Question

The mean free path of neon is 6.4 x 10^-2 m at a pressure of 1.00 x 10^-4 atm and 298 K. A neon atom has a mass of 3.351 x 10^-26 kg. Calculate the collision cross section and the collision frequency of neon under these conditions.

Answer 1

Pressure , P = 1.00 X 10^-4 atm = 10.1325 Pa = 10.1325 Jm^-3

Temperature , T = 298 K

Step 1: to calculate number of atoms per unit volume(number density), n_d

Using Gas law : PV = nRT

PV = (N/N_A)RT , [ no. of moles, n = no. of atoms, N / avogadro's no., N_A]

R = 8.314 JK^-1mol^-1   N_A  = 6.022X10²³mol^-1

n_d = N/V = PN_A/RT

    n_d = (10.1325 Jm^-3)(6.022X10²³mol^-1) / (8.314 JK^-1mol^-1)(298K)

   n_d  = 2.4628 X 10²¹ m^-3

Step 2: To calculate the collision cross section ,

mean free path, = 6.4 x 10^-2m (given)

mean free path, = 1 / n_d

so,   = 1/ n_d

    = 1/(2.4628 X 10²¹ m^-3)(6.4X10^-2m) = 4.486 X 10^-21 m²

The collision cross section of neon = 4.486 X 10^-21 m² =

step 3: To calculate the mean velocity , <v>

P = (1/3) n_d × m× <v²> , m = mass of 1 neon atom , <v²> = mean square velocity

<v²> = (3 X P)/(n_d × m) = (3 X 10.1325 Jm^-3)/(2.4628 X 10²¹ m^-3)(3.351x10^-26kg)

<v²> = 3.6833X10⁵m²s^-2

<v> = 6.069 X 10²m/s

Step 4: to calculate collision frequency , z

z =    <v> n_d

z =   (4.486 X 10^-21 m²)( 6.069 X 10²m/s)(2.4628 X 10²¹ m^-3)

z = 9482.4 s^-1

the collision frequency of neon = 9482.4 s^-1