Chapter 15, Problem 15.27P

Interpretation Introduction

Interpretation:

The surface energy required to change a spherical drop of water of $2\text{mm}$ diameter into 2 spheres of equal size has to be determined.

Concept Introduction:

A molecule inside liquid feels attraction from all directions uniformly, but surface molecules only feel force from inward direction. So, surface molecules of a liquid feel a net inward force and therefore tends to minimize their surface area to minimize the number of molecules on the liquid surface. This inward force is surface tension.

Higher the attractive force on molecules, higher will be the surface tension of liquid. The surface tension of a liquid holds the liquid droplet in spherical shape because a sphere has the lowest surface area. For a spherical drop, high surface tension in necessity.

Answer to Problem 15.27P

The surface energy needed to change a spherical drop of water of $2\text{mm}$ diameter into 2 spheres of equal size is $3.346\times {10}^{-4}\text{mJ}$.

Explanation of Solution

The diameter of bigger sphere is $2\text{mm}$.

The radius of bigger sphere is calculated as follows:

  $\begin{array}{c}\text{Radius}=\frac{\text{diameter}}{2}\\ =\frac{2\text{mm}}{2}\left(\frac{0.001\text{m}}{1\text{mm}}\right)\\ =0.001\text{m}\end{array}$

The formula for surface area of sphere is as follows:

  $\text{surface}\text{area}=4\text{π}{r}^2$        (1)

Here,

$r$ is radius.

Substitute $0.001\text{m}$ for $r$ in equation (1) to determine the surface area of bigger sphere.

  $\begin{array}{c}\text{surface}\text{area}=\left(4\right)\left(3.14\right){\left(0.001\text{m}\right)}^2\\ =1.256\times {10}^{-5}{\text{m}}^2\end{array}$

The formula to calculate the surface energy of bigger sphere is as follows:

  $\text{surface}\text{energy}=\left(\text{surface}\text{area}\right)\left(\text{surface}\text{tension}\right)$        (2)

Substitute $1.256\times {10}^{-5}{\text{m}}^2$ for surface area and $72\text{mJ}\cdot {\text{m}}^{-\text{2}}$ for $\text{surface}\text{tension}$ in equation (2) to calculate the surface energy of bigger sphere.

  $\begin{array}{c}\text{surface}\text{energy}=\left(1.256\times {10}^{-5}{\text{m}}^2\right)\left(72\text{mJ}\cdot {\text{m}}^{-\text{2}}\right)\\ =9.0432\times {10}^{-4}\text{mJ}\end{array}$

The formula to determine volume of bigger sphere is as follows:

  $\text{volume}=\frac{4}{3}\text{π}{r}^3$        (3)

Substitute $0.001\text{m}$ for $r$ in equation (3) to determine volume of bigger sphere.

  $\begin{array}{c}\text{volume}=\left(\frac{4}{3}\right)\left(3.14\right){\left(0.001\text{m}\right)}^3\\ =4.1866\times {10}^{-9}{\text{m}}^3\end{array}$

The bigger sphere is divided into 2 smaller spheres of equal size so the volume of both smaller spheres will be same. Consider volume of smaller sphere be $V$ then the volume of 2 spheres will be $2V$ and it will be equal to the volume of bigger sphere shown as follows:

  $2V=\text{volume}\text{of}\text{bigger}\text{sphere}$        (4)

Rearrange equation (4) to calculate $V$.

  $V=\frac{\text{volume}\text{of}\text{bigger}\text{sphere}}{2}$        (5)

Substitute $4.1866\times {10}^{-9}{\text{m}}^3$ for volume of bigger sphere in equation (5) to calculate $V$.

  $\begin{array}{c}V=\frac{4.1866\times {10}^{-9}{\text{m}}^3}{2}\\ =2.093\times {10}^{-9}{\text{m}}^3\end{array}$

Rearrange equation (3) to calculate radius of smaller sphere.

  $r=\sqrt[3]{\frac{\text{volume}}{\frac{4}{3}\text{π}}}$        (6)

Substitute $2.093\times {10}^{-9}{\text{m}}^3$ for volume in equation (6) to calculate radius of smaller sphere.

  $\begin{array}{c}r=\sqrt[3]{\frac{2.093\times {10}^{-9}{\text{m}}^3}{\frac{4}{3}\left(\text{3}\text{.14}\right)}}\\ =7.937\times {10}^{-4}\text{m}\end{array}$

Substitute $7.937\times {10}^{-4}\text{m}$ for $r$ in equation (1) to determine the surface area of inner sphere.

  $\begin{array}{c}\text{surface}\text{area}=\left(4\right)\left(3.14\right){\left(7.937\times {10}^{-4}\text{m}\right)}^2\\ =7.9123\times {10}^{-6}{\text{m}}^2\end{array}$

Substitute $7.9123\times {10}^{-6}{\text{m}}^2$ for surface area and $72\text{mJ}\cdot {\text{m}}^{-\text{2}}$ for $\text{surface}\text{tension}$ in equation (2) to determine surface energy of smaller sphere.

  $\begin{array}{c}\text{surface}\text{energy}=\left(7.9123\times {10}^{-6}{\text{m}}^2\right)\left(72\text{mJ}\cdot {\text{m}}^{-\text{2}}\right)\\ =5.6968\times {10}^{-4}\text{mJ}\end{array}$

So the difference in surface energy of smaller and bigger sphere is calculated as follows:

  $\begin{array}{c}\text{change in surface}\text{energy}=\left[\begin{array}{c}\left(\text{surface}\text{energy for large sphere}\right)-\\ \left(\text{surface}\text{energy for small sphere}\right)-\end{array}\right]\\ =9.0432\times {10}^{-4}\text{mJ}-5.6968\times {10}^{-4}\text{mJ}\\ =3.346\times {10}^{-4}\text{mJ}\end{array}$

So the surface energy needed to change a spherical drop of water of $2\text{mm}$ diameter into 2 spheres of same size is $3.346\times {10}^{-4}\text{mJ}$.