3 moles of an ideal monoatomic gas performs A B C D A cyclic...

Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of form $y=m x$. Also, $p=\frac{\mu R}{V} T(\mu=3)$ $p \propto T$ So, volume remains constant for the graphs $A B$ and $C D$. So, no work is done during processes for $A$ to $B$ and $C$ to $D$. $W_{A B}=W_{C D}=0$ and $\quad W_{B C}=p_2\left(V_C-V_B\right)$=\mu R\left(T_C-T_B\right)$=3 R(2400-800)$ $=3 R \times 1600$ $=4800 R$ $W_{D A}=p_1\left(V_A-V_D\right)$=\mu R\left(T_A-T_D\right)$=3 R(400-1200)$ $=-2400 R$ Work done in complete cycle $W=W_{A B}+W_{B C}+W_{C D}+W_{D A}$ $=0+4800 R+0+(-2400) R$ $=2400 R$ $=19.944 \mathrm{~J}=20 \mathrm{~kJ}$