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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 vaporization for water $Lv = 540 k cal kg^{-1}$ , the mechanical equivalent of heat $J = 4.2 J cal^{-1}$ , density of water $\rho w = 103 kg l^{-1}$ , Avagadro’s number $N_{A} = 6.0 \times 10^{26} k mole ^{-1}$ and the molecular weight of water $M_{A} = 18 kg$ 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
$d=\left [ \frac{M_{A}}{N_{A}}\times \frac{1}{\rho _{w}} \right ]^{\frac{1}{3}}$
(c) 1 g of water in the vapor state at 1 atm occupies 1601 cm³. 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 where d=3.1x10^-10m
(e) Calculate $\frac{F}{d}$ , which is a measure of the surface tension.

Answers (1)

a) $Lv = 540 k cal kg^{-1}$ $= 540 \times 4.2 \times 1000 J/Kg$

For evaporation of 1 kg water, energy required = Lv K Cal

For evaporation of Ma kg of water, La Ma K Cal

Number of molecules in Ma kg of water = Na

Hence, for the evaporation of 1 molecule, the energy required (U)= (Lv Ma/Na) K. Cal

$U = \frac{Ma Lv \times 1000 \times 4.2}{Na} = \frac{18 \times 540 \times 1000 \times 4.2}{6 \times 10 ^{26}} = 6.8 \times 10 ^{-20} J$

b)

water molecules are assumed to be of a point size and have distance d between each other

total volume of Na molecules of water = mass/density

$=\frac{M_{a}}{p}$

volume occupied by 1 molecule =

$d^{3} = \frac{M_{a}}{N_{a}}\times p$

$d = \left [ \frac{M_{a}}{N_{a}} \times p \right ]^{\frac{1}{3}}$

hence, the volume occupied by 1kg vapour $= 1601 cm^3 = 1601 \times 10^{-6} m^3$

c)

the volume occupied by 18kg water vapour $= 18\times 1601 \times 10^{-6} cm^3$

number of molecules in 18 kg water = $6\times 10^{26}$

so, the volume occupied by 1 molecule = $\frac{18 \times 1601 \times 10^{-6} cm^{3}}{6 \times 10 ^{26}}$

hence $d^{'3}=\frac{18 \times 1601 \times 10^{-6} cm^{3}}{6 \times 10 ^{26}}$

$d^{'} = 36.3 \times 10 ^{-10} m$

d) work done to change the distance to d’ from d = F (d’-d)

$6.8 \times 10^{-20}= F (36.3 \times 10^{-10} - 3.1 \times 10^{-10}) = F \times 33.2 \times 10^{-10}$

Hence, $F = \frac{6.8 \times 10^{-20}}{33.2 \times 10^{-10}}$

$F = 2.05 \times 10 ^{-11} N$

e)

from the formula of surface tension, we know that, surface tension = F/d

Surface tension = $\frac{2.05 \times 10 ^{-11} N}{3.1 \times 10^{-10}}$

Surface tension = $6.6 \times 10^{-2} N/m$

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