The mean free path of molecules of a gas (rad | vedclass

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(B) The mean free path $\lambda$ of a gas molecule is given by the formula:$\lambda = \frac{1}{\sqrt{2} n \pi d^2}$where $n$ is the number density of

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$r^2$

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(B) The mean free path $\lambda$ of a gas molecule is given by the formula:$\lambda = \frac{1}{\sqrt{2} n \pi d^2}$where $n$ is the number density of molecules and $d$ is the diameter of the molecule.Since the radius of the molecule is $r$,the diameter is $d = 2r$.Substituting this into the formula:$\lambda = \frac{1}{\sqrt{2} n \pi (2r)^2} = \frac{1}{4\sqrt{2} n \pi r^2}$From this expression,it is clear that $\lambda \propto \frac{1}{r^2}$.Therefore,the mean free path is inversely proportional to $r^2$.

The mean free path of molecules of a gas (radius $r$) is inversely proportional to

Similar Questions

The mean free path of gas molecules depends on $(d =$ molecular diameter$)$

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What is mean free path?

Statement-$1$: Real gas approaches ideal gas behaviour for low pressures and high temperatures.
Statement-$2$: At low pressure,the density of a gas is very low.

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