Problem 14: A monodisperse suspension of latex particles can be produced in water through a radical polymerization of styrene monomers. The generated particles are typically called polystyrene latex (PS) particles or latex particles, for short. The particles are spherical, with a uniform diameter. These dispersions are often used as a type of standard to calibrate scientific instruments. Calculate the following quantities of the aqueous PS particle dispersion.
- Terminal settling velocity of PS particles with a diameter of 1.0 micrometers in water under gravity. Assume the specific gravity of PS particles to be 1.05.
- Decay distance of the particle concentration profile in water in the vertical direction (away from gravity). Assume a Boltzmann distribution is formed. The decay distance means the distance over which the value of the number concentration of PS particles decreases by a factor of 1/e.
- How will these values change if the diameter of the PS particle decreases to 0.1 micrometers.
Given data:
- Diameter of PS particle, d = 1.0 micrometers = 1.0 × 10^(-6) m
- Radius of PS particle, r = d/2 = 0.5 × 10^(-6) m
- Density of PS particle, ρ_p = 1.05 × 10^3 kg/m^3
- Density of water, ρ_f = 1.0 × 10^3 kg/m^3
- Dynamic viscosity of water at room temperature, η ≈ 1.0 × 10^(-3) Pa·s
- Acceleration due to gravity, g = 9.81 m/s^2
Using Stokes' law for small Reynolds numbers (laminar flow): [ v_t = \frac{2 (ρ_p - ρ_f) g r^2}{9 η} ]
For diameter d = 1.0 micrometers: [ v_t = \frac{2 (1.05 × 10^3 - 1.0 × 10^3) × 9.81 × (0.5 × 10^(-6))^2}{9 × 1.0 × 10^(-3)} ] [ v_t ≈ 1.09 × 10^(-7) m/s ]
Therefore, the terminal settling velocity is approximately 1.09 × 10^(-7) m/s.
Using the Boltzmann distribution: [ h = \frac{k_B T}{mg} ]
Mass of a single particle: [ m = \frac{4}{3} π r^3 ρ_p ] [ m ≈ 5.5 × 10^(-19) kg ]
Assuming the temperature T is 298 K (25°C): [ h = \frac{1.38 × 10^(-23) × 298}{5.5 × 10^(-19) × 9.81} ] [ h ≈ 0.763 mm ]
Therefore, the decay distance is approximately 0.763 mm.
For diameter d = 0.1 micrometers: [ v_t = \frac{2 (1.05 × 10^3 - 1.0 × 10^3) × 9.81 × (0.05 × 10^(-6))^2}{9 × 1.0 × 10^(-3)} ] [ v_t ≈ 2.72 × 10^(-10) m/s ]
Therefore, the terminal settling velocity is approximately 2.72 × 10^(-10) m/s.
For the new particle mass: [ m ≈ 5.5 × 10^(-22) kg ]
Using the decay distance formula: [ h = \frac{1.38 × 10^(-23) × 298}{5.5 × 10^(-22) × 9.81} ] [ h ≈ 0.76 m ]
Therefore, the decay distance is approximately 0.76 m.
-
Terminal Settling Velocity:
- For 1.0 micrometers diameter: 1.09 × 10^(-7) m/s
- For 0.1 micrometers diameter: 2.72 × 10^(-10) m/s
-
Decay Distance:
- For 1.0 micrometers diameter: 0.763 mm
- For 0.1 micrometers diameter: 0.76 m