Question

The pressure exerted by a non-reactive gaseous mixture of $6.4\mathrm{g}$ of methane and $8.8\mathrm{g}$ of carbon dioxide in a $10\mathrm{L}$ vessel at $27°\mathrm{C}$ is _____ kPa. (Round off to the Nearest Integer). Assume gases are ideal, $\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$ Atomic masses: $\mathrm{C}:12.0\mathrm{u},\mathrm{H}:1.0\mathrm{u},\mathrm{O}:16.0\mathrm{u}$

Answer 1

Step 1: Given data

• Volume$\left(\mathrm{V}\right)=10\mathrm{L}$
• Temperature$\left(\mathrm{T}\right)={27}^{\circ }\mathrm{C}$
• Methane in mixture$=6·4\mathrm{g}$
• Carbon in mixture$=8·8\mathrm{g}$

Step 2: Formula Used

• The pressure exerted by pressure can be computed using the following formula:
• $\mathrm{PV}=\mathrm{nRT}$
• Where $\mathrm{P}=$Pressure, $\mathrm{V}=$Volume, $\mathrm{n}=$Amount of substance, $\mathrm{R}=$gas constant and $\mathrm{T}=$Temperature

Step 3: Compute the Total moles of gases

Total moles of gases in the mixture is,

$\begin{array}{rcl}\mathrm{n}& =& \frac{\mathrm{Mass}\mathrm{of}\mathrm{methane}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{methane}}+\frac{\mathrm{Mass}\mathrm{of}\mathrm{carbon}}{\mathrm{Molar}\mathrm{maas}\mathrm{of}\mathrm{carbon}}\\ & =& \frac{6.4}{16}+\frac{8.8}{44}\\ & =& 0.6\end{array}$

Step 4: Compute the exerted pressure

Substitute the known values in the formula $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$

$\begin{array}{rcl}\mathrm{P}& =& \frac{0·6\times 8·314\times \left(273+27\right)}{10}\\ & =& 149.652\\ & =& 150\mathrm{kPa}\end{array}$

Hence, the pressure exerted by the pressure is $150\mathrm{kPa}$.