4. (6 points) In an ideal gas, a thermodynamic cycle consisting of two isochoric and two adiabatic processes is carried out. The maximum \( T_{\max} = 900 \, \text{K} \) and minimum \( T_{\min} = 350 \, \text{K} \) absolute temperatures of the gas in the cycle are known, as well as the efficiency of the cycle \( \eta = 0.4 \). Determine the ratio of the final and initial absolute temperatures in the isochoric heating process.

## Possible solution.

In the graph of the cyclic process, it is clear that the absolute temperatures \( T_{1} = T_{\max} \), and \( T_{3} = T_{\min} \). By definition, the efficiency is the ratio of the work done by the gas in the cycle to the amount of heat transferred to the gas. In the studied cycle, work is performed in adiabatic processes, while energy is transferred in the form of heat in isochoric processes. Using the first law of thermodynamics, we obtain for the work done by the gas in adiabatic processes:

![](https://cdn.mathpix.com/cropped/2024_05_06_0593b734296e341e16d2g-12.jpg?height=574&width=632&top_left_y=198&top_left_x=1363)

$$
\begin{aligned}
A_{12} & = -\Delta U_{12} = \frac{i}{2} v R \left( T_{\max} - T_{2} \right) > 0 \\
A_{34} & = -\Delta U_{34} = \frac{i}{2} v R \left( T_{\min} - T_{4} \right) < 0
\end{aligned}
$$

where \( \Delta U \) is the change in internal energy of the gas, \( i \) is the number of degrees of freedom of the gas molecule, \( v \) is the amount of substance, and \( R \) is the universal gas constant.

The work done by the gas in the cycle:

$$
A = A_{12} + A_{34} = \frac{i}{2} v R \left[ \left( T_{\max} - T_{2} \right) - \left( T_{4} - T_{\min} \right) \right]
$$

The amount of heat transferred to the gas in the isochoric process \( 4-1 \):

$$
Q_{1} = \Delta U_{41} = \frac{i}{2} v R \left( T_{\max} - T_{4} \right)
$$

The efficiency

$$
\eta = \frac{\left( T_{\max} - T_{4} \right) - \left( T_{2} - T_{\min} \right)}{T_{\max} - T_{4}} = 1 - \frac{T_{2} - T_{\min}}{T_{\max} - T_{4}}
$$

In accordance with Poisson's equation for adiabatic processes:

$$
\begin{aligned}
p_{1} V_{1}^{\gamma} & = p_{2} V_{2}^{\gamma} \\
p_{4} V_{1}^{\gamma} & = p_{3} V_{1}^{\gamma} \\
\frac{p_{1}}{p_{4}} & = \frac{p_{2}}{p_{3}}
\end{aligned}
$$

Here, \( \gamma \) is the adiabatic index. According to Charles's law:

$$
\begin{aligned}
& \frac{p_{1}}{p_{4}} = \frac{T_{\max}}{T_{4}} \\
& \frac{p_{2}}{p_{3}} = \frac{T_{2}}{T_{\min}}
\end{aligned}
$$

We obtain that the required temperature ratio is

$$
\frac{T_{\max}}{T_{4}} = \frac{T_{2}}{T_{\min}} = k
$$

Considering the obtained equality, we can write:

$$
\eta = 1 - \frac{T_{2} - T_{\min}}{T_{\max} - T_{4}} = 1 - \frac{T_{\min} (k - 1)}{T_{\max} \left( 1 - \frac{1}{k} \right)} = 1 - k \frac{T_{\min}}{T_{\max}}
$$

Then

$$
k = \frac{T_{\max}}{T_{\min}} (1 - \eta) = \frac{900}{350} \cdot (1 - 0.4) \approx 1.54
$$
Answer: $k=\frac{T_{\max }}{T_{\text {min }}}(1-\eta) \approx 1.54$.

## Evaluation Criteria

| Performance | Score |
| :--- | :---: |
| The participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| The expression for work in the cycle is written | $\mathbf{1}$ |
| The expression for the amount of heat received is written | $\mathbf{1}$ |
| The expression for the efficiency coefficient in terms of temperatures in the cycle is written | $\mathbf{1}$ |
| The expression for the temperature ratios in isochoric processes is obtained | $\mathbf{1}$ |
| Necessary transformations are performed and the answer is obtained in general form | $\mathbf{1}$ |
| The numerical answer is obtained | $\mathbf{1}$ |
| Total points | $\mathbf{6}$ |