Question

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).

Answer 1

Here,
Pressure ( P ) = 2 atm = 2 × 1.013 × 10^5 N/m²
Temperature ( T ) = 17 + 273.15 = 290.15 K
Molecular weight ( M) = 28 g
= 28 × 10^-3 Kg

Root mean square velocity = √{3RT/M}
= √{ 3 × 8.314 × 290/28 × 10^-3}
= 508.26 m/s

For mean free path ( L) on using ,
L = 1/√2πnd²
Where,
n = number of molecules per unit volume
d = diameter of moecule
Use PV = n.Kb.T
Where , n = no of morcules
Kb = Boltzmann's constant .
n/V = PV/Kb.T /V = P/Kb.T
Then,
Mean free path ( L) = KT/√2πd² × P
L = 1.38 × 10^-23 × 290.15/√2 × 3.14 × (2 × 10^-10)² × 2 × 1.013 × 10^5
= 1.11 × 10^-7 m

Collision frequency = no of collision per second
= Vrms/mean free path
= 508.26/1.11 × 10^-7
= 5.1 × 10^9 /sec

Time taken for collision = diameter of a moecule /Vrms
= 2 × 10^-10/508.26
= 4 × 10^-13 sec

Hence, time taken between two successive collision = L/Vrms
= 1.11 × 10^-7/508.26
= 2 × 10^-10 sec

Hence, time taken between two successive collision is 500 times the time taken for a collision . thus , a molecule in a gas moves essentially free for most of the time .