Question

Smoke particles in the air typically have masses of the order of 10−16kg. The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope.
Find the root-mean-square speed of Brownian motion for a particle with a mass of 3.00×10−16kg in air at 300K.

Answer 1

Answer:

V_rms = 6.437 x 10^(-3) m/s

Explanation:

We are given;

Mass; m = 3 x 10⁻¹⁶Kg

Temperature; T = 300K

Now, the formula for root mean squares velocity is given as;

V_rms = √(3KT/m)

K is a constant known as Boltzmann's constant and it has a value of 1.381 x 10⁻²³J/mol.K

Thus, plugging in the relevant values;

V_rms = √[(3x1.381x10⁻²³x 300)/3x10⁻¹⁶]

V_rms = 64366 x 10^(7)

V_rms = 6.437 x 10^(-3) m/s

Answer 2

Answer:

6.44*10⁻³m/s

Explanation:

mass = 3*10⁻¹⁶Kg

T = 300K

K = 1.381*10⁻²³J/mol.K = Boltzmann's constant

Using the formula below,

Vrms = √(3KT/m)

Vrms = √[(3*1.381*10⁻²³*300)/3.0*10⁻¹⁶]

Vrms = 6.44*10^-3 m/s

The root mean square velocity of the gas is 6.44*10^-3 m/s.