The pressure exerted by a non- reactive gaseous mixture of 6-4 g of methane and 8-8 g of carbon dioxide in a 10 L vessel at 27-C is kPaRound off to the Nearest Integer-Assume gases are ideal- R-8-314 J mol-1 K-1 Atomic masses - C-12-0 u- H - 1-0 u- O-16-0 u-

Number of moles of CH_4=6.416=0.4 mol nbsp; Number of moles CO_2=8.844=0.2 mol Total number of moles of the mixture=0.6 mol Pressure of the mixture of gases i ...

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Standard XII

Chemistry

Law of Radioactivity

The pressure ...

Question

The pressure exerted by a non- reactive gaseous mixture of $6.4 g$ of methane and $8.8 g$ of carbon dioxide in a $10 L$ vessel at $27^{\circ}$ C is ____ $k P a$
(Round off to the Nearest Integer).
[Assume gases are ideal, $R = 8.314 J m o l^{- 1} K^{- 1}$
Atomic masses : $C = 12.0 u, H : 1.0 u, O : 16.0 u$ ]

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Solution

$Number of moles of C H_{4} = \frac{6.4}{16} = 0.4 m o l$
$Number of moles C O_{2} = \frac{8.8}{44} = 0.2 m o l$
$Total number of moles of the mixture = 0.6 m o l$
Pressure of the mixture of gases in $10 L (0.01 m^{3})$ vessel at $300 K$ is given as:
$P = \frac{n R T}{V} = \frac{0.6 \times 8.314 \times 300}{0.01} = 149652 P a \approx 150 k P a$

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The pressure exerted by a non- reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27∘C is ____kPa (Round off to the Nearest Integer). [Assume gases are ideal, R=8.314 J mol−1 K−1 Atomic masses : C=12.0 u, H:1.0 u, O:16.0 u]

Number of moles of CH4=6.416=0.4 mol Number of moles CO2=8.844=0.2 mol Total number of moles of the mixture=0.6 mol Pressure of the mixture of gases in 10 L (0.01 m3) vessel at 300 K is given as: P=nRTV=0.6×8.314×3000.01=149652 Pa≈150 kPa