To solve this problem, we need to calculate the work done by the gas during the expansion.$\\$Given:$\\$• Initial volume $V_1 = 0.1$ L$\\$• Final volume $V_2 = 0.25$ L$\\$• External pressure $P_{ext} = 2$ bar$\\$• Conversion factor: $1$ L bar $= 100$ J$\\$The work done by the gas during expansion against a constant external pressure is given by:$\\ W = -P_{ext} \times \Delta V \\$where $\Delta V = V_2 - V_1$ is the change in volume.$\\$Calculate $\Delta V: \\ \Delta V = 0.25$ L$- 0.1$ L$= 0.15$ L$\\$Substitute the values into the work formula:$\\ W = -2$ bar$\times 0.15$ L$\\$Convert the work from L bar to J using the conversion factor:$\\ W = -2 \times 0.15 \times 100$ J$\\ W = -30$ J$\\$Therefore, the work done by the gas is $-30$ J, which corresponds to Option 1.