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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).

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  1. JamesJ
    • 4 years ago
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    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.

  2. S
    • 4 years ago
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    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?

  3. S
    • 4 years ago
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    So iv V = 1, V of nitrogen is 0.84, and V of Oxygen is 0.16...

  4. JamesJ
    • 4 years ago
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    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 \[ \rho = \frac{m}{V} = \frac{(m_O + m_N) \ kg }{1 \ m^3} = (m_O + m_N) \ kg/m^3 \]

  5. JamesJ
    • 4 years ago
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    (Notice in this last equation the m^3 is a measure of volume, not the mass cubed.)

  6. S
    • 4 years ago
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    Okay this makes perfect sense! Great explanation! Now i get it! Thank you very very much!

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