Calculating Total Mass of Nitrogen in Atmosphere

[gunnar]

Hi. Can someone give me a hint to how to calculate total mass of nitrogen in the atmosphere. I've calculated the mass of all the atmosphere, I got the percent by volume of nitrogen to be 78.03 %. I converted the mass of the atmosphere to Liters. I got the volume of nitrogen to be 3.18e21 L and converted to 3,099 e18 kg. The correct answear is 3,96e18 kg can't see what I doing wrong. To get the Kg of nitrogen I used the partial pressure of N2 in O°C and the gas equation to get the moles of N2. Maybe I'm getting the partial pressure wrong. I did 78.03/100 = 0,7803*1 atm

 

[gschjetne]

I'm not completely sure, but I'd suggest using the old $PV=nRT$, using the volume of the atmosphere, along with the average partial pressure of Nitrogen.

 

[dextercioby]

U need the density of air and the density of gaseous nitrogen under normal conditions of pressure & temperature.

100l air------------------------x Kg air
100 l air----------------------78.03 l N_{2}
78.03 l N_{2}----------------- y Kg N_{2}

Therefore the massic concentration of N_{2} is y/x...

Multiply the massic concentration with the mass of the atmosphere & u'll find the mass of N_{2}

Daniel.

 

[gunnar]

Thanks Daniel. I did what you told me, if I got it right I still didn't get the right answear. Calculated the density of air and N2 in STP multiplied by 100 and 78.03 respectively. the difference was about 0.5 something. multiplied with the mass of air whis is 5,266e18 Kg. Should get 3,96e18 Kg got 3,01e18 Kg

 

[dextercioby]

The density of air under normal conditions of pressure & temp.is

$$\rho_{air}^{STP}\simeq \frac{28.89}{22.41} Kg \ m^{-3}$$

The density of nitrogen under normal conditions of temp & pressure is

$$\rho_{N_{2}}^{STP} \simeq \frac{28}{22.41} Kg \ m^{-3}$$

So 22.41 l of air contain $28.89 \cdot 10^{-3} \ Kg$ of air.Now,22.41 l of air contain 78.03% N_{2} which means exactly

$$\frac{22.41\cdot 78.03}{100} \ l _{N_{2}}$$

The mass of N_{2} in 22.41 l air is

$$mass_{N_{2}}=\frac{22.41\cdot 10^{-3} m^{-3} \cdot 78.03}{100}\cdot \frac{28}{22.41}Kg \ m^{-3}=\frac{28 \cdot 78.03 \cdot 10^{-3}}{100} \ Kg$$

The massic concentration is

$$n_{massic}^{N_{2}}=\frac{28\cdot 78.03 \ 10^{-3} \ Kg}{100} \frac{1}{28.89 \cdot 10^{-3} \ Kg} =\frac{78.03}{100}\cdot \frac{28}{28.89}$$

Then the mass of N_{2} in the atmosphere under normal cond-s of temp & press. is

$$m_{N_{2}}^{atmosphere,STP} =m_{atmosphere} \cdot \frac{78.03}{100}\cdot\frac{28}{28.89}$$

Daniel.

 

[gunnar]

It works. Thanks a lot