The pressure exerted by 6-0 g of methane gas in a 0-03 m3 vessel at 129-C is Atomic masses - C-12-01- H-1-01 and R-8-314 JK-1 mol-1

P=nRTV=6× 8.314× 40216.05× 0.03=41647 Pa ...

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Standard XII

Chemistry

Law of Radioactivity

The pressure ...

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 $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ $)$

31684 Pa

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215216 Pa

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13409 Pa

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41647 Pa

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Solution

The correct option is D $41647 Pa$
$P = \frac{n R T}{V} = \frac{6 \times 8.314 \times 402}{16.05 \times 0.03} = 41647 Pa$

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The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129∘C is (Atomic masses : C=12.01,H=1.01 and R=8.314 JK−1 mol−1)

The correct option is D 41647 PaP=nRTV=6×8.314×40216.05×0.03=41647 Pa

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 $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ $)$

The correct option is D $41647 Pa$
$P = \frac{n R T}{V} = \frac{6 \times 8.314 \times 402}{16.05 \times 0.03} = 41647 Pa$