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JEE Main 202116 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2021 · 16 March · Evening Shift · Physics · Kinetic Theory of Gases

Calculate the value of the mean free path (λ) for oxygen molecules...

Calculate the value of the mean free path (λ)\left(\lambda\right) for oxygen molecules at temperature 27°C27 ° C and pressure 1.01×105Pa1 . 01 \times {10}^{5}Pa. Assume the molecular diameter 0.3nm0 . 3nm and the gas is ideal. (k=1.38×10−23JK−1)\left(k = 1 . 38 \times {10}^{- 23}J{K}^{- 1}\right)

Options

  1. A58nm58nm
  2. B32nm32nm
  3. C86nm86nm
  4. D102nm102nm

Correct answer

D. 102nm102nm

Step-by-step solution

λ=RT2πd2NAP\lambda = \dfrac{RT}{\sqrt{2}\pi{d}^{2}{N}_{A}P}

λ=8.314×3002×3.14×(0.3×10−9)2×6.023×1023×1.01×105\lambda = \dfrac{8 . 314 \times 300}{\sqrt{2} \times 3 . 14 \times {\left(0 . 3 \times {10}^{- 9}\right)}^{2} \times 6 . 023 \times {10}^{23} \times 1 . 01 \times {10}^{5}}

λ=102nm\lambda = 102nm

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