Question -
Answer -
Mean free path = 1.11 x 10-7 m
Collision frequency = 4.58 x 109 s-1
Successive collision time ≅ 500 x(Collision time)
Pressure inside the cylinder containing nitrogen, P = 2.0 atm =2.026 x 105 Pa
Temperature inside the cylinder, T = 170 C= 290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 x 1010 m
Diameter, d = 2 x 1 x 1010 =2 x 1010 m
Molecular mass of nitrogen, M = 28.0 g = 28 x 10-3 kg
The root mean square speed of nitrogen is given by the relation:
Vrms=√3RT / M
Where,
R is the universal gas constant = 8.314 J mol-1 K-1
Hence,
Vrms=3 x 8.314 x 290 / 28 x 10-3
On calculation, we get,
= 508.26 m/s
The mean free path (l) is given by relation:
l = KT / √2 x π x d2 xP
Where,
k is the Boltzmann constant = 1.38 x 10-23 kgm2 s-2 K-1
Hence,
l = (1.38 x 10-23 x290) / (√2 x 3.14 x (2 x 10-10)2 x 2.026 x 105
We get,
= 1.11 x 10-7 m
Collision frequency = Vrms / l
= 508.26 / 1.11 x 10-7
On calculation, we get,
= 4.58 x 109 s-1
Collision time is given as:
T = d / Vrms
= 2 x 10-10 /508.26
On further calculation, we get
= 3.93 x 10-13 s
Time taken between successive collisions:
T’ = l / Vrms =1.11 x 10-7 / 508.26
We get,
= 2.18 x 10-10
Hence,
T’ / T = 2.18 x 10-10 /3.93 x 10-13
On calculation, we get,
= 500
Therefore, the time taken between successive collisions is 500times the time taken for a collision