0
$\begingroup$

When humans take in air to their lungs, we capture about 5% of the total quantity of the air as oxygen (which in turn equals about 24% of the available oxygen in the air)

Is there an equivalent rate for plants when photosynthesising (or a rate for any one specific plant?). I.e. if a plant had a given amount of air pass through their stomata, what percentage of the available CO2 is used by the chloroplasts?

$\endgroup$
4
  • 1
    $\begingroup$ Stomata should be compared to human alveoli, not the lungs. Also the lungs are pumps with heavy circulation whereas the plant heomoglobin is passive. The oxygen will tend to be carried by a metabolite whereas the CO2 can diffuse straight into the transportation medium, and the exchange depends on the gaseous gradient near the internal stomata surface. $\endgroup$ Jan 13 at 8:46
  • 1
    $\begingroup$ I understand that trying to compare leaves/stomata to lungs/alveoli is problematic. However, I guess it would be possible to measure the amount of air that passes through a plant organism in a given period of time, and to measure the CO2 levels before and after in a sealed environment, and calculate from there? I guess there must be a lot more variables given it is a passive process in a non-homeostatic environment (like the lungs are) - but I maybe an efficiency range would still be possible? $\endgroup$
    – Amphibio
    Jan 13 at 9:54
  • 1
    $\begingroup$ I've started towards a way to figure it out - 1 litre of CO2 weighs 1.964g at NTP, and the amount of CO2 in one litre of air is about 0.04%. So one litre of NTP air contains 0.0007856g of CO2. $\endgroup$
    – Amphibio
    Jan 13 at 13:25
  • $\begingroup$ An adult tree (yes, very generic!) can take in about 21kg of CO2 per year, or 21,000 grams. 21000 / 0.0007856 = 26731160. So an adult tree must process at least 26 million litres of air a year (and that is assuming it is being 100% efficient!). Please let me know if I have done my calculations right $\endgroup$
    – Amphibio
    Jan 13 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.