States of Matter - Result Question 38

####38. The pressure exerted by $6.0 g$ of methane gas in a $0.03 m^{3}$ vessel at $129^{\circ} C$ is (Atomic masses : $C=12.01, H=1.01$ and $R=8.314 JK^{-1} mol^{-1}$ )

(a) $31684 Pa$

(b) $215216 Pa[2010]$

(c) $13409 Pa$

(d) $41648 Pa$

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Solution:

(d) $P=\frac{n R T}{V}=\frac{w}{m} \frac{R T}{V}$

$=\frac{6}{16.05} \frac{\times 8.314 \times 402}{0.03} \simeq 41648 Pa$



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