Question

(a) Show that the number of collisions a molecule makes per second, called the collision frequency, f is given by f= v/lₘ and thus f=4√(2) πr²v N/V.

(b) What is the collision frequency for N₂ molecules in air at T=20 °C and P=1.0x10⁻² atm?

Answer 1

47149435.4884 /s

Step-by-step explanation:

`k` = Boltzmann constant =
`1.38* 10^(-23)\ J/K`

T = Temperature = 20 °C

P = Pressure =
`1* 10^(-2)\ atm`

Average time between collision is given by

`\Delta t=(l_m)/(v)`

Frequency is given by

`f=(1)/(\Delta t)`

Mean free path is given by

`l_m=(1)/(4\pi√(2)* r^2* (N)/(V))`

`\Delta t=((1)/(4\pi√(2)* r^2* (N)/(V)))/(v)`

`f=(1)/(((1)/(4\pi√(2)* r^2* (N)/(V)))/(v))\\\Rightarrow f=(v4\pi√(2)* r^2* N)/(V)`

Hence, proved.

Average velocity is given by

`v=(8kT)/(\pi m)`

`PV=NkT`

`f=\frac{4\pi√(2)* r^2* ((8kT)/(\pi m))^{(1)/(2)}* P}{kT}`

`f=16r^2P((\pi)/(kTm))^{(1)/(2)}\\\Rightarrow f=16(1.5* 10^(-10))^2* 1* 10^(-2)* 101325((\pi)/(1.38* 10^(-23)* (20+273.15)* 28* 1.66* 10^(-27)))^{(1)/(2)}\\\Rightarrow f=47149435.4884\ /s`

The collision frequency is 47149435.4884 /s