Question

If we approximate the atmosphere to be 84% Nitrogen and 16% Oxygen estimate the density of air (kg/m^3) at STP conditions (0c, 1atm).

Answer 1

By the perfect gas law,
\[ \frac{PV}{T} = nR \]
where here \( n \) is the weighted sum of the nitrogen and oxygen. Let \( V = 1 \ m^3 \) and now you can figure out how many moles of O of N there must be in that space; hence the weight; hence the density.

Answer 2

Thank you but that's still not ringing a bell. This is for my environmental engineering class and I took chem and physics like 3 years ago so I can't remember much =\ Could you please tell me how to go on from here?

Answer 3

So iv V = 1, V of nitrogen is 0.84, and V of Oxygen is 0.16...

Answer 4

Using
T = 0 C = 273 K,
V = 1 m^3
P = 1 atm
R = 8.314 J/(K.mole)

you can solve for n, the number of moles.

Then in that space there are
0.84n moles of N
0.16n moles of O

From that you can calculate the mass of the N and O, write them as \( m_O \) and \( m_N \).

Thus the total mass is \( m = m_O + m_N \) and hence the density
\[ \rho = \frac{m}{V} = \frac{(m_O + m_N) \ kg }{1 \ m^3} = (m_O + m_N) \ kg/m^3 \]

Answer 5

(Notice in this last equation the m^3 is a measure of volume, not the mass cubed.)

Answer 6

Okay this makes perfect sense! Great explanation! Now i get it! Thank you very very much!