Q11 / 120 · VITEEE 2012 · Chemistry
The pressure exerted by $6.0 \mathrm{~g}$ of methane gas in $\mathrm{a} 0.03 \mathrm{~m}^{3}$ vessel at $129^{\circ} \mathrm{C}$ is
(Atomic masses: $\mathrm{C}=12.01, \mathrm{H}=1.01$ and $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )
- A$215216 \mathrm{~Pa}$
- B$13409 \mathrm{~Pa}$
- Correct answerC$41648 \mathrm{~Pa}$
- D$31684 \mathrm{~Pa}$
Solution
$\left[\begin{array}{l}\text { w(given mass of methane) }=6 \mathrm{~g} \\ \text { temperature, } T=129+273=402 \mathrm{~K} \\ \text { mol mass of methane, } M=12.01+4 \times 1.01 \\ =16.05\end{array}\right]$
From, ideal gas equation,
$$
\begin{array}{l}
p V=n R T \Rightarrow P=\frac{n R T}{v} \\
\mathrm{p}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}=41648 \mathrm{~Pa}
\end{array}
$$
Hence, the correct answer is option: 'C'.