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Question:

A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid then:

Energy=3VT(1r+1R)is released

Energy=4VT(1r−1R)is released

Energy is neither released nor absorbed

Energy=3VT(1r−1R)is released

Solution:

ΔU=(T)(ΔA)A(initial)=(4πr²)nA(final)=4πR²ΔA=(4πr²)n−4πR²(4/3πr³)n=4/3πR³n=R³/r³ΔA=4π[R³/r³⋅r²−R²]=4π[R³/r−R²]=(4πR³/3)3[1/r−1/R]ΔA=3V[1/r−1/R]ΔU=3VT[1/r−1/R]