Isothermal and Adiabatic Expansion Plots for Ideal Gas

To determine the correct graphical representations of the given state transitions, let us analyze the changes in the thermodynamic variables ($p, V, T, H, S$) for each step:

---

1. Process $\mathbf{I} \to \mathbf{II}$: Reversible Isothermal Expansion

• Temperature ($T$): Remains constant, so $T_{\mathbf{I}} = T_{\mathbf{II}}$.
• Volume ($V$) and Pressure ($p$): Expansion implies $V_{\mathbf{II}} > V_{\mathbf{I}}$. Since $p \propto \frac{1}{V}$ at constant temperature, pressure decreases ($p_{\mathbf{II}} < p_{\mathbf{I}}$).
• Enthalpy ($H$): For an ideal gas, enthalpy depends only on temperature ($H = n C_p T + \text{constant}$). Since $T$ is constant, $\Delta H = 0 \implies H_{\mathbf{I}} = H_{\mathbf{II}}$.
• Entropy ($S$): The entropy change during a reversible isothermal expansion is given by: $\Delta S = n R \ln \left( \frac{V_{\mathbf{II}}}{V_{\mathbf{I}}} \right) > 0 \implies S_{\mathbf{II}} > S_{\mathbf{I}}$

---

2. Process $\mathbf{II} \to \mathbf{III}$: Reversible Adiabatic Expansion

• Entropy ($S$): For a reversible adiabatic process, $dq_{\text{rev}} = 0$, which means it is an isentropic process ($\Delta S = 0 \implies S_{\mathbf{II}} = S_{\mathbf{III}}$).
• Volume ($V$) and Work ($W$): Expansion implies $V_{\mathbf{III}} > V_{\mathbf{II}}$, so work done by the gas is positive ($W > 0$).
• Internal Energy ($U$) and Temperature ($T$): By the First Law of Thermodynamics, $\Delta U = q - W = -W < 0$. Since internal energy decreases, the temperature drops: $T_{\mathbf{III}} < T_{\mathbf{II}}$
• Pressure ($p$): As volume expands adiabatically, pressure decreases further ($p_{\mathbf{III}} < p_{\mathbf{II}}$).
• Enthalpy ($H$): Since $H \propto T$ for an ideal gas and $T_{\mathbf{III}} < T_{\mathbf{II}}$, the enthalpy decreases: $H_{\mathbf{III}} < H_{\mathbf{II}}$

---

3. Evaluation of the Options

1. Option (A) — $p$ vs $V$ Plot:

   • $\mathbf{I} \to \mathbf{II}$: Isothermal expansion shows a curve going downwards to the right ($p \propto V^{-1}$).
   • $\mathbf{II} \to \mathbf{III}$: Adiabatic expansion shows a curve going downwards to the right with a steeper slope ($p \propto V^{-\gamma}$, where $\gamma > 1$).
   • Volume continuously increases from $\mathbf{I}$ to $\mathbf{III}$.
   • (A) is CORRECT.

2. Option (B) — $p$ vs $T$ Plot:

   • $\mathbf{I} \to \mathbf{II}$: Temperature is constant while pressure decreases. This corresponds to a vertical line pointing downwards.
   • $\mathbf{II} \to \mathbf{III}$: Both pressure and temperature decrease. This corresponds to a curve moving down and to the left.
   • (B) is CORRECT.

3. Option (C) — $H$ vs $S$ Plot:

   • $\mathbf{I} \to \mathbf{II}$: $H$ is constant while $S$ increases (horizontal line to the right).
   • $\mathbf{II} \to \mathbf{III}$: $S$ is constant while $H$ decreases due to cooling (vertical line pointing downwards).
   • Option (C) incorrectly shows $H$ increasing (vertical line pointing upwards from $\mathbf{II}$ to $\mathbf{III}$).
   • (C) is INCORRECT.

4. Option (D) — $T$ vs $S$ Plot:

   • $\mathbf{I} \to \mathbf{II}$: Temperature is constant while entropy increases (horizontal line to the right).
   • $\mathbf{II} \to \mathbf{III}$: Entropy is constant while temperature decreases (vertical line pointing downwards).
   • (D) is CORRECT.

---

Correct Options:

(A), (B), (D)