**Collision Diameter, Collision Frequency, Collision Number and Mean Free Path**

In the kinetic molecular theory of gases, we come to know that the molecules of a gas aremoving randomly, colliding among themselves and with the walls of the vessel. During the collisions, they suffer a change in their directions and also change their velocities. The time for which the two molecules are in contact at the time of collision is called compression time.

Since, we are going to discuss the frequency of the collisions and the free path in between the collisions, so first of all we should know about the nature of the collisions.

The collisions are of three types.

**(i) Grazing Collision or Glancing Collision**

“In these collisions, the molecules are moving just parallel to each other, with the average velocity and their outer boundaries touch each other.”

**(ii) Head on Collision**

“When two molecules approach each other on a straight line, then they collide head to head and the collision is head on.” The approaching molecules retrace the straight line path in the reverse direction. The relative speed becomes 2.

**(iii) Right Angled Collision**

When two molecules approach each other and their approaching lines are approximately 90 to each other. Then the collision is right angled. The relative speed is.

**Elastic Collision**

During the collisions there happens a change in the direction of the motion but total energy remains the same. The collisions in which there is no net loss or gain of energy are called elastic collisions.

**Collision Diameter**

In order to do the collision, the molecules approach each other. At the time of contact of the outer boundaries, there is a limit beyond which they cannot come close to each other. This is called the distance of closest approach. “The closest distance between the centers of two molecules taking place during collision is called collision diameter.

**Collision Frequency (Z _{I})**

“If we follow a single molecule in one cm^{3} of the gas, and want to observe that, how many collisions are being faced by this molecule in one second, then it will be collision frequency of that molecule.” It is denoted by Z_{I}. Its value should depend upon the velocity of gas molecules, sizes of molecules and closeness of the molecules in the vessel.

**Mean Free Path (λ)**

When a molecule travels and collides with various molecules, then it travels free path in the vessel. Anyhow, all the free paths of the molecules arc not equal. So, if we take the average ofall these free paths, then we get the mean free path. “Hence, mean free path is average distance covered by a molecule between two successive collisions.” It is denoted by ‘λ’.

Mathematically, ‘λ’is related to the mean distance travelled by the molecule in one second and its number of collisions per second.

‘λ =

‘λ = (1)

**Collision Number (Z _{II})**

It is the number of collisions happening in all the molecules in 1 cm^{3} of the gas in one second. It is denoted by ‘Z_{II}‘. Mathematically, we can say that,

Z_{II} = (2)

‘nZ_{I}‘, is divided by 2, so that each collision may not be counted twice.

**Derivation for the Expression of Collision Frequency (Z _{I})**

Consider a molecule ‘A’ which is travelling in the center of imaginary cylinder from left the right. This imaginary cylinder is supposed to be present in the vessel of a gas. Diameter of this cylinder is ‘2’. The diameter of the molecule itself is”. It means that two molecules can travel in this imaginary cylinder simultaneously. The average velocity of the molecule travelling in the center of the cylinder is ” ms^{-1}. If the length of the imaginary cylinder is supposed to be”meters, then it means that the molecule ‘A’ will approach the other end of the cylinder in one second.

There are many molecules present in this imaginary cylinder. The categories of molecules are ‘B’, ‘C’, ‘D’ and ‘E’. The molecule ‘A’ can collide with ‘B’ type molecules doing head-on collisions, with ‘C’ type, right-angled collisions with ‘D’ grazing collisions and therewill be no collision with molecules of the type ‘E’. Keep it in mind, that there are many molecules of each category in this cylinder.

Now, the question arises, that how many collisions this molecule ‘A’ will face while moving from one end of the cylinder to the other end.

Its answer is that number of collisions depend upon the number of molecules in this imaginary cylinder.

How to count the number of molecules? It is very easy.

Let the radius of imaginary cylinder is =

Base area of this imaginary cylinder =

Length of the cylinder in meters =

Volume of the imaginary cylinder =

Let the number of molecules in one cm^{3}=n

Number of molecules in imaginary cylinder= n

So, the number of collisions ‘which the molecule ‘A’will experience in one second should be

n

Be careful, that the number of right-angled collision is much greater than grazing collisions and head-on collisions. So, the exact number of such collisions should be n.We have multiplied with ”, because the relative speed of two molecules approaching each other to do right-angled collision, is” and not. It means that considering the right-angled collisions, the length of the imaginary cylinder is proposed to be ”

Hence, Z_{I} = n (3)

This is the collision frequency of the molecule.

**Formula of Mean Free Path (****λ****)**

As we have previously explained that mean free path is the ratio of and Z_{I}

‘λ = (4)

Putting value of Z_{I} from equation (3) into equation (4)

λ =

λ = cm collision (5)

According to the equation (5), the mean free path of the gas molecule is inversely proportional to the square of the collision diameter and inversely proportional to the number of molecules per unit volume. It means, greater the number of molecules per unit volume, smaller the distance the molecule has to the cover between successive collisions.

**Formula of Collision Number (Z _{II})**

Collision number is the number of collisions happening in all the molecules in one second in 1 cm^{3} of the gas.

We know that number of collisions of a single molecule, called the collision frequency is Z_{I}. If we multiply ‘Z_{I}‘ with ‘n’ and divided with 2, then we get the total number of collisions in one second in one cm^{3}.

Z_{II} =

We have divided it with two, because each collision involves two molecules and collision is not the property of a single molecule. The division with two, is to be done so that each collision not be counted twice.

Z_{II} = =

Z_{II}= (6)

According to the equation (6), the collision number depends upon the collision diameter, number of collisions cm^{-3} and the averagevelocity.

The expression for average velocity is

=

It means that, average velocity depends upon the temperature and the molar mass indirectly, we can say that the collision number Z_{II} is,

(i) Directly proportional to square root of temperature.

(ii) Inversely proportional to square root of molar mass.

(iii) Directly proportional to square of collision diameter.

(iv) Directly proportional to square of number of molecules per unit volume.

**Example (4) **

Oxygen is maintained at 1 atm. pressure and 25C. Calculate:**(i)** Number of collisions s^{-1} molecule^{-1}, **(ii)** Number of collisions s^{-1} m^{-3}. The collision diameter of oxygen molecule is 3.60 x 10^{-10}m.

**Solution**

First of all we calculate average velocity,

**Data**

Molar mass of O^{2} = 32 x 10^{-3} kg mol^{-1}

Temperature= 25C +273 = 298K

R = 8.3143 JK^{-1}mol^{-1}

=

By putting values

=

= 4.44 x 10^{2}ms^{-1}

Number of molecules = Number of moles x Avogadro’s number

= n x N_{A}= x N_{A}Since n =

V = 1 dm^{3}

Putting values

Number of molecules = =0.246 10^{23}

= 2.46 x 10^{25}m^{-3} = 2.46 x 10^{22} dm^{-3}

= 2.46 x 10^{25} m^{-3}

The collision frequency is the number of collisions per second.

Since 1 dm^{3} = 10^{-3} m^{3}

(i) Z_{I}=

Z_{I} = 1.414 x 3.14 x (3.60 x 10^{-10} m^{2})

x (4.44 x 10^{2} ms^{-1}) x (2.64 x 10^{25} m^{-3})

Z_{I} = 6.74 X 10^{9} s^{-1}Ans.

The collision number is the number of collision per second per m^{3}

Z_{II }=

Z_{II}=

Z_{II} =7.73 x 10^{34} collisions s^{-1} m^{-3}Ans.