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)

Match the thermodynamics processes given under column I with expression given under column II

Column I
(A) Freezing water at 273 K and 1 atm
(B) Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions.
(C) Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container.
(D) Reversible heating of H₂(g) at 1 atm from 300K to 600K, followed by reversible cooling to 300K at 1 atm

Column II
(p) q = 0
(q) w = 0
(r) $$\Delta S_{sys}$$ < 0
(s) $$\Delta U$$ = 0
(t) $$\Delta G$$ = 0

For the process
H₂O(l) $$\to$$ H₂O(g)
at T = 100^oC and 1 atmosphere pressure, the correct choice is