The standard state Gibbs free energies of formation of graphite and diamond at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite to diamond reduces its volume by $\pu{2e-6 m3 mol-1}$.
If graphite is converted to diamond isothermally at $T = \pu{298 K}$, the pressure at which graphite is in equilibrium with diamond, is
(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$
I applied
$$\Delta G_{(p,T)} =\Delta_\mathrm{f}G^\circ + \int_{p_1}^{p_2}V\,\mathrm dp,$$
and since the system is at equilibrium,
$$\Delta_\mathrm{f}G^\circ = -\int_{p_1}^{p_2}V\,\mathrm dp.$$
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?