What is the mean free path in m of a gas of randomly-moving hard spheres of radius 0.5 × 10-9 m when the number density of spheres is 4.3 × 1019 per cubic meter?
Question
What is the mean free path in m of a gas of randomly-moving hard spheres of radius 0.5 × 10-9 m when the number density of spheres is 4.3 × 1019 per cubic meter?
Solution 1
The mean free path (λ) of a gas can be calculated using the formula:
λ = 1 / ( √2 * π * d² * n)
where:
- d is the diameter of the gas molecules, which is twice the radius. So, d = 2 * 0.5 × 10^-9 m = 1 × 10^-9 m.
- n is the number density of the gas molecules, which is given as 4.3 × 10^19 per cubic meter.
Substituting these values into the formula, we get:
λ = 1 / ( √2 * π * (1 × 10^-9 m)² * 4.3 × 10^19 m^-3)
Solving this will give you the mean free path of the gas in meters.
Solution 2
The mean free path (λ) of a gas can be calculated using the formula:
λ = 1 / ( √2 * π * d² * n)
where:
- d is the diameter of the gas molecules, which is twice the radius. So, d = 2 * 0.5 × 10^-9 m = 1 × 10^-9 m.
- n is the number density of the gas molecules, which is given as 4.3 × 10^19 per cubic meter.
Substituting these values into the formula, we get:
λ = 1 / ( √2 * π * (1 × 10^-9 m)² * 4.3 × 10^19 m^-3)
Solving this equation will give you the mean free path of the gas in meters.
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