To solve this problem, we will use the ideal gas equation:$\\ PV = nRT \\$Given:$\\$• Volume $V = 10.0$ L$\\$• Mass of oxygen $= 64$ g$\\$• Temperature $T = 27^\circ$C$\\$• Universal gas constant $R = 0.0831$ L bar K$^{-1}$ mol$^{-1} \\$First, convert the temperature from Celsius to Kelvin:$\\ T = 27 + 273.15 = 300.15$ K$\\$Next, calculate the number of moles of $O_2: \\$Molar mass of $O_2 = 32$ g/mol$\newline n = \frac{mass}{molar \ mass} = \frac{64 g}{32 g/mol} = 2$ mol$\\$Now, substitute the values into the ideal gas equation:$\\ P \cdot 10.0 = 2 \cdot 0.0831 \cdot 300.15 \\$Simplify the equation:$\\ P \cdot 10.0 = 49.86 \\ P = \frac{49.86}{10.0} \\ P = 4.986$ bar$\\$Rounding to one decimal place, the pressure inside the flask is $4.9$ bar.$\\$This corresponds to Option 4.