SATHEE: States of Matter - Result Question 38

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 $R = 8.314 J K^{- 1} m o l^{- 1}$ )

(a) $31684 P a$

(b) $215216 P a [2010]$

(c) $13409 P a$

(d) $41648 P a$

Show Answer

Solution:

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

$= \frac{6}{16.05} \frac{\times 8.314 \times 402}{0.03} ≃ 41648 P a$

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