Question

Please answer all questions.

a) With the use of the appropriate equations and diagram, develop a relationship between the collision diameter and the collision frequency. (10) b) What is the mean free path of a gas, show it is derived from the collision frequency? (5) c) Given that the collision cross section of a gas is 0.43nm2, what is the mean free path when the temperature is 217K and pressure is 0.0Satm? (10)

Answer 1

a) Collision frequency is defined as the average rate at which two reactants collide for a given system.This is generally used to denote the average number of collisions per unit time, in a well defined system.Now, in order to establish the relation between collision frequency and collision diameter as well as other factors, we need to undertake certain assumptions. They are as follows:

i)We treat the molecules as hard spheres of certain diameter(=d₁, say).

ii)The molecules are assumed to travel through straight lines in free space.

iii)For two molecules to collide,their centers must come together to within a distance 'd' of each other.

iv) The two molecules don't interact with each other,except for the collision

no collision position att 0 r* position att t collision rel

So the above cylinder represents the cylindrical path swept by the molecules in a time t.Where the collision cross-section is given by (pi)d² .

This cylinder is also known as the collisional cylinder

The basis of the equation is given by

f_i = (Volume of the collisional cylinder) * (Density) / (Time)

= (root 2) pi * d² * <c> * $\Delta$t * (N/V) / $\Delta$t .................(1)

= (root 2) pi * d² * <c> * (N/V) , Where <c> = mean speed of the molecules = root(8 * K_B * T / pi * m )

Hence, collision frequency depends upon two factors :

i) Temperature : Increase in temperature leads to an increase in the collision frequency

ii) Size(diameter) of the reactants : Increasing the size of the molecules renders an increase in the collision frequency.

iii) Density : Upon increasing density, the frequency does increase, since the number of molecules per unit volume also increases.

b)Mean free path is defined as the average distance a particle would cover, before getting into or experiencing a collision.Now we would derive it from the average collision frequency

Since the average collision frequency is defined to be the no of collisions per unit time, we can write it as : N_{total collisions} /$\Delta$ t = A_i * N₀ * l / $\Delta$ t = A_i * N₀ * <c> ...................(2)

Where, A_i = Individual cross section of the particles

N₀ = Number density of the system and

l = distance moved by the particle in time $\Delta$ t

<c> = mean velocity of the particles

And since, Mean free path ($\lambda$) is equal to the product of the speed of the particles and time between the collisions, $\Rightarrow$$\lambda$ = <c> * root(2)/ A_i * N₀ * <c> = <c> / root(2) * f_i , .........................(3)

Where f_i = average collision frequency for the molecules.

c) From the above equation 3 ,

We can write $\lambda$ = [1/ root(2)] * 1/  A_i * N₀

And since Number density N₀ = N/V = P/kT = 0.05 * 10⁵ Pa / (1.38 * 10^-23 J/K* 217 K ) =  and  A_i = 0.43 nm² (as given) = 0.43 * 10^-18 m²

We get $\lambda$ = (1/1.414) * 5000 / ( 16.7 * 10²³ * 0.43 * 10^-18 ) = 696.28 * 10^-5 /1.414 = 4.92 mm