91Ó°ÊÓ

1. The diameter of the molecule,$d=2.9×{10}^{−10}\text{″¾}$
2. The molecules collide with oxygen gas at temperature,$T=400\text{‿é}$
3. The pressure at which the collision occurs,$P=2.02×{10}^3\text{ p²¹}$
4. Molar mass of oxygen,$M=0.032\text{kg/mol}$
5. Molar gas constant value,$R=8.31\text{J/mol.K}$

The mean free path of a gas is defined as its average distance covered by the moving particle before undergoing any substantial change in direction or energy due to one or more collisions with the present particles in the system. The molecules will continuously collide with other particles and randomly change their direction of motion. Similarly, here, when the molecule comes into a collision with the oxygen gas, the interaction between them gives rise to a collision frequency as the molecule repeatedly collides with the molecules of the oxygen gas.

Formula:

Frequency of molecular collision,${\text{Z}}_{frequency}=\frac{}{\mathrm{λ}}$ (i)

where,$V$is the average speed of the molecule,$\mathrm{λ}$is the mean free path of the gas molecule.

The average molecular speed of any gas, $\stackrel{¯}{v}=\sqrt{\frac{8RT}{\mathrm{π}M}}$(ii)

where, R is the gas constant, M is the molecular or molar mass of the gas, T is the temperature of the gas.

The mean free path of a gas,role="math" localid="1661918785796" $\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{d}^2\left(\frac{N}{V}\right)}$ (iii)

where, N is the number of molecules,is the diameter of the gas,V is the volume of the gas.

The ideal gas equation, $PV=NkT$(iv)

where,P is the pressure applied on the gas, V is the volume of the gas, n is the number of moles of the gas, k is the Boltzmann’s constant, T is the temperature of the gas.

Using the given data in equation (ii), the average speed of the gas can be given as follows:

$\begin{array}{c}\stackrel{¯}{v}=\sqrt{\frac{8(8.31{\text{m}}^3.\text{Pa/K.mol})\left(400\text{K}\right)}{\left(3.14\right)\left(0.032\text{kg/mol}\right)}}\\ =514.44\text{″¾}/\text{s}\end{array}$

Now, the mean free path of the gas can be given using equation (iv) in equation (iii) with the given data as follows:

$\begin{array}{c}\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{d}^2\left(\frac{P}{kT}\right)}\\ =\frac{kT}{\sqrt{2}\mathrm{π}{d}^{2}P}\\ =\frac{(1.38×{10}^{−23}{\text{m}}^2.{\text{kg/s}}^2\text{.K})\left(400\text{K}\right)}{\sqrt{2}\left(3.14\right){(290×{10}^{−12}\text{m})}^2(1.01×{10}^5\text{Pa})}\\ =1.46×{10}^{−7}\text{m}\end{array}$

Now, using the above calculated values, the frequency of the molecules for the given collision can be given using equation (i) as follows:

$\begin{array}{c}{Z}_{frequency}=\frac{514.44\text{m/s}}{1.46×{10}^{−7}\text{″¾}}\\ =3.5×{10}^9\text{collision}/\text{sec}\end{array}$

Hence, the value of frequency of collision is.$3.5×{10}^9\text{collision}/\text{sec}$