In a flask of volume V litres- 0-2 mol of oxygen- 0-4 mol of nitrogen- 0-1 mol of ammonia and 0-3 mol of helium are enclosed 27-circ-C- If the total pressure exerted by these non-reacting gases is one atmosphere- the partial pressure exerted by nitrogen is -0-1 atmosphere0-3 atmosphere0-4 atmosphere0-2 atmosphere

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Toppr Admin · Accepted Answer

As we know-Partial pressure - Mole fraction x total pressure-x2234-PN2-XN2-xD7-PtotalPN2-0-40-2-0-4-0-1-0-3-xD7-1-0-4-xA0-atm

As we know,Partial pressure = Mole fraction x total pressure∴PN2=XN2×PtotalPN2=0.40.2+0.4+0.1+0.3×1=0.4 atm

As we know,
Partial pressure = Mole fraction x total pressure
$∴ P_{N_{2}} = X_{N_{2}} \times P_{t o t a l}$
$P_{N_{2}} = \frac{0.4}{0.2 + 0.4 + 0.1 + 0.3} \times 1 = 0.4 a t m$