Question

Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decrease with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporization for water L₀ = kcl kg^-1, the mechanical equivalent of heat J = 4.2 J cal^-1, density of water ρ_w = 10³ kgm^-3, Avagadro’s No N_A = 6.0 × 10²⁶ k mole^-1 and the molecular weight of water M_A = 18 kg for 1 mole.

(a) Estimate the energy required for one molecule of water to evaporate.

(b) Show that the inter-molecular distance for water is d = [\(\frac{M_A}{N_A}\) × \(\frac{1}{ρ_w}\)]^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 vaporization a molecule overcome a force F< assumed constant to go from an intermolecular distance d to d’. Estimate the value of F, where d = 3.1 × 10^-10 m.

(e) Calculate F/d which is a measure of the surface tension.

Answer 1

Given, L_v = 540\(\frac{kcal}{Kg}\)

= 540 × 10³ × 4.2 J/Kg

(a) Such that, 1kg of water requires energy for evaporation = L_v k cal

∴ M_A kg of water requires M_AL_vk cal

Since there are N_A molecules to evaporates is

\(u=\frac{M_AL_v}{N_A}J\)        [Here, N_A = 6 × 10²⁶ = Avogadro number]

= \(\frac{18×540×4.2×10^3}{6×10^{26}}\)

= 90 × 18 × 4.2 × 10^-23 J

= 6.8 × 10^-20 J

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

Volume of N_A molecules = \(\frac{M_A}{ρ_w}l\)

Thus, the volume around one molecule is \(\frac{M_A}{N_Aρ_w}l\)

The volume around one molecule is d³ = (\(\frac{M_A}{N_Aρ_w}\))

∴ d = \((\frac{M_A}{N_Aρ_w})^\frac{1}{3}\) = \((\frac{18}{6×10^{26}×10^3})^\frac{1}{3}\)

= (3 × 10⁻³⁰)^{\(\frac{1}{3}\)} m ≃ 3.1 × 10⁻¹⁰ m

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

∴ 18 kg of vapour occupies = 18 × 1601 × 10^-3 m³

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

∴ 1 molecule occupies = \(\frac{18×1601×10^3}{6×10^{26}}m^3\)

If d₁ is the inter molecular distance.

Then

\(d^3_1\) = (3 × 1601 × 10^-29) m³

∴ d₁ = (30 × 1601)^{\(\frac{1}{3}\)} × 10^-10 m

= 36.3 × 10^-10 m

(d) To change the distance from d to d₁,

Work done = F(d₁ − d)

Work done = energy required to evaporate 1 molecule

Or F = \(\frac{6.8×10^{−20}}{d_1−d}\)

= \(\frac{6.8×10^{−20}}{ (36.3−3.1)×10^{−10}}\)

= 0.2048 × 10⁻¹⁰ N

(e) Surface Tension = \(\frac{F}{d}\)

= \(\frac{2.05×10^{−20}}{3.1×10^{−10}}\)

= 6.6 × 10^-2 Nm^-1