Answer :

(a) Given:

Latent heat of vaporisation of water = L_{v} = 540 k Cal kg^{-1} = 540 ×10^{3} ×4.2 J = 2268×10^{3}J

Energy required to evaporate 1kmole of water is given by,

E

We know,

The number of molecules in 1kmole of water are N_{A}.

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 = d^{3}

Also, we know, volume of N_{A} whose mass is M_{A} is given by,

So, volume of 1 molecule =

Equating we get,

(c) Volume of 1g of water =1601 cm^{3} = 1.601 × 10^{-3}m^{3}

There are N_{A} 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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