Given that the density of air at 0.987 bar and 27°C is 1.146 kg m 3 , calculate

Question

Given that the density of air at 0.987 bar and 27°C is 1.146 kg m3, calculate the mole fraction and partial pressure of nitrogen and oxygen assuming the following conditions.

(a) air consists only of these two gases
mole fraction nitrogen
  
mole fraction oxygen
  
partial pressure nitrogen
  bar
partial pressure oxygen
  bar

(b) air also contains 1.0 mole percent Ar
mole fraction nitrogen
  
mole fraction oxygen
  
partial pressure nitrogen
  bar
partial pressure oxygen
  bar
partial pressure argon
  bar

Explanation / Answer

Given that the density of air at 0.987 bar and 27°C is 1.146 kg m3, calculate the mole fraction and partial pressure of nitrogen and oxygen assuming the following conditions.

For air:

mol fraction of O2 = 0.21

mol fraction of N2 = 0.79

for partial pressures:

P-O2 = x-O2 * Ptotal = 0.21*0.987 = 0.20727 bar

P-N2 = = x-N2 * Ptotal = 0.79*0.987 = 0.77973 bar

b)

(b) air also contains 1.0 mole percent Ar
x-N2 = 0.79

x-O2 = 0.20

x-Ar = 0.01

so..

P-O2 = x-O2 * Ptotal = 0.20*0.987 = 0.1974bar

P-N2 = = x-N2  * Ptotal = 0.79*0.987 = 0.77973 bar

P-Ar= x-Ar* Ptotal = 0.01*0.987 = 0.00987 bar

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