How Does the Beer-Lambert Law Apply to Gas Absorbance Calculations?

[cmoral1265]

Hi,

This isn't a homework question but it is quite basic so thought I should post it here.

I am trying to confirm the calculation of absorbance of a gas using the Beer-Lambert law and the gas's absorption cross-section.

Is the equation?

A = 2.303 * absorption cross-section * pathlength * number of molecules/cm3

i.e. a gas with 2.478 e17 molecules/cm3 * 1.4e-19 * 7.5 cm give an absorbance of 0.599 or does it give A = 0.260 (without the 2.303 multiplier). I am confused about the multiplication of the 2.303, which is the difference between the base e and base 10 of the Beer-Lambert law.

I am used to calculating absorbance for liquids which uses the base 10 in the formula.

Thanks for your help.

 

[BigJDubs]

Yes, the equation you have provided is correct. The 2.303 multiplier is necessary because it is the conversion factor from a base 10 logarithm to a base e logarithm. In other words, it converts the absorbance (A) calculated using the Beer-Lambert law from a base 10 logarithm to a base e logarithm. This is necessary because the Beer-Lambert law uses a base 10 logarithm while the absorption cross-section of gases is usually given in terms of a base e logarithm.