The mean free path of a molecule of diameter | vedclass

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Question

(C) The formula for the mean free path $\lambda$ is given by $\lambda = \frac{k_{B}T}{\sqrt{2}\pi\sigma^{2}P}$.Given values are:$k_{B} = 1.38 	imes 1

Vedclass · Accepted Answer

$2\sqrt{2}	imes10^{-8}$

Vedclass · Answer

(C) The formula for the mean free path $\lambda$ is given by $\lambda = \frac{k_{B}T}{\sqrt{2}\pi\sigma^{2}P}$.Given values are:$k_{B} = 1.38 	imes 10^{-23} \ J/K$$T = 41^{\circ}C = 41 + 273 = 314 \ K$$P = 1.38 	imes 10^{5} \ Pa$$\sigma = 5 	imes 10^{-10} \ m$Substituting these values into the formula:$\lambda = \frac{1.38 	imes 10^{-23} 	imes 314}{\sqrt{2} 	imes 3.14 	imes (5 	imes 10^{-10})^{2} 	imes 1.38 	imes 10^{5}}$$\lambda = \frac{1.38 	imes 10^{-23} 	imes 314}{\sqrt{2} 	imes 3.14 	imes 25 	imes 10^{-20} 	imes 1.38 	imes 10^{5}}$$\lambda = \frac{314 	imes 10^{-23}}{\sqrt{2} 	imes 3.14 	imes 25 	imes 10^{-15}}$$\lambda = \frac{100 	imes 10^{-23}}{\sqrt{2} 	imes 25 	imes 10^{-15}} = \frac{4 	imes 10^{-8}}{\sqrt{2}} = 2\sqrt{2} 	imes 10^{-8} \ m$.

The mean free path of a molecule of diameter $5\times10^{-10} \ m$ at the temperature $41^{\circ}C$ and pressure $1.38\times10^{5} \ Pa$ is given as . . . . . . $m$. (Given $k_{B}=1.38\times10^{-23} \ J/K$).

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