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).
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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.
Step 1: Understanding the Concept:
The total work done in a cyclic process is the sum of the work done in each individual process that makes up the cycle. The work done is the area enclosed by the cycle on a P-V diagram. We need to calculate the work for each segment: AB, BC, and CA. Step 2: Key Formula or Approach:
- Work done during an isothermal process (constant T): \(W = nRT \ln(V_f/V_i)\).
- Work done during an isobaric process (constant P): \(W = P(V_f - V_i)\).
- Work done during an isochoric process (constant V): \(W = 0\).
- Total work \(W_{cycle} = W_{AB} + W_{BC} + W_{CA}\). Step 3: Detailed Explanation: Process AB:
The temperature is constant at \(T_1\), so this is an isothermal expansion from \(V_1\) to \(3V_1\).
\[ W_{AB} = nRT_1 \ln\left(\frac{V_B}{V_A}\right) = 3RT_1 \ln\left(\frac{3V_1}{V_1}\right) = 3RT_1 \ln(3) \]
Process BC:
The pressure is constant at \(P_2\), so this is an isobaric compression from \(3V_1\) to \(V_1\).
\[ W_{BC} = P_2(V_C - V_B) = P_2(V_1 - 3V_1) = -2P_2V_1 \]
Process CA:
The volume is constant at \(V_1\), so this is an isochoric process.
\[ W_{CA} = 0 \]
Total Work Done:
\[ W_{total} = W_{AB} + W_{BC} + W_{CA} = 3RT_1 \ln(3) - 2P_2V_1 \]
To get the answer in the desired form, we need to express \(P_2V_1\) in terms of \(R\) and \(T_1\). We can use the ideal gas law for state B: \((P_2, 3V_1, T_1)\).
\[ P_B V_B = nRT_B \implies P_2 (3V_1) = 3RT_1 \]
Dividing by 3, we get:
\[ P_2V_1 = RT_1 \]
Now, substitute this back into the total work equation:
\[ W_{total} = 3RT_1 \ln(3) - 2(RT_1) = RT_1[3\ln(3) - 2] \]
Step 4: Final Answer:
The total work done in the cycle is \(RT_1[3\ln(3)-2]\). Therefore, option (D) is correct.
