Question

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

Answer 1

It is known that,
$p=\frac{m}{M}=\frac{RT}{V}$
For methane $\left(C{H}_4\right)$,
$P_{C{H}_4}=\frac{3.2}{16}\times \frac{8.314\times 300}{9\times {10}^{-3}}\left[\begin{array}{c}Since9d{m}^3=9\times {10}^{-3}{m}^3\\ {27}^{\circ }=300K\end{array}\right]=5.543\times {10}^{4}Pa$
For carbon dioxide $\left(C{O}_2\right)$,
$p_{C{O}_2}=\frac{4.4}{44}\times \frac{8.314\times 300}{9\times {10}^{-3}}=2.771\times {10}^{4}PA$
Total pressure exerted by the mixture can be obtained as:
$p=P_{C{H}_4}+P_{C{O}_2}=(5.543\times {10}^4+2.771\times {10}^4)Pa=8.314\times {10}^{4}Pa$
Hence, the total pressure exerted by the mixture is $8.314\times {10}^{4}Pa$.