Question

The surface tension of a soap solution is 0.03 nm-1.How much work is done to produce a soap bubble of radius 0.05m?

Answer 1

Here T = 0.03 Nm-1

r = 0.05 m

Surface area = 4πr2

= 4×22/7×(0.05)2

No. of surface area of a soap bubble = 2 (outer and inner area)

Surface energy is given by T∆S per surface.

Total surface energy = 2 T∆S

E = 2×0.03× 4×22/7×(0.05)2 J

Answer 2

Given:

• The surface tension of a soap bubble = 0.03 N/m
• Radius of the soap = 0.05 m

To Find:

• The work done in producing a soap bubble of 0.05 radius.

Solution:

The work done is given by,

W = 2×surface tension×surface area = 2×T×A → {equation 1}

The surface area of the soap is 4π$r^2$  since bubbles are circular in shape.

Substitute the values in equation 1 then we get,

W = 2×0.03×(4×3.14×0.05×0.05)

Multiplying all the values in the above formula.

W = 0.00184 J or 18.4×$10^-4$

∴ The work done to produce a soap bubble of 0.05 radius = 0.00184 J or 18.4×$10^-4$ J.