Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0$ atm and temperature $17^{\circ} \mathrm{C}$. Take the radius of a nitrogen molecule to be roughly $1.0$ A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\mathrm{N}_{2}=28.0 \mathrm{u}$ ).

Home » Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at atm and temperature . Take the radius of a nitrogen molecule to be roughly A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of ).

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0$ atm and temperature $17^{\circ} \mathrm{C}$ . Take the radius of a nitrogen molecule to be roughly $1.0$ A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\mathrm{N}_{2}=28.0 \mathrm{u}$ ).

CBSE | Class 11 | Kinetic Theory | NCERT | Physics

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0$ atm and temperature $17^{\circ} \mathrm{C}$ . Take the radius of a nitrogen molecule to be roughly $1.0$ A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\mathrm{N}_{2}=28.0 \mathrm{u}$ ).

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,
$\mathrm{I}=\left(1.38 \times 10^{-23} \times 290\right) /\left(\sqrt{2} \times 3.14 \times\left(2 \times 10^{-10}\right)^{2} \times 2.026 \times 10^{5}\right.$