A mixture of 8 g oxygen ,14 g nitrogen and 22g carbon dioxide is contained in a vessel of volume 4L What will be the pressure of the gas mixture if the temperature of the mixture is `27^@C`?

Explanation

We know ,pV=nRT [n=number of moles]
or, `p=(nRT)/V=g/M.(RT)/V`[g= mass of the gas, M=atomic weight of the gas]
Pressure exerted by `O_2` is `p_(O_2)`= `8/32,(RT)/V=1/4(RT)/V`
Pressure exerted by `N_2` is `p_(N_2)` =`14/28,(RT)/V=1/(RT)/V`
and pressure exerted by `CO_2` is `p_(CO_2)`=`22/44.(RT)/V=1/2(RT)/V`
From Dalton.s law pressure of the gas mixture
`p=p_(O_2)+P_(N_2)+P_(CO_2)=(RT)/V(1/4+1/2+1/2)=5/4(RT)/V`
`5/4times(8.315times300)/(4times10^-3)=7.795 times 10^5 N//m^2`