# Several drops of radius r coalesce to form a drop of radius R . if the tension is T and the volume under consideration is V , what is the energy released during the process ?

1
by namku

2015-04-24T14:07:16+05:30

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Volume of small drop of radius r = V1 = 4/3  π r³
volume of the big drop of radius R = V = 4/3 π R³

number of small drops combined to form the big drop = N = V / V1
N = R³ / r³

Surface tension of liquid = surface energy / surface area = T
surface energy = surface tension * surface area

Total surface energy stored in N small drops = 4 π r² * T * N
= 4π T R³ / r

Total surface energy stored in 1 big drop = 4 π R² * T

surface Energy released during the merging process of all drops :
ΔE =  4 π T R³ /r - 4 π R² T
= 4 π R³ T [ 1/r - 1/R]
= 3 V T  [ 1/r - 1/R ]

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Here we are taking into account only the surface tension related energy.  We have ignored gravitational potential energy and the changes in the potential energy of the molecules when they merge.  Also the kinetic energy of the drops is not considered.