How much will the atmospheric carbon dioxide change?

[unscientific]

Homework Statement

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 $5.7 \times 10^{-4} kg/kg$.

Homework Equations

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?

 

[Chestermiller]

Please show us how you calculated the mass of the atmosphere.

Chet

 

[unscientific]

Please show us how you calculated the mass of the atmosphere.

Chet

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.

 

[SteamKing]

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.

And (4/3) π R_E³ represents what geometrically? Think about this carefully.

 

[Chestermiller]

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.

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

 

[unscientific]

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

Sorry I meant $4\pi R_E^2 P_0$.

 

[Chestermiller]

Sorry I meant $4\pi R_E^2 P_0$.

You forgot to divide by g.

Chet

 

[theodoros.mihos]

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.

 

[Chestermiller]

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.

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