Question

Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a temperature of 298 K. Assume that nitrogen comprises 78% of air by volume and that oxygen comprises 21%.

Answer 1

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