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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 $27^{\circ} \mathrm{C}$ is $\mathrm{kPa}$. (Round off to the Nearest Integer).
[Assume gases are ideal, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ Atomic masses : $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}]$
Solution:
Total moles of gases, $\mathrm{n}=\mathrm{n}_{\mathrm{CH}_{4}}+\mathrm{n}_{\mathrm{CO}_{2}}$
$=\frac{6.4}{16}+\frac{8.8}{44}=0.6$
Now, $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{0.6 \times 8.314 \times 300}{10 \times 10^{-3}}$
$=1.49652 \times 10^{5} \mathrm{~Pa}=149.652 \mathrm{kPa}$
$\approx 150 \mathrm{kPa}$
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