The work done in a reversible isothermal process can be calculated using the formula:

$$ W = -nRT \ln\left(\frac{V_f}{V_i}\right) $$

Here:

• $ W $ is the work done by the gas.
• $ n $ is the number of moles of the gas.
• $ R $ is the universal gas constant.
• $ T $ is the temperature in Kelvin.
• $ V_i $ and $ V_f $ are the initial and final volumes of the gas, respectively.

However, since the volumes are not directly provided but the pressures are given, we use the ideal gas law, $ PV = nRT $, to relate pressures and volumes at the same temperature and amount of gas:

For an ideal gas undergoing a change at constant temperature, we can also write the work done in terms of the initial and final pressures:

$$ W = -nRT \ln\left(\frac{P_i}{P_f}\right) $$

where:

• $ P_i $ and $ P_f $ are the initial and final pressures, respectively.

Given:

• $ n = 1 $ (one mole of hydrogen)
• $$ T = 25^{\circ} \mathrm{C} = 25 + 273.15 = 298.15 \, \mathrm{K} $$
• $$ R = 2.0 \, \mathrm{cal} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $$
• $$ P_i = 20 \, \mathrm{atm} $$
• $$ P_f = 10 \, \mathrm{atm} $$

Substituting all values into the formula:

$$ W = -1 \cdot 2.0 \cdot 298.15 \ln\left(\frac{20}{10}\right) $$

$$ W = -2.0 \cdot 298.15 \cdot \ln(2) $$

Now, solving this using the value of $ \ln(2) \approx 0.693 $:

$$ W \approx -2.0 \cdot 298.15 \cdot 0.693 $$

$$ W \approx -413.14 \, \text{calories} $$

Thus, the work done during the process is about $$ -413.14 \, \text{calories} $$, indicating that this amount of energy was done by the system (expansion work being done by the gas against external pressure), and hence is negative as it indicates work done by the system. Therefore, the correct option is:

Option B: $$ -413.14 \, \text{calories} $$