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The pressure exerted by $6.0 \mathrm{~g}$ of methane gas in a $0.03 \mathrm{~m}^3$ vessel at $129^{\circ} \mathrm{C}$ is (Atomic masses : $\mathrm{C}=12.01, \mathrm{H}=1.01$ and $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )
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Verified Answer
The correct answer is:
$41648 \mathrm{~Pa}$
Given,
volume, $\mathrm{V}=0.03 \mathrm{~m}^3$
temperature, $\mathrm{T}=129+273=402 \mathrm{~K}$
mass of methane, $w=6.0 \mathrm{~g}$
mol. mass of methane, $M=12.01+4 \times 1.01$
$=16.05$
From, ideal gas equation,
$$
\begin{aligned}
& \mathrm{pV}=\mathrm{nRT} \\
& \mathrm{p}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}=41648 \mathrm{~Pa}
\end{aligned}
$$
volume, $\mathrm{V}=0.03 \mathrm{~m}^3$
temperature, $\mathrm{T}=129+273=402 \mathrm{~K}$
mass of methane, $w=6.0 \mathrm{~g}$
mol. mass of methane, $M=12.01+4 \times 1.01$
$=16.05$
From, ideal gas equation,
$$
\begin{aligned}
& \mathrm{pV}=\mathrm{nRT} \\
& \mathrm{p}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}=41648 \mathrm{~Pa}
\end{aligned}
$$
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