Analyzing Entropy Change During Water Evaporation

The question asks for the change in entropy ($\Delta S$) for the system (water) and the surroundings during evaporation at $100^{\circ}\text{C}$ and 1 atm pressure.

System Entropy Change ($\Delta\text{S}_{\text{system}}$)

Evaporation is a phase transition from liquid water to gaseous steam. Gases have molecules that are much more disordered and spread out than liquids. Therefore, the transition from a more ordered state (liquid) to a less ordered state (gas) results in an increase in entropy.

• Phase transition: Liquid $\rightarrow$ Gas
• Increase in disorder
• Conclusion: $\Delta\text{S}_{\text{system}} > 0$

Surroundings Entropy Change ($\Delta\text{S}_{\text{surrounding}}$)

Evaporation at $100^{\circ}\text{C}$ and 1 atm is the normal boiling point of water. At this temperature and pressure, the process is reversible. The system (water) absorbs heat ($q_{\text{system}} > 0$) from the surroundings to undergo evaporation.

The heat absorbed by the system comes from the surroundings. Thus, the heat change for the surroundings is $q_{\text{surrounding}} = -q_{\text{system}}$, which means $q_{\text{surrounding}} < 0$.

The change in entropy for the surroundings is calculated as $\Delta\text{S}_{\text{surrounding}} = \frac{q_{\text{surrounding}}}{\text{T}}$.

• Heat absorbed by system: $q_{\text{system}} > 0$
• Heat lost by surroundings: $q_{\text{surrounding}} < 0$
• Temperature: $\text{T} = 100^{\circ}\text{C} = 373.15 \text{ K}$ (positive value)
• $\Delta\text{S}_{\text{surrounding}} = \frac{\text{Negative value}}{\text{Positive value}} < 0$
• Conclusion: $\Delta\text{S}_{\text{surrounding}} < 0$

Overall Entropy Change Conclusion

Based on the analysis:

• System entropy change: $\Delta\text{S}_{\text{system}} > 0$
• Surroundings entropy change: $\Delta\text{S}_{\text{surrounding}} < 0$

This corresponds to Option 4.