Chemistry - Thermodynamics - II Question with Solution | TestHub

The standard state Gibb's free energies of formation of$\mathrm{C}$(graphite) and$\mathrm{C}$(diamond) at$\mathrm{T}=298\mathrm{ }\mathrm{K}$are
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\mathrm{o}}\left[\mathrm{C} (\mathrm{graphite}\right]=0\mathrm{ }\mathrm{kJ}\mathrm{ }{\mathrm{mol}}^{-1}$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\mathrm{o}}\left[\mathrm{C} \left(\mathrm{diamond}\right)\right]=2.9\mathrm{ }\mathrm{kJ}\mathrm{ }{\mathrm{mol}}^{-1}$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [$\mathrm{C}$(graphite)] to diamond [$\mathrm{C}$(diamond)] reduces its volume by$2\times {10}^{-6}\mathrm{ }{\mathrm{m}}^3 {\mathrm{mol}}^{-1}$. If$\mathrm{C}$(graphite) is converted to$\mathrm{C}$(diamond) isothermally at$\mathrm{T}=298 \mathrm{K}$, the pressure at which$\mathrm{C}$(graphite) is in equilibrium with$\mathrm{C}$(diamond), is
[Useful information:$1\mathrm{ }\mathrm{J}=1\mathrm{ }\mathrm{kg}\mathrm{ }{\mathrm{m}}^2 {\mathrm{s}}^{-2},\mathrm{ }1\mathrm{ }\mathrm{Pa}=1 \mathrm{kg}\mathrm{ }{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}; 1\mathrm{bar}={10}^5\mathrm{Pa}$]