Q1. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in 9 dm3 flask at 27 degree celsius?
Q2. What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of O2 at 0.7 bar are introduced in a 1 L vessel at 27 degree celsius?

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Solution

1. Molar mass of methane = 12 + 4 x 1 = 16 g/mol
Molar mass of carbon dioxide = 12 + 2 x 16 = 44 g/mol

Number of moles of methane in 3.2 g = 3.2/16 = 0.2
Number of moles of carbondioxide in 4.4 g = 4.4/44 = 0.1

Total number of moles of gases in the mixture = 0.2 + 0.1 = 0.3

Using ideal gas equation:
PV = nRT
P = nRT/V
= 0.3 x 0.08206 x 300 K) / 9 L (1L = 1dm³)
= 0.82 atm

2. Applying ideal gas equation to find the number of moles of H₂ and O₂ gases in the mixture:

$n_{H_{2}} = \frac{P V}{R T} = \frac{0.8 b a r X 0.1 L}{0.08206 L \cdot b a r \cdot m o l^{- 1} \cdot K^{- 1} x 300 K} (\sin c e 1 b a r ≃ 1 a t m) n_{O_{2}} = \frac{0.7 b a r X 2 L}{0.08206 L \cdot b a r \cdot m o l^{- 1} \cdot K^{- 1} x 300 K} N o w, P_{m i x t u r e} = P_{H_{2}} + P_{O_{2}} = (n_{H_{2}} + n_{O_{2}}) \frac{R T}{V} = \frac{0.8 b a r x 0.1 L + 0.7 b a r x 2 L}{0.08206 L \cdot b a r \cdot m o l^{- 1} \cdot K^{- 1} x 300 K} x \frac{0.08206 L \cdot b a r \cdot m o l^{- 1} \cdot K^{- 1} x 300 K}{I L} = (0.4 + 1.4) b a r = 1.8 b a r$

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H_{2}

27^{\circ} C

0.5 L

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3.2 g

4.4 g

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R = 8.314 {J K}^{- 1} {mol}^{- 1}

What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a $9 d m^{3}$ flask at $27^{\circ} C$ ?

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