A vessel is filled with a mixture of oxygen and nitrogen. At what ratio of partial pressures will the mass of gases be identical?

P (O_{2}) = 0.785 P (N_{2})

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P (O_{2}) = 8.75 P (N_{2})

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P (O_{2}) = 11.4 P (N_{2})

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P (O_{2}) = 0.875 P (N_{2})

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Solution

The correct option is D $P (O_{2}) = 0.875 P (N_{2})$
As we know,
$P V = n R T$
$P V = \frac{w}{m} R T$
$P_{O_{2} V} = \frac{w}{32} R T$ ...(i)
$P_{O_{2} V} = \frac{w}{28} R T$ ...(ii)
$\frac{P_{O_{2}}}{P_{N_{2}}} = \frac{28}{32}$
$P_{O_{2}} = 0.875 P_{N_{2}}$

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Q. A vessel is filled with a mixture of oxygen and nitrogen. At what ratio of the partial pressure will the masses of the gases be identical?

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A vessel is filled with a mixture of oxygen and nitrogen. At what ratio of partial pressures will the mass of gases be identical?

The correct option is D P(O2)=0.875P(N2)As we know,PV=nRTPV=wmRTPO2V=w32RT ...(i)PO2V=w28RT ...(ii)PO2PN2=2832PO2=0.875PN2

The correct option is D $P (O_{2}) = 0.875 P (N_{2})$
As we know,
$P V = n R T$
$P V = \frac{w}{m} R T$
$P_{O_{2} V} = \frac{w}{32} R T$ ...(i)
$P_{O_{2} V} = \frac{w}{28} R T$ ...(ii)
$\frac{P_{O_{2}}}{P_{N_{2}}} = \frac{28}{32}$
$P_{O_{2}} = 0.875 P_{N_{2}}$