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Gas laws from KTG

A closed vess...

Question

A closed vessel contains a mixture of one mole of oxygen $(O_{2})$ and two moles of Nitrogen $(N_{2})$ at $300 K$ . Find the ratio of partial pressure exerted by $O_{2}$ to the total pressure exerted by mixture on the wall of vessel?

1 : 3

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3 : 1

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2 : 3

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1 : 2

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Solution

The correct option is A $1 : 3$
From Dalton's law of partial pressures,
$P_{total} = P_{O_{2}} + P_{N_{2}}$
We know,
$P_{total} V = n_{total} \times R \times T . . . (i)$
where, $n_{total} = n_{O_{2}} + n_{N_{2}}$
Also,
$P_{O_{2}} V = n_{O_{2}} R T . . . . (i i)$

From equation $(i)$ and $(i i)$ , we get
$\frac{P_{O_{2}}}{P_{total}} = \frac{n_{O_{2}}}{n_{O_{2}} + n_{N_{L}}} = \frac{1}{1 + 2} = \frac{1}{3}$

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Gas laws from KTG

Standard XI Physics

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A closed vessel contains a mixture of one mole of oxygen (O2) and two moles of Nitrogen (N2) at 300 K. Find the ratio of partial pressure exerted by O2 to the total pressure exerted by mixture on the wall of vessel?

The correct option is A 1:3From Dalton's law of partial pressures, Ptotal=PO2+PN2 We know, PtotalV=ntotal×R×T...(i) where, ntotal=nO2+nN2 Also, PO2V=nO2RT....(ii) From equation (i) and (ii), we get PO2Ptotal=nO2nO2+nNL=11+2=13

A closed vessel contains a mixture of one mole of oxygen $(O_{2})$ and two moles of Nitrogen $(N_{2})$ at $300 K$ . Find the ratio of partial pressure exerted by $O_{2}$ to the total pressure exerted by mixture on the wall of vessel?