Question

For water at 100°C and 1 bar, Δ_vap H – Δ_vap U = ..... x 10² J mol^–1 . (Round off to the Nearest Integer) [Use : R=8.31 J mol^–1 K ^–1] [Assume volume of H₂O(l) is much smaller than volume of H₂O(g). Assume H₂O(g) treated as an idea gas]

Answer 1

Answer is : 31

H₂O_(l) ⇌ H₂O_(v)

ΔH = ΔU + Δn_gRT

for 1 mole waters; 

Δn_g = 1

Δn_gRT = 1 mol x 8.31 J/mol-k x 373 K

= 3099.63 J = 31 x 10² J