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Standard XII

Chemistry

Dalton's Law of Partial Pressures

Air contains ...

Question

Air contains 79% of $N_{2}$ and 21% of $O_{2}$ by volume. If the barometric pressure is 750 mm of Hg, the partial pressure of the oxygen is:

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Solution

Sijnce the percentage composition of air in the mixture is $21$ %, mole fraction of air is equal to, $\frac{21}{21 + 79} = 0.21$

According to Dalton's theory of partial pressure,

$p_{O_{2}} = χ_{O_{2}} p_{t o t a l}$

$\Rightarrow p_{O_{2}} = 0.21 \times 750 = 157.5 m m o f H g$

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Air contains 79% of N2 and 21% of O2 by volume. If the barometric pressure is 750 mm of Hg, the partial pressure of the oxygen is:

Sijnce the percentage composition of air in the mixture is 21 %, mole fraction of air is equal to, 2121+79=0.21According to Dalton's theory of partial pressure, pO2=χO2ptotal⇒pO2=0.21×750=157.5 mm of Hg

Air contains 79% of $N_{2}$ and 21% of $O_{2}$ by volume. If the barometric pressure is 750 mm of Hg, the partial pressure of the oxygen is:

Sijnce the percentage composition of air in the mixture is $21$ %, mole fraction of air is equal to, $\frac{21}{21 + 79} = 0.21$

According to Dalton's theory of partial pressure,

$p_{O_{2}} = χ_{O_{2}} p_{t o t a l}$

$\Rightarrow p_{O_{2}} = 0.21 \times 750 = 157.5 m m o f H g$