What would the pressure 𝑃 of an ideal gas be if the mean free path was 142 cm? Assume that the gas is at room temperature, 𝑇=22.6 ∘C, and that the diameter of the molecule is 𝑑=1.25×10^-10 m.

Question

What would the pressure 𝑃 of an ideal gas be if the mean free path was 142 cm? Assume that the gas is at room temperature, 𝑇=22.6 ∘C, and that the diameter of the molecule is 𝑑=1.25×10^-10 m.

Answer 1

Take k below as Boltzmann's Constant equal to 1.38065812E-23 Joules Per Kelvin.

Take T below as (22.6 + 273.15) or 295.75 Kelvin.

Take b below as (1.25E-10 ÷ 2) or 6.25E-11 Meter.

Take l below as 1.42 Meters.

The mean free path of a gas molecule or "l" is the average distance traveled by that molecule between collisions with other molecules.  

For an ideal gas of spherical molecules with radius b, l is equal to 1/[4π(2^0.5)b²(N/V)] with N/V (also known as n_u or number density) the number of molecules per unit volume.

To consider the pressure, write l = 1/[4π(2^0.5)b²n_u] with n_u in place of (N/V). Guide on 

PV = nRT or P = (n/V)RT and note that (n/V) equals (n_u/N_A) and that R/N_A gives k.

Then construct P = (n_u/N_A)RT equal to (n_u/N_A)(N_Ak)T or n_ukT. Then n_u = P/kT.

Next, l = 1/[4π(2^0.5)b²(P/kT)] enables:

P = kT/[4π(2^0.5)b²l].

Then P here is [(1.38065812E-23 J/K)(295.75 K)] divided by 

[4π(2^0.5)(6.25E-11 M)²(1.42 M)] which goes to 0.04142262945

Newton-Dot-Meter-Per-Cubic-Meter or 0.04142262945 Pascal.