Identification of Heat Absorption Steps in Four Step Gas Cycle

To determine the steps in which heat is absorbed by the ideal gas, we use the First Law of Thermodynamics: $Q = \Delta U + W$ where:

• $Q$ is the heat added to the system,
• $\Delta U = n C_v \Delta T$ is the change in internal energy,
• $W = \int P \, dV$ is the work done by the gas.

From the ideal gas law $PV = nRT$, the temperature $T$ is directly proportional to the product $PV$.

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Analysis of Each Step:

1. Step 1: Isobaric Expansion

   • Pressure ($P$): Constant at a higher value $P_{\text{high}}$
   • Volume ($V$): Increases ($\Delta V > 0$), so the work done is positive: $W_1 = P_{\text{high}} \Delta V > 0$.
   • Temperature ($T$): Since $P$ is constant and $V$ increases, $T = \frac{PV}{nR}$ increases ($\Delta T > 0$).
   • Internal Energy ($\Delta U$): $\Delta U_1 = n C_v \Delta T > 0$.
   • Heat ($Q$): $Q_1 = \Delta U_1 + W_1 > 0$
   • Thus, heat is absorbed in step 1.

2. Step 2: Isochoric Pressure Drop

   • Volume ($V$): Constant at $V_{\text{high}}$ ($\Delta V = 0 \implies W_2 = 0$).
   • Pressure ($P$): Decreases ($\Delta P < 0$).
   • Temperature ($T$): Since $V$ is constant and $P$ decreases, $T$ decreases ($\Delta T < 0$).
   • Internal Energy ($\Delta U$): $\Delta U_2 = n C_v \Delta T < 0$.
   • Heat ($Q$): $Q_2 = \Delta U_2 + 0 < 0$
   • Thus, heat is released in step 2.

3. Step 3: Isobaric Compression

   • Pressure ($P$): Constant at a lower value $P_{\text{low}}$
   • Volume ($V$): Decreases ($\Delta V < 0$), so $W_3 = P_{\text{low}} \Delta V < 0$.
   • Temperature ($T$): Since $P$ is constant and $V$ decreases, $T$ decreases ($\Delta T < 0$).
   • Internal Energy ($\Delta U$): $\Delta U_3 = n C_v \Delta T < 0$.
   • Heat ($Q$): $Q_3 = \Delta U_3 + W_3 < 0$
   • Thus, heat is released in step 3.

4. Step 4: Isochoric Pressure Increase

   • Volume ($V$): Constant at $V_{\text{low}}$ ($\Delta V = 0 \implies W_4 = 0$).
   • Pressure ($P$): Increases ($\Delta P > 0$).
   • Temperature ($T$): Since $V$ is constant and $P$ increases, $T$ increases ($\Delta T > 0$).
   • Internal Energy ($\Delta U$): $\Delta U_4 = n C_v \Delta T > 0$.
   • Heat ($Q$): $Q_4 = \Delta U_4 + 0 > 0$
   • Thus, heat is absorbed in step 4.

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Conclusion:

Heat is absorbed by the gas in steps 1 and 4.

Correct Option: (C)