To find the partial pressure of $N_2,$ we will use the concept of partial pressures and mole fractions.$\\$Given:$\\$• Mass of $N_2 = 7$ g$\\$• Mass of Ar$= 8$ g$\\$• Total pressure $P_{total} = 27$ bar$\\$Atomic masses:$\\$• $N = 14$ g/mol$\\$• $Ar = 40$ g/mol$\\$First, calculate the number of moles of each gas:$\\$Moles of $N_2 = \frac{mass \ of \ N_2}{molar \ mass \ of N_2} = \frac{7}{28} = 0.25$ mol$\\$Moles of$Ar = \frac{mass \ of \ Ar}{molar \ mass \ of \ Ar} = \frac{8}{40} = 0.2$ mol$\\$Total moles of gas in the mixture:$\\ n_{total} = n_{N_2} + n_{Ar} = 0.25 + 0.2 = 0.45$ mol$\\$Mole fraction of $N_2: \\ \chi_{N_2} = \frac{n_{N_2}}{n_{total}} = \frac{0.25}{0.45} \\$Calculate $\chi_{N_2}: \\ \chi_{N_2} = \frac{0.25}{0.45} = \frac{5}{9} \\$Partial pressure of $N_2$ is given by:$\\ P_{N_2} = \chi_{N_2} \times P_{total} \\ P_{N_2} = \frac{5}{9} \times 27 \\ P_{N_2} = 15$ bar$\\$Therefore, the partial pressure of $N_2$ is $15$ bar, which corresponds to Option 2.