**Question:**

**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.**

**Solution:**

(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