Let's say I have a gas in a fixed volume being heated by a uniform, monochromatic light source.

I'd like to know how the radiative heating rate per unit mass of the gas in J/kg/s scales with the density of the gas.

If I wanted to know how the peak temperature of the gas scales with density, then I suppose I'd just do an energy-balance problem given some specification of the input source and properties of the gas. Would a similar approach be used to get at the heating rate that I am after?

I am attempting to model the radiative heating of an air parcel in the atmosphere containing some density of an aerosol, by incident radiation, without actually modeling the chemistry or radiation itself.


1 Answer 1


That would depend on whether it was optically thick (absorbs all radiation incident upon it) or optically thin (allows almost all radiation through), or somewhere in between.

If the gas is thin and let's radiation through, it is attenuated by a fraction $\exp(-\kappa \rho x)$, where $\kappa$ is the opacity in units of m$^2$/kg, $\rho$ is the mass density and $x$ is the distance travelled through the gas. If $I_0$ is the incident light intensity in W/m$^2$, then the power absorbed per kg will be $$\frac{{\rm W}}{\rm kg} = \frac{I_0(V/x)(1-\exp[-\kappa\rho x])}{\rho V}=\frac{I_0(1-\exp[-\kappa\rho x])}{\rho x}$$ and the result depends on the path length through the gas as well as the density. If the gas is very thin, so that $\kappa\rho x \ll 1$, this approximates to $I_0 \kappa$ and is independent of density.

If the gas is optically thick and $\kappa\rho x \gg 1$, then the absorbed power per kg will be $I_0/\rho x$. This is power per unit area divided by mass per unit area. Most of the power is absorbed in the first layer of gas of thickness $(\kappa \rho)^{-1}$.

  • $\begingroup$ Hmm, this could be very useful. Can you provide a source or link some material that might help me see how to derive what you state in your second paragraph? i.e. I'd like to see the density drop out when applying a condition of an optically thick gas $\endgroup$ Commented Oct 31, 2022 at 7:27
  • $\begingroup$ Also, for the intermediate case that the gas is thin but not "very thin", I am not sure I'm understanding this form. Does it really say that, in this regime, the power absorbed per unit mass decreases with density? $\endgroup$ Commented Oct 31, 2022 at 7:28
  • $\begingroup$ @pretzlstyle The exponential decay term is very standard radiation transfer theory. See any textbook. As for you first comment, well where else can the absorbed energy go? The density will merely determine whether the gas is optically thick or not. If it is, then it absorbs all the energy. I suppose at that point you could say that the W/kg is $I_0/\rho x$. $\endgroup$
    – ProfRob
    Commented Oct 31, 2022 at 8:38
  • $\begingroup$ @pretzlstyle ... i.e. there is a density dependence. There would be no density dependence if considering a fixed mass of gas, rather than a fixed volume. $\endgroup$
    – ProfRob
    Commented Oct 31, 2022 at 8:47