1. The problem statement, all variables and given/known data Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar How can diamond ever be more stable than graphite, when it has less entropy 2. Relevant equations 3. The attempt at a solution (dG/dP)=V, T and N are fixed. dP=15 kbar. V((diamond)=3.42e-6 m^3 V((graphite))=5.31e-6 m^3 dG(diamond)=15 kbar(3.42e-6) dG(graphite)=15 kbar(5.31e-6) Do my calculations above represent the the procedure I would go through to prove that Diamond is more stable than graphite? the answer to the second question is, . A gas has a much higher entropy than any solid, but at low temperatures solids and liquids are more favorable gibbs free energy G= U + PV - TS Diamond has smaller radius than graphite, therefore it has a smaller radius . U(graphite)>U(diamond) . Atoms inside graphite have the autonomy to move around its lattice, therefore it has more entropy and therefore has higher temperatures.