A gaseous mixture of two substances A and B , under a total pressure of 0 . 8   atm is in equilibrium…

Raoult’s law states that a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.

Raoult’s law is valid only in the case of ideal solutions

${\mathrm{P}}_{\mathrm{A}}={{\mathrm{P}}^{\mathrm{o}}}_{\mathrm{A}}{\mathrm{\chi }}_{\mathrm{A}}\left(\mathrm{According} \mathrm{to} \mathrm{Raoult}\text{'}\mathrm{s} \mathrm{law}\right)$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{T}}{\mathrm{Y}}_{\mathrm{A}}\left(\mathrm{According} \mathrm{to} \mathrm{Dalton}\text{'}\mathrm{s} \mathrm{law}\right) \\ {\mathrm{P}}_{\mathrm{T}}=\mathrm{Total} \mathrm{pressure} \\ {\mathrm{P}}_{\mathrm{A}}=\mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{A} \mathrm{in} \mathrm{solution} \end{gathered}$$

${{\mathrm{P}}^{\mathrm{o}}}_{\mathrm{A}}=\mathrm{Vapour} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{liquid} \mathrm{A}$

${\mathrm{\chi }}_{\mathrm{A}}=\mathrm{Mole} \mathrm{fraction} \mathrm{of} \mathrm{component} \mathrm{A} \mathrm{in} \mathrm{liquid} \mathrm{phase}$

${\mathrm{Y}}_{\mathrm{A}}=\mathrm{Mole} \mathrm{fraction} \mathrm{of} \mathrm{component} \mathrm{A} \mathrm{in} \mathrm{gaseous} \mathrm{phase}$

So, ${\mathrm{P}}_{\mathrm{A}}={{\mathrm{P}}^{\mathrm{o}}}_{\mathrm{A}}{\mathrm{\chi }}_{\mathrm{A}}={\mathrm{Y}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{T}}$

${\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}\left(0.2\right)=0.5\times 0.8$

${\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}=2 \mathrm{atm}$