = \frac{W}{Q_{a}} = \frac{2}{5}(1 - \mathcal{l}n2)\ \text{~Option~}(c) \\ & Q_{\text{rej~}} = Q_{AB} + Q_{CA} = - \frac{P_{0}M}{\rho_{0}}\left( \mathcal{l}n2 + \frac{3}{2} \right) \end{matrix}$$