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.

neela | Student

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.

giorgiana1976 | Student

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