Question

Three moles of an ideal gas undergoes a cyclic process ABCA as shown in the figure. The pressure, volume and absolute temperature at points A, B and C are respectively \((P_1, V_1, T_1)\), \((P_2, 3V_1, T_1)\) and \((P_2, V_1, T_2)\). Then the total work done in the cycle ABCA is (R- Universal gas constant).

• A. \(RT_1[3\ln(3)+2]\)
• B. \(RT_1[3\ln(2)]\)
• C. \(3RT_1(\ln 3)\)
• D. \(RT_1[3\ln(3)-2]\)

Hint:

For any cyclic process on a P-V diagram, the sign of the work done can be determined by the direction of the cycle. A clockwise cycle represents net work done by the gas (positive work), while a counter-clockwise cycle (like this one) represents net work done on the gas (negative work). Since \(\ln(3) \approx 1.0986\), \(3\ln(3)-2 \approx 3.29 - 2>0\). Wait, let me recheck the sign. The diagram shows the cycle is A → B → C → A. This is a clockwise cycle. A→B is expansion (+ve work). B→C is compression (-ve work). The area under AB is larger than the area under BC, so net work should be positive. My formula gives \(3RT_1 \ln(3) - 2RT_1 = RT_1(3.29-2)>0\). The reasoning is correct.

Answer 1

The Correct Option is D

Step 1: Identifying the Processes: - Process A \(\to\) B: Temperature is constant (\(T_1\)). This is an Isothermal Expansion. - Process B \(\to\) C: Pressure is constant (\(P_2\)). This is an Isobaric Compression. - Process C \(\to\) A: Volume is constant (\(V_1\)). This is an Isochoric Pressurization.
Step 2: Work Done in Each Step: Given \( n = 3 \) moles. 1. Work done in A \(\to\) B (Isothermal): \[ W_{AB} = nRT \ln\left(\frac{V_f}{V_i}\right) \] \[ W_{AB} = 3RT_1 \ln\left(\frac{3V_1}{V_1}\right) = 3RT_1 \ln(3) \] 2. Work done in B \(\to\) C (Isobaric): \[ W_{BC} = P\Delta V = P_2(V_C - V_B) \] \[ W_{BC} = P_2(V_1 - 3V_1) = -2P_2V_1 \] From the state at B \((P_2, 3V_1, T_1)\), using Ideal Gas Law \( PV = nRT \): \[ P_2(3V_1) = nRT_1 = 3RT_1 \] \[ P_2V_1 = RT_1 \] Substituting \( P_2V_1 = RT_1 \) into the work equation: \[ W_{BC} = -2(RT_1) = -2RT_1 \] 3. Work done in C \(\to\) A (Isochoric): Since \( \Delta V = 0 \), work done is zero. \[ W_{CA} = 0 \]
Step 3: Total Work Done: Total Work \( W_{\text{cycle}} = W_{AB} + W_{BC} + W_{CA} \) \[ W_{\text{cycle}} = 3RT_1 \ln(3) - 2RT_1 + 0 \] \[ W_{\text{cycle}} = RT_1 [3 \ln(3) - 2] \]
Step 4: Final Answer: The total work done is \( RT_1[3\ln(3)-2] \).