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Dalton's Law of Partial Pressures

If a mixture ...

Question

If a mixture of $C O_{2} a n d N_{2}$ in equal amount exert a total pressure of 1 atm, the partial pressure of $N_{2}$ in the mixture is –

0.56 atm

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1 atm

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1.52 atm

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0.75 atm

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Solution

The correct option is A
0.56 atm

Applying Dalton’s Law of partial pressure :

$p = P_{T} . χ$

Less mass of $N_{2}$ and $C O_{2}$ be 1 g each

$χ (m o l e f a c t i o n) o f N_{2} = \frac{n N_{2}}{n N_{2} + n C O_{2}} = \frac{(\frac{1}{28})}{\frac{1}{28} () + (\frac{1}{44})} = \frac{0.0357}{0.0357 + 0.0227} = 0.56 ∴ P = 1 \times 0.56 = 0.56 a t m$

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If a mixture of $C O_{2} a n d N_{2}$ in equal amount exert a total pressure of 1 atm, the partial pressure of $N_{2}$ in the mixture is –

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If a mixture of CO2 and N2 in equal amount exert a total pressure of 1 atm, the partial pressure of N2 in the mixture is –

The correct option is A 0.56 atm Applying Dalton’s Law of partial pressure : p=PT.χ Less mass of N2 and CO2 be 1 g each χ (mole faction) of N2=nN2nN2+nCO2=(128)128()+(144)=0.03570.0357+0.0227=0.56∴P=1×0.56=0.56 atm

If a mixture of $C O_{2} a n d N_{2}$ in equal amount exert a total pressure of 1 atm, the partial pressure of $N_{2}$ in the mixture is –