Mean Free Path and Pressure of a Gas

Mean Free Path:
The mean free path of a molecule in a gas is the average distance traveled by the molecule between two successive collisions with other molecules. It is given by the equation:

Mean Free Path (λ) = (1/√2) * (1/(π * d^2 * N/V))

Where:
- λ is the mean free path
- d is the diameter of the molecule
- N is the Avogadro's number
- V is the volume of the gas

Given:
Temperature (T) = 25°C = 298 K
Mean Free Path (λ) = 2.63 × 10^-5 m
Radius of the molecule (r) = 2.56 × 10^-10 m

Calculating Diameter and Volume:
The diameter (d) of the molecule can be calculated from the radius (r) using the formula:

d = 2 * r

The volume (V) of the gas can be calculated from the ideal gas equation:

V = N * k * T / P

Where:
- k is the Boltzmann constant
- P is the pressure of the gas

Calculating Pressure:
We can rearrange the formula for mean free path to solve for pressure (P):

λ = (1/√2) * (1/(π * d^2 * N/V))
P = (1/λ) * (√2) * (π * d^2 * N/V)

Substituting the values and solving:

d = 2 * r = 2 * 2.56 × 10^-10 m = 5.12 × 10^-10 m

Using the value of the Boltzmann constant (k = 1.38 × 10^-23 J/K) and Avogadro's number (N = 6.022 × 10^23 mol^-1), we can calculate the volume (V) of the gas:

V = N * k * T / P
V = (6.022 × 10^23 mol^-1) * (1.38 × 10^-23 J/K) * (298 K) / P

Simplifying the equation:

P = (1/λ) * (√2) * (π * d^2 * N/V)
P = (1 / (2.63 × 10^-5 m)) * (√2) * (π * (5.12 × 10^-10 m)^2 * (6.022 × 10^23 mol^-1) / ((6.022 × 10^23 mol^-1) * (1.38 × 10^-23 J/K) * (298 K))

Evaluating the expression:

P ≈ 134 N/m^2

Therefore, the pressure of the gas is approximately 134 N/m^2.