What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g

It is known that,

$p=\frac{m}{M} \frac{\mathrm{RT}}{V}$

For methane (CH­4),

$p_{\mathrm{CH}_{4}}=\frac{3.2}{16} \times \frac{8.314 \times 300}{9 \times 10^{-3}}\left[\begin{array}{l}\text { Since } 9 \mathrm{dm}^{3}=9 \times 10^{-3} \mathrm{~m}^{3} \\ 27^{\circ} \mathrm{C}=300 \mathrm{~K}\end{array}\right]$

$=5.543 \times 10^{4} \mathrm{~Pa}$

For carbon dioxide (CO2),

$p_{\mathrm{CO}_{2}}=\frac{4.4}{44} \times \frac{8.314 \times 300}{9 \times 10^{-3}}$

$=2.771 \times 10^{4} \mathrm{~Pa}$

Total pressure exerted by the mixture can be obtained as:

$p=p_{\mathrm{CH}_{4}}+p_{\mathrm{CO}_{2}}$

$=\left(5.543 \times 10^{4}+2.771 \times 10^{4}\right) \mathrm{Pa}$

$=8.314 \times 10^{4} \mathrm{~Pa}$

Hence, the total pressure exerted by the mixture is 8.314 × 104 Pa.