4. (6 points) In an ideal gas, a thermodynamic cycle consisting of two isochoric and two adiabatic processes is carried out. The ratio of the initial and final absolute temperatures in the isochoric cooling process is \( k = 1.5 \). Determine the efficiency of this cycle, given that the efficiency of the Carnot cycle with the same ratio of maximum to minimum temperature as in the considered cycle is \( 50\% \).

## 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 the adiabatic processes, while energy is transferred in the form of heat in the isochoric processes. Using the first law of thermodynamics, we obtain for the work of

![](https://cdn.mathpix.com/cropped/2024_05_06_3672d113fcc2ad54f09dg-44.jpg?height=566&width=626&top_left_y=1116&top_left_x=1363)
the gas in the adiabatic processes:

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

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

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

According to 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. By 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

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

Considering the obtained equality, we can write:

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

For the Carnot cycle

$$
\eta_{\mathrm{K}} = 1 - \frac{T_{\min}}{T_{\max}}
$$

Then

$$
\eta = 1 - k \left( 1 - \eta_{\mathrm{K}} \right) = 0.25 = 25\%.
$$
Answer: $\eta=1-k\left(1-\eta_{C}\right)=0.25=25 \%$.

## Evaluation Criteria

| Performance | Score |
| :--- | :---: |
| Participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for work in the cycle is written | $\mathbf{1}$ |
| Expression for the amount of heat received is written | $\mathbf{1}$ |
| Expression for the efficiency coefficient in terms of temperatures in the cycle states is written | $\mathbf{1}$ |
| Expression for the temperature ratios in isochoric processes is obtained | $\mathbf{1}$ |
| Expression for the efficiency of the Carnot cycle is written | $\mathbf{1}$ |


| Necessary transformations are performed and a numerical answer is obtained | $\mathbf{1}$ |
| :---: | :---: | :---: |
| Total points | $\mathbf{6}$ |