8g oxygen, 14 g nitrogen and 22 g carbon dioxide are mixed in a container of volume 4 l. Find out the pressure of the gas mixture at `27^(@)C`. Given `R = 8.315 J cdot mol^(-1) cdot K^(-1)`.

Explanation

If n = number of moles in a gas, then pV = nRT.
So, `p = (nRT)/(V) = m/V (RT)/(V)`, where m = mass of the gas and M = molecular weight.
Then the pressure due to oxygen, nitrogen and carbon dioxide gases, respectively, are
`p_(1) = 8/32 (RT)/V cdot p_(2) = 14/28 (RT)/V cdot p_(3) = 22/44 (RT)/V`
`:.` Net pressure of the gas mixture is
`p = p_(1) + p_(2) + p_(3)`
`=(8/32 + 14/28 + 22/44) RT/V = (1/4 + 1/2 + 1/2) (RT)/V = 5/4 (RT)/V`
`=5/4 xx (8.315 xx 300)/(4 xx 10^(-3))`
[Here, `T = 27^(@)C = 300K, V = 4 l = 4 xx 10^(-3) ^(3)]`
`= 7.795 xx 10^(5) N cdot m^(-2)`.