Moles of N_2=0.4 and moles of O_2=0.1 find p_ N_2 (Partial...

Key Idea According to Raoult's law, "at a given temperature, for a solution of volatile liquids, the partial vapour pressure of each component of the solution is direcently proportional to its mole fraction present in solution. According to Raoult's law, $\mathrm{p}_{\mathrm{A}}=\mathrm{x}_{\mathrm{A}} \mathrm{p}_{\mathrm{T}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}}} \mathrm{p}_{\mathrm{T}}$ where, $\mathrm{P}_{\mathrm{A}}=$ partial pressure $\mathrm{p}_{\mathrm{T}}=$ total pressure $\mathrm{n}_{\mathrm{A}}$ and $\mathrm{n}_{\mathrm{B}}=$ moles Given, ${ }^{\mathrm{n}} \mathrm{N}_2=0.4,{ }^{\mathrm{n}} \mathrm{O}_2=0.1, \mathrm{p}_{\mathrm{T}}=1 \mathrm{~atm}$$\begin{aligned} \therefore \quad \mathrm{P}_{\mathrm{N}_2} & =\frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{N}_2}+\mathrm{n}_{\mathrm{O}_2}} \mathrm{P}_{\mathrm{T}}=\frac{0.4}{0.4+0.1} \times 1 \mathrm{~atm} \\ & =\frac{0.4}{0.5}=0.8 \mathrm{~atm} \end{aligned}$$$