Question

A cell is roughly spherical with a radius of 20.0 micro meters. If the cell doubles in volume, calculate the work required to expand the cell surface against a surface tension of 72 mN/m.

Answer 1

Area of a cell = area of sphere = 4r²

Initially, area of cell = A₁ = 4r₁² = 4*3.14*(20.0*10^-6)² = 5.03*10^-9 m²

After expansion, the volume became double. This means that, V₂ = 2V₁

So, (4/3)r₂³ = 2(4/3)r₁³

So, r₂ = (2)^1/3r₁ = 1.26*20.0*10^-6 = 25.2*10^-6 m

Here Area of cell = A2 = 4r₂² = 4*3.14*(25.2*10^-6)² = 7.98*10^-9 m²

Work done in expansion against surface tension = -(A₂-A₁) = -72*10^-3[(7.98*10^-9)-(5.03*10^-9)]

= -212.4*10^-12 J = -2.12*10^-10 J

Work done in expansion = -2.12*10^-10 J