Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I recently saw cm^3/g as a unit for amount adsorbed. Usually, you see either kg adsorbate/kg adsorbent or mole adsorbate/kg adsorbent. Does anyone know the meaning of this unit?

share|cite|improve this question

Indeed, when you just want to measure how much material has been adsorbed, it's more natural to use the same units for both materials.

I think that 1 cubic centimeter per gram is not a unit of "adsorption" per se - adsorption is a process, not a quantity, after all - but it's a unit of "specific pore volume". You take one gram of a material and measure the volume of the pores inside this material in cubic centimeters - and you get "specific pore volume" in cubic centimeters per gram. Note that pores - holes - are empty so they may only be measured by a unit of volume, and not a unit of mass or moles. ;-)

Gas adsorption is a major method to measure "specific pore volume". In some sense, the "specific pore volume" also does quantify adsorption because it tells you at how many microscopic places the gas may get adsorbed inside a material. See some papers about the two concepts to get a flavor:

share|cite|improve this answer

I've only ever seen those units described as "specific surface area". It does have implications for adsorption, but I don't think it is the be all and end all.

share|cite|improve this answer

In the case of cm^3/g, the quantity of gas adsorbed (the adsorbate) is expressed as its volume in gaseous form at STP. These units are still commonally used although the International Union of Pure and Applied Chemistry (IUPAC) recommends that the quantity adsorbed be described in units of mol/g. The quantity of adsorbate in cm^3 units occurs from the manner by which the absolute adsorbed quantity is calculated in the volumetric adsorption analytical technique.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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