The mean free path of a molecule of diameter 5 x 10-10 m at the temperature 41 and pressure 1.38 x 105 Pa, is given as ____ m. (Given~kB = 1.38x 10-23 J / K).

where, lambda = mean free path kB = Boltzmann constant T = absolute temperature in Kelvin sigma = diameter of the molecule P = pressure of the gas d = sigma = 5 x 10-10 m T = 41^ = 273 + 41 = 314 K P = 1.38 x 105 Pa kB = 1.38 x 10-23 J/K Cancel out 1.38 from numerator and denominator: The temperature 314 K and pi approx 3.14 are intentionally matched to cancel out and give a factor of 100. 1.38 cancels completely. The rest is simply handling powers of 10. Class 11 Physics: Kinetic Theory

The mean free path of a molecule of diameter 5 x 10-10 m at the temperature 41 and pressure 1.38 x 105 Pa, is given as ____ m. (Given~kB = 1.38x 10-23 J / K).

where, lambda = mean free path kB = Boltzmann constant T = absolute temperature in Kelvin sigma = diameter of the molecule P = pressure of the gas d = sigma = 5 x 10-10 m T = 41^ = 273 + 41 = 314 K P = 1.38 x 105 Pa kB = 1.38 x 10-23 J/K Cancel out 1.38 from numerator and denominator: The temperature 314 K and pi approx 3.14 are intentionally matched to cancel out and give a factor of 100. 1.38 cancels completely. The rest is simply handling powers of 10. Class 11 Physics: Kinetic Theory

The mean free path of a molecule of diameter 5 x 10-10 m at the temperature 41 and pressure 1.38 x 105 Pa, is given as ____ m. (Given~kB = 1.38x 10-23 J / K).

where, lambda = mean free path kB = Boltzmann constant T = absolute temperature in Kelvin sigma = diameter of the molecule P = pressure of the gas d = sigma = 5 x 10-10 m T = 41^ = 273 + 41 = 314 K P = 1.38 x 105 Pa kB = 1.38 x 10-23 J/K Cancel out 1.38 from numerator and denominator: The temperature 314 K and pi approx 3.14 are intentionally matched to cancel out and give a factor of 100. 1.38 cancels completely. The rest is simply handling powers of 10. Class 11 Physics: Kinetic Theory

The mean free path of a molecule of diameter 5 x 10-10 m at the temperature 41 and pressure 1.38 x 105 Pa, is given as ____ m. (Given~kB = 1.38x 10-23 J / K).

where, lambda = mean free path kB = Boltzmann constant T = absolute temperature in Kelvin sigma = diameter of the molecule P = pressure of the gas d = sigma = 5 x 10-10 m T = 41^ = 273 + 41 = 314 K P = 1.38 x 105 Pa kB = 1.38 x 10-23 J/K Cancel out 1.38 from numerator and denominator: The temperature 314 K and pi approx 3.14 are intentionally matched to cancel out and give a factor of 100. 1.38 cancels completely. The rest is simply handling powers of 10. Class 11 Physics: Kinetic Theory

The mean free path of a molecule of diameter $5 \times 10⁻¹⁰ m$ $5 \times 10^{-10} \text{ m}$ at the temperature $41°C$ $41^{\circ}\text{C}$ and pressure $1.38 \times 10⁵ Pa$ $1.38 \times 10^{5} \text{ Pa}$, is given as ____ m. $( Given~kB = 1.38\times 10⁻²³ J / K).$

A. $2\sqrt(2) \times 10⁻¹⁰$ $2\sqrt{2} \times 10^{-10}$
B. $10\sqrt(2) \times 10⁻⁸$ $10\sqrt{2} \times 10^{-8}$
C. $2\sqrt(2) \times 10⁻⁸$ $2\sqrt{2} \times 10^{-8}$
D. $2 \times 10⁻⁸$ $2 \times 10^{-8}$

Solution & Explanation

Related Formula

λ = kB T\sqrt(2) π σ² P

where, $λ$ $\lambda$ = mean free path $kB$ $k_B$ = Boltzmann constant $T$ $T$ = absolute temperature in Kelvin $σ$ $\sigma$ = diameter of the molecule $P$ $P$ = pressure of the gas

Step 1: Extract and Convert Given Data

$d = σ = 5 \times 10⁻¹⁰ m$ $T = 41^ = 273 + 41 = 314 K$ $P = 1.38 \times 10⁵ Pa$ $P = 1.38 \times 10^5 \text{ Pa}$ $kB = 1.38 \times 10⁻²³ J/K$

Step 2: Substitution and Calculation

λ = 1.38 \times 10⁻²³ \times 314\sqrt(2) \times π \times (5 \times 10⁻¹⁰)² \times 1.38 \times 10⁵

Cancel out $1.38$ $1.38$ from numerator and denominator:

λ = 10⁻²³ \times 314\sqrt(2) \times 3.14 \times 25 \times 10⁻²⁰ \times 10⁵

λ = 314 \times 10⁻²³\sqrt(2) \times 3.14 \times 25 \times 10⁻¹⁵

λ = 100 \times 10⁻²³\sqrt(2) \times 25 \times 10⁻¹⁵

λ = 4 \times 10⁻⁸\sqrt(2) = 2\sqrt(2) \times 10⁻⁸ m

Pattern Recognition

The temperature $314 K$ $314 \text{ K}$ and $π \approx 3.14$ $\pi \approx 3.14$ are intentionally matched to cancel out and give a factor of 100. $1.38$ $1.38$ cancels completely. The rest is simply handling powers of 10.

Chapter Mix

Class 11 Physics: Kinetic Theory

Reference Study Guides

Kinetic Theory

The mean free path of a molecule of diameter $5 \times 10⁻¹⁰ m$ $5 \times 10^{-10} \text{ m}$ at the temperature $41°C$ $41^{\circ}\text{C}$ and pressure $1.38 \times 10⁵ Pa$ $1.38 \times 10^{5} \text{ Pa}$, is given as ____ m. $( Given~kB = 1.38\times 10⁻²³ J / K).$

Solution & Explanation

Related Formula

Step 1: Extract and Convert Given Data

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Kinetic Theory Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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Solution & Explanation

Related Formula

Step 1: Extract and Convert Given Data

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Kinetic Theory Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Kinetic Theory Questions — jee_main_2026_28_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

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