A reversible cyclic process for an ideal gas is shown below. Here, $P, V$ and $T$ are pressure, volu

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A reversible cyclic process for an ideal gas is shown below. Here, $P, V$ and $T$ are pressure, volume and temperature, respectively. The thermodynamic parameters $q, w, H$ and $U$ are heat, work, enthalpy and internal energy, respectively.\ The correct option(s) is (are)\

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(b, c) : A-C (Isochoric process)\ $\Rightarrow w_{A C}=0$ and $\Delta U_{A C}=q_{A C}$\ $B-C$ (Isobaric process)\ $\Rightarrow \quad \Delta U_{B C}=q_{B C}+w_{B C}$ $$ w_{B C}=-P_{2}\left(V_{1}-V_{2}ight)=P_{2}\left(V_{2}-V_{1}ight) $$ $$ \Rightarrow \quad q_{B C}=\Delta H_{B C} $$ $\because \quad(\Delta T)_{A-C}=(\Delta T)_{B-C}$\ $	herefore \quad \Delta U_{B C}=\Delta U_{A C}=q_{A C}$\ $\Delta H_{B C}=\Delta H_{A C}=q_{B C} \quad\left(\because T_{2}>T_{1}ight)$\ $\Delta H_{C A}$ and $\Delta U_{C A}$ are negative.\ $\Delta H_{C A}=\Delta U_{C A}+V \Delta P$\ (--ve)\ $	herefore \quad \Delta H_{C A}<\Delta U_{C A}$\ $A-B$ (Isothermal process)\ $\Delta U_{A B}=\Delta H_{A B}=0$\ $w_{A B}=-n R T_{1} \ln \frac{V_{2}}{V_{1}}$\