Question

A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a mass of 1.9 x 10^−14 kg. How long does it take for all of the particles to settle to the bottom of the container?

Answer 1

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = $\dfrac{4}{3}\pi r^3$

                              =$\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3$

Density =$\dfrac{mass}{volume}$

Density =$\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}$

ρ = 2322 kg/m³

terminal velocity

$v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}$

$v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}$

v_t = 6693 x 10⁻⁷ m/s

$t = \dfrac{d}{v_t}$

$t = \dfrac{0.15}{6693 \times 10^{-7}}$

t = 224 s