Question

Define mean free path of a molecule in a gas. Derive the law of distribution of free paths

Answer 1

Answer on Question #66115, Physics / Mechanics | Relativity

Define mean free path of a molecule in a gas. Derive the law of distribution of free paths.

Answer:

A moving molecule the average distance travels the mean free path between successive collisions, which change its direction or energy or other properties of particles.

Consider a large number of molecules at a certain time. As they travel, they will collide with each other and with other molecules. Let the number of molecules that survive a collision at a distance $r$ of the path be $r$. If each molecule is allowed to move to a larger distance $dx$, more collisions will occur. Suppose that the number of collisions is proportional to the number of molecules $N$ and specific $dx$. That is, the number of molecules removed by these collisions will be proportional to. Since their number decreases with increasing area, we can write

$$dN = -P_c N dx$$

where $P_c$ is a constant of proportionality and is called the Collision probability.

One can rewrite the above equation as

$$dN / N = -P_c dx$$

This can be integrated to $N = N_0 e^{-P_c x}$

where $N_0$ is the number of molecules at $x = 0$.

From the equation we find that number of molecules surviving, a collision decreases exponentially. Further, the probability that a gas molecule will cover a distance $x$ without making any collision is

$$N / N_0 = f(x) = e^{-P_c x}.$$

This is the law of distribution of free paths. $P_c = 1 / l$,

where $l$ is mean free path.

Thus

$$N / N_0 = e^{-x / l}$$

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