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 \bigg( \frac{1}{r} - \frac{1}{R} \bigg) $ is released
• B. energy = $ 3VT \bigg( \frac{1}{r} + \frac{1}{R} \bigg) $ is absorbed
• C. energy = $ 3VT \bigg( \frac{1}{r} - \frac{1}{R} \bigg) $ 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 $)=4 \pi R ^{2}$
$\Delta A =\left(4 \pi r ^{2}\right) n -4 \pi R ^{2}$
$\left(\frac{4}{3} \pi r^{3}\right) n =\frac{4}{3} \pi R ^{3}$
$n =\frac{ R ^{3}}{ r ^{3}}$
$\Delta A =4 \pi\left[\frac{ R ^{3}}{ r ^{3}} \cdot r ^{2}- R ^{2}\right]=4 \pi\left[\frac{ R ^{3}}{ r }-\frac{ R ^{3}}{ R }\right]=\left(\frac{4 \pi R ^{3}}{3}\right) 3 \left[\frac{1}{ r }-\frac{1}{ R }\right]$
$\Delta A = 3 V \left[\frac{1}{ r }-\frac{1}{ R }\right]$
$\Delta U = V V T \left[\frac{1}{ r }-\frac{1}{ R }\right]$