Question

Spherical particles of a protein of density 1.50 g/cm are shaken up in a solution of 20°C water. The solution is allowed to stand for 1.00 h. If the depth of the water in the tube is 6.00 cm, find the radius of the largest particles that remain in solution (i.e. haven’t yet reached the bottom of the tube) at the end of the hour. The viscosity of 20°C water is 1.00 × 10−3 Pa⋅s.

Answer 1

Spherical particles of a protein of density 1.50 g/cm are shaken up in a solution of 20°C water. The solution is allowed to stand for 1.00 h. If the depth of the water in the tube is 6.00 cm, find the radius of the largest particles that remain in solution (i.e. haven’t yet reached the bottom of the tube) at the end of the hour. The viscosity of 20°C water is 1.00 × 10?3 Pa?s.

Density 1.50 g/cm

T = 20°C

Depth 6 cm

Viscosity of water 1.00 × 10?3 Pa?s.

Where

? = viscosity (kg/m s), uT = terminal velocity (m/s), ?s = density of the particle

(kg/m3), ?i = density of the liquid medium (kg/m3)

The density of water at 20 degrees centigrade is 1 g/cm3 while its viscosity is 1

centipoise (= 0.01 poise, g/cm s).

The terminal velocity of the particles is (6/3600)(depth/time in sec)

cm/s = 0.00166 cm/s.

The acceleration due to gravity is 981 cm/s²

Using equation (1):

re = [(9x0.01x0.00166 cm) / (2 x 981 x (1.5-1)]^1/2 = 7.61 x 10^-8 cm

Therefore, the diameter is 7.61 x 10^-8 cm and radius is 3.807 x 10^-8