Published by:
CGP EDU Academic Team
Published on: August 14, 2026
At 5 ×10 5 bar pressure, density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain temperature T. Find the value of 
for the conversion of 1 mole of graphite to 1 mole of diamond (in KJ) at temperature T. (1L.atm = 100 J)
Text Solution
Verified by ExpertsThe correct answer is:
5
(5)
C (graphite) ⎯→ C (diamond)
1 mole 1 mole
ρ = 2 g/cm 3 ρ = 3 g/cm 3
V = 
ml V = 
ml
Δ H = Δ U + P Δ V
Δ U – Δ H = –P Δ V
= –5×10 5 
× 10 –3 L.atm
= – 5× 10 5 × 
×10 –1 J= 100 KJ/mole
= 10×10 4 J = 100KJ/mole
= 5 Ans.
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