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

List-I	List-II
(P) Physisorption	(1) $\Delta H > 0$ and $\Delta S > 0$
(Q) Diamond $\longrightarrow$ Graphite	(2) $\Delta H < 0$ and $\Delta S < 0$
(R) Denaturation of protein	(3) $\Delta H < 0$ and $\Delta S = 0$
(S) Propene $\longrightarrow$ Cyclopropane	(4) $\Delta H > 0$ and $\Delta S < 0$
	(5) $\Delta H < 0$ and $\Delta S > 0$

(P) Physisorption is an exothermic process because attractive forces are formed between the adsorbate and the adsorbent, so $\Delta H < 0$. The gas molecules are restricted on the solid surface, decreasing randomness, so $\Delta S < 0$. This matches (2).

(Q) Diamond $\longrightarrow$ Graphite is an exothermic process because graphite is the thermodynamically more stable allotrope of carbon at standard conditions ($\Delta H < 0$). Graphite has a layered structure and is less dense, so it has higher entropy than the rigid 3D network of diamond ($\Delta S > 0$). This matches (5).

(R) Denaturation of protein involves the breaking of hydrogen bonds and the unfolding of its specific 3D structure into a more random coil. Breaking these interactions requires energy ($\Delta H > 0$) and the unfolding increases randomness ($\Delta S > 0$). This matches (1).

(S) Propene $\longrightarrow$ Cyclopropane is an endothermic process because cyclopropane has significant ring strain compared to the open-chain propene, making it less stable ($\Delta H > 0$). The formation of a rigid ring restricts the degrees of freedom (such as rotation around single bonds), decreasing entropy ($\Delta S < 0$). This matches (4).

Thus, P $\rightarrow$ 2, Q $\rightarrow$ 5, R $\rightarrow$ 1, S $\rightarrow$ 4.

Answer: P $\rightarrow$ 2; Q $\rightarrow$ 5; R $\rightarrow$ 1; S $\rightarrow$ 4