Calculate the entropy change involved in the conversion of one mole of liquid water at 373 K to vapour at the same temperature (Latent heat of vaporization of water ∆vapH = 2.257 kJ/g).

Question

Calculate the entropy change involved in the conversion of one mole of liquid water at 373 K to vapour at the same  temperature (Latent heat of vaporization of water ∆vapH = 2.257 kJ/g).

Shiksha Nation · Accepted Answer

For the conversion of water→vapour, the entropy change is given by Δ_vapS =Δ_vapH /T_b

Here, Δ_vapH = 2.257 kJ / kg = 2.257 × 18 kJ / mol = 40.626 kJ / mol, T_b = 373 K

∴ Δ_vapS =40.626 kJmol⁻¹/373 K

= 0.1089 kJK⁻¹mol⁻¹

Calculate the entropy change involved in the conversion of one mole of liquid water at 373 K to vapour at the same temperature (Latent heat of vaporization of water ∆_vapH = 2.257 kJ/g).

Verified Answer