6. (15 points) The operation of a heat engine is based on a cycle consisting of an isochore, an isotherm, and a process with a directly proportional dependence of pressure on volume (see figure). An ideal monatomic gas is used as the working substance. It is known that the maximum and minimum temperatures differ by a factor of two. Determine the efficiency of this heat engine.

![](https://cdn.mathpix.com/cropped/2024_05_06_96efd573cfdf000c103ag-04.jpg?height=346&width=388&top_left_y=2023&top_left_x=914)
Answer: $8.8\%$.

Solution. Since the temperature changes by a factor of two during the process 3-1, it follows from the equation of state of an ideal gas $p V=\vartheta R T$ that the volume changes by a factor of $\sqrt{2}$.

The magnitude of the work done by the gas in the process 3-1 is equal to the area under the graph:

$A_{3-1}=\frac{1}{2}\left(p_{1}+p_{3}\right)\left(V_{3}-V_{1}\right)=\frac{1}{2}\left(p_{1} V_{3}-p_{1} V_{1}+p_{3} V_{3}-p_{3} V_{1}\right)$.

Since: $p_{1} V_{3}=p_{3} V_{1}$,

we get: $A_{3-1}=\frac{1}{2}\left(-p_{1} V_{1}+p_{3} V_{3}\right)=\frac{1}{2} v R\left(T_{3}-T_{1}\right)$.

The amount of heat in the processes: $Q_{12}=\Delta U_{12}=\frac{3}{2} \vartheta R\left(T_{2}-T_{1}\right)=\frac{3}{2} \vartheta R T_{1}$.

(2 points)

$Q_{23}=A_{23}=\vartheta R T_{2} \ln \left(\frac{V_{3}}{V_{1}}\right)=2 \vartheta R T_{1} \ln \sqrt{2}$.

$Q_{31}=\Delta U_{31}+A_{31}=\frac{3}{2} \vartheta R\left(T_{1}-T_{3}\right)+\frac{1}{2} \vartheta R\left(T_{1}-T_{3}\right)=2 \vartheta R\left(T_{1}-T_{3}\right)=-2 \vartheta R T_{1}$.

(2 points)

The efficiency of the heat engine:

$\eta=\frac{Q_{\mathrm{H}}-Q_{\mathrm{X}}}{Q_{\mathrm{H}}}=\frac{Q_{12}+Q_{23}-\left|Q_{31}\right|}{Q_{12}+Q_{23}}=\frac{\frac{3}{2} \vartheta R T_{1}+2 \vartheta R T_{1} \ln \sqrt{2}-2 \vartheta R T_{1}}{\frac{3}{2} \vartheta R T_{1}+2 \vartheta R T_{1} \ln \sqrt{2}}=\frac{1.5+2 \ln \sqrt{2}-2}{1.5+2 \ln \sqrt{2}}$.

As a result: $\eta=0.088=8.8\%$.