Verifying That Diamond Is More Stable At Certain Conditions

1. The problem statement, all variables and given/known data
Verify that diamond becomes more stable than graphite at approximately 15 kbars

2. Relevant equations

3. The attempt at a solution

So, I understand that to show one substance is more stable than another at particular environmental conditions, I just have to show that the one with a lower Gibbs free energy is more stable.

In the section, the formula [itex]\left( \frac{\partial G}{\partial P} \right)_{T,N} = V[/itex]. Since the partial derivative is given approximately by [itex]\frac{\partial G}{\partial P} = \frac{\Delta G}{\Delta P}[/itex], I can write [itex]\frac{\Delta G}{\Delta P} = V[/itex].

In the back of the textbook I am using is a table containing the Gibbs free energy of formation of substances at at temperature of T=293 K and a pressure of P = 1 bar, call it [itex]G_1[/itex] Using this information, I figured I could obtain the Gibbs free energy at P = 15 kbar, call it [itex]G_{15}[/itex]:

[itex]\frac{\Delta G}{\Delta P} = V \implies[/itex]

[itex]\Delta G = \Delta P V \implies[/itex]

[itex]G_{15} = \Delta P V + G_1[/itex].

Now, if we assume that the change in volume of the substance is negligible as the pressure changes from P = 1 bar to P = 15 kbar, which is an assumption the author of the textbook uses, then the formula above is applicable. The volume of diamond at T=293 and P=1 bar is 5.31 x 10^-6. Calculating the Gibbs free energy at 15 kbars of the diamond:

[itex]G_{15,D} = (15 \times 10^{3}~bar -1~bar)(5.31 \times 10^{-6} ~m^3)+ 2.900 \times 10^{3} ~J = 2900 ~J[/itex]

And for graphite, where the volume is 3.42 x 10^-6

[itex]G_{15,G} = (15 \times 10^{3}~bar -1~bar)(3.42 \times 10^{-6}) + 0 = 0.0513 ~J[/itex].

Clearly the Gibbs free energy is less for graphite. What did I do incorrectly?

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