The respective mole fractions of `N_(2) and O_(2)` is dry air are 0.78 and 0.21 gt if the atmospheric presure and temperature are 740 torr and `20^(@)C` respectively, then what will be the mass of `N_(2) and O_(2)` present in a room of volume of `3000ft^(3)`? (Assuming the relative humidity of air as zero)

Explanation

`P_(N_(2))=0.78xx740` torr `=0.78xx(740)/(760)=0.76` atm
`3000ft^(3)=3000xx(30.48)^(3)" "[because 1ft=30.48cm]`
`=84.95xx10^(6)` cc `=84.95xx10^(3)L`
`therefore P_(N_(2)) xxV=n_(N_(2))xxRT`
or, `0.76xx84.95xx10^(3)=n_(N_(2))xx0.0821xx(273+20)`
or, `n_(N_(2))=2.683xx10^(3)mol` ltbgt `therefore`Mass of `N_(2)=2.683xx10^(3)xx28g=75.149kg`
In calculation, it can be shown that mass of `O_(2)=23.107 kg`.