Question

At what pressure does the mean free path of Argon at 20 oC become comparable to 10 times the diameters of the molecules themselves. Assume the collision cross-section (σ) of Argon is 0.36 nm2 . Hint: The mean free path of Argon is comparable to 10 times of its diameter basically means that = 10, where is the diameter of argon gas molecules. To calculate , you will need the definition of the collision cross section area.

Answer 1

Using the definition of mean free path and collision cross-section, we calculate the pressure required for the mean free path of Argon to be 10 times the molecule's diameter by setting up an equation based on the kinetic theory of gases.

Step-by-step explanation:

To find at what pressure does the mean free path of Argon at 20 °C become comparable to 10 times the diameter of the molecules, we must use the definition of mean free path (λ) and the collision cross-section (σ) provided. If λ = 10d, where d is the diameter of an argon atom, and knowing that the collision cross-section (σ) is related to the diameter by the formula σ = πd2, we can set up an equation to solve for the pressure required using the formula derived from kinetic theory of gases for the mean free path, λ = kT / (√2 σ P), where k is the Boltzmann constant, T is the temperature in kelvins, and P is the pressure.

Given that the collision cross-section (σ) of Argon is 0.36 nm2 (or 3.6 x 10-19 m2), we can calculate the diameter (d = √(σ / π)) of an argon atom. With both the value of σ and the temperature (converted to kelvins), we can solve for the pressure (P) that gets us a mean free path 10 times the diameter of the molecule.