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A molecule is travelling in air at $300 \mathrm{~K}$ and $1 \mathrm{~atm}$, and the radius of the molecule is $0.6 \times 10^{-10} \mathrm{~m}$. Calculate the approx. mean free path of the molecule. (The number density is $2.44 \times 10^{25}$ molecules $/ \mathrm{m}^3$ )
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Verified Answer
The correct answer is:
$\frac{0.2}{\pi} \times 10^{-5} \mathrm{~m}$
Given that, temperature, $T=300 \mathrm{~K}$
Pressure, $p=1$ atm $=1.01 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
Radius, $r=0.6 \times 10^{-10} \mathrm{~m}$, diameter $d=1.2 \times 10^{-10} \mathrm{~m}$
Number density, $\rho=2.44 \times 10^{25}$ molecules $/ \mathrm{m}^3$
Boltzmann constant, $K=1.38 \times 10^{-23} \mathrm{JK}^{-1}$
Now, mean free path, $\lambda=\frac{K T}{\sqrt{2} \pi d^2 p}$
$$
\begin{aligned}
\lambda & =\frac{1.38 \times 10^{-23} \times 300}{1.414 \pi \times\left(1.2 \times 10^{-10}\right)^2 \times 1.01 \times 10^5} \\
& =\frac{0.2}{\pi} \times 10^{-5} \mathrm{~m}
\end{aligned}
$$
Pressure, $p=1$ atm $=1.01 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
Radius, $r=0.6 \times 10^{-10} \mathrm{~m}$, diameter $d=1.2 \times 10^{-10} \mathrm{~m}$
Number density, $\rho=2.44 \times 10^{25}$ molecules $/ \mathrm{m}^3$
Boltzmann constant, $K=1.38 \times 10^{-23} \mathrm{JK}^{-1}$
Now, mean free path, $\lambda=\frac{K T}{\sqrt{2} \pi d^2 p}$
$$
\begin{aligned}
\lambda & =\frac{1.38 \times 10^{-23} \times 300}{1.414 \pi \times\left(1.2 \times 10^{-10}\right)^2 \times 1.01 \times 10^5} \\
& =\frac{0.2}{\pi} \times 10^{-5} \mathrm{~m}
\end{aligned}
$$
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