Two separate experiments were carried out involving adiabatic reversible expansion, taking monatomic gas in one case and a diatomic gas in the second case. In each case, the initial temperature was${\mathrm{T}}_{\mathrm{i}}$ and expansion was from volume $\mathrm{V}{ }_1$ to volume ${\mathrm{V}}_2$. The final temperatures attained were ${\mathrm{T}}_{\mathrm{M}}$ for monatomic gas and ${\mathrm{T}}_{\mathrm{D}}$ for diatomic gas. Then

Consider the following liquid-vapour equilibrium

$\mathrm{Liquid} \rightleftharpoons \mathrm{Vapour}$

Which of the following relations is correct.

The standard state Gibbs free energies of formation of $\mathrm{C}\left(\mathrm{graphite}\right)+\mathrm{C}\left(\mathrm{diamond}\right) \mathrm{at} \mathrm{T}=298 \mathrm{K}$ are 

 ${∆}_{\mathrm{f}}\mathrm{G}°[\mathrm{C} (\mathrm{graphite}\left)\right] = 0 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${∆}_{\mathrm{f}} \mathrm{G}°[\mathrm{C} (\mathrm{diamond}\left)\right] = 2.9 \mathrm{kJ} {\mathrm{mol}}^{-1}$

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

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