6. (15 points) The operation of a heat engine is based on a cycle consisting of an isobaric process, an isothermal process, 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-11.jpg?height=384&width=452&top_left_y=1855&top_left_x=879)
Answer: $6.1\%$.

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

(1 point)

The magnitude of the work done by the gas in the process 2-3 is equal to the area under the graph:
$A_{2-3} = \frac{1}{2} \left(p_{2} + p_{3}\right) \left(V_{2} - V_{3}\right) = \frac{1}{2} \left(p_{2} V_{2} - p_{2} V_{3} + p_{3} V_{2} - p_{3} V_{3}\right)$.

Since $p_{2} V_{3} = p_{3} V_{2}$

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

The amount of heat in the processes

$$
\begin{aligned}
& Q_{12} = \Delta U_{12} + A_{12} = \frac{3}{2} \vartheta R \left(T_{2} - T_{1}\right) + \vartheta R \left(T_{2} - T_{1}\right) = \frac{5}{2} \vartheta R T_{1} \\
& Q_{31} = A_{31} = \vartheta R T_{1} \ln \left(\frac{p_{3}}{p_{1}}\right) = -\vartheta R T_{1} \ln \sqrt{2} \\
& Q_{23} = \Delta U_{23} + A_{23} = \frac{3}{2} \vartheta R \left(T_{3} - T_{2}\right) + \frac{1}{2} \vartheta R \left(T_{3} - T_{2}\right) = 2 \vartheta R \left(T_{3} - T_{2}\right) = -2 \vartheta R T_{1}
\end{aligned}
$$

(2 points)

The efficiency of the heat engine:

$\eta = \frac{Q_{\mathrm{H}} - Q_{\mathrm{x}}}{Q_{\mathrm{H}}} = \frac{Q_{12} - \left|Q_{31} + Q_{23}\right|}{Q_{12}} = \frac{\frac{5}{2} \vartheta R T_{1} - \vartheta R T_{1} \ln \sqrt{2} - 2 \vartheta R T_{1}}{\frac{5}{2} \vartheta R T_{1}} = \frac{2.5 - \ln \sqrt{2} - 2}{2.5}$.

As a result: $\eta = 0.061 = 6.1\%$.