1. Process I $\to$ II: reversible isothermal expansion

For an ideal gas in an isothermal process, $T=\text{constant}, \qquad pV=\text{constant}$ So during $I \to II$:

• $V$ increases
• $p$ decreases
• $T$ remains constant
• For an ideal gas, enthalpy depends only on temperature, so $H=H(T)$ hence $H$ remains constant.
• Entropy increases because reversible isothermal expansion gives $\Delta S = nR\ln\frac{V_2}{V_1}>0$

2. Process II $\to$ III: reversible adiabatic expansion

For a reversible adiabatic process of an ideal gas, $q=0, \qquad pV^\gamma=\text{constant}, \qquad TV^{\gamma-1}=\text{constant}$ During expansion:

• $V$ increases
• $p$ decreases
• $T$ decreases
• Since $H$ depends only on $T$ for an ideal gas, $H=nC_pT$ so $H$ decreases.
• For a reversible adiabatic process, $\Delta S=0$ Therefore entropy remains constant from state II to III.

3. Net change from I to III

Combining both steps:

• Volume: increases throughout.
• Pressure: decreases throughout.
• Temperature: constant first, then decreases. Hence from I to III, final temperature is lower than initial temperature.
• Enthalpy: constant first, then decreases. So final enthalpy is lower than initial enthalpy.
• Entropy: increases in the isothermal step, then remains constant in the adiabatic step. Thus $S_3>S_1$

4. Shape/plot conclusions

From the above behavior:

• In a $p$ vs $V$ plot, first an isotherm, then a steeper adiabatic curve. This is correct.
• In a $T$ vs $V$ plot, first horizontal (isothermal), then decreasing curve (adiabatic). This is correct.
• In an $H$ vs $S$ plot, first horizontal to the right (since $H$ constant, $S$ increases), then vertical downward (since $S$ constant, $H$ decreases). This is correct.
• Any plot showing entropy decreasing, or enthalpy increasing, or temperature remaining constant in the adiabatic step would be incorrect.

Therefore the correct options are: $\boxed{A,\ B,\ D}$

5. Comparison with stored answer

Stored correct answer: $A, B, D$

My derived answer matches the stored answer exactly.