thermodynamics - Pressure at which graphite and diamond are in equilibrium

I came across this problem :

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.

The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $\pu{2e-6 m3 mol-1}$.

If C(graphite) is converted to C(diamond) isothermally at $\pu{T = 298 K}$, the pressure at which C(graphite) is in equilibrium with C(diamond), is

(A) $\pu{14501 bar}$

(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$

I applied : $$\Delta G_{(P,T)} =\Delta_f G^{\circ}+\int_{p_1}^{p_2}Vdp$$.

Since the system is at equilibrium,$$ \Delta_f G^{\circ}= -\int_{p_1}^{p_2}Vdp$$

Now I am stuck. I have not been given any relation between Pressure and Volume. Is there any assumption i have to make to solve this integral ?

Answer

For each phase (graphite or diamond) you can write

$$ \mu = \mu^\circ+\int_{P^\circ}^{P} V_mdP $$ at constant T.

We are asked to find the pressure $P=P_{eq}$ at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that

$$ \mu(diamond) = \mu(graphite)$$

This leads to

$$\Delta \mu^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

and, since $\Delta \mu^{\circ}= \Delta_f G_m^{\circ}$, ultimately to the expression you provided:

$$\Delta_f G_m^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

The approximation you are allowed to make at this point is that the solids are incompressible, such that their $V_m$ are constant with change of pressure. It follows that

$$\Delta_f G_m^{\circ}=-\Delta V_m\int_{P^\circ}^{P_{eq}}dP=-\Delta V_m(P_{eq}-P^\circ)$$

which can be solved for $P_{eq}$:

$$P_{eq}=-\frac{\Delta_f G_m^{\circ}}{\Delta V_m}+P^\circ$$