Question

Calculate the total number of molecular collisions that occur per second in 1 cm³ of air ( 80% N₂ and 20% O₂ by number ) at 1 atm and 298K.

(No more information provided)

Answer 1

We can calcualte the collisions per second by the following formula

Collisions / second = root mean square velocity / mean free path

Rms for nitrogen = 515.23 m/s

RMS for oxygen = 481.96 m/s

The mean free path is calcualted as

$\lambda = \frac{RT}{\sqrt{2}\pi d^{2}Na P}$

Where

T = 298K , Pressure = 1atm = 760mmHg

Na = 6.023 X 10^23 R = 0.0821

(i) for nitrogen , molecular diameter = 3.64 X 10^-10 meters

mean free path = 0.689 X 10^-7 meters

Collisions per second = 515.23 m/s / 0.689 X 10^-7 meters = 747.46 X 10^7 collisions per second

(ii) for oxygen

Molecular diameter= 1.2 x 10^-10 meters

Mean free path = 0.634 X 10^-6 meters

So collisions per second = 481.96 m/s / 0.634 X 10^-6 meters = 760.189 X 10^6 collisions per second