Question

3. Calculate the collision frequency, binary collision number, mean free path, and viscosity for Ar gas at 298 K and 1.0 atm pressure. The collision diameter of Ar is 3.64 Å. Assume that Ar behaves ideally under these conditions.

Answer 1

Average speed = sqrt ( 8RT / pi x M )

so

MW of Ar = 39.948 g/mol = 39.948*10^-3 kg/mol

average speed of Argon = sqrt ( 8*8.314*298 / (3.1416**0.039)) =

average speed of nitrogen (V) = 402.2 m/s

now

collision frequency = sqrt(2) *pi *d^2 * V * * (N/V)

note that N/V must be calculated via gas aw

PV = NRT

n/V = P/(RT) = (1)/(0.082*298) = 0.04092 mol

1 mol = 6.022*10^23

0.04092 mol --> 0.04092*6.022*10^23 = 2.464202*10^22 mol/L ---> 2.464202*10^19 mol/m3

d = 3.64*10^-10 m

substitute in

collision frequency = sqrt(2) *3.1416 * ( 3.64*10^-10 m )^2 * (402.2 m/s ) (2.464202*10^19 mol/m3 )

collision frequency = 5834.2767 = 5.8*10^3 collision / s

binary collsion number = collison frequency x ( N/V ) / 2

binary collsion number = (5.8*10^3) (2.464202*10^19 ) / 2

binary collsion number = 7.14*10^22 1/m3s

For mean free Path (MFP)

mean free path = average velocity / collision frequency

mean free path = 402.2 /(5834.2767 )

= 0.06893 m3