Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of soap solution is 30 dyne/cm. (π=3.142).
Answer
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Hint: Generally the property of liquid is surface tension. All the molecules inside the liquid experience force from all the 360 degrees direction. Hence net force will be zero on those molecules. The molecules which are on the liquid surface will be experiencing force only from below molecules as there are no molecules above them. So they have some surface energy.
Formula used:
$S = 2T\Delta A$
$\Delta A = {A_f} - {A_i}$
Complete step by step answer:
When the soap bubble’s radius is increased, the area of that bubble gets increased and then the surface energy of that bubble gets increased. Energy cannot be created nor destroyed. Hence the extra energy generated must be coming from somewhere. The work we do will be converted to energy and stored as that extra energy. Now change in the energy will give us the work done. Change in surface energy will be $2T\Delta A$. Where T is the surface tension and $\Delta A$ is the change in surface area.
The reason why we multiplied with number 2 is because soap bubble will be having two surfaces and hence surface tension will act on two surfaces
$\Delta A = {A_f} - {A_i}$
$A = 4\pi {r^2}$
$\eqalign{
& \Rightarrow \Delta A = {A_f} - {A_i} = (r_f^2 - r_i^2) \times 4\pi \cr
& \Rightarrow \Delta A = {A_f} - {A_i} = ({2^2} - {1^2}) \times 4 \times 3.142 \cr
& \Rightarrow \Delta A = 37.704c{m^2} \cr} $
So the work done in increasing the radius of soap bubble is
$S = 2T\Delta A$
$\eqalign{
& \Rightarrow S = 2 \times 30 \times 37.704 \cr
& \therefore S = 2262.24ergs \cr} $
Hence the work required to convert the soap bubble from 1cm radius to 2cm radius is 22262.24 ergs.
Note:
Same work can be expressed in terms of joules too. Erg is the CGS unit of work done while joule is the SI unit of work done. Since every information is given in CGS we had calculated work done in CGS system. One joule will be equal to ten power seven times of one erg.
Formula used:
$S = 2T\Delta A$
$\Delta A = {A_f} - {A_i}$
Complete step by step answer:
When the soap bubble’s radius is increased, the area of that bubble gets increased and then the surface energy of that bubble gets increased. Energy cannot be created nor destroyed. Hence the extra energy generated must be coming from somewhere. The work we do will be converted to energy and stored as that extra energy. Now change in the energy will give us the work done. Change in surface energy will be $2T\Delta A$. Where T is the surface tension and $\Delta A$ is the change in surface area.
The reason why we multiplied with number 2 is because soap bubble will be having two surfaces and hence surface tension will act on two surfaces
$\Delta A = {A_f} - {A_i}$
$A = 4\pi {r^2}$
$\eqalign{
& \Rightarrow \Delta A = {A_f} - {A_i} = (r_f^2 - r_i^2) \times 4\pi \cr
& \Rightarrow \Delta A = {A_f} - {A_i} = ({2^2} - {1^2}) \times 4 \times 3.142 \cr
& \Rightarrow \Delta A = 37.704c{m^2} \cr} $
So the work done in increasing the radius of soap bubble is
$S = 2T\Delta A$
$\eqalign{
& \Rightarrow S = 2 \times 30 \times 37.704 \cr
& \therefore S = 2262.24ergs \cr} $
Hence the work required to convert the soap bubble from 1cm radius to 2cm radius is 22262.24 ergs.
Note:
Same work can be expressed in terms of joules too. Erg is the CGS unit of work done while joule is the SI unit of work done. Since every information is given in CGS we had calculated work done in CGS system. One joule will be equal to ten power seven times of one erg.
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