Use the first law of thermodynamics

For any process, $Q = \Delta U + W$ where:

• $Q$ = heat absorbed by the gas,
• $\Delta U$ = change in internal energy,
• $W$ = work done by the gas.

For an ideal gas, internal energy depends only on temperature, and $T \propto PV$

Interpret the rectangular $P$-$V$ cycle

In the standard four-step rectangular cycle:

• Step 1: vertical upward line (constant volume, pressure increases)
• Step 2: horizontal rightward line (constant pressure, volume increases)
• Step 3: vertical downward line (constant volume, pressure decreases)
• Step 4: horizontal leftward line (constant pressure, volume decreases)

Check heat transfer in each step

Step 1: Constant volume, pressure increases

Since $V$ is constant and $P$ increases, $PV \text{ increases} \Rightarrow T \text{ increases}$ so $\Delta U > 0$.

Also, at constant volume, $W = 0$ Hence, $Q = \Delta U > 0$ So heat is absorbed in step 1.

Step 2: Constant pressure, volume increases

Since $P$ is constant and $V$ increases, $PV \text{ increases} \Rightarrow T \text{ increases}$ so $\Delta U > 0$.

Also, expansion occurs, so $W = P\Delta V > 0$ Therefore, $Q = \Delta U + W > 0$ So heat is absorbed in step 2.

Step 3: Constant volume, pressure decreases

Here $PV$ decreases, so temperature decreases: $\Delta U < 0, \quad W = 0$ Therefore, $Q = \Delta U < 0$ Heat is released in step 3.

Step 4: Constant pressure, volume decreases

Here $PV$ decreases, so temperature decreases: $\Delta U < 0$ Compression occurs, so $W = P\Delta V < 0$ Hence, $Q = \Delta U + W < 0$ Heat is released in step 4.

Conclusion

Heat is absorbed during: $\boxed{\text{steps 1 and 2}}$

So the correct option is: $\boxed{A}$

Comparison with stored answer

The stored correct answer is C (steps 1 and 4), but from thermodynamic analysis, step 4 is a compression with decreasing temperature, so heat cannot be absorbed there.

Therefore, the stored answer appears incorrect.