Analysis of the Thermodynamic Processes

The cyclic process for an ideal gas consists of three reversible steps: A → B, B → C, and C → A. Let's analyze each step using the first law of thermodynamics, $\Delta U = q + w$, where $w$ is the work done on the system ($w = -\int P_{ext} dV$). For a reversible process, $P_{ext} = P$. We use standard relations for an ideal gas: $\Delta U = nC_V\Delta T$ and $\Delta H = nC_P\Delta T = \Delta U + \Delta(PV)$.

1. Process A → B (Isobaric Expansion):

   • Pressure is constant at $P_2$. Volume changes from $V_1$ to $V_2$.
   • Work done on the system: $w_{AB} = -\int_{V_1}^{V_2} P_2 dV = -P_2(V_2 - V_1)$.
   • Heat added: $q_{AB} = \Delta H_{AB} = nC_P(T_B - T_A)$.

2. Process B → C (Isochoric Cooling):

   • Volume is constant at $V_2$. Pressure changes from $P_2$ to $P_1$.
   • Work done on the system: $w_{BC} = -\int_{V_2}^{V_2} P dV = 0$.
   • Heat exchanged: $q_{BC} = \Delta U_{BC} = nC_V(T_C - T_B)$.

3. Process C → A:

   • Both pressure and volume change. This is a compression from $V_2$ to $V_1$.
   • Work done on the system, $w_{CA}$, is the negative of the area under the curve C-A. Since it's a straight line, the area is a trapezoid: $w_{CA} = - \text{Area} = - \frac{1}{2}(P_1+P_2)(V_1-V_2) = \frac{1}{2}(P_1+P_2)(V_2-V_1)$. Since $V_2>V_1$, work is done on the gas, so $w_{CA} > 0$.
   • Change in internal energy: $\Delta U_{CA} = nC_V(T_A - T_C)$.
   • Change in enthalpy: $\Delta H_{CA} = nC_P(T_A - T_C)$. We also know $\Delta H_{CA} = \Delta U_{CA} + \Delta(PV)_{CA} = \Delta U_{CA} + (P_2V_1 - P_1V_2)$.

Evaluation of the Options

Option A: ${q_{Ac}} = \Delta {U_{BC}}\,\,$ and ${w_{AB}} = {P_2}\left( {{V_2} - {V_1}} \right)$

• The term $q_{Ac}$ is ambiguous. If it's a typo for $q_{BC}$, the statement becomes $q_{BC} = \Delta U_{BC}$, which is true for an isochoric process.
• The second part is $w_{AB} = P_2(V_2 - V_1)$. The work done on the system is $w_{on, AB} = -P_2(V_2 - V_1)$. The work done by the system is $w_{by, AB} = P_2(V_2 - V_1)$. The statement is correct if $w$ denotes work done by the system, which contradicts the standard chemistry convention for the first law as stated above. Assuming chemistry convention, this part is incorrect due to the sign. Thus, option A is incorrect.

Option B: ${w_{BC}} = {P_2}\left( {{V_2} - {V_1}} \right)\,\,$ and ${q_{BC}} = \Delta {H_{AC}}$

• The first part states $w_{BC} = P_2(V_2 - V_1)$. As established, process B → C is isochoric ($V=constant$), so $w_{BC} = 0$. Since $P_2 > 0$ and $V_2 > V_1$, the term $P_2(V_2 - V_1)$ is non-zero. Thus, the first statement is false.
• Since one part of the option is false, the entire option is incorrect.

Option C: $\Delta {H_{CA}} \lt \Delta {U_{CA}}\,\,$ and $\,\,{q_{AC}} = \Delta {U_{BC}}$

• The first inequality, $\Delta H_{CA} < \Delta U_{CA}$, implies $\Delta U_{CA} + (P_2V_1 - P_1V_2) < \Delta U_{CA}$, which simplifies to $P_2V_1 < P_1V_2$. Using the ideal gas law ($PV=nRT$), this is equivalent to $T_A < T_C$. This is not a general property of the cycle; it depends on the specific values of pressures and volumes.
• The second part, $q_{AC} = \Delta U_{BC}$, is ambiguous due to the subscript $AC$. No reasonable interpretation makes this a generally true statement.

Option D: $\,\,{q_{BC}} = \Delta {H_{AC}}\,\,$ and $\Delta {H_{CA}} \gt \Delta {U_{CA}}$

• The first part, $q_{BC} = \Delta H_{AC}$, is not a general thermodynamic identity. It would only hold if the cycle satisfies a specific condition.
• The second part, $\Delta H_{CA} > \Delta U_{CA}$, simplifies to $P_2V_1 > P_1V_2$, or $T_A > T_C$. Again, this is not generally true.

Contradiction between Options B and C

Let's analyze the consequence of the statement $q_{BC} = \Delta H_{AC}$ (from options B and D). $q_{BC} = \Delta U_{BC} = nC_V(T_C - T_B)$. $\Delta H_{AC} = nC_P(T_C - T_A)$. So, $nC_V(T_C - T_B) = nC_P(T_C - T_A)$. $C_V T_C - C_V T_B = C_P T_C - C_P T_A$ $C_P T_A = (C_P - C_V) T_C + C_V T_B = R T_C + C_V T_B$. Dividing by $\gamma = C_P/C_V$: $\frac{C_P}{\gamma}T_A = \frac{R}{\gamma}T_C + \frac{C_V}{\gamma}T_B \implies C_V T_A = \frac{R}{\gamma}T_C + \frac{C_V}{\gamma}T_B$. This is getting complicated. Let's do it differently: $T_A - T_C = \frac{1}{C_P} ( (C_P - C_V)T_C + C_V T_B ) - T_C = (1-\frac{C_V}{C_P}-1)T_C + \frac{C_V}{C_P}T_B = \frac{1}{\gamma}(T_B-T_C)$. In process B→C, pressure decreases at constant volume, so the gas cools down. Thus, $T_B > T_C$, which means $T_B - T_C > 0$. Since $\gamma > 1$, we get $T_A - T_C > 0$, or $T_A > T_C$. This implies $\Delta H_{CA} > \Delta U_{CA}$.

Now, let's look at the options:

• Option C states $\Delta H_{CA} < \Delta U_{CA}$ (i.e., $T_A < T_C$).
• Option B contains the statement $q_{BC} = \Delta H_{AC}$, which implies $T_A > T_C$.

These two conditions are mutually exclusive. Therefore, it is impossible for both options B and C to be correct.

Conclusion

The question is flawed as presented.

1. Option B is demonstrably false because its first statement, $w_{BC} = P_2(V_2-V_1)$, is incorrect as $w_{BC}=0$.
2. Options B and C contain statements that are mutually exclusive.

This question was part of the JEE Advanced 2018 paper and was declared a "Bonus" question, meaning it was recognized as flawed and all candidates were awarded marks for it. Based on a rigorous analysis, none of the options are correct. My derived answer is that there are no correct options.

Since the stored answer is (B, C), which is logically impossible as shown, I must disagree with it.