Two moles of an ideal gas undergo free expansion from 10 L to 100 L...

For the free expansion of an ideal gas, the external pressure is zero, so the work done $W = 0$.

Since the process is isothermal at $300$ K, the change in internal energy $\Delta U = 0$.

From the first law of thermodynamics, $q = \Delta U - W = 0$.

Since no heat is exchanged with the surroundings, the change in entropy of the surroundings is:

$\Delta S_{\text{surroundings}} = \dfrac{q_{\text{surr}}}{T} = 0$

Entropy is a state function, so the change in entropy of the system for an isothermal expansion is calculated using a reversible path between the same initial and final states:

$\Delta S_{\text{system}} = nR \ln\left(\dfrac{V_2}{V_1}\right)$

Substituting $n = 2$ moles, $V_1 = 10$ L, and $V_2 = 100$ L:

$\Delta S_{\text{system}} = 2R \ln\left(\dfrac{100}{10}\right) = 2R \ln(10)$

Using $\ln(10) = 2.303 \log_{10}(10) = 2.303$:

$\Delta S_{\text{system}} = 2 \times 2.303 R = 4.606 R$

Answer: $\Delta S_{\text{system}}=4.606$ R; $\Delta S_{\text{surroundings}}=0$