Step 1: Recall the four processes that make up a Carnot cycle:
A Carnot cycle is built from exactly two isothermal processes and two adiabatic processes, arranged so the cycle can be reversed without any loss.
Isothermal processes happen at constant temperature, with heat flowing in or out of the system, while adiabatic processes happen with no heat transfer at all.
A process with no heat transfer, carried out reversibly, is called isentropic, because entropy stays constant along it.
Step 2: Identify which of the four processes in the figure are adiabatic:
From the P-V diagram, state 1 sits at the highest pressure, and the cycle moves 1 to 2 to 3 to 4 and back to 1.
Process 1-2 is the isothermal expansion, where the gas expands at the higher constant temperature while absorbing heat.
Process 2-3 continues the expansion, but now with no heat exchange, so the temperature itself falls, this is the adiabatic expansion.
Process 3-4 is the isothermal compression, where the gas is compressed at the lower constant temperature while rejecting heat.
Process 4-1 completes the loop with an adiabatic compression, raising the temperature back to the starting value with no heat exchange.
Step 3: Match adiabatic with isentropic:
Since the Carnot cycle is fully reversible, both adiabatic legs, 2-3 and 4-1, are also isentropic, entropy does not change along them.
The two isothermal legs, 1-2 and 3-4, are not isentropic because heat is transferred during them, and entropy changes whenever heat crosses the system boundary.
Step 4: Check each option:
Option (A), 1-2, is isothermal, not isentropic, so it is wrong.
Option (B), 2-3, is the adiabatic expansion, so it is isentropic, correct.
Option (C), 3-4, is isothermal, not isentropic, so it is wrong.
Option (D), 4-1, is the adiabatic compression, so it is isentropic, correct.
Final Answer:
The isentropic legs of the Carnot cycle are the two adiabatic processes, 2-3 and 4-1.
\[ \boxed{\text{(B) and (D)}} \]
