Question

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 L_v = 540 k cal kg^–1, the mechanical equivalent of heat J = 4.2 J cal^–1, density of water ρ_w = 10³ kg l^–1, Avagadro’s No N_A = 6.0 × 10²⁶ k mole ^–1 and the molecular weight of water M_A = 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

(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.

Answer 1

(a) 1 kg of water requires L_v k cal 

M_A kg of water requires M_AL_v k cal 

Since there are N_A molecules in M_A kg of water the energy required for 1 molecule to evaporate is

(b) Consider the water molecules to be points at a distance d from each other.

(c) 1 kg of vapour occupies 1601 × 10^-3 m³.

18 kg of vapour occupies 18 × 1601 × 10^–3 m³

= 6 × 10²⁶ molecules occupies 18 × 1601 × 10^–3 m³ 

1 molecule occupies (18 x 1601 x 10^-3 / 6 x 10²⁶)m³ 

If d' is the inter molecular distance, then 

d'³ = (3 × 1601 × 10^-29)m³ 

d' = (30 × 1601)^1/3 × 10^-10 m = 36.3 × 10^-10 m