Question

A cell is roughly spherical with a radius of 30.5×10−6m. The surface tension of water is 71.99 mN⋅m−1. If the cell grows to double its volume, calculate the magnitude of the work required to expand the cell surface. Assume the cell is surrounded by pure water and that T=298.15K. Express your answer with the appropriate units. Please explain how you arrived to your answer.

Answer 1

Answer:

To calculate the magnitude of work required to expand the cell surface

consider,

Area of a cell = area of sphere = 4$\pi$r²

A₁ = 4$\pi$r₁²

= 4 *3.14 *(30.5*10^-6)²

= 1.17*10^-8 m²

Here after the expansion, the volume became double.

i.e.,

V₂ = 2V₁

So, (4/3)$\pi$r₂³ = 2(4/3)$\pi$r₁³

So, r₂ = (2)^1/3r₁

= 1.26*30.5*10^-6

= 38.43*10^-6 m

Here Area of cell = A2 = 4$\pi$r₂²

= 4 *3.14 *(38.43*10^-6)²

= 1.855*10^-8 m²

Work done in expansion against surface tension = -$\gamma$(A₂-A₁)

= -71.99*10^-3[(1.855*10^-8 )-( 1.17*10^-8)]

= - 4.93*10^-10 J

Work done in expansion = - 4.93*10^-10 J