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Kinetic Theory - Mean Free Path

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition: The mean free path is the average distance traveled by a gas molecule between two successive collisions. It is denoted by the symbol λ\lambda.

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Molecular Motion: According to the kinetic theory of gases, molecules move randomly and undergo elastic collisions. Between two collisions, a molecule moves in a straight line with constant velocity.

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Dependence on Number Density (nn): The mean free path is inversely proportional to the number of molecules per unit volume (n=N/Vn = N/V). If the density increases, the probability of collisions increases, thus decreasing λ\lambda.

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Dependence on Molecular Size (dd): λ\lambda is inversely proportional to the square of the diameter (dd) of the gas molecule. A larger molecular size increases the collision cross-section πd2\pi d^2.

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Dependence on Pressure (PP) and Temperature (TT): Using the ideal gas relation P=nkBTP = n k_B T, the mean free path can be expressed as λ=kBT2πd2P\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}. This shows that λ∝T\lambda \propto T at constant pressure and λ∝1P\lambda \propto \frac{1}{P} at constant temperature.

📐Formulae

λ=12πnd2\lambda = \frac{1}{\sqrt{2} \pi n d^2}

λ=kBT2πd2P\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}

n=NV=PkBTn = \frac{N}{V} = \frac{P}{k_B T}

Collision Frequency (f)=vˉλ\text{Collision Frequency } (f) = \frac{\bar{v}}{\lambda}

💡Examples

Problem 1:

Calculate the mean free path of a gas molecule with a diameter of 3×10−10 m3 \times 10^{-10} \text{ m} at a number density of 2×1025 m−32 \times 10^{25} \text{ m}^{-3}.

Solution:

Given: d=3×10−10 md = 3 \times 10^{-10} \text{ m} and n=2×1025 m−3n = 2 \times 10^{25} \text{ m}^{-3}. \n Using the formula: λ=12πnd2\lambda = \frac{1}{\sqrt{2} \pi n d^2} \n λ=11.414×3.1415×(2×1025)×(3×10−10)2\lambda = \frac{1}{1.414 \times 3.1415 \times (2 \times 10^{25}) \times (3 \times 10^{-10})^2} \n λ=11.414×3.1415×2×1025×9×10−20\lambda = \frac{1}{1.414 \times 3.1415 \times 2 \times 10^{25} \times 9 \times 10^{-20}} \n λ=179.97×105≈1.25×10−7 m\lambda = \frac{1}{79.97 \times 10^5} \approx 1.25 \times 10^{-7} \text{ m}.

Explanation:

The mean free path is calculated by substituting the molecular diameter and the number density into the standard Maxwellian distribution formula for mean free path.

Problem 2:

How does the mean free path λ\lambda change if the absolute temperature TT of a gas is doubled and the pressure PP is quadrupled?

Solution:

The formula for mean free path in terms of TT and PP is λ∝TP\lambda \propto \frac{T}{P}. \n Let the initial mean free path be λ1=CT1P1\lambda_1 = C \frac{T_1}{P_1}. \n New temperature T2=2T1T_2 = 2T_1 and new pressure P2=4P1P_2 = 4P_1. \n New mean free path λ2=CT2P2=C2T14P1=12(CT1P1)=λ12\lambda_2 = C \frac{T_2}{P_2} = C \frac{2T_1}{4P_1} = \frac{1}{2} \left( C \frac{T_1}{P_1} \right) = \frac{\lambda_1}{2}.

Explanation:

Since λ\lambda is directly proportional to temperature and inversely proportional to pressure, doubling the temperature increases it by a factor of 22, but quadrupling the pressure decreases it by a factor of 44, resulting in a net decrease by half.

Mean Free Path - Revision Notes & Key Formulas | CBSE Class 11 Physics