= Solution
At fixed <temperature>, a phase's <Gibbs free energy> satisfies $\mathrm dG=V\,\mathrm dp$. For the metallic-minus-molecular difference,
$$
\left(\frac{\partial\Delta G}{\partial p}\right)_T=\Delta V<0.
$$
\b[Increasing <pressure> favors the smaller-volume metallic phase.] With a pressure-independent <volume> approximation, $\Delta G(p)\simeq\Delta G(p_0)+(p-p_0)\Delta V$, so <pressure> can overcome an initially positive free-energy difference. The <volume> comparison explains the direction of stabilization; it does not by itself determine the transition <pressure> or the electronic origin of <metallic bonding>.
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