Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water Lv = 540 k cal kg^-1, the mechanical equivalent of heat J = 4.2 J cal^-1, density of water ρ_w = 103 kg/m³, Avagadro’s No N_A = 6.0 × 10²⁶ k mole^-1 and the molecular weight of water MA = 18 kg for 1 k mole.

(a) Estimate the energy required for one molecule of water to evaporate.

(b) Show that the inter–molecular distance for water is d= ${\left[\frac{{\mathit{M}}_{\mathit{A}}}{{\mathit{N}}_{\mathit{A}}}\times \frac{1}{{\mathit{\rho }}_{\mathit{w}}}\right]}^{1/3}$  and find its value.

(c) 1 g of water in the vapor state at 1 atm occupies 1601cm³. Estimate the intermolecular distance at boiling point, in the vapour state.

(d) During vaporisation a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to d′. Estimate the value of F.

(e) Calculate F/d, which is a measure of the surface tension.