Application of Beer-Lambert Law to Absorptivity

Question

The Beer-Lambert law gives a linear relationship between the concentration of a solute $c$ and the absorbance $A$, with absorbance defined as the logarithm of the ratio between the transmitted radiant power to the incident one: $$A = \epsilon lc$$ $$A = -\log\left(\frac{I_t}{I_i}\right) = -\log\left(\frac{I_i - I_a}{I_i}\right) = -\log{\left(1 -\frac{I_a}{I_i}\right)} $$

In a simplified model of the effect of CO2 on the atmosphere I am playing around with, the absorptivity $a$ in the Stefan-Boltzmann law of radiation is used to quantify the aomunt of CO2 present, and I found that $a$ is defined as the ratio between the absorbed radiant power to the incident one. $$j = a\sigma T^4$$ $$ a = \frac{I_a}{I_i} $$

I am wondering if it is plausible to say, from the Beer-Lambert law, that in this case the absorptivity is related to the concentration of CO2 in the atmosphere by the following, considering an atmosphere of uniform properties and thickness (which is certainly an overly ideal model). That is: $$ -\log(1-a) = A \propto \ce{ [CO_2] }$$

P.S. I am not sure if this question is more suited for Physics Stackexchange, but since it's about the Beer-Lambert law, I'll post it here first.

Edit: as the commenters helpfully pointed out, my original definitions for the terms were awfully incorrect.

Answer 1

Absorbance $A$ of atmosphere, caused by $\ce{CO2}$, can be for the given relevant IR absorption wavelength $\lambda$ formulated as:

$$A(\lambda) = \int_{h_1}^{h_2}{(\epsilon(\lambda)c(h)\mathrm{d}h}$$

where

• $A(\lambda)$ is absorbance at the given wavelength
• $h$ is altitude.
• $\epsilon(\lambda)$ is the $\ce{CO2}$ (mass/molar) extinction coefficient for the given wavelength
• $c(h)$ is (mass/molar) $\ce{CO2}$ concentration at the given altitude.

Transmittance $T$ for the given wavelength is then $$T(\lambda) = \frac{I(\lambda)}{I_0(\lambda)}= 10^{-A(\lambda)}$$

The relative emissivity, referring to the emissivity of a black body at the given wavelength, is $\frac{I(\lambda,T)}{I_\mathrm{BB}(\lambda,T)}=1-T(\lambda) $

If we switch from the wav length to frequency by $\nu = \frac{c}{\lambda}$, then from Planck's law page we can use the relation of the Plancks' law and Stefan-Boltzmann law:

$$P = \int_0^\infty d\nu \int_0^{\tfrac{\pi}{2}} d\theta \int_0^{2\pi}d\phi \, B_\nu(T) \cos(\theta)\sin(\theta) = \sigma T^4$$ where $$\sigma = \frac{2k_\mathrm{B}^4\pi^5}{15c^2h^3}\approx 5.670 400 \times 10^{-8}\, \mathrm{J\, s^{-1} m^{-2} K^{-4}}$$ is known as the Stefan–Boltzmann constant.

and further

$$B_\nu(\nu, T) =\frac{ 2 h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 }$$

For the thermally emitted power at given frequency $\nu$, temperature $T$ and transmittance $T(\lambda=\frac{c}{\nu})$(Be aware of clash of quantity symbols):

$$dP(\nu,T,T(\lambda) = d\nu (1-T(\lambda))\int_0^{\tfrac{\pi}{2}} d\theta \int_0^{2\pi}d\phi \, B_\nu(T) \cos(\theta)\sin(\theta) \\ = d\nu (1-T(\lambda)) \int_0^{\tfrac{\pi}{2}} d\theta \int_0^{2\pi}d\phi \, \frac{ 2 h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } \cos(\theta)\sin(\theta) \\ = d\nu (1-T(\lambda)) \int_0^{\tfrac{\pi}{2}} d\theta \frac{ 4 \pi h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } \cos(\theta)\sin(\theta) \\ = d\nu (1-T(\lambda)) \frac{ 4 \pi h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } \int_0^{\tfrac{\pi}{2}} d\theta \cos(\theta)\sin(\theta) \\ = d\nu (1-T(\lambda)) \frac{ 2 \pi h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } $$