Q22 of 24 Page 75

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 ρ w = 103 kg l –1, Avogadro’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 []1/3 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.


(a) Given:


Latent heat of vaporisation of water = Lv = 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 NA.


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 NA whose mass is MA 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 NA 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,



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