Standard enthalpy of vapourisation ΔvapH° for water at 100°C is 40.66 kJ mol⁻¹. The internal energy of vapourisation of water at 100°C (in kJ mol⁻¹) is

Question

Dr. Abubacker · Accepted Answer

[Solution from source] For H₂O(l) → H₂O(g), Δn_g = 1. ΔU = ΔH − Δn_gRT = 40.66 − (8.314 × 10⁻³ × 373) = 40.66 − 3.10 = 37.56 kJ mol⁻¹. 43.76 comes from adding RT instead of subtracting it. Answer (2).

Q. Standard enthalpy of vapourisation ΔvapH° for water at 100°C is 40.66 kJ mol⁻¹. The internal energy of vapourisation of water at 100°C (in kJ mol⁻¹) is

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?