Question

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given that the collision cross-section for the molecules of that gas is 2.0 × 10-20 m2? Boltzmann's constant is k = 1.38 × 10-23 J/K.

Answer 1

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

Given:

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

To determine:

The mean free path, λ

Calculation:

The mean free path is related to the collision cross section by the following equation:

$\lambda =\frac{1}{n\sigma }------(1)$

where n = number density

$n = \frac{P}{kT}-----(2)$

Substituting for P, k and T in equation (2) gives:

$n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\ m^{3}$

Next, substituting for n and σ in equation (1) gives:

$\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m$