Answer:
The concentration of nitrogen per liter of air is 3.189 × 10⁻² moles
while the concentration of oxygen per liter of air is 8.59 × 10⁻³ moles
Explanation:
From Dalton's law of partial pressure, we have the pressure of the air is due to the partial pressures of the constituent gases
Therefore from the ideal gas law we have
PV = nRT, or P = nRT/V which for nitrogen is with volume = 78 % of the total volume giving us
Pressure of nitrogen = nRT/(0.78V) and oxygen gives
nRT/(0.21V) while the remaining contituent 1 % is
nRT/(0.01V)
Therefore we have nRT/(0.78V) + nRT/(0.21V) + nRT/(0.01V) = Ptotal
however by Avogadro's law, equal volume of all gases at the same temperature and pressure contain equal number of molecules
and the partial pressure of a gas =mole fraction × total pressure
From which we have the mole fraction of nitrogen given as
nRT/(0.78V) + nRT/(0.21V) + nRT/(0.01V) = nRT/V
cancelling like terms gives
n(nitrogen)/0.78 +n(oxygen)/0.21 +n(others)/0.01 =n/1
therefore the mole fraction of nitrogen = 0.78
and the mole fraction of oxygen = 0.21
or n = PV/RT = 1 * 0.78/(0.082057* 298) = 3.189 × 10⁻² moles per liter
Also we have for oxygen =1*0.21/(0.082057* 298) = 8.59 × 10⁻³ moles per liter
