What will be the pressure of the gas mixture of $$3.2\,g$$  methane and $$4.4\,g$$  carbon dioxide contained in a $$9\,d{m^3}$$  flask at $$27{\,^ \circ }C?$$

Solution :
$$P = \frac{n}{V}RT$$
$$n\,\,{\text{for}}\,\,C{H_4} = \frac{{3.2}}{{16}} = 0.2;$$     $$n\,\,{\text{for}}\,\,C{O_2} = \frac{{4.4}}{{44}} = 0.1$$
$${n_{{\text{total}}}} = 0.2 + 0.1 = 0.3;$$     $$P = \frac{{0.3 \times 0.0821 \times 300}}{9} = 0.82\,\,atm$$