Question

an oxygen .molecule is traveling in air at 300 k and 1 atm and the diameter of oxygen molecules is 1.2×10^-10m.calculate the mean free path of oxygen molecule?​

Answer 1

Mean free path will be $76.93\times 10^{-13}m$

Explanation:

We have given temperature T = 300 K

Pressure P = 1 atm

Diameter of oxygen molecule $d=1.2\times 10^{-10}m$

Boltzmann constant $k=1.38\times 10^{-23}$

Formula of mean free path is $\lambda =\frac{kT}{\sqrt{2}\pi d^2P}$

So mean free path $\lambda =\frac{1.38\times 10^{-23}\times 300}{1.414\times 3.14\times 1.2\times 10^{-10}\times 1.01\times 10^5}=76.93\times 10^{-13}m$

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The mean free path of benzene

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Answer 2

Answer:

The given answer is correct and solution also given