
The process $\mathrm{BC}$ is an isochoric process since the volume is constant
The process $\mathrm{CA}$ is isothermal since, temperature remains constant.
The only $P-v$ graph that matches the above criteria is option $\mathrm{C}$
The correct option ($s$) is (are)
$(A)$ $q_{A C}=\Delta U_{B C}$ and $W_{A B}=P_2\left(V_2-V_1\right)$ $(B)$ $W _{ BC }= P _2\left( V _2- V _1\right)$ and $q _{ BC }= H _{ AC }$ $(C)$ $\Delta H _{ CA }<\Delta U _{ CA }$ and $q _{ AC }=\Delta U _{ BC }$ $(D)$ $q_{B C}=\Delta H_{A C}$ and $\Delta H_{C A}>\Delta U_{C A}$
