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CGP EDU Academic Team
Published on: August 14, 2026
The boiling point of a substance X at 1 atm pressure is 500 K. The enthalpy of vapourisation at the boiling point of X( λ ) is 80 kJ/mol. The molar specific heat of X ( λ ) = 5 × 10 –3 kJ/mol and C P of X(g) = 5 × 10 –4 kJ/mol.What is the molar latent heat of vapourisation of X( λ ) at 800 K and at 1 atm ?
Text Solution
Verified by ExpertsThe correct answer is:
A
Δ H 1 = 5 × 10 –3 × (500 – 800) × 10 –3 kJ;
Δ H 2 = 80 kJ;
Δ H 3 = 5 × 10 –4 × (800 – 500) × 10 –3 kJ
Δ H 1 = – 0.0015 kJ ; Δ H 2 = 80 kJ,
Δ H 3 = 0.00015 kJ ;
Δ H = Δ H 1 + Δ H 2 + Δ H 3 ;
Hence
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