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).

8g oxygen, 14 g nitrogen and 22 g carbon dioxide are mixed in a contai

1.	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).
Answer» Solution :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)`.