How can one find the rate of collision per unit area with the walls of a gas container given the distribution as $$d N\left(v_{x}\right)=N\left(\frac{m}{2 \pi k T}\right)^{1 / 2} e^{-m v_{x}^{2} / 2 k T} d v_{x} \quad.$$ I tried to find it but I end up with an extra factor of $1/2$ in the end. The correct expression being $f=\frac{1}{4} n\langle v\rangle$. Where $\langle v\rangle$= average velocity of gas molecules and $n$= concentration of gas molecules.


1 Answer 1


Since the Maxwell-Boltzmann propability distribution, for particles moving in only one direction - e.g., the $x$-direction here, $f(v_x)$ is given as

\begin{align} f(v_x) = \left( \frac{m}{2 \pi k_B T} \right)^{1/2} \cdot e^{- m \, v_x^2 / 2 k_B T} \quad, \end{align}

we can further deduce that $f(v_x) dv_x$ represents the fraction of molecules/particles with a velocity component between [$v_x$, $v_x+dv_x$]. Consequently we can infer1 that with

$$N_c(total) = n \,dA\, dt\, \int^\infty_0 v_x\, f(v_x)\, dv_x \tag{1}\label{eq:NumOfCollisions}$$

we compute the total number of collisions $N_c(total)$ that occur at one specific wall (hence, we only integrate in one - the positive - direction of the $x$-axis for the particle velocities), with each particle velocity weigthed by the propability distribution function $f(v_x)$, per time interval $dt$ and area interval $dA$.

The collision flux, often denoted as $Z_w$ in pertinent literature, and is defined as the number of collisions per unit area, per unit time, i.e. we divide Eq.\eqref{eq:NumOfCollisions} by $dA$ and $dt$,

\begin{align} Z_w := n \, \int^\infty_0 v_x \, f(v_x) \, dv_x \tag{2}\label{eq:CollisionFluxInt} \quad, \end{align}

and after evaluation of Eq.\eqref{eq:CollisionFluxInt}, one gets

\begin{align} Z_w := n \left( \frac{k_B T}{2 \pi m} \right)^{1/2} \tag{3}\label{eq:CollisionFluxAnaly} \quad. \end{align}

Furthermore, we know the mean velocity $\langle v \rangle$ (as can be derived through the Maxwellian distribution above) is given as

\begin{align} \langle v \rangle := \sqrt{ \frac{8 k_B T}{\pi m} } \tag{4}\label{eq:MeanVel} \quad. \end{align}

Now we can use both equations, \eqref{eq:CollisionFluxAnaly} and \eqref{eq:MeanVel} and write,

\begin{align} \frac{2 \, Z_w^2}{n^2} = \color{orange}{\frac{k_B T}{\pi m}} \qquad \text{and} \qquad \frac{\langle v \rangle^2}{8} = \color{orange}{\frac{k_B T}{\pi m}} \quad, \end{align}


\begin{align} \frac{2 \, Z_w^2}{n^2} \equiv \frac{\langle v \rangle^2}{8} \quad, \end{align}

and after some minor rearrangements of the expression above we arrive at the equation you referred to through the Maxwell-Boltzmann distribution,

\begin{align} \boxed{ Z_w \equiv f = \frac{1}{4} n \, \langle v \rangle \;} \quad. \end{align}

I hope this line of thought and deduction is helpful.

Addendum & Footnotes:

1 In order to motivate on how Eq.\eqref{eq:NumOfCollisions} is set up here we reflect about what variables or quantities directly influence the number of collisions $N_c$ against, e.g. on one wall of a cube where the gas is being studied.

First we consider one particular group of particles, namely the particles that travel with one specific numerical value of the velocity $v_x$ along the positive $x$-axis. The number of particles with velocity $v_x$ colliding with the wall/area perpenticular to $\vec{v}_x$ can be linked to the following quantites:

\begin{align} N_c(v_x) \propto n \, dA \, dt \, v_x \end{align}

because the number of collisions against the wall for any particle with a positive $v_x$ component, i.e. $N_c(v_x)$, increases with the number density of particles in the volume $n$, the size of the area the particles are hitting $dA$, the length of the timeperiod $dt$ we observe (and count) the collisions of course and how fast the particles are moving towards the wall, i.e. $v_x$ (the faster particles are moving towards the wall the more particles will hit with wall, and in the opposite: If the particles are moving really slowly or are not moving at all in our gedankenexperiment very few or no collisions take place.) Consequently the proportionality relation above is well-founded.

However, we are missing one crucial factor in order to convert the above expression into an equality '=': The Maxwell-Boltzmann distribution $f(v_x)$ allows that each value of $v_x$ is being weighted accordingly by the probability of its occurence,

\begin{align} N_c(v_x) = n \, dA \, dt \, f(v_x) \, v_x \quad. \end{align}

Finally, we do not only like to limit our study to a group of particles which one single value of the velocity $v_x$ but rather all particles moving towards the wall along the positive $x$-axis. This is eventually achieved by taking the integral over all positive $v_x$,

\begin{align} \int^\infty_0 N_c(v_x) \, dv_x \equiv N_c(total) = n \, dA \, dt \, \int^\infty_0 v_x \, f(v_x) \, dv_x \equiv \eqref{eq:NumOfCollisions} \quad. \end{align}

  • $\begingroup$ Thank you but i fail to understand how equation$1$ gives us the collision frequency. Could you please explain it further.? $\endgroup$
    – Kashmiri
    Commented Aug 30, 2020 at 14:29
  • $\begingroup$ @YasirSadiq You mean the number of collisions with Eq.(1), right? Because the collision frequency (i.e.per unit of area and unit of time) is given in Eq.(2) as $Z_w$. Thank you for the additional info on what part of the deduction is most elusive to you. I try to improve my line of reasoning. Stay tuned. $\endgroup$ Commented Aug 30, 2020 at 14:37
  • $\begingroup$ Thank you buddy $\endgroup$
    – Kashmiri
    Commented Aug 30, 2020 at 14:49
  • 2
    $\begingroup$ All the molecules with velocity component $v_x$ that will hit wall area $dA$ in a time $dt$ will be contained within a distance $v_x\ dt$ of the wall, and so in a volume $dA\ v_x\ dt$ (assuming $v_x << \sqrt A$). $\endgroup$ Commented Aug 30, 2020 at 14:53
  • $\begingroup$ great. got it. thanks Mr Phillip . $\endgroup$
    – Kashmiri
    Commented Aug 31, 2020 at 14:47

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