Mean free path is given as $1.11\times10^{-7}$
Collision frequency is given as $4.58\times10^{9}s^{-1}$
Successive collision time ≅ 500 x (Collision time)
Pressure inside the cylinder containing nitrogen will be,
P = $2.0 atm = 2.026\times10^{5} Pa$
Temperature inside the cylinder is given as T = 170 C = 290 K
Radius of a nitrogen molecule is given as r = 1.0 Å = 1 x 1010 m
Diameter will be $d = 2\times 1\times 10^{10} = 2\times10^{10} m$
Molecular mass of nitrogen is known as $M = 28.0 g = 28\times 10^{-3} kg$
The root mean square speed of nitrogen is given by the expression,
$V_{rms}= √3RT / M$
Where,
R = universal gas constant having value $8.314 J mol^{-1} K^{-1}$
Hence,
$Vrms= \frac{3\times 8.314\times 290}{28\times10^{-3}}$
On calculation, we get,
= 508.26 m/s
The mean free path (l) is given by relation:
$l = KT / √2\times π\times d2\times P$
Where,
$\mathrm{k}$ is the Boltzmann constant having value $1.38 \times 10^{-23} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} \mathrm{~K}^{-1}$
So,
$\mathrm{I}=\left(1.38 \times 10^{-23} \times 290\right) /\left(\sqrt{2} \times 3.14 \times\left(2 \times 10^{-10}\right)^{2} \times 2.026 \times 10^{5}\right.$
We get,
$=1.11 \times 10^{-7} \mathrm{~m}$
Collision frequency $=\mathrm{V}_{\text {rms }} / 1$
$=508.26 / 1.11 \times 10^{-7}$
On calculation,
$=4.58 \times 10^{9} \mathrm{~s}^{-1}$
On calculation, we get
$4.58 \times 10^{9} \mathrm{~s}^{-1}$
$\begin{array}{l}
\mathrm{T}=\mathrm{d} / \mathrm{V}_{\mathrm{rms}} \\
=2 \times 10^{-10} / 508.26
\end{array}$
On further calculation, we get
$3.93 \times 10^{-13} \mathrm{~s}$
Time taken between successive collisions:
$T’ = \frac{l}{V_{rms}}=\frac{1.11\times10^{-7}}{508.26}$
$= 2.18\times 10^{-10}$
Hence,
$T’ / T = 2.18\times 10^{-10} / 3.93 \times 10^{-13}$
On calculation, we get,
= 500
As a result, the time taken between successive collisions is 500 times the time taken for a collision.