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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 ? |
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Answer» of small drop of radius R = V1 = 4/3 π r³volume of the BIG drop of radius R = V = 4/3 π R³number of small drops combined to FORM the big drop = N = V / V1 N = R³ / r³Surface tension of liquid = surface energy / surface area = T surface energy = surface tension * surface areaTotal surface energy stored in N small drops = 4 π r² * T * N = 4π T R³ / rTotal surface energy stored in 1 big drop = 4 π R² * Tsurface 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 ]===================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. |
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