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

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Decreases

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(A) The mean free path $\lambda$ of a gas molecule is given by the formula: $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the diameter of the molecule and $n$ is the number density (number of molecules per unit volume).From the formula,it is clear that the mean free path $\lambda$ is inversely proportional to the number density $n$ $(\lambda \propto \frac{1}{n})$.Therefore,if the number density $n$ increases,the mean free path $\lambda$ decreases because the frequency of collisions between molecules increases.

On increasing the number density of a gas in a vessel,the mean free path of the gas:

Similar Questions

What is free path? And what is mean free path?

The mean free path of electrons in a metal is $4 \times 10^{-8} \;m$. The electric field which can give on an average $2 \;eV$ energy to an electron in the metal will be in units of $V/m$.

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion,the average time of collision between molecules increases as $V^q$,where $V$ is the volume of the gas. The value of $q$ is $\left( \gamma = \frac{C_P}{C_V} \right)$.

If the diameter of a gas molecule is very small,what will be the mean free path?

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