Import Question JSON

$\text{Hint: } W = -2.303 \text{ nRT}\log \frac{V_2}{V_1}$ $\text{Step 1: Identify the given values}$ $\text{The given values are as follows:}$ $V_1 = 10 \text{ L (initial volume)}$ $V_2 = 20 \text{ L (final volume)}$ $n = 1 \text{ mol (number of moles)}$ $T = 25^{\circ}\text{C} = 273 + 25 = 298 \text{ K (temperature)}$ $R = 8.314 \text{ J mol}^{-1}\text{K}^{-1} \text{ (gas constant)}$ $\text{Conversion factor: } 1 \text{ J} = 10^7 \text{ erg}$ $\text{Therefore, } R = 8.314 \times 10^7 \text{ erg mol}^{-1}\text{K}^{-1}$ $\text{Step 2: Apply the formula for work done in reversible isothermal expansion}$ $\text{For a reversible isothermal expansion of an ideal gas, the work done is given by:}$ $W = -nRT\ln\frac{V_2}{V_1}$ $\text{Using base-10 logarithm: } W = -2.303 \times nRT\log\frac{V_2}{V_1}$ $W = -2.303 \times 1 \times 8.314 \times 10^7 \times 298 \times \log\frac{20}{10}$ $W = -2.303 \times 8.314 \times 10^7 \times 298 \times \log 2$ $\text{Rearranging: } W = -298 \times 10^7 \times 8.314 \times 2.303 \times \log 2$ $\text{Step 3: Verify the answer}$ $\text{The negative sign indicates that the gas is doing work on the surroundings during expansion.}$ $\text{The answer matches option 2: } -298 \times 10^7 \times 8.31 \times 2.303 \log 2$ $\text{Note: There's a slight difference in the value of R (8.31 vs. 8.314), but this is within acceptable rounding for multiple-choice options.}$