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What is mean free path? Derive an expression for mean free path. |
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Answer» Solution :Let us CONSIDER a system of molecules each with diameter d. Let n be the number of molecules per unit volume . It is assumed that only one molecule is in motion and all other are at rest as shown in the figure.  `implies` Mean free path If a molecule moves with average speed v in a timet , the distance travelled is vt. In this time t, let us consider the molecule to move in an imaginary cylinder of volume `pid^2vt`. It collides with any molecule whose centre is within this cylinder . Hence , the number ofcollisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equation of `pid^2vtn.` The total path length DIVIDED by the number of collision in time t is the mean free path . Mean free path, `lamda = ("distance travelled")/("Number of collisions")` `lamda=(vt)/(npid^2vt) =1/(npid^2) "" ....(1)` It is assumed that only one molecule is MOVING at a time and other molecules are at rest. But in actual practice all the molecules are in random motion . Hence the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct. expression for mean free path, `lamda=1/(sqrt2npid^2)""..(2)` It is implied from the equation that the mean free path is inversely PROPORTIONAL to number density. When the number density increases the molecular collisions increases. Hence it decreases the distance travelled bu the molecule before collisions. Rearranging the equation (2) using 'm' (mass of the molecule) `:. lamda=m/(sqrt2pid^2mn)` But mn= mass per unit volume = `rho` (density of the gas) `lamda=m/(sqrt2pid^2rho) ""...(3)` Also we know that PV = NkT `P = N/V kT = nkT "" [:.n=N/v]` `:. n=P/(kT)` Substituting `n= P/(kT)` in equation (2), we get `lamda(kT)/(sqrt2pid^2P)` |
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