If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most probable, the average, and the root mean square speeds, respectively, is

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