Given one mole of diamond vs one mole of graphite,
G S V
1 mol Graphite 0J 5.74 J/K 5.30 cm^3
1 mol Diamond 2900J 2.38 J/K 3.42 cm^3
(Assume S and V are constant over the temperatures and pressures we’re considering, which is a reasonable approximation.)
(a) At room temperature 300 K, at what pressure does diamond become more stable?
(b) At standard pressure 10^5 Pa, at what temperature does diamond become more stable?

Correct Data:

                                      G                       S                 V   

1  mol Graphite             0 J               5.74 J/K     5.30cm³

1 mol Diamond           2900 J          2.38 J/K     3.42cm³

a) P = -639 MPa

b) T =1218.63 K

Explanation:

a) For stability of diamond, the gibb's free energy, G = 0 J

The Gibb's free energy is given by G = E + PV - TS

E + PV - TS = 0

where, E = enthalpy, P = pressure, S = entropy

E = 2900 J

T = 300 K

S = 2.38 J/K

V = 3.42 cm³ = 3.42 * 10⁻⁶ m³

2900 + P3.42 * 10⁻⁶  - 300(2.38) = 0

P3.42 * 10⁻⁶  = -2186

P = 2186/(3.42 * 10⁻⁶)

P = -639* 10⁶ Pa

P = -639 MPa

b) At P = 10⁵

E + PV - TS = 0

2900 + 10⁵(3.42 * 10⁻⁶) - T(2.38) = 0

T =1218.63 K