A vessel contains a mixture of `7 g` of nitrogen and `11 g` carbon dioxide at temperature ` T = 290 K`. If pressure of the mixure is `1 atm (= 1.01 xx 10^5 N//m^2)`, calculate its density `(R = 8.31 J//mol - K)`.

Explanation

We know, for the mixture, `N_2` and `CO_2` (being regarded as ideal gases, their mixture too behaves like an ideal gas)
`p V = v R T`, so `p_0 V = v R T`
Where, `v` is the total number of moles of the gases (mixture) present and `V` is the volume of the vessel. If `v_1` and `v_2` are number of moles of `N_2` and `CO_2` respectively present in the mixture, then
`v = v_1 + v_2`
Now number of moles of `N_2` and `CO_2` is, by definition, given by
`v_1 = (m_1)/(M_1)` and, `v_2 = (m_2)/(M_2)`
where, `m_1` is the mass of `N_2` (Moleculer weight `= M_1`) in the mixture and `m_2` is the mass of `CO_2` (Molecular weight `= M_2`) in the mixture.
Therefore density of the mixture is given by
`rho = (m_1 + m_2)/(V) = (m_1 + m_2)/((v RT//P_0))`
=`(p_0)/(RT).(m_1 + m_2)/(v_1 + v_2) = (p_0(m_1 + m_2)M_1 M_2)/(RT(m_1 M_2 +m_2 M_1))`
=`1.5 kg//m^3` on substitution.