1
$\begingroup$

One day as I carpooled back home from work, our backseat passenger cut loose a particularly unpleasant wind, forcing the driver to open the car's windows to the freezing cold. Our offending passenger remarked "I swear, the smell must travel faster than the sound." Of course, as engineers, we took to analyzing this claim as if it were serious, but soon found ourselves stumped. We've all had a basic physics series, have taken a semiconductor physics class, and work in an RF focused industry, so we're familiar with the concepts of diffusion and wave propagation, but not enough to answer this question: how is the rate of diffusion in air related to the speed of sound? What factors are at play here (mass of particles, temperature, pressure...)? What does it even mean to ask how quickly a smell travels given the long tail on a distribution of particle velocities in a gas?

$\endgroup$
1
$\begingroup$

This is an interesting question, and I hope to shed some light on it.

The sound speed and the diffusion coefficient are indeed conceptually different quantities, but we may study the relation between them. In a simple gas, we may demonstrate that the ratio between these two quantities depends basically on the collision frequency. The sound speed and the diffusion coefficient are given by the following expressions \begin{equation} c = \sqrt{\frac{\gamma p}{nm}} = \sqrt{\frac{kT}{m}} \end{equation}
\begin{equation} D = \frac{kT}{m\nu} \end{equation} where $p$ is the gas pressure, $n$ is the gas density in atoms per unit volume, $m$ is the particle mass, $T$ is the gas temperature, $\gamma$ is the heat capacity ratio, $\nu$ is the collision frequency and $k$ is the Boltzmann constant. In the first expression above we considered $\gamma = 1$ and the ideal gas equation of state $p = n k T$. We may readily write the relation between $D$ and $c^2$ as \begin{equation} D = \frac{c^2}{\nu} \end{equation} We see that the lower the collision frequency is, the higher is the Diffusion coefficient. In a rarefied gas, the sound velocity does not depend on the collision frequency, so this holds whatever the sound velocity is. Depending on the initial conditions, such as in a free expansion of a gas, the expansion velocity can be supersonic.

In your case, if we consider that the unpleasant wind from your fellow is a gas with a mass very similar to the mass of the surrounding gas this analysis applies. However, the collision frequency at atmospheric pressure is very high and the diffusion flux will occur at velocities much lower than the sound speed. If your fellow generated a powerful high pressure blow, it could well develop into a supersonic expansion and reach you at supersonic speed...

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The two speeds are entirely unrelated. The smell travels through diffusive/convective mass transfer, while speed of sound is the speed with which pressure fluctuation propagate. In your example the speed of sound is somewhere between one and two orders of magnitude faster than the convective speed. In short: Sound travels much, MUCH faster than smell.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How do you explain the long tail on a distribution of particle velocities? Why does smell not travel nearly instantaneously? $\endgroup$ – Void Star Dec 11 '16 at 5:38
  • $\begingroup$ That would be because it must be transported by the macroscopic velocities of fluid flow in the cabin, which are fairly small. Molecular velocities are of the order of the speed of sound, yes, but molecules don't travel far; the mean free path is tiny. $\endgroup$ – Pirx Dec 11 '16 at 5:56
  • $\begingroup$ Your claim then is that smell travels primarily through convection and not diffusion. Can you substantiate that? $\endgroup$ – Void Star Dec 11 '16 at 23:17
  • 1
    $\begingroup$ For typical ventilation in cars, diffusion is indeed much slower than convection. The diffusion coefficient of a gas in air is typically of the order of $10^{-5}$m$^2$/s. With a bit of dimensional analysis you'll arrive at characteristic speeds of concentration fronts of the order of cm/s or so. The process is similar to diffusion of heat (conduction). If you'd have to rely on conduction to get your car warm in the winter, you would not be happy. Like I said, the main claim is that either of these processes (convection or diffusion) result in signals traveling MUCH slower than pressure waves $\endgroup$ – Pirx Dec 11 '16 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.