An ideal monoatomic gas is carried along the cycle ABCDA as shown in the figure. The total heat absorbed during this process is

Heat absorbed means heat is extracted from source, i.e. Q must be positive. The occurs along path
$A\to B\to C$  
 $\therefore$Heat added = $\Delta Q_{ABC}=\Delta U_{ABC}+\Delta w_{ABC}$
 $$\begin{gathered} =C_v\left(T_C-T_A\right)+2p_0\left(2V_0-V_0\right) \\ =n\frac{3}{2}R\left(T_C-T_A\right)+3p_0{V}_0=\frac{3}{2}\left(nR{T}_C-nR{T}_A\right)+3p_0-V_0 \\ =\frac{3}{2}\left(p_C{V}_C-p_A{V}_A\right)+3p_0{V}_0 \\ =\frac{3}{2}\left(3p_{0}2V_0-p_0{V}_0\right)+3p_0{V}_0 \\ \frac{3}{2}\times {}^{3}P_0{V}_0+{}^{3}P_0{V}_0=\frac{21}{2}{p}_0{V}_0=10.5p_0{V}_0 \end{gathered}$$