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Kinetic Theory

JEE Main level, single correct, medium

51 of 168

Question 51 of 168

Calculate the value of the mean free path $\left(\lambda \right)$ for oxygen molecules at temperature $27^{\circ}C$ and pressure $1.01\times {10}^{5}Pa$ . Assume the molecular diameter $0.3nm$ and the gas is ideal. $\left(k=1.38\times {10}^{-23}J{K}^{-1}\right)$

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Answer: (D) $102nm$

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

\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}}

\lambda =102nm

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Kinetic Theory (Speed of Gas), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.