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?

(a) $0.8\times {10}^{5}atm$ (b) $0.008N{m}^{-2}$ (c) $8\times {10}^{4}N{m}^{-2}$ (d) 0.25 atm

Answer 1

Answer: (c) $8\times {10}^{4}N{m}^{-2}$

Given that, pressure of a 1:4 mixture of $H_2$ and $O_2$ enclosed ia a vessel is one atmosphere. This suggests that the moles ratio of $H_2$ and $O_2$ is 1:4. Thus, partial pressure of dioxygen $\left(O_2\right)$ is given as

$therefore$ Partial pressure of $O_2$= Mole fraction of $O_2$ $\times$ total pressure of mixture

$=\frac{4}{1+4}\times \frac{4}{5}\times 1atm$

$=0.8atm=0.8\times {10}^{5}N{m}^{-2}=8\times {10}^{4}N{m}^{-2}$