First slide

Gaseous state

Question

The pressure of a 1: 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?

Moderate

A

$0 .8 \times 10^{5} atm$
B

$0 .008 {Nm}^{- 2}$
C

$8 \times 10^{4} {Nm}^{- 2}$
D

$0 .25 atm$

Solution

Given that, pressure of a 1 : 4 mixture of H₂ and O₂ enclosed in a vessel is one atmosphere. This suggests that the moles ratio of H₂ and O₂ is 1:4. Thus, partial pressure of dioxygen (O₂ ) is given as
$∴ Partial pressure of O_{2} = mole fraction of O_{2} \times total pressure of mixrure$
$= \frac{4}{1 + 4} \times 1 = \frac{4}{5} \times 1 atm = 0 .8 atm = 0 .8 \times 10^{5} {Nm}^{- 2} = 8 \times 10^{4} {Nm}^{- 2}$

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