Calculate the value of mean free path (lambda | vedclass

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(D) The formula for the mean free path $(\lambda)$ is given by: $\lambda = \frac{k T}{\sqrt{2} \pi d^2 P}$.Given values are:Temperature $T = 27^{\circ

Vedclass · Accepted Answer

$102$

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(D) The formula for the mean free path $(\lambda)$ is given by: $\lambda = \frac{k T}{\sqrt{2} \pi d^2 P}$.Given values are:Temperature $T = 27^{\circ} C = 27 + 273 = 300 K$.Pressure $P = 1.01 	imes 10^{5} Pa$.Diameter $d = 0.3 nm = 0.3 	imes 10^{-9} m$.Boltzmann constant $k = 1.38 	imes 10^{-23} J K^{-1}$.Substituting these values into the formula:$\lambda = \frac{1.38 	imes 10^{-23} 	imes 300}{\sqrt{2} 	imes 3.14 	imes (0.3 	imes 10^{-9})^2 	imes 1.01 	imes 10^{5}}$.$\lambda = \frac{4.14 	imes 10^{-21}}{1.414 	imes 3.14 	imes 0.09 	imes 10^{-18} 	imes 1.01 	imes 10^{5}}$.$\lambda = \frac{4.14 	imes 10^{-21}}{0.403 	imes 10^{-13}} \approx 1.027 	imes 10^{-7} m$.Converting to $nm$: $\lambda \approx 102.7 nm \approx 102 nm$.

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

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