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

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(C) The number of moles of oxygen $(n_{O_2})$ is $8/32 = 0.25\,mol$.The number of moles of nitrogen $(n_{N_2})$ is $7/28 = 0.25\,mol$.According to Dal

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$5$

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(C) The number of moles of oxygen $(n_{O_2})$ is $8/32 = 0.25\,mol$.The number of moles of nitrogen $(n_{N_2})$ is $7/28 = 0.25\,mol$.According to Dalton's law of partial pressures,the total pressure $P_{total} = (n_{O_2} + n_{N_2}) \frac{RT}{V} = 10\,atm$.Since $n_{O_2} = n_{N_2} = 0.25\,mol$,the partial pressure of each gas is equal.$P_{O_2} = P_{N_2} = \frac{10}{2} = 5\,atm$.When oxygen is removed,only nitrogen remains in the vessel.Therefore,the new pressure in the system will be equal to the partial pressure of nitrogen,which is $5\,atm$.

$A$ closed vessel contains $8\,g$ of oxygen and $7\,g$ of nitrogen. The total pressure is $10\,atm$ at the given temperature. If all the oxygen is removed from the system without permitting any change in temperature,then the pressure will become (in atmosphere):

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