At sea level, the composition of dry air is approximately `N_(2)=75.5%`, `O_(2)=23.2%`, and `Ar=1.3%` by mass. If the total pressure at sea level is `1 "bar"`, what is the partial pressure of each component?

Answer

`p(N_(2))="0.781 bar"`
`p(O_(2))="0.210 bar"`
`p(Ar)="0.009 bar"`

Explanation

Let us first calculate the mole fraction of each component.
`"Moles of "N_(2)=(75.5)/(28)=2.696`,
`"Moles of "O_(2)=(23.3)/(32)=0.725`
`"Moles of Ar"=(1.3)/(40)=0.0325`
`"Total moles"=2.696+0.725+0.0325=3.4535`
`"Mole fraction of "N_(2), x_(N_(2))=(2.696)/(3.4535)=0.781,`
`"Mole fraction of "O_(2), x_(O_(2))=(0.725)/(3.4535)=0.210`
`"Mole fraction of Ar, x"_(Ar)=(0.0325)/(3.4535)=0.0094`
`p_((N_(2)))=x_(N_(2))xxp=0.781xx1="0.781 bar"`
`p_((O_(2)))=x_(O_(2))xxp=0.210xx1="0.210 bar"`
`p_((Ar))=x_(Ar)xxp=0.009xx1="0.0094 bar."`