The work done ($w$) during the reversible isothermal expansion of an ideal gas is given by the formula: $w = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$ Since $P_1V_1 = P_2V_2$ for an isothermal process, we can also write this as: $w = -2.303 nRT \log_{10}\left(\frac{P_1}{P_2}\right)$

Given the values: Number of moles, $n = 1$ mol Universal gas constant, $R = 2.0 \text{ cal K}^{-1} \text{mol}^{-1}$ Temperature, $T = 25^\circ\text{C} = 25 + 273 = 298 \text{ K}$ Initial pressure, $P_1 = 20 \text{ atm}$ Final pressure, $P_2 = 10 \text{ atm}$

Substituting these values into the formula: $w = -2.303 \times 1 \times 2.0 \times 298 \times \log_{10}\left(\frac{20}{10}\right)$ $w = -2.303 \times 596 \times \log_{10}(2)$ $w = -2.303 \times 596 \times 0.3010$ $w \approx -413.14 \text{ calories}$