An ideal mono-atomic gas is expanded adiabatically from A to B. It is then compressed in an isobaric process from B to C. Finally, the pressure is increased in an isochoric process from C to A. The cyclic process is shown in the figure below. For this system, which of the following is(are) correct?
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For any cyclic process on a P-V diagram, remember these two fundamental rules:
1. The net change in internal energy (\(\Delta U_{cycle}\)) is always zero.
2. The net heat transferred (\(Q_{cycle}\)) is equal to the net work done (\(W_{cycle}\)). The net work done is the area enclosed by the cycle on the P-V diagram (clockwise is positive work done by the gas, counter-clockwise is negative).
Work done along the path AB is \(\frac{3}{2}(P_1V_1 - P_2V_2)\)
Total work done during the entire process is \(\frac{3}{2}(P_1V_1 - P_2V_2) + P_2(V_1 - V_2)\)
Total heat absorbed during the entire process is \(\frac{3}{2}(P_1 - P_2)V_1\)
Total change in internal energy during the entire process is \(\frac{5}{2}P_2(V_2 - V_1)\)
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The Correct Option isB
Solution and Explanation
Step 1: Understanding the Concept:
This problem involves calculating work done, heat transfer, and change in internal energy for a cyclic process involving an ideal mono-atomic gas. The process consists of three parts: adiabatic expansion (A to B), isobaric compression (B to C), and isochoric heating (C to A). Step 2: Key Formula or Approach:
1. Work Done (W): \(W = \int P dV\).
- Adiabatic process: \(W = \frac{P_iV_i - P_fV_f}{\gamma - 1}\). For a mono-atomic gas, \(\gamma = C_p/C_v = (5/2 R)/(3/2 R) = 5/3\).
- Isobaric process: \(W = P \Delta V = P(V_f - V_i)\).
- Isochoric process: \(W = 0\) since \(dV=0\).
2. First Law of Thermodynamics: \(\Delta U = Q - W\).
3. Internal Energy (U): For an ideal gas, \(U\) depends only on temperature. \(\Delta U = nC_v \Delta T\). For a mono-atomic gas, \(C_v = \frac{3}{2}R\). So \(\Delta U = \frac{3}{2} nR \Delta T = \frac{3}{2} \Delta(PV)\).
4. Cyclic Process: For any complete cycle, the total change in internal energy is zero (\(\Delta U_{cycle} = 0\)). This implies \(Q_{cycle} = W_{cycle}\). Step 3: Detailed Explanation: Let's analyze each path:
- Path A \(\rightarrow\) B (Adiabatic Expansion):
Work done \(W_{AB} = \frac{P_A V_A - P_B V_B}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{5/3 - 1} = \frac{P_1V_1 - P_2V_2}{2/3} = \frac{3}{2}(P_1V_1 - P_2V_2)\).
- Path B \(\rightarrow\) C (Isobaric Compression):
Work done \(W_{BC} = P_B (V_C - V_B) = P_2(V_1 - V_2)\). Note that since \(V_1<V_2\), this work is negative (work done on the gas).
- Path C \(\rightarrow\) A (Isochoric Heating):
Work done \(W_{CA} = 0\) since volume is constant (\(V_1\)).
Now let's evaluate the options: (A) Work done along the path AB is \(\frac{3}{2}(P_1V_1 - P_2V_2)\).
From our calculation above, \(W_{AB} = \frac{3}{2}(P_1V_1 - P_2V_2)\). This matches the calculation. (B) Total work done during the entire process is \(\frac{3}{2}(P_1V_1 - P_2V_2) + P_2(V_1 - V_2)\).
Total work done \(W_{total} = W_{AB} + W_{BC} + W_{CA}\).
\[ W_{total} = \frac{3}{2}(P_1V_1 - P_2V_2) + P_2(V_1 - V_2) + 0 \]
This expression exactly matches option (B). Thus, statement (B) is correct. (C) Total heat absorbed during the entire process is \(\frac{3}{2}(P_1 - P_2)V_1\).
For a cyclic process, \(Q_{total} = W_{total}\).
So, \(Q_{total} = \frac{3}{2}(P_1V_1 - P_2V_2) + P_2(V_1 - V_2)\).
Let's expand this: \(\frac{3}{2}P_1V_1 - \frac{3}{2}P_2V_2 + P_2V_1 - P_2V_2 = \frac{3}{2}P_1V_1 + P_2V_1 - \frac{5}{2}P_2V_2\).
The expression in option (C) is \(\frac{3}{2}(P_1 - P_2)V_1 = \frac{3}{2}P_1V_1 - \frac{3}{2}P_2V_1\).
These two expressions are not equal. Statement (C) is incorrect. (D) Total change in internal energy during the entire process is \(\frac{5}{2}P_2(V_2 - V_1)\).
For any cyclic process, the system returns to its initial state. Since internal energy U is a state function, the total change in internal energy over a complete cycle is always zero. \(\Delta U_{total} = 0\).
The expression in (D) is non-zero. Statement (D) is incorrect.
Step 4: Final Answer:
Based on the analysis of each step of the cycle, the expression for the total work done in option (B) is correct. All other options are incorrect.
