Problem 4. One mole of an ideal gas undergoes a closed cycle, in which:

$1-2$ - isobaric process, during which the volume increases by 4 times;

$2-3$ - isothermal process, during which the pressure increases;

$3-1$ - a process in which the gas is compressed according to the law $T=\gamma V^{2}$.

Find how many times the volume in state 3 exceeds the initial volume in state 1.
Solution. Let the initial volume and pressure be denoted as $\left(V_{0} ; P_{0}\right)$. Then $V_{2}=4 V_{0}$.

From the Mendeleev-Clapeyron law, we have three relationships:

$$
P_{0} V_{0}=R T_{1}, P_{0} V_{2}=R T, P_{3} V_{3}=R T
$$

Dividing the third relationship by the second, we get: $\frac{P_{3}}{P_{0}}=\frac{V_{2}}{V_{3}}$.

From the dependence of temperature on volume in the process $3-1$, there is a linear relationship between pressure and volume: $P=\gamma V$. Then we can write the relationship between the parameters corresponding to states 1 and $3: \frac{P_{3}}{P_{0}}=\frac{V_{3}}{V_{0}}$.

From the two obtained relationships, it follows that:

$$
\frac{V_{2}}{V_{3}}=\frac{V_{3}}{V_{0}} \Rightarrow V_{3}^{2}=V_{2} V_{0} \Rightarrow V_{3}=2 V_{0}
$$

Answer: 2.

Answer to variant 212: 2.