2
$\begingroup$

So recently I was thinking gases (at least colourless ones) are more transparent in their gaseous state than liquid state.

And as we talk of continuity in liquid and gaseous state (fluids) is it possible that gases become more and more opaque as their number density (number of gas molecules per unit volume) increases (by cooling, compression, etc) increases.

Consider a ideally cubical transparent container with a gas in it and compress it continuously. Assume a monochromatic light source behind the container giving out light with constant intensity with all light rays parallel to each other. Would intensity of transmitted light decrease as number density increases?

It would seem true intuitively to me I guess but I cannot figure out why?

So my question:

Is it possible to relate the intensity of transmitted light with number density of gas (Mathematically)?

Please provide validation too for your formula or reasoning.

$\endgroup$
  • $\begingroup$ I have simplified a practical problem to a idealised theoretical one so that it is easier to solve. Please help me point out irrelevant assumptions. $\endgroup$ – Hrishabh Nayal Mar 27 at 13:35
  • $\begingroup$ If it remains a gas, the absorption per molecule should not change. Once you get condensation you may have droplets (scattering) or liquids (density fluctuations on shorter length scales also leading to scattering) and life gets more difficult. $\endgroup$ – Jon Custer Mar 27 at 14:35
  • $\begingroup$ @JonCuster But take the example of water boiling you can see water vapours as they rise. But as they continue to move upward they become less and less opaque(and less denser) and eventually above a certain point we can no longer $see$ them. So if there is no change on absorption why do we observe this? $\endgroup$ – Hrishabh Nayal Mar 27 at 15:30
  • 2
    $\begingroup$ Because what you see is small drops of condensed water formed when the boiling vapor hits the cooler air, and then as the column of air continues to mix with more air the drops evaporate. $\endgroup$ – Jon Custer Mar 27 at 16:21
  • $\begingroup$ Ah! You see right where we started. I read here(britannica.com/science/gas-state-of-matter/…) that > there is no fundamental distinction between a gas and a liquid. So how can we say that what is condensed drops and what are not? This is what prompted me to ask this question in first place. $\endgroup$ – Hrishabh Nayal Mar 28 at 7:34
2
$\begingroup$

Relation between transmission and density

You ask for a mathematical relation between the intensity $I$ of the transmitted light and the number density $n$ of gas molecules. It is $$ I = I_0 \, e^{-n\sigma r}, $$ where $I_0$ is the emitted intensity (i.e. before the beam enters the gas), $\sigma$ is the (wavelength-dependent) cross section of individual particles, and $r$ is the distance traveled through the gas.

Optical depth

The quantity $\tau \equiv nr\sigma$ is called the optical depth of the gas. We sometimes speak of the two different regimes $\tau \ll 1$ and $\tau \gg 1$ as optically thin and optically thick, respectively. In the optically thin regime, "most" light is transmitted. In the optically thick regime, "most" light is absorbed.

Derivation

Consider a light beam of cross-sectional area $A$, traveling a small distance $dr$ (so small that the particles don't "cover" each other). The volume covered is $V=A\,dr$, and the total area of all particles in this volume is $\Sigma = nV\sigma$.

The transmitted fraction $I/I_0$ is equal to the fractional area covered by particles: $$ \frac{dI}{I} = -\frac{\Sigma}{A} = -n\sigma \,dr. $$ If there are multiple species of particles, you use $(n_1\sigma_1 + n_2\sigma_2 + \cdots)dr$.

Integrating on both sides (i.e. along the path of the beam), $$ \begin{array}{rcl} \int_{_{I_0}}^{^{I}}\frac{dI'}{I} & = & -\int_0^r n\sigma\,dr'\\ \ln I - \ln I_0 & = & n\sigma r\\ \frac{I}{I_0} & = & e^{-n \sigma r}. \end{array} $$

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

Clear air is colorless in the visible spectrum. Yet consider what happens when the Sun changes its altitude from high in the sky to just over horizon: intensity of solar radiation decreases for an earthly observer. That's due to scattering (Rayleigh scattering, in particular, is relevant here) from a larger amount of atmospheric gases.

Also, when you go higher above the ground (e.g. in an airplane), you get more solar radiation from the Sun at the same zenith angle. That's due to lower atmospheric density at high altitudes, which results in less scattering along the path between the Sun and the airborne observer.

So yes, due to scattering, gases do become less transparent with increase of molecule number density.

Is it possible to relate the intensity of transmitted light with number density of gas (Mathematically)?

Yes, you want to learn the Beer–Lambert law.

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

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.