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S
 4 years ago
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).
S
 4 years ago
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).

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1By 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.

S
 4 years ago
Best ResponseYou've already chosen the best response.0Thank 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?

S
 4 years ago
Best ResponseYou've already chosen the best response.0So iv V = 1, V of nitrogen is 0.84, and V of Oxygen is 0.16...

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Using 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 \[ \rho = \frac{m}{V} = \frac{(m_O + m_N) \ kg }{1 \ m^3} = (m_O + m_N) \ kg/m^3 \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1(Notice in this last equation the m^3 is a measure of volume, not the mass cubed.)

S
 4 years ago
Best ResponseYou've already chosen the best response.0Okay this makes perfect sense! Great explanation! Now i get it! Thank you very very much!
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