Question

Calculate the work done in increasing the radius of a soap bubble in air from $1cm$ to $2cm$. The surface tension of soap solution is $30dyne/cm$. ($\pi =3.142$)

Answer 1

Whenever a liquid film is expanded, the work done gets stored as energy in the film. Hence the increase in energy of the film is equal to the work done.

The soap bubble expands hence the surface area increases.

The increment in the surface area$\times$ the surface tension $T$ is the extra energy and hence equal to the work done.

Note that the bubble has two areas, inner and outer. Since the thickness of the film is negligible, both the radii can be assumed as equal.

Area of sphere$=4\pi r^2$

Previous area $=A_1=8\pi r_{{}_1}^2$

Final area $=A_2=8\pi r_{{}_2}^2$

Increment $=\Delta A=A_2-A_1=(r_{{}_2}^2-r_{{}_1}^2)\times 8\pi$

Work done $W=\Delta A\times T=(r_{{}_2}^2-r_{{}_1}^2)\times 8\pi \times T$

$\therefore W=(2^2-1^1)\times 8\times 3.142\times 30=2262.24$ $ergs.$

CGS unit of work is $erg.$