Step 1: Use the shape of the curves instead of memorising the cycle sequence:
On a P-V diagram, an easy way to tell an isothermal curve from an adiabatic curve is by how steep it looks near the same point.
For an ideal gas, an isothermal curve follows $PV = constant$, while an adiabatic curve follows $PV^{\gamma} = constant$, and since $\gamma$ is always greater than 1, the adiabatic curve falls faster, so it looks steeper.
Step 2: Look at the four curves drawn between states 1, 2, 3 and 4:
Going around the loop from 1 to 2 to 3 to 4, two of the curves are flatter (these carry heat at a fixed temperature) and two are steeper (these carry no heat at all).
The curve from 1 to 2 and the curve from 3 to 4 are the flatter, constant temperature curves.
The curve from 2 to 3 and the curve from 4 to 1 are the steeper curves, where the gas changes temperature but exchanges no heat with the surroundings.
Step 3: Connect steepness to entropy behaviour:
A process with zero heat transfer, done reversibly, keeps entropy fixed, and that is exactly the definition of isentropic.
So the two steeper curves, 2-3 and 4-1, are the isentropic ones, while the two flatter curves, 1-2 and 3-4, involve heat transfer and therefore a change in entropy.
Step 4: Match this back to the given options:
(A) 1-2 and (C) 3-4 belong to the flatter, isothermal curves, so they are ruled out.
(B) 2-3 and (D) 4-1 belong to the steeper, adiabatic curves, so these are the correct picks.
Final Answer:
Reading the steepness of the curves confirms that 2-3 and 4-1 are the isentropic processes.
\[ \boxed{\text{(B) and (D)}} \]
