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Standard XII

Physics

In a mixture ...

Question

In a mixture of $N_{2}$ and $C O_{2}$ gases, the partial pressure of $C O_{2}$ is 1.25 atm. The total pressure of the mixture is 5 atm.Find the mole fraction of $N_{2}$ in the mixture.

0.82

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0.75

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0.80

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0.65

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Solution

The correct option is B
0.75

$P_{N_{2}} = X_{N_{2}} \times P_{T}$

$P_{T} = P_{C O_{2}} + P_{N_{2}}$

5 = 1.25 + $P_{N_{2}}$

$P_{N_{2}}$ = 3.75

3.75 = $X_{N_{2}} \times 5$

$X_{N_{2}} = \frac{3.75}{5} \Rightarrow 0.75$

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In a mixture of N2 and CO2­ gases, the partial pressure of CO2 is 1.25 atm. The total pressure of the mixture is 5 atm.Find the mole fraction of N2 in the mixture.

The correct option is B 0.75 PN2=XN2×PT PT=PCO2+PN2 5 = 1.25 + PN2 PN2 = 3.75 3.75 = XN2×5 XN2=3.755⇒0.75

In a mixture of $N_{2}$ and $C O_{2}$ gases, the partial pressure of $C O_{2}$ is 1.25 atm. The total pressure of the mixture is 5 atm.Find the mole fraction of $N_{2}$ in the mixture.

The correct option is B
0.75

$P_{N_{2}} = X_{N_{2}} \times P_{T}$

$P_{T} = P_{C O_{2}} + P_{N_{2}}$

5 = 1.25 + $P_{N_{2}}$

$P_{N_{2}}$ = 3.75

3.75 = $X_{N_{2}} \times 5$

$X_{N_{2}} = \frac{3.75}{5} \Rightarrow 0.75$