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 Lv = 540 k cal kg-1, the mechanical equivalent of heat J = 4.2 J cal-1, density of water ρw = 103 kg/m3, Avagadro's No NA = 6.0 * 1026 k mole-1 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= MANA*1ρw1/3  and find its value.

(c) 1 g of water in the vapor state at 1 atm occupies 1601cm3. 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.

6 Views|Posted 8 months ago
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8 months ago

This is a long answer type question as classified in NCERT Exemplar

(a) Lv=540 kcal kg-1

= 540 * 10 3 c a l kg-1 = 540 * 10 3 * 4.2jkg-1

Energy required to evaporate 1kg of water = Lv kcal

And MA kg of water requires MALV kcal

Since there are NA molecules in MA kg of water the energy required for 1 molecule to evaporate

Is

U=

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Physics NCERT Exemplar Solutions Class 11th Chapter Ten 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Ten 2025

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