Surface tension is exhibited by liquids due to the force of attraction between molecules of the liquid. The surface tension decreases with increase in
temperature and vanishes at boiling point. Give that the latent heat of vaporisation for water Lv = 540 k cal/kg, the mechanical equivalent of heat J = 4.2 J/cal,
density of water ρw = 103 kg/l, Avogadro’s number NA = 6.0 1026 k/mole, 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
$d=\left[\frac{M_{A}}{N_{A}} \times \frac{1}{\rho_{w}}\right]^{\frac{1}{3}}$ 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 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.
(a) According to the given problem,
Latent heat of vaporisation for water is = 2268 × 103 J
The energy required to evaporate 1 k mol water = 4.0824 × 107 J
The energy required for evaporation of 1 molecule is, U = 6.8 × 10-20 J
(b) Let the distance between the water molecules be d
Then the volume around one molecule is:
$\frac{\text { volume of } 1 \mathrm{kmol}}{\text { number of molecules } / \mathrm{kmol}}=\frac{M_{A}}{N_{A} \rho_{w}}$
Therefore, the volume around one molecule = d3
Which is given as:
$d^{3}=\frac{M_{A}}{N_{A} \rho_{w}}$
d = 3.1 × 10-10m
(c) Volume occupied by 1 kmol of water molecules = 28818 × 10-3 m3
Volume occupied by 1 molecule of water = 48030 × 10-30 m3
Let d’ be the intermolecular distance, then
(d’)3 = 48030 × 10-30 m3
d’ = 36.3 × 10-10 m
(d) Work done to change the distance from d to d’ U = F(d’-d)
F = 2.05 × 10-11 N
(e) Surface tension = F/d = 6.6 × 10-2 N/m