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

• A. energy = $4VT(\frac{1}{r}-\frac{1}{R})$ is released
• B. energy = $3VT(\frac{1}{r}+\frac{1}{R})$ is absorbed
• C. energy = $3VT(\frac{1}{r}-\frac{1}{R})$ is released
• D. energy is neither released nor absorbed.

Answer 1

Answer: (C)

$\Delta U=(T)(\Delta A)$
$A($ initial $)=\left(4\pi r^2\right)n$
$A($ final $)=4\pi R^2$
$\Delta A=\left(4\pi r^2\right)n-4\pi R^2$
$\left(\frac{4}{3}\pi r^3\right)n=\frac{4}{3}\pi R^3$
$n=\frac{R^3}{r^3}$
$\Delta A=4\pi \left[\frac{R^3}{r^3}\cdot r^2-R^2\right]=4\pi \left[\frac{R^3}{r}-\frac{R^3}{R}\right]=\left(\frac{4\pi R^3}{3}\right)3\left[\frac{1}{r}-\frac{1}{R}\right]$
$\Delta A=3V\left[\frac{1}{r}-\frac{1}{R}\right]$
$\Delta U=VVT\left[\frac{1}{r}-\frac{1}{R}\right]$