List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy( ΔH )and entropy( ΔS). Match each entry in List-I to the appropriate entry in List-II and choose the correct option.

The correct option is: P → 2; Q → 5; R → 1; S → 4

To determine the correct matching between the physical/chemical processes in List-I and the thermodynamic state function changes in List-II, let us analyze each process step-by-step:

1. (P) Physisorption:
Physisorption involves the accumulation of gas molecules on a solid surface via weak van Der Waals interactions.
• Since attractive forces are established, energy is released into the surroundings, making the process exothermic. Thus, .
• The gas molecules lose their translational freedom upon being adsorbed, leading to a decrease in molecular randomness/disorder. Thus, .
Therefore, P matches with 2 ( and ).

2. (Q) Diamond → Graphite:
Graphite is the thermodynamically more stable allotrope of carbon compared to diamond at standard temperature and pressure.
• The conversion from diamond to graphite is an exothermic process, which means enthalpy decreases: .
• Diamond has a very rigid, highly ordered 3D tetrahedral network structure, whereas graphite has a layered hexagonal structure with greater degree of freedom/disorder between layers. Consequently, entropy increases: $S\; >\; 0$.
Therefore, Q matches with 5 ( and $S\; >\; 0$).

3. (R) Denaturation of protein:
Denaturation is the process in which proteins lose their quaternary, tertiary, and secondary structures due to external stress or chemicals.
• Breaking the weak hydrogen bonds, ionic interactions, and hydrophobic interactions present in the native folded structure requires heat input (endothermic process), so $H\; >\; 0$.
• The highly ordered folded structure unfolds into a randomized, uncoiled polypeptide chain, drastically increasing randomness, so $S\; >\; 0$.
Therefore, R matches with 1 ($H\; >\; 0$ and $S\; >\; 0$).

4. (S) Propene → Cyclopropane:
This ring-closure (cyclization) converts an open-chain alkene into a strained three-membered cycloalkane.
• Ring formation introduces significant ring strain (angle strain), rendering cyclopropane higher in enthalpy than propene. Thus, the conversion is endothermic: $H\; >\; 0$.
• The open-chain molecule possesses internal rotational freedoms about single bonds that become restricted upon forming a rigid cyclic structure, leading to a decrease in entropy: .
Therefore, S matches with 4 ($H\; >\; 0$ and ).

Summary of Matches:
• P → 2
• Q → 5
• R → 1
• S → 4