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.

1. Estimate the energy required for one molecule of water to evaporate.
2. Show that the inter–molecular distance for water is `d = [M_A/N_A xx 1/ρ_w]^(1/3)` and find its value.
3. 1 g of water in the vapor state at 1 atm occupies 1601 cm³. Estimate the intermolecular distance at boiling point, in the vapour state.
4. 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.
5. Calculate F/d, which is a measure of the surface tension.

Answer 1

a. According to the problem, latent heat of vaporisation for water

`(L_v) = 540  k cal = 540 xx 10^3 xx 4.2  J = 2268 xx 10^3  J`

Therefore, energy required to evaporate 1 kmol (18 kg) of water

= `(2268 xx 10^3  J)(18)`

= `40824 xx 10^3  J`

= `4.0824 xx 10^7  J`

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

`U = (M_AL_v)/N_A  J`  .....[Where NA = 6 × 10²⁶ = Avogadro number]

`U = (4.0824 xx 10^7)/(6 xx 10^26)  J`

= `0.68 xx 10^-19  J`

= `6.8 xx 10^-20  J`

b. Let the water molecules to be at points and are placed at a distance d from each other,

The volume of Na molecule of water = `M_A/ρ_w`

Thus, the volume around one molecule is = `"Volume of 1 kmol"/"Number of molecules/kmol"`

= `M_A/(N_Aρ_w)`

And also volume around one molecule = d³

Thus, by equating these, we get

`d^3 = M_A/(N_Aρ_w)`

∴ `d = (M_A/(N_Aρ_w))^(1/3)`

= `(18/(6 xx 10^26 xx 10^3))^(1/3)`

= `(30 xx 10^-30)^(1/3) m`

= `3.1 xx 10^-10  m`

c. The volume occupied by 18 kg water vapour = `18 xx 1601 xx 10^-6  cm^3`

Number of molecules in 18 kg water = `6 xx 10^26`

Since `6 xx 10^26` molecules occupies `18 xx 1601 xx 10^-3  m^3`

∴ Volume occupied by 1 molecule = `(28818 xx 10^-3 m^3)/(6 xx 10^26)`

= `48030 xx 10^-30  m^3`

If d' is the intermolecular distance, then

`(d)^3 = 48030 xx 10^-30 m^3`

So, `d^' = 36.3 xx 10^-10 m = 36.3 xx 10^-10 m`

d. Work done to change the distance from d to d' is `U = F(d^' - d)`,

This work done is equal to energy required to evaporate 1 molecule.

∴ `F(d^' - d) = 6.8 xx 10^-20`

or `F = (6.8 xx 10^-20)/(d^' - d)`

= `(6.8 xx 10^-20)/((36.3 xx 10^-10 - 3.1 xx 10^-10))`

= `2.05 xx 10^-11 N`

e. Surface tension = `F/d`

= `(2.05 xx 10^-11)/(3.1 xx 10^-10)`

= `6.6 xx 10^-2`  N/m