Question

What will be the pressure exerted by a mixture of $3.2\text{g}$ of methane and $4.4\text{g}$ of carbon dioxide contained in a $9{\text{dm}}^3$ flask at ${27}^{\circ }\text{C}$?
(Given : $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$)

• A. $1.326\times {10}^6\text{Pa}$
• B. $8.314\times {10}^4\text{Pa}$
• C. $8.314\times {10}^6\text{Pa}$
• D. $3.751\times {10}^4\text{Pa}$

Answer 1

The correct option is B $8.314\times {10}^4\text{Pa}$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$
$\text{Number of moles of Methane}=\frac{3.2}{16}=0.2$
$\text{Number of moles of Carbon dioxide}=\frac{4.4}{44}=0.1$
$\text{Total number of moles}=0.1+0.2=0.3$
$\text{V}=9{\text{dm}}^3=9\times {10}^{-3}{\text{m}}^3$
$\text{T}=273+27=300K$
We have,
$PV=nRT$
$\Rightarrow P=\frac{nRT}{V}=\frac{0.3\times 8.314\times 300}{9\times {10}^{-3}}=8.314\times {10}^4\text{Pa}$