Energy absorbed by the gas from a PV-diagram
I suspect you forgot to take the work done by the gas into account. From the first law of thermodynamics:

$$Q = \Delta U - W \label{eq:1}\tag{1}$$

For the monoatomic gas $U=\frac{3}{2}PV$, so

$$\Delta U = U_2 - U_1 = \frac{3}{2}(P_2V_2 - P_1V_1) = \frac{3}{2}(3P_0 \cdot 2V_0 - 2P_0 \cdot V_0) = 6P_0V_0 \tag{2}$$

On the other hand, since the gas expanded ($V_2 &gt; V_1$), the work is negative and can be found geometrically as the area under the line:



$$W = -(\color{blue}{2P_0(2V_0-V_0)} + \color{orange}{0.5 \cdot (3P_0 - 2P_0)(2V_0 - V_0)}) = -2.5P_0V_0 \tag{3}$$

Now back to \eqref{eq:1}:

$$Q = 6P_0V_0 -(-2.5P_0V_0) = 8.5P_0V_0$$