Question

$n$ drops of a liquid, each with surface energy $E$, join to form a single drop. What happens?

• A. Some energy will be released in the process.
• B. Some energy will be absorbed in the process.
• C. The energy released or absorbed will be $E(n-n^{\frac{2}{3}})$.
• D. The energy released or absorbed will be $nE[2^{\frac{2}{3}}-1]$.

Answer 1

Answer: (A), (C)

Some energy will be released in the process.
The energy released or absorbed will be $E(n-n^{\frac{2}{3}})$.

Let us say $r$ is the radius of each single drop before these coalesce and $R$ is the radius of the new drop after these coalesce. Since volume of liquid remains same, we have
$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow R=n^{\frac{1}{3}}r$ .... $\left(I\right)$
Initial surface energy of single drop $=E=\left(4\pi r^2\right)\times T$
Initial surface energy of $n$ drops $=nE=n\left(4\pi r^2\right)\times T$
final surface energy of big single drop $=\left(4\pi R^2\right)\times T=\left(4\pi {\left(n^{\frac{1}{3}}r\right)}^2\right)\times T=n^{\frac{2}{3}}(4\pi r^2\times T)=n^{\frac{2}{3}}E$ from $\left(I\right)$
Clearly, $n^{\frac{2}{3}}E<nE$
i.e. final surface energy < initial surface energy
That means some energy will be released in this process.
And the energy released in this process is given by $nE-n^{\frac{2}{3}}E=E(n-n^{\frac{2}{3}})$