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How much will the atmospheric carbon dioxide change?

  1. Jun 10, 2015 #1
    1. The problem statement, all variables and given/known data

    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##.

    2. Relevant equations


    3. The attempt at a solution

    Using density of air as around ##29 g/mol##, the weight of the atmosphere is about ##10^{18} kg##. The percentage change is ##\frac{10^{14}}{10^{18}} \times 100 = 0.01%##? Where does the mixing ratio come in?
     
  2. jcsd
  3. Jun 10, 2015 #2
    Please show us how you calculated the mass of the atmosphere.

    Chet
     
  4. Jun 11, 2015 #3
    Taking pressure to be ##p_0 = 10^5 Pa##, I used ##\frac{4}{3}\pi R_E^3 p_0 \sim 10^{18} kg##. Sorry the ##29g/mol## bit was misleading.
     
    Last edited: Jun 11, 2015
  5. Jun 11, 2015 #4

    SteamKing

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    And (4/3) π RE3 represents what geometrically? Think about this carefully.
     
  6. Jun 11, 2015 #5
    You can't just multiply the volume of the entire earth by the air pressure at the surface of the earth and think that you are getting the mass of the atmosphere. The units don't even match, and the earth is not filled with air. What is the air density at the surface of the earth? Are you familiar with the barotropic equation?

    Chet
     
  7. Jun 11, 2015 #6
    Sorry I meant ##4\pi R_E^2 P_0##.
     
  8. Jun 12, 2015 #7
    You forgot to divide by g.

    Chet
     
  9. Jun 13, 2015 #8
    The value ##29 (g/mol)## is equivalent to ## 1.29 (kg/m^3) ##. Taking the same density for about 3km and the earth surface we have:
    ## \pi\times(6.4\times10^6)^2\times3000\times1.3 \approx 5\times10^{17} (kg) ##

    For better answer we must use exponential law for pressure.
     
  10. Jun 13, 2015 #9
    The correct so-called scale height of the atmosphere is close to 7 km, rather than 3 km. But an easier way of doing this problem is to determine the weight of a column of air above each square meter. That would be 105 N. Assuming that the gravitational acceleration does not change much over the first few km, the mass of air over each square meter of surface would be 104 kg. Multiplying this by the area of the earth, one obtains 1.3 x 1018 kg. This compares with a value of 1.2 x 1018 obtained by taking your answer and multiplying it by 7/3.

    Chet
     
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