To solve this problem, we will use the formula for work done during a reversible isothermal expansion of an ideal gas.$\\$The formula for work done $(W)$ during isothermal expansion is:$\\ W = -nRT \ln \left(\frac{P_{final}}{P_{initial}}\right) \\$Given:$\\$• $n = 1$ mole (since we have one mole of hydrogen gas)$\\$• $R = 2.0$ cal K$^{-1}$ mol$^{-1} \\$• $T = 25^\circ$C$= 298$ K (since temperature must be in Kelvin)$\\$• $P_{initial} = 20$ atm$\\$• $P_{final} = 10$ atm$\\$Substitute these values into the formula:$\\ W = -1 \times 2.0 \times 298 \ln \left(\frac{10}{20}\right) \\$Simplify the expression:$\\ W = -596 \ln \left(\frac{1}{2}\right) \\$Recall that $\ln \left(\frac{1}{2}\right) = -\ln(2) \\$Thus, $W = -596 \times (-\ln(2)) \\ W = 596 \ln(2) \\$Using the approximate value $\ln(2) \approx 0.693: \\ W = 596 \times 0.693 \\ W \approx 413.148$ calories$\\$Since the work done is negative (indicating work done by the system),$\\ W \approx -413.148$ calories$\\$Therefore, the correct option is Option 4: $-413.14$ calories.