Question

Evaporation of water taking place at $100^\circ\text{C}$ at 1 atm. pressure, the correct choice of change in $\Delta S$ for system and surrounding would be :

• A. $\Delta S_{\text{system}} > 0, \Delta S_{\text{surrounding}} > 0$
• B. $\Delta S_{\text{system}} < 0, \Delta S_{\text{surrounding}} < 0$
• C. $\Delta S_{\text{system}} < 0, \Delta S_{\text{surrounding}} > 0$
• D. $\Delta S_{\text{system}} > 0, \Delta S_{\text{surrounding}} < 0$

Hint:

Whenever an endothermic phase change occurs (like melting or boiling), the system's entropy always increases ($>0$) and the surroundings' entropy always decreases ($<0$). For exothermic phase changes (freezing, condensation), the exact opposite is true.

Answer 1

The Correct Option is D

Step 1: Concept:
The question explores the thermodynamic entropy changes ($\Delta S$) during a phase transition, specifically the evaporation of water at its standard boiling point ($100^\circ\text{C}$, 1 atm).
We must evaluate the sign of entropy change for both the system (the water undergoing evaporation) and its surroundings.

Step 2: Key Formula or Approach:
Entropy ($S$) is a measure of randomness or disorder.
For the system undergoing a phase change from liquid to gas:
\[ \Delta S_{\text{system}} = S_{\text{gas}} - S_{\text{liquid}} \]
For the surroundings, the entropy change is related to the heat exchanged:
\[ \Delta S_{\text{surrounding}} = \frac{q_{\text{surrounding}}}{T} = \frac{-q_{\text{system}}}{T} \]

Step 3: Step-by-step Explanation:

• System: The process is the evaporation of water ($H_2O_{(l)} \rightarrow H_2O_{(g)}$).

• Gases have significantly higher entropy than liquids because the molecules have far greater freedom of motion and occupy a much larger volume.

• Therefore, the entropy of the system increases, making $\Delta S_{\text{system}} > 0$ (positive).

• Surroundings: Evaporation is an endothermic process. The system must absorb heat from its surroundings to overcome the intermolecular forces in the liquid phase.

• Since the system absorbs heat ($q_{\text{system}} > 0$), the surroundings lose an equal amount of heat ($q_{\text{surrounding}} < 0$).

• Because the surroundings lose thermal energy, their thermal disorder decreases. Therefore, $\Delta S_{\text{surrounding}} = -q_{\text{system}} / T$, which means $\Delta S_{\text{surrounding}} < 0$ (negative).

• Additionally, at the normal boiling point, the phase change is a reversible equilibrium process where the total entropy change of the universe is zero ($\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0$). Thus, $\Delta S_{\text{sys}} = -\Delta S_{\text{surr}}$.

Step 4: Final Answer:
The entropy of the system increases while the entropy of the surroundings decreases. This matches option (D).