The process described is free expansion of an ideal gas. Free expansion occurs against zero external pressure ($P_{ext} = 0$). Since the gas expands into a vacuum, no work is done ($W = 0$).
For an ideal gas, the change in entropy ($\Delta S_{system}$) for an isothermal process is given by the formula: $$ \Delta S_{system} = nR \ln\left(\frac{V_f}{V_i}\right) $$ Given $n = 2$ moles, $V_i = 10$ L, and $V_f = 100$ L: $$ \Delta S_{system} = 2 \times R \times \ln\left(\frac{100}{10}\right) $$ $$ \Delta S_{system} = 2 \times R \times \ln(10) $$ Using $\ln(10) \approx 2.303$: $$ \Delta S_{system} = 2 \times R \times 2.303 = 4.606 R $$
In free expansion, the gas expands into a vacuum. Since $P_{ext} = 0$, the work done by the system on the surroundings is zero. Because the process is isothermal and no work is exchanged, there is no heat exchange with the surroundings ($q_{surr} = 0$). Therefore: $$ \Delta S_{surroundings} = \frac{-q_{sys}}{T} = 0 $$
Final Answer: Option (A)
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