Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decrease with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporization for water L0 = kcl kg-1, the mechanical equivalent of heat J = 4.2 J cal-1, density of water ρw = 103 kgm-3, Avagadro’s No NA = 6.0 × 1026 k mole-1 and the molecular weight of water MA = 18 kg for 1 mole.
(a) Estimate the energy required for one molecule of water to evaporate.
(b) Show that the inter-molecular distance for water is d = [\(\frac{M_A}{N_A}\) × \(\frac{1}{ρ_w}\)]1/3 and find its value.
(c) 1 g of water in the vapour state at 1 atm occupies 1601 cm3. Estimate the intermolecular distance at boiling point, in the vapour state.
(d) During vaporization a molecule overcome a force F< assumed constant to go from an intermolecular distance d to d’. Estimate the value of F, where d = 3.1 × 10-10 m.
(e) Calculate F/d which is a measure of the surface tension.