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.

Explanation

(a) The mixture contains `v_1, v_2` and `v_3` moles of `O_2, N_2` and `CO_2` respectively. Then the total number of moles of the mixture
`v = v_1 + v_2 + v_3`
We know, ideal gas equation for the mixture
`p V = vRT` or `p = (v RT)/(V)`
or, `p = ((v_1 + v_2 + v_3)RT)/(V) = 1.968 atm` on substitution.
(b) Mass of oxygen `(O_2)` present in the mixture : `m_1 = v_1 M_1`
Mass of nitrogen `(N_2)` present in the mixture : `m_2 = v_2 M_2`
Mass of carbon dioxide `(CO_2)` present in the mixture : `m_3 = v_3 M_3`
So, mass of the mixture
`m = m_1 + m_2 + m_3 = v_1 M_1 + v_1 M_2 + v_3 M_3`
Molecular mass of mixture : `M = ("mass of the mixture")/("total number of moles")`
=`(v_1 M_1 + v_2 M_2 + v_3 M_3)/(v_ 1+ v_2 + v_3) = 36.7 g//mol`. on substitution.