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surface tension is exhibited by liquids due to the force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Give that the latent heat of vaporisation for water Lv = 540 k cal/kg, the mechanical equivalent of heat J = 4.2 J/cal, density of water ρw = 103 kg/l, Avogadro’s number NA = 6.0 1026 k/mole, and the molecular weight of water MA = 18 kg for 1 k mole.
surface tension is exhibited by liquids due to the force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Give that the latent heat of vaporisation for water Lv = 540 k cal/kg, the mechanical equivalent of heat J = 4.2 J/cal, density of water ρw = 103 kg/l, Avogadro’s number NA = 6.0 1026 k/mole, and the molecular weight of water MA = 18 kg for 1 k mole.
CBSE | Class 11 | Mechanical Properties of Fluids | NCERT Exemplar | Physics
surface tension is exhibited by liquids due to the force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Give that the latent heat of vaporisation for water Lv = 540 k cal/kg, the mechanical equivalent of heat J = 4.2 J/cal, density of water ρw = 103 kg/l, Avogadro’s number NA = 6.0 1026 k/mole, 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

Answer :

a) According to the question, the latent heat of vaporization for water is given equal to 2268 × 103 J

4.0824 × 107 J is the energy required to evaporate 1 k mol water and the energy required for evaporation of 1 molecule is:

U = 6.8 × 10-20 J

b) Let d represent the distance between the water molecules. Then the volume around one molecule can be written as:

$ \frac{volume(1kmol)}{number\_of\_molecules}=\frac{{{M}_{A}}}{{{N}_{A}}{{\rho }_{w}}}$

Therefore, d3 represents the volume around one molecule

$ {{d}^{3}}=\frac{{{M}_{A}}}{{{N}_{A}}{{\rho }_{w}}}$

Therefore, upon substituting the known values, we get

d = 3.1 × 10-10m

 

i

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