NEETChemistryThermodynamicsMCQ+4 / −1
The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm3, respectively. If the standard free energy difference is equal to 1895 J mol1, the pressure at which graphite will be transformed into diamond at 298 K is
- A11.14 108 Pa
- B11.14 107 Pa
- C11.14 106 Pa
- D11.14 105 Pa
View written solutionFree
Correct answer: A
C(graphite) C(diamond)
Volume of graphite = cm3
mol–1
Volume of diamond = cm3
mol–1
V = Vgraphite – Vdiamond
= 1.70 cm3
mol–1
= 1.70 × 10–3 L mol–1
So, Go = PV
1895 J mol–1 = P(1.70 × 10–3 L mol–1)
1114.70 × 103
J/L = P
= P
P = 11000.69 atm × 1.013 × 105
Pa
= 11143.69 × 105
Pa
= 11.14 × 108
Pa
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