- #1

Fabrizio Vassallo

- 17

- 6

Molar mass of N2: 28.0134 g/mole

1 mol = 6.022x10

^{23}molecules/mole

Therefore 1 molecule = (28.0134 g/mole) / (6.022x10

^{23}molecules/mole) = 4.6517x10

^{-23}grams = 4.6517x10

^{-26}kg

And then I took the average density of the atmosphere between 0 and 80 km that I calculated using a 1976 Technical Report by NASA, which resulted to be 0.13523 kg/m

^{3}and multiplied that density by the percentage of nitrogen present in the atmosphere, which is 78.084% to get the density of N

_{2}in the atmosphere, which resulted in 0.10559 kg/m

^{3}. I then divided the mass in kg of a single molecule that I calculated before (4.6517x10

^{-26}) by that density of N

_{2}to get the volume occupied by one molecule of N

_{2}.

I know that this result is wrong, somewhere, because what I'm seeking to know is what is the ratio of occupied space to free space in the atmosphere, and when I calculate the amount of space that the N

_{2}molecules would occupy in total, it gives me just a bit less than the total volume of the atmosphere between 0 and 80km, and that can't be the case because you still have to fit the +20% of oxygen and other gases, and still should have plenty of free space. I know that if you put a gas in any place it will occupy all available space, but my aim is to work out how much free space there is between each of those molecules that form that gas.

Just to be clear, I'm considering the atmosphere to be at all times comprehended by the space there is between 0 km above the surface and 80 km above the surface. I know there's other ways of defining the atmosphere but this is the one that I chose for this exercise.

Thanks for reading, and if I wasn't very clear in my explanation please let me know and I'll try to rephrase or explain what I'm trying to do better so you can understand it. Thanks again :)