91Ó°ÊÓ

We can choose the reference value because the Gibbs free energy of one mole of diamond is 2900 J higher than the Gibbs free energy of one mole of graphite. Let the graphite's Gibbs free energy be 0 for the sake of simplicity.

Therefore the two equations of the lines drawn in fig 5.15 are

G_g=V_gP

G_d=V_dP+2.9kJ

Where,

G_gis Gibbs energy of graphite

G_d is Gibbs energy of diamond

V_gis volume of graphite

V_d is volume of diamond

We need to prove that the diamond becomes more stable than graphite at pressures of 15 kbar, and to do so, we need to identify the intersection of the two curves, set G_g equals to G_d to get

V_gP=V_dP+2.9kJ

Solve for P to get,

P(V_g-V_d)=2.9 kJ

P=$\frac{2.9kJ}{V_g-V_d}$

the molar volume of graphite is V_g=$5.31×{10}^{-6}{m}^3$ and the molar volume of diamond is V_d=$3.42×{10}^{-6}{m}^3$

Substitute the values and solve
$$\begin{gathered} P=\frac{2900J}{5.31×{10}^{-6}{m}^3-3.42×{10}^{-6}{m}^3} \\ P=1.534×{10}^{9}Pa \\ P=1.534×{10}^{4}bar \\ P=15.34kbar \end{gathered}$$

We used $1bar={10}^{4}kbar$