Energy absorbed by the gas from a PV-diagram


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> 10. The $PV$-graph for a monoatomic gas is shown in the figure. Find the energy absorbed by the gas during this process.
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> [![enter image description here](https://i.stack.imgur.com/17EOc.png)](https://i.stack.imgur.com/17EOc.png)
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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](https://en.wikipedia.org/wiki/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 ($V\_2 > V\_1$), the work is negative and can be found geometrically as the area under the line:


[![enter image description here](https://i.stack.imgur.com/g8SQQ.png)](https://i.stack.imgur.com/g8SQQ.png)


$$W = -(\color{blue}{2P\_0(2V\_0-V\_0)} + \color{orange}{0.5 \cdot (3P\_0 - 2P\_0)(2V\_0 - V\_0)}) = -2.5P\_0V\_0 \tag{3}$$


Now back to \eqref{eq:1}:


$$Q = 6P\_0V\_0 -(-2.5P\_0V\_0) = 8.5P\_0V\_0$$


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