Spherical particles of a protein of density 2.0 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 3.7 cm, find the radius of the largest particles that remain in solution at the end of the hour.
ρ =2 g/cm³= 2•10³kg/m³
The terminal speed of the particle is
v<h/t =0.037/3600=1.0228•10^-5 m/s.
The terminal speed of the particle of density ρ and radius R which is falling down in the water od density ρ1 =1000 kg/m³ and viscosity η is
v=2•R²•g• (ρ - ρ1)/9•η.
R=sqrt[9• η•v/2•g•(ρ - ρ1)] =
= sqrt[9•0.001•1.0228•10^-5/2•9.8•1•10³]=
=2.17•10^-6 m =2.17 μm
To find the radius of the largest particles remaining in solution, we need to consider the concept of sedimentation and its relation to the buoyant force and gravitational force.
We can start by calculating the buoyant force acting on the particles using Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
The buoyant force (F_b) can be calculated using the formula:
F_b = V × ρ_f × g
Where:
V is the volume of the particle,
ρ_f is the density of the fluid (water in this case), and
g is the acceleration due to gravity.
Given that the density of the protein particles (ρ_p) is 2.0 g/cm^3 and the density of water (ρ_f) is 1.0 g/cm^3, we can calculate the mass of the particles (m_p) using the formula:
m_p = V × ρ_p
Next, we need to determine the gravitational force (F_g) acting on the particles, which can be calculated using the formula:
F_g = m_p × g
Now, we can set up an equilibrium condition, where the buoyant force is equal to the gravitational force:
F_b = F_g
Substituting the formulas, we can solve for V (the volume of the particle) as follows:
V × ρ_f × g = V × ρ_p × g
Canceling out the g and rearranging the equation, we have:
V = ρ_f / ρ_p
Now, we can determine the radius of the particles using the formula for the volume of a sphere:
V = (4/3) × π × r^3
We already know the value of V from the previous equation, so we can solve for r (the radius of the particle). Rearranging the equation, we get:
r = (3 × V / (4 × π))^(1/3)
Now, let's plug in the values and calculate:
ρ_f = 1.0 g/cm^3, ρ_p = 2.0 g/cm^3
From the given information, we can assume g ≈ 9.8 m/s^2.
V = (1.0 g/cm^3) / (2.0 g/cm^3)
V = 0.5
r = (3 × 0.5 / (4 × π))^(1/3)
r ≈ 0.423 cm
Therefore, the radius of the largest particles that remain in solution after 1.0 hour is approximately 0.423 cm.
To find the radius of the largest particles that remain in solution at the end of the hour, we can use the concept of sedimentation.
The sedimentation velocity of a spherical particle can be calculated using Stokes' Law:
v = (2/9) * (r^2) * g * (d_p - d_f) / η
Where:
v = sedimentation velocity
r = radius of the particle
g = acceleration due to gravity (approximately 9.8 m/s^2)
d_p = density of the particle
d_f = density of the fluid (water in this case)
η = dynamic viscosity of the fluid (water in this case)
We are given the following information:
Density of the particle (d_p) = 2.0 g/cm^3
Density of water (d_f) = 1.0 g/cm^3
Depth of water in the tube = 3.7 cm (0.037 m)
Time = 1.0 h (3600 seconds)
First, we need to convert the densities from g/cm^3 to kg/m^3 and the depth from cm to meters:
d_p = 2.0 * 1000 = 2000 kg/m^3
d_f = 1.0 * 1000 = 1000 kg/m^3
Depth = 0.037 m
Time = 3600 seconds
Next, let's rearrange the formula to solve for the radius (r):
r = √((v * η) / ((2/9) * g * (d_p - d_f)))
To calculate the sedimentation velocity (v), we need to find the height (h) the particles fall through. This can be calculated using the depth (d) and time (t):
h = (1/2) * g * t^2
h = (1/2) * 9.8 * (3600^2) = 63504000 m^2
Now, plug in the given values:
v = h / t = 63504000 / 3600 = 17640 m/s
Substituting the known values into the formula for r:
r = √((v * η) / ((2/9) * g * (d_p - d_f)))
= √(((17640) * (0.001)) / ((2/9) * (9.8) * (2000 - 1000)))
= √(17.64 / 3520)
= √0.005
Therefore, the radius of the largest particles that remain in solution at the end of the hour is approximately 0.0707 m.