For increasing to N=8.0 times the volume of a bubble soap with the radius R=10mm, what kind of mechanical work has to be done against stress forces?The stress forces are surface kind.
A soap bubble has a cavity and therefore has two surfaces.
The surface energy required to maintain the surface of the bubble, is therefore= 2 times spherical suface*surface energy per unit area.=2*4pi*R^2* gamma = 8pi*R^2 * gamma
When the volume is incresed 8 times,the radius r also icreases to the cube root of 8R^3 to 2R
Therefore, there will be inrease in the surface area, and the new suface area is 2*4pi(2R)^2=32pi*R^2.
and the energy of the surface =16pi*R^2*gamma.
So, the mechanical work done to increse the surface enrgy is : 32pi*R^2.gamma-8pi*R^2*gamma= 24pi*R^2*gamma.
=24*pi*(.01)^2*gamma, as 10mm=0.01 meter.
=7.5398 *10^-3 gamma Joules.
Hope this helps.
The mechanical work is necessary for increasing the superficial energy of the bubble soap. We note that the soap pellicle is limited, from inside and outside, by the coat of soap-water solution, so the surface of both superficial layers, has to be increased.
L=2*sigma[4pi*R'^2- 4 pi*R^2], but (4pi/3)*R'^3=N*(4pi/3)*R^3, so R'=R* cubic root N
L=2*sigma*4*pi*R^2[(cubic root N^2)-1]=0.32mJ