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5.8 What will be the pressure of the gaseous mixture when $0.5\: L$ of $H_{2}$ at $0.8$ bar and $2.0\: L$ of dioxygen at $0.7$ bar are introduced in a 1L vessel at $27^{o}C$ ?

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Pressure of the gas mixture will be the sum of partial pressure of $H_{2}$ gas and partial pressure of $O_{2}$ gas.

So, calculating the partial pressures of each gas.

Partial pressure of $H_{2}$ in a 1L volume vessel.

$P_{1} = 0.8\ bar$ , $P_{2},\ V_{1} = 0.5L,\ V_{2} =1L$

As the temperature remains constant, $P_{1}V_{1} =P_{2}V_{2}$

$\implies (0.8\ bar)(0.5L) = P_{2}(1L)$

or $P_{2} = 0.4\ bar$ or $P_{H_{2}} = 0.4\ bar$

Now, calculating the partial pressure of $O_{2}$ gas in 1L vessel.

$P'_{1}V_{1} = P'_{2} V'_{2}$

$\implies (0.7\ bar)(2L) = P_{2}(1L)$

$\implies P'_{2} = 1.4\ bar$ or $P_{O_{2}} = 1.4\ bar$

Therefore the total pressure $= P_{H_{2}} +P_{O_{2}} = 0.4\ bar+1.4\ bar = 1.8\ bar$

Posted by

Divya Prakash Singh

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