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Mean free path of gas molecules

This type of question, asked 4 times: 2021 Paper 6, 13 Apr 2023, 5 Apr 2024, 28 Jan 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 6

    10
    Calculate the value of mean free path (λ)(\lambda) for oxygen molecules at temperature 27∘C27^{\circ}C and pressure 1.01×105 Pa1.01 \times10^{5}\text{ }Pa. Assume the molecular diameter is 0.3 nm0.3\text{ }nm and the gas is ideal. (k=1.38×10−23 J K−1)\left( k = 1.38 \times 10^{-23}\ J\,K^{-1} \right)

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  2. Question 2

    13 Apr 2023

    110
    The mean free path of molecules of a certain gas at STP is 1500d1500d, where dd is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K373\text{ }K is approximately:

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  3. Question 3

    5 Apr 2024

    96
    If nn is the number density and dd is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by :

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  4. Question 4

    28 Jan 2026 S2

    4
    The mean free path of a molecule of diameter 5×10−10 m5 \times 10^{- 10}\text{ }m at the temperature 41∘C41^{\circ}C and pressure 1.38×105 Pa1.38 \times 10^{5}\text{ }Pa, is given as ____\_\_\_\_ m. (Given kB=1.38×10−23 J/Kk_{B} = 1.38 \times 10^{- 23}\text{ }J/K).

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