Suppose $10^{14} kg$ of carbon dioxide was released into the atmosphere and absorbed completely, what is the percentage change of carbon dioxide concentration? Take initial atmospheric mass mixing ratio to be $5.7 \times 10^{−4} kg/kg$.

Using density of air as around $29g/mol$, the weight of the atmosphere is about $10^{18}kg$. The percentage change is $10^{14} \times 10^{18} \times 100=0.01$? Where does the mixing ratio come in?

  • 2
    $\begingroup$ This looks like a homework question... the "mixing ratio" tells you how much carbon dioxide was already there; when you ask about the percentage change, it is relative to this initial ratio. So if you go from 0.01 to 0.02, that would be a 100% change. Is that enough to get you going? $\endgroup$ – Floris Jun 11 '15 at 1:29
  • 1
    $\begingroup$ So the i find the new ratio by using $\frac{10^{14}}{10^{18} + 10^{14}}$ which is about $10^{-4}$. So the percentage change is $18%$? $\endgroup$ – user44840 Jun 11 '15 at 1:32

Following up on your comment:

If you initially have $5.7 \cdot 10^{-4} \mathrm{kg/kg}$ of carbon dioxide, and the total mass of the atmosphere is $10^{18} \mathrm{kg}$, you can compute the mass of $\mathrm{CO_2} = 5.7 \cdot 10^{14} \mathrm{kg}$. Adding $10^{14}\mathrm{kg}$ to that does indeed give a $\frac{1}{5.7} = 18\%$ change as you calculated in your comment.

  • $\begingroup$ Sorry, which method are you referring to? Is the percentage change 18%? $\endgroup$ – user44840 Jun 11 '15 at 1:45
  • $\begingroup$ Yes the percentage is 18%. $\endgroup$ – Floris Jun 11 '15 at 2:02

Your Answer

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

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