Determine partial pressures of `O_(2) and N_(2)` in air at `0^(@)C` and 760 mm Hg pressure. Air contains 78% `N_(2) and 22%` `O_(2)` by volume.

Explanation

Mole-fraction of
`N_(2)(x_(N_(2)))=("Partial volume of "N_(2))/("Total volume")=(78)/(100)=0.78`
Similarly, mole-fraction of `O_(2)(x_(O_(2)))=(22)/(100)=0.22`
`therefore p_(N_(2))=x_(N_(2)) xxP=0.78xx760 mm Hg=592.8mm Hg`
`p_(O_(2))=x_(O_(2)) xxP=0.22xx760mm" "Hg=167.2mm" "Hg`