Energy absorbed by the gas from a PV-diagram The $PV$-graph for a monoatomic gas is shown in the figure. Find the energy absorbed by the gas during this process.



I’m having trouble with this question because both pressure and volume are changing and I cannot calculate a value for $T$. So far, I ended up with an answer using the ratio of $C_p$ and $C_v$, and got $6.25P_0V_0$. The answer is $8.5P_0V_0$.
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}:

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