One mole of an ideal gas at $350 \mathrm{~K}$ is in a $2.0 \mathrm{~L}$ vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0 L to $3.0 \mathrm{~L}$ against a constant pressure of $4 \mathrm{~atm}$. The change in entropy of the surroundings ( $\Delta \mathrm{S})$ is ___________ $\mathrm{J} \mathrm{K}^{-1}$ (Nearest integer)

Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

<p>For an isothermal, reversible process, the change in entropy (ΔS) of the system can be calculated using the formula:</p> <p>$ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) $</p> <p>where (n) is the number of moles, (R) is the gas constant, and (V_2) and (V_1) are the final and initial volumes, respectively.</p> <p>Substituting the given values:</p> <p>$ \Delta S_{\text{system}} = 1 \times 8.314 \, \ln\left(\frac{3}{2}\right) = 3.37 \, \text{J K}^{-1} $</p> <p>Since the process is reversible and the total entropy change in the universe should be zero for reversible processes, the change in entropy of the surroundings is equal to the negative change of the system&#39;s entropy:</p> <p>$ \Delta S_{\text{surroundings}} = - \Delta S_{\text{system}} = -3.37 \, \text{J K}^{-1} $</p> <p>But considering the heat transfer from the system to the surroundings, the sign should be positive, which means the entropy of the surroundings also increases:</p> <p>$ \Delta S_{\text{surroundings}} = 3.37 \, \text{J K}^{-1} $</p> <p>So, rounded to the nearest integer, </p> <p>$ \Delta S_{\text{surroundings}} = 3 \, \text{J K}^{-1} $</p>