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) $$\Delta H > 0 \text{ and } \Delta S > 0$$
(Q) Diamond $\rightarrow$ Graphite	(2) $$\Delta H < 0 \text{ and } \Delta S < 0$$
(R) Denaturation of protein	(3) $$\Delta H < 0 \text{ and } \Delta S = 0$$
(S) Propene $\rightarrow$ Cyclopropane	(4) $$\Delta H > 0 \text{ and } \Delta S < 0$$
	(5) $$\Delta H < 0 \text{ and } \Delta S > 0$$

The standard state Gibbs free energies of formation of $$C$$(graphite) and $$C$$(diamond) at $$T=298$$ $$K$$ are

$${\Delta _f}{G^0}$$ [$$C$$(graphite)] $$ = 0kJmo{l^{ - 1}}$$

$${\Delta _f}{G^0}$$ [$$C$$(diamond)] $$ = 2.9kJmo{l^{ - 1}}$$

The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C$$(graphite)] to diamond [$$C$$(diamond)] reduces its volume by $$2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$$ If $$C$$(graphite) is converted to $$C$$(diamond) isothermally at $$T=298$$ $$K,$$ the pressure at which $$C$$(graphite) is in equilibrium with $$C$$(diamond), is

[Useful information : $$1$$ $$J=1$$ $$kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};$$ $$1$$ bar $$ = {10^5}$$ $$Pa$$]

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding ($$\Delta$$S_surr)in JK^–1 is (1L atm = 101.3 J)