Surface tension i

(a) Given:

Latent heat of vaporisation of water = L_v = 540 k Cal kg^-1 = 540 ×10³ ×4.2 J = 2268×10³J

Energy required to evaporate 1kmole of water is given by,

E

We know,

The number of molecules in 1kmole of water are N_A.

Energy required for evaporation of 1 molecule is given by,

(b) Assume molecules to be at a distance d from each other.

Volume around 1 molecule = d³

Also, we know, volume of N_A whose mass is M_A is given by,

So, volume of 1 molecule =

Equating we get,

(c) Volume of 1g of water =1601 cm³ = 1.601 × 10^-3m³

There are N_A molecules in 18kg of water. So, volume occupied by 1 molecule is given by,

If d^’ is the intermolecular distance and d^’3 is the volume of one molecule, then

(d) The work done by force in move a molecule from distance d to d^’ is equal to the energy required to evaporate one molecule.

(e) Here,