   Chapter 9, Problem 57P

Chapter
Section
Textbook Problem

Spherical panicles of a protein of density 1.8 g/cm3 are shaken up in a solution of 20°C water. The solution is allowed to stand for 1.0 h. If the depth of water in the tube is 5.0 cm, find the radius of the largest particles that remain in solution at the end of the hour.

To determine
The radius of the largest particles that remain in solution at the end of the hour.

Explanation
The terminal velocity of the particle must be less than, if it is still in suspension after 1 hour. The terminal velocity is vt=2r2g(ρρf)/9ηd/t=2r2g(ρρf)/9η and now the maximum radius of the particle is rmax=9ηwater(vt)max/2g(ρsphereρwater) .

Given info: The solution allowed to stand for 1.0h , depth of water in the tube is 5.0cm , density of water is 1000kg/m3 , density of the spherical particles is 1.8g/cm3 , acceleration due to gravity is 9.8m/s2 , and coefficient of viscosity is 1.00×103Ns/m2 .

The formula for the radius of the largest particles that remain in solution is,

rmax=9ηwater(d/t)2g(ρsphereρwater)

• ηwater is coefficient of viscosity of water.
• d is depth of the water in the tube.
• g is acceleration due to gravity.
• t is time for the particle to reach the water depth.
• ρsphere is density of sphere

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