The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm-3, respectively. If the standard free energy difference ( G ) is equal to 1895 J mol-1, the pressure at which graphite will be transformed into diamond at 298 K is
- 11.14 108 Pa
- 11.14 107 Pa
- 11.14 106 Pa
- 11.14 105 Pa
Answer
(1) 11.14 108 Pa
Solution
For 1 mol of carbon (12 g):
Vgraphite = 122.25 = 5.333 cm3
Vdiamond = 123.31 = 3.625 cm3
⇒ V = 3.625 - 5.333 = -1.708 cm3 = -1.708 10-6 m3 mol-1
∵ At constant temperature dG = V dP, the extra pressure P must make the work term equal to G :
P | V| = G
⇒ P = 1895 J mol-11.708 10-6 m3 mol-1
⇒ P 1.11 109 Pa = 11.1 108 Pa
∴ The closest value is 11.14 108 Pa.
Answer: (1)