Thermodynamic State Functions for Ideal Gas in Reversible Cyclic Process

To determine the correct option(s), we analyze the thermodynamic properties for each step of the reversible cyclic process undergone by the ideal gas.

1. Identification of the Processes

From the given $V\text{--}T$ diagram:

• Process $\text{A} \to \text{B}$: The temperature remains constant at $T = T_1$, while the volume decreases from $V_1$ to $V_2$. This is a reversible isothermal compression.
• Process $\text{A} \to \text{C}$: The volume remains constant at $V = V_1$, while the temperature increases from $T_1$ to $T_2$. This is a reversible isochoric heating.
• Process $\text{B} \to \text{C}$: Since the state $\text{B}$ is $(P_2, V_2, T_1)$ and state $\text{C}$ is $(P_2, V_1, T_2)$, the pressure is constant at $P = P_2$ (as $V \propto T$). This is a reversible isobaric expansion.

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2. Evaluation of Thermodynamic Quantities

(i) Process $\text{A} \to \text{C}$ (Isochoric):

Since the volume is constant ($\Delta V = 0$): $w_{AC} = 0$ From the First Law of Thermodynamics: $q_{AC} = \Delta U_{AC} = n C_v (T_2 - T_1)$ The enthalpy change for this step is: $\Delta H_{AC} = n C_p (T_2 - T_1)$

(ii) Process $\text{B} \to \text{C}$ (Isobaric at pressure $P_2$):

The internal energy change depends only on the initial and final temperatures: $\Delta U_{BC} = n C_v (T_2 - T_1)$ Comparing with $q_{AC}$: $q_{AC} = \Delta U_{BC}$

The work done on/by the gas (using the standard IUPAC chemistry convention $w = -\int P_{\text{ext}} dV$): $w_{BC} = -\int_{V_2}^{V_1} P_2 \, dV = -P_2(V_1 - V_2) = P_2(V_2 - V_1)$

The heat absorbed at constant pressure is equal to the enthalpy change: $q_{BC} = \Delta H_{BC} = n C_p (T_2 - T_1)$ Since $\Delta H_{AC} = n C_p (T_2 - T_1)$, we have: $q_{BC} = \Delta H_{AC}$

(iii) Process $\text{A} \to \text{B}$ (Isothermal):

For a reversible isothermal process of an ideal gas: $w_{AB} = -\int_{V_1}^{V_2} P \, dV = -n R T_1 \ln\left(\frac{V_2}{V_1}\right) \neq P_2(V_2 - V_1)$

(iv) Process $\text{C} \to \text{A}$:

Here, the temperature decreases from $T_2$ to $T_1$ ($\Delta T = T_1 - T_2 < 0$). $\Delta H_{CA} = n C_p (T_1 - T_2)$ $\Delta U_{CA} = n C_v (T_1 - T_2)$ For an ideal gas, $C_p > C_v > 0$. Since $(T_1 - T_2)$ is negative: $n C_p (T_1 - T_2) < n C_v (T_1 - T_2) \implies \Delta H_{CA} < \Delta U_{CA}$

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3. Analysis of Options

• Option (A): $q_{AC} = \Delta U_{BC}$ is correct, but $w_{AB} = P_2(V_2 - V_1)$ is incorrect.
• Option (B): $w_{BC} = P_2(V_2 - V_1)$ and $q_{BC} = \Delta H_{AC}$ are both correct.
• Option (C): $\Delta H_{CA} < \Delta U_{CA}$ and $q_{AC} = \Delta U_{BC}$ are both correct.
• Option (D): $q_{BC} = \Delta H_{AC}$ is correct, but $\Delta H_{CA} > \Delta U_{CA}$ is incorrect.

Thus, the correct options are B and C.