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  • Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K^–1 mol^–1.

5.15     Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of  1\: dm^{3} at 27°C. R = 0.083\: bar \: dm^{3} K^{-1} mol^{-1}.

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Given the mass of oxygen gas and mass of hydrogen gas.

Molar mass of O_{2} =32g\ mol^{-1}

Therefore 8g\ O_{2} = \frac{8}{32}mol = 0.25\ mol

Molar mass of H_{2} = 2g\ mol^{-1}

Therefore 4g H_{2} =\frac{4}{2} =2\ mol

\Rightarrow Total number of moles of mixture n = 2+0.25 =2.25

and V = 1\ dm^3,\ T = 27^{\circ }C = 300K,\ R=0.083\ bar\ dm^{3}K^{-1}\ mol^{-1}

So, Ideal gas equation; PV = nRT

or P = \frac{nRT}{V} = \frac{(2.25mol\times) (0.083 bar\ dm^3K^{-1}mol^{-1})(300K)}{1 dm^3}

Therefore Total pressure is = 56.025\ bar

Posted by

Divya Prakash Singh

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