A vessel of volume `V = 7.51` contains a mixture of ideal gases at a temperature `T = 300 K : v_1 = 0.10` mole of oxygen, `v_2 = 0.20` mole of nitrogen, and `v_3 = 0.30` mole of carbon dioxide. Assuming the gases to be ideal, find :
(a) the pressure of the mixture ,
(b) the mean molar mass `M` of the given mixture which enters its equation of state `p V = (m//M) RT`, where `m` is the mass of the mixture.

Answer

(a)2.32 (b)1.7r

Explanation

(a) Let `p_(1),P_(2) and P_(3)` be the partical pressures of the 3 gases Then `P_(1) v= n_(1)RT,P_(2) = n_(2) RT and P_(3) V=n_(3)RT`
`therefore P_(mix) = P_(1)+P_(2)+P_(3)=(n_(1)+n_(2)+n_(3))(RT)/(V)`
`=(0.1+0.2+0.3)xx(8.3xx300)/(7.5xx10^(-3)) = 1.992xx10^5Nm^(-2) = 2 atm`
(b) `M_(mix) = (n_(1)M_(1)+n_(2)M_(2)+n_(3)M_(3))/(n_(1)+n_(2)+n_(3))=(0.1xx32xx0.2xx28xx0.3xx44)/(0.1+0.2+0.3)=36.7`