Graphite is more compressible than diamond.

(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?

(b) The isothermal compressibility of graphite is about 3 x 10^-6 bar^-1, while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

When the compressibility factor is ignored, the volume of the substance can be used to derive the pressure dependence of Gibbs free energy at constant temperature and is expressed as

${\left(\frac{\partial G}{\partial P}\right)}_T=V$

Where,

G is the Gibbs energy

P is the pressure

V is the volume of the substance

The Gibbs free energy for one mole of diamond is greater than the Gibbs free energy of one mole of graphite by 2900 J.

(a) When the compressibility factor is taken into account, the pressure dependence of the Gibbs free energy (G) can be stated as follows:

${\left(\frac{\partial G}{\partial P}\right)}_T=V_o\left(1-{\kappa }_{T}P\right)$

Where,

k_T is the isothermal compressibility factor

V_o is the change in the volume of the substance

It can be seen from the preceding relationship that as pressure rises, volume decreases and G rises more slowly.

As a result, the graphite-to-diamond transition will occur at a little higher pressure.

The pressure at which the diamond is more stable than graphite may be estimated by ignoring the compressibility of both material:

$$\begin{gathered} G_g=V_{g}P \\ {G}_d=V_{d}P+2.9\mathrm{kJ} \end{gathered}$$

Where,

g, d are for graphite and diamond respectively

G is the free energy

V is the volume

P is the pressure

The intersection of two lines will give the value for pressure

$$\begin{gathered} V_{g}P=V_{d}P+2.9\mathrm{kJ} \\ \mathrm{P}=\frac{2.9\mathrm{kJ}}{{\mathrm{V}}_{\mathrm{g}}-{\mathrm{V}}_{\mathrm{d}}} \end{gathered}$$

Substituting the value of $V_g-V_d=0.189kJ/kbar$and solve for P

$$\begin{gathered} P=\frac{2.9\mathrm{kJ}}{0.189\mathrm{kJ}/\mathrm{kbar}} \\ =15.3\mathrm{kbar} \end{gathered}$$

When the compressibility factor is taken into account, the pressure dependence of Gibbs free energy at constant temperature is

${\left(\frac{\partial G}{\partial P}\right)}_T=V_o\left(1-{\kappa }_{T}P\right)$

For a finite value of Gibbs free energy change at constant temperature,

$$\begin{gathered} \Delta G={\int }_0^P{V}_o\left(1-{\kappa }_T·P\right)\Delta P \\ ={\int }_0^P\left(V_o-V_o·{\kappa }_T·P\right)\Delta P \end{gathered}$$

On integrating further

$\Delta G=P·V_o-\frac{V_o·{\kappa }_T·P^2}{2}$

Substituting $∆G$as $G_{od}-G_{og}$and localid="1646814662046" $V_o=V_g-V_d$in the equation

$$\begin{gathered} G_{od}-G_{og}=P\left(V_g-V_d\right)-\frac{1}{2}\left(V_g-V_d\right){\kappa }_T{P}^2 \\ P\left(V_g-V_d\right)=\left(G_{od}-G_{og}\right)+\frac{1}{2}\left(V_g-V_d\right){\kappa }_T{P}^2 \end{gathered}$$

$V_d$can be neglected as compressibility of diamond is negligible as compared to graphite.

localid="1646814925186" $$\begin{gathered} P\left(V_g-V_d\right)=\left(G_{od}-G_{og}\right)+\frac{1}{2}{V}_g{\kappa }_T{P}^2 \\ P=\frac{1}{\left(V_g-V_d\right)}\left[\left(G_{od}-G_{og}\right)+\frac{1}{2}{V}_g{\kappa }_T{P}^2\right] \end{gathered}$$

Substituting the value of $G_{od}-G_{og}=2.9kJ$

$P=\frac{1}{\left(V_g-V_d\right)}\left[2.9\mathrm{kJ}+\frac{1}{2}{V}_g{\kappa }_T{P}^2\right]$

Substituting the values in the above equation

localid="1646815170817" $$\begin{gathered} P=\frac{1}{(0.189\mathrm{kJ}/\mathrm{kbar})}\left[2.9\mathrm{kJ}+\frac{1}{2}(0.531\mathrm{kJ}/\mathrm{kbar})\left(3\times {10}^{-3}{\mathrm{kbar}}^{-1}\right)(15.3\mathrm{kbar}{)}^2\right] \\ P=\frac{1}{(0.189\mathrm{kJ}/\mathrm{kbar})}[2.9\mathrm{kJ}+0.186\mathrm{kJ}] \\ P=\frac{3.086\mathrm{kJ}}{(0.189\mathrm{kJ}/\mathrm{kbar})} \\ P=16.33\mathrm{kbar} \end{gathered}$$

Hence, the pressure will be $16.77kbar$.