Mean Free Path

The mean free path is the average distance a molecule in a gas travels between collisions with other molecules. Let us crudely approximate the molecules in the gas as hard spheres of diameter $R$. Any two molecules whose centers are less than a distance $R$ apart will collide. Suppose that one molecule is moving with velocity ${\bf v}$, whereas the other molecules are stationary. The moving molecule sweeps out a cylindrical volume $\pi\,R^{\,2}\,\langle v\rangle$ in one second. Any other molecule whose center lies in this volume will collide with the moving molecule. There are $n\,\pi\,R^{\,2}\,\langle v\rangle$ such molecules, where $n$ is the number density of molecules. Hence, the number of collisions per second is

$$f= \pi\,R^{\,2}\,n\,\langle v\rangle.$$ (5.209)

Thus, the mean distance that the molecule travels between collisions, which is the mean free path, is

$$l = \frac{\langle v\rangle}{f} =\frac{1}{\pi\,R^{\,2}\,n}.$$ (5.210)

If we now take into account the fact that all of the molecules in the gas are moving then it is clear that the previous two equation generalize to give

$$f = \pi\,R^{\,2}\,n\,\langle V\rangle,$$ (5.211)

$$l = \frac{\langle v\rangle}{\pi\,R^{\,2}\,n\,\langle V\rangle},$$ (5.212)

where ${\bf V}$ is the relative velocity between molecules. Consider two molecules of velocities ${\bf v}_1$ and ${\bf v}_2$. The relative velocity of the molecules is

$${\bf V} = {\bf v}_1-{\bf v}_2.$$ (5.213)

Now,

$$V^{\,2} = v_1^{\,2} + v_2^{\,2} -2\,{\bf v}_1\cdot{\bf v}_2,$$ (5.214)

which implies that

$$\left\langle V^{\,2}\right\rangle = \left\langle v_1^{\,2}\right ... ...e v_2^{\,2} \right\rangle - 2\left\langle {\bf v}_1\cdot{\bf v}_2\right\rangle.$$ (5.215)

However, $\left\langle {\bf v}_1\cdot{\bf v}_2\right\rangle=0$, because the cosine of the angle subtended between ${\bf v}_1$ and ${\bf v}_2$ is just as likely to be positive as to be negative. Thus, we deduce that

$$\left\langle V^{\,2}\right\rangle = \left\langle v_1^{\,2}\right \rangle + \left\langle v_2^{\,2} \right\rangle =2\left\langle v^2\right\rangle.$$ (5.216)

Assuming, as seems reasonable, that

$$\frac{\langle V\rangle}{\langle v \rangle} = \sqrt{\frac{\left\langle V^{\,2}\right\rangle}{\left\langle v^2\right\rangle}},$$ (5.217)

we obtain

$$\langle V\rangle= \sqrt{2}\,\langle v\rangle.$$ (5.218)

Hence, Equation (5.212) yields

$$l = \frac{1}{\sqrt{2}\,\pi\,R^{\,2}\,n}.$$ (5.219)

Let us estimate the mean free path in air at standard temperature ( $T=15^\circ$ C) and pressure ( $p=10^5\,{\rm N\,m}^{-2}$.) From the idea gas law, (5.97),

$$n = \frac{\nu\,N_A}{V} = \frac{p}{k_B\,T} = \frac{10^5}{(1.381\times 10^{-23})\,(288)}= 2.5\times 10^{25}\,{\rm m}^{-3}.$$ (5.220)

Now, $R=2\times 10^{-10}\,{\rm m}$ is a typical diameter of an air molecule. Thus, we obtain

$$l = \frac{1}{\sqrt{2}\,\pi\,(2\times 10^{-10})^2\,(2.5\times 10^{25})}=2\times 10^{-7}\,{\rm m}.$$ (5.221)

Consider a molecule moving along the $x$-axis. The molecule is subject to random collisions. Thus, the probability that the molecule undergoes a collision between moving a distance $x$ and moving a distance $x+dx$ is $\alpha\,dx$, where $\alpha$ is a constant. Let $P(x)$ be the probability that the molecule moves a distance $x$ without undergoing a collision. It is evident that the probability that the molecule's first collision occurs between moving a distance $x$ and a distance $x+dx$ is $-(dP/dx)\,dx$. However, this probability is also equal to the probability that the molecule does not undergo a collision in moving a distance $x$, and then undergoes a collision between moving a distance $x$ and a distance $x+dx$. In other words,

$$-\frac{dP}{dx}\,dx = [P(x)]\,[\alpha\,dx],$$ (5.222)

or

$$\frac{dP}{dx} = -\alpha\,P.$$ (5.223)

The previous equation can be integrated to give

$$P(x) = p_0\,{\rm e}^{-\alpha\,x},$$ (5.224)

where $p_0$ is an arbitrary constant. However, $P(0)=0$, because the molecule has no chance of undergoing a collision in moving zero distance. Hence, the probability that the molecule moves a distance $x$ without undergoing a collision is

$$P(x) = {\rm e}^{-\alpha\,x}.$$ (5.225)

Moreover, the probability that the molecule undergoes its first collision between moving a distance $x$ and moving a distance $x+dx$ is $f(x)\,dx = [P(x)]\,[\alpha\,dx]$ (i.e., the molecule needs to not undergo a collision in moving a distance $x$, and then undergo a collision between moving a distance $x$ and a distance $x+dx$), so

$$f(x) = \alpha\,{\rm e}^{-\alpha\,x}.$$ (5.226)

The mean distance that the molecule travels before undergoing its first collision is (see Section 5.1.6)

$$\langle x\rangle = \int_0^\infty f(x)\,x\,dx= \int_0^\infty \alph... ...\,x\,dx= \frac{1}{\alpha} \int_0^\infty y\,{\rm e}^{-y}\,dy = \frac{1}{\alpha}.$$ (5.227)

However, $\langle x\rangle$ is equivalent to the mean free path, $l$. Hence, the probability density for a molecule to move a distance $x$ between collisions is

$$f(x) = \frac{{\rm e}^{-x/l}}{l}.$$ (5.228)

Moreover, the probability that the molecule moves a distance $x$ without undergoing a collision is

$$P(x)= {\rm e}^{-x/l}.$$ (5.229)