Equal amounts of two gases of molecular weigh | vedclass

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(D) Given, total pressure of the mixture $(P_{total}) = 1.1 \ atm$. Let the mass of each gas be $m$. Moles of lighter gas $(n_1) = \frac{m}{4}$. Moles

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(D) Given, total pressure of the mixture $(P_{total}) = 1.1 \ atm$. Let the mass of each gas be $m$. Moles of lighter gas $(n_1) = \frac{m}{4}$. Moles of heavier gas $(n_2) = \frac{m}{40}$. Total moles $(n_{total}) = \frac{m}{4} + \frac{m}{40} = \frac{10m + m}{40} = \frac{11m}{40}$. Mole fraction of lighter gas $(x_1) = \frac{n_1}{n_{total}} = \frac{m/4}{11m/40} = \frac{m}{4} 	imes \frac{40}{11m} = \frac{10}{11}$. Partial pressure of lighter gas $= x_1 	imes P_{total} = \frac{10}{11} 	imes 1.1 \ atm = 1 \ atm$.

Equal amounts of two gases of molecular weights $4$ and $40$ are mixed. The pressure of the mixture is $1.1 \ atm$. What will be the partial pressure of the lighter gas in the mixture (in $atm$)?

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