What is mean free path? Derive an expression for mean free path.

Answer

A gas is colliection of a large number of molecules, which are in continuous random motion. During their random motion, the molecules collide against one another. Between two collisions, a molecule moves along a straight line and the distance covered between the successive collisions is called the free path of the molecule. due to random nature of the motion of the gas molecules, thefree paths of a molecule during the collisions are bound to be of differnet lengths and zig-zag as shown in Fig. ''The mean free path of a molecule is defined as the avergae two succesive collisions and is measured as the mean of large number of free paths. Consider that a molecule `O` undegoes collisions at points `A,B,C,...` so that thefree paths of the molecule are of lengths `lambda_(1),lambda_(2),lambda_(3)........`, respectively. Then, mean free path of themolecule is given by
`barlambda=(lambda_(1)+lambda_(2)+lambda_(3)+...)/("total number of collisions")`
`(##RES_HAT_PHY_XI_C01_E01_035_A01##)`
Expression for mean free path: In order to derive the expanssion for mean free path of gas molecule, the following assumptions are made:
1. A gas molecule is perfectly spherical in shape.
2. All the gas molecules of the gas expect the one under considration are at rest. Consider a sample of gas, which constains n molecules per unit volume of the gas. Let d be the diameter of each gas molecule. Further consider a molecule `O` moving with the velocity v (fig.) The molecule `O` in motion wills trike with all those molecules, whose centree lie within a distance d from its path. Obviously, all such molecules with which the molecules `O` wills trike in time t, lie inside a cylinder of length `v t` and area of corss-section `pid^(2)` as shown in Fig. Thus, the number of collisions suffered by the molecule `O` in time `t`,
`(##RES_HAT_PHY_XI_C01_E01_035_A02##)`
`N =` volume of cylinder of length `vt` and area of cross-section
`pid^(2) xx` number of molecules per unit volume
`=(pid^(2) xx vt) xxn`
or `N = pid^(2) nvt`
Now, mean free path of a gas molecule,
`bar lambda = ("distance convered by the molecules in time t)/(number of collisions sufferend in time t)`
`bar lambda = (vt)/(pid^(2)nvt)`
`bar lambda = (1)/(pid^(2)n)`
The expression for mean free path has been obtained by assuming that all the gas molecules except the one under consideration, are at rest. In fact, all the gas molecules are in random motion, their velocities being governed by Maxwell's law of distribition of velocities. On taking the motion of all the gas molecules into consideration, the mean free path comes out to be:
`bar lambda = (1)/(sqrt(2)pid^(2)n)`
If `m` is the mass of each gas molecule, then density of the gas,
`rho = mn or n = (rho)/(m)`
In the equation, substituting for `n`, we have
`bar lambda = (m)/(sqrt(2)pid^(2) rho)`
From the equation, if follows that the mean free path of a gas molecule is:
(i) directly proportional to the mass of the gas molecule.
(ii) inversely proportional to the density of the gas
(iii) inversely proportional to the square of the molecule diameter.