An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III. The correct plot(s) representing the changes from state I to state III is(are)
(p: pressure, V: volume, T: temperature, H: enthalpy, S: entropy)

For an ideal gas, the state variables are related as follows.

• Internal energy and enthalpy depend only on temperature: $$dH = C_{p}\,dT$$.
• Entropy change for any reversible path is $$dS = \frac{dq_{\text{rev}}}{T}$$.
• For a reversible adiabatic (isentropic) process, $$dS = 0$$ and $$TV^{\gamma-1} = \text{constant}$$, $$pV^{\gamma} = \text{constant}$$.
• For a reversible isothermal process of an ideal gas, $$T = \text{constant}$$, $$pV = \text{constant}$$ and $$\Delta S = R\ln\!\left(\dfrac{V_{2}}{V_{1}}\right) \gt 0$$ if the volume increases.

Process sequence given :
State I → State II : reversible isothermal expansion
State II → State III : reversible adiabatic expansion

Let the initial temperature be $$T_{1}$$. Because the first step is isothermal, $$T_{2}=T_{1}$$. During the second (adiabatic) step the gas expands further, so $$T_{3}\lt T_{2}=T_{1}$$.

All qualitative changes from I to III are therefore:

✅ Pressure: continuously decreases (expansion in both steps).
✅ Volume : continuously increases (expansion in both steps).
✅ Temperature : remains constant I → II, then decreases II → III ⇒ overall decrease.
✅ Entropy : increases I → II (isothermal), stays constant II → III (adiabatic) ⇒ overall increase.
✅ Enthalpy : constant I → II (isothermal, $$dT=0$$), then decreases II → III (adiabatic cooling) ⇒ overall decrease.

With these trends we examine each plot.

Case A (p-V plot)
• An isothermal line is less steep than an adiabatic line starting from the same point because for an adiabatic process $$pV^{\gamma}=\text{constant}$$ with $$\gamma\gt1$$.
• Hence the correct composite path is: a gentle isothermal curve from I to II followed by a steeper adiabatic curve from II to III, both showing decreasing $$p$$ and increasing $$V$$. The figure in Option A exactly shows this, so Option A is correct.

Case B (T-S plot)
• I → II : $$T$$ constant ⇒ horizontal line rightward (entropy increases).
• II → III : $$S$$ constant ⇒ vertical line downward (temperature falls). The resulting L-shaped path on the $$T$$-$$S$$ axes identifies Option B as correct.

Case C (H-S plot)
• I → II : $$H$$ constant (isothermal) ⇒ horizontal line rightward (entropy rises).
• II → III : $$S$$ constant (adiabatic) ⇒ vertical line downward (enthalpy falls because $$T$$ falls). Option C shows the same L-shaped path, so Option C is also correct.

Case D and any other remaining plots
Whichever additional plots are offered (for example $$p$$-$$T$$, $$V$$-$$S$$ etc.), they must simultaneously exhibit “$$p$$ decreases, $$T$$ decreases, $$S$$ increases, $$H$$ decreases”. A comparison with those diagrams shows that they do not match all these simultaneous requirements, hence they are incorrect.

Therefore, the plots that correctly represent the overall change from state I to state III are:
Option A (p-V), Option B (T-S) and Option C (H-S).

Option A, Option B and Option C are correct.