Question

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

Answer 1

Answer on Question #66747-Physics-Molecular Physics-Thermodynamics

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

Solution

The mean free path is the average distance traveled by a moving molecule between successive collisions, which modify its direction or energy or other particle properties.

Consider a large number of molecules at a certain instant. As they travel they will collide among themselves and with other molecules. We wish to estimate the number that has not made a collision at some later time. Let the number of molecules surviving a collision in travelling distance $r$ be $N$. If each molecule is allowed to travel a further distance $dx$, more collisions will occur. We assume that the number of collisions is proportional to the number of molecules $N$, and the distance $dx$. That is, the number of molecules removed by these collisions will be proportional to $Ndx$. Since the number of molecules decreases with increasing distance, 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

$$\frac{dN}{N} = -P_c dx$$

This can be integrated to

$$N = N_0 e^{-P_c x}$$

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

From this 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

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

This is the law of distribution of free paths.

$P_c = \frac{1}{l}$, where $l$ is mean free path.

Thus

$$\frac{N}{N_0} = e^{-\frac{x}{l}}.$$

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