A closed vessel contains 8,g of oxygen and 7, | vedclass

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(D) The number of moles of oxygen $(n_{O_2})$ is $n_{O_2} = \frac{8\,g}{32\,g/mol} = 0.25\,mol$.The number of moles of nitrogen $(n_{N_2})$ is $n_{N_2

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$5$

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(D) The number of moles of oxygen $(n_{O_2})$ is $n_{O_2} = \frac{8\,g}{32\,g/mol} = 0.25\,mol$.The number of moles of nitrogen $(n_{N_2})$ is $n_{N_2} = \frac{7\,g}{28\,g/mol} = 0.25\,mol$.According to Dalton's law of partial pressures,the total pressure $P_{total}$ is proportional to the total number of moles $n_{total} = n_{O_2} + n_{N_2} = 0.25 + 0.25 = 0.50\,mol$.Given $P_{total} = 10\,atm$,the partial pressure of each gas is proportional to its mole fraction.Since $n_{O_2} = n_{N_2}$,the partial pressure of nitrogen is $P_{N_2} = \frac{n_{N_2}}{n_{O_2} + n_{N_2}} 	imes P_{total} = \frac{0.25}{0.50} 	imes 10\,atm = 5\,atm$.When oxygen is removed,only nitrogen remains in the vessel. Therefore,the pressure of the remaining gas is $5\,atm$.

$A$ closed vessel contains $8\,g$ of oxygen and $7\,g$ of nitrogen. The total pressure is $10\,atm$ at a given temperature. If oxygen is now absorbed by introducing a suitable absorbent,the pressure of the remaining gas in $atm$ will be:

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