Answer :

(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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