The figure shows the $P$-$V$ plot of an ideal gas taken through a cycle $ABCDA$. The part $ABC$ is a semi-circle and $CDA$ is half of an ellipse. Then,

Step 1: Analyze the path $A \to B$.

For an isothermal process of an ideal gas, the product of pressure and volume must remain constant.

$$PV = \text{constant}$$

At the endpoints $A(2,3)$ and $B(3,2)$, the product is equal.

$$P_A V_A = P_B V_B = 6$$

However, along the semi-circular path connecting $A$ and $B$, the product $PV$ varies and reaches a maximum value of approximately $7.33$.

$$PV \neq \text{constant}$$

Therefore, the path $A \to B$ is not isothermal.

Step 2: Analyze the path $B \to C \to D$.

Determine the sign of the heat exchange using the First Law of Thermodynamics.

$$Q = \Delta U + W$$

Along this path, both temperature and volume decrease, meaning both the change in internal energy and work done are negative.

$$\Delta U < 0$$ $$W < 0$$

Since both terms are negative, the net heat exchange is negative, meaning heat flows out of the gas.

$$Q < 0$$

Step 3: Calculate the work done along the path $A \to B \to C$.

The work done is represented by the area under the path on the $P$-$V$ diagram.

$$W_{A \to B \to C} = \int P \, dV$$

This integral corresponds to the area of the right semi-circle.

$$W_{A \to B \to C} = \frac{\pi}{2} \neq 0$$

Step 4: Determine the net work done in the cycle $ABCDA$.

The cycle is traversed in a clockwise direction.

$$W_{\text{cycle}} = \oint P \, dV$$

For a clockwise cycle, the net work done by the gas is positive and equal to the enclosed area of the semi-circle and semi-ellipse.

$$W_{\text{cycle}} = \frac{\pi}{2} + \frac{\pi a}{2} > 0$$

Answer: The correct statements are:

• Heat flows out of the gas during path $B \to C \to D$.
• Positive work is done by the gas in the cycle $ABCDA$.