### Related Formula W = -2.303 \, nRT log left( fracV_2V_1 right)$W = -2.303 \, nRT \log \left( \frac{V_2}{V_1} \right)$ ### Core Logic For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: n = 5text moles$n = 5\text{ moles}$ R = 8.314text J K^-1textmol^-1$R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$ T = 300text K$T = 300\text{ K}$ V_1 = 10text L$V_1 = 10\text{ L}$ V_2 = 100text L$V_2 = 100\text{ L}$ ### Step 1: Calculating Work Done W = -2.303 times 5 times 8.314 times 300 times logleft( frac10010 right)$W = -2.303 \times 5 \times 8.314 \times 300 \times \log\left( \frac{100}{10} \right)$ W = -2.303 times 5 times 8.314 times 300 times log(10)$W = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$ W = -2.303 times 12471 times 1$W = -2.303 \times 12471 \times 1$ W = -28720.713text J$W = -28720.713\text{ J}$ ### Step 2: Final Formatting The question asks for work w = -xtext J$w = -x\text{ J}$. So x = 28720.713$x = 28720.713$, which rounds to 28721$28721$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics