Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2017-Paper-2 The standard state Gibbs free energies of fo…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2017-Paper-2 Single Correct MCQ
Published on: August 13, 2026

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are

7 G° [C(graphite)] = 0 kJmol -1

7 G° [C(diamond)] = 2.9 kJmol -1

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 x 10 -6 m 3 mol -1 . If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is

[Useful information: 1 J = 1 kg m 2 s -2 ; 1 Pa = 1 kg m -1 s -2 ; 1 bar = 10 5 Pa]

A
14501 bar
B
58001 bar
C
1450 bar
D
29001 bar

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Text Solution

Verified by Experts
The correct answer is:
A

G = PdV

2.9x10 3 Jmol -1 = Px2x10 -6 m 3 mol -1

P = 1.45x10 9 Pa

P = 1.45x10 9 x10 -5 bar

P=1.45x 10 4 bar

P = 14500 bar

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