Question

The boiling point of a substance X at 1 atm pressure is 500 K. The enthalpy of vapourisation at the boiling point of X(l) is 80 kJ/mol.

The molar specific heat of X (l) = 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(l) at 800 K and at 1 atm ?

• A. Lesser than 80 kJ
• B. Greater than 80 kJ
• C. Equal to 80 kJ
• D. Since 800 K is the higher temperature than the boiling point of X(l) ; therefore, enthalpy of vapourisation is not applicable

Answer 1

Answer: (A)

Lesser than 80 kJ

Answer 2

DH₁ = 5 × 10^–3 × (500 – 800) × 10^–3 kJ;

DH₂ = 80 kJ;

DH₃ = 5 × 10^–4 × (800 – 500) × 10^–3 kJ

DH₁ = – 0.0015 kJ ; DH₂ = 80 kJ,

DH₃ = 0.00015 kJ ;

DH = DH₁ + DH₂ + DH₃ ;

Hence (1)