Question

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 $\left.R=8.314\, JK ^{-1} \,mol ^{-1}\right)$

• A. 215216 Pa
• B. 13409 Pa
• C. 41648 Pa
• D. 31684 Pa

Answer 1

Answer: (C)

41648 Pa

Answer 2

Given, volume, $V=0.03 \,m ^{3}$
temperature, $T=129+273=402\, K$
mass of methane, $W=6.0\, g$
mol mass of methane, $M=12.01+4 \times 1.01$
$=16.05$
From, ideal gas equation,
$p V=n R T$
$p=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}$
$=41648\, Pa$