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Thermodynamics question

2003 · Q114
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Thermodynamics question

2003 · Q114

NEETChemistryThermodynamicsMCQ+4 / −1
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 (ΔGo)\left( {\Delta {G^o}} \right)(ΔGo) is equal to 1895 J mol−-−1, the pressure at which graphite will be transformed into diamond at 298 K is
  1. A
    11.14 ×\times× 108 Pa
  2. B
    11.14 ×\times× 107 Pa
  3. C
    11.14 ×\times× 106 Pa
  4. D
    11.14 ×\times× 105 Pa
View written solutionFree

Correct answer: A

C(graphite) →\to→ C(diamond)

Volume of graphite = 122.25=5.33{{12} \over {2.25}} = 5.332.2512​=5.33 cm3 mol–1

Volume of diamond = 123.31=3.63{{12} \over {3.31}} = 3.633.3112​=3.63 cm3 mol–1

Δ\Delta ΔV = Vgraphite – Vdiamond

= 1.70 cm3 mol–1

= 1.70 × 10–3 L mol–1

So, Δ\Delta ΔGo = PΔ\Delta ΔV

⇒\Rightarrow⇒ 1895 J mol–1 = P(1.70 × 10–3 L mol–1)

⇒\Rightarrow⇒ 1114.70 × 103 J/L = P

⇒\Rightarrow⇒ 1114.7×103101.33{{1114.7 \times {{10}^3}} \over {101.33}}101.331114.7×103​ = P

⇒\Rightarrow⇒ P = 11000.69 atm × 1.013 × 105 Pa

= 11143.69 × 105 Pa

= 11.14 × 108 Pa

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