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For the process H(2)O(l) to H(2)O(g) at ...

For the process `H_(2)O(l) to H_(2)O(g)` at `T = 100^(@)C` and 1 atmosphere pressure, the correct choice is

A

`DeltaS_("system") gt 0 and DeltaS_("surroundings") gt 0`

B

`DeltaS_("system") gt 0 and DeltaS_("surroundings") lt 0`

C

`DeltaS_("system") lt 0 and DeltaS_("surroundings") gt 0`

D

`DeltaS_("system") lt 0 and DeltaS_("surroundings") lt 0`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of the process \( H_2O(l) \) to \( H_2O(g) \) at \( T = 100^\circ C \) and 1 atmosphere pressure, we can analyze the thermodynamic aspects of the phase change from liquid to gas. ### Step-by-Step Solution: 1. **Identify the Process**: The process described is the phase change of water from liquid to gas (evaporation) at a specific temperature and pressure. 2. **Understand the Conditions**: The temperature is \( 100^\circ C \) and the pressure is 1 atmosphere. At this temperature and pressure, water is at its boiling point, meaning that liquid water can coexist with water vapor in equilibrium. 3. **Equilibrium Condition**: At equilibrium, the process of evaporation and condensation occurs simultaneously. This means that the system is in a state where the rate of evaporation equals the rate of condensation. 4. **Entropy Change**: For the process of evaporation, the entropy of the system (water turning into vapor) increases because gas has higher entropy than liquid. Thus, \( \Delta S_{system} > 0 \). 5. **Surroundings Entropy Change**: When the system's entropy increases, the surroundings' entropy will decrease due to the heat absorbed by the system during the phase change. Therefore, \( \Delta S_{surroundings} < 0 \). 6. **Total Entropy Change**: According to the second law of thermodynamics, the total entropy change of the universe (system + surroundings) must be zero for a reversible process at equilibrium. Thus, we can write: \[ \Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} = 0 \] This implies: \[ \Delta S_{surroundings} = -\Delta S_{system} \] 7. **Conclusion**: Since the process is at equilibrium, the correct choice must reflect that the entropy of the system increases while the entropy of the surroundings decreases, maintaining the total entropy change at zero. ### Final Answer: The correct choice for the process \( H_2O(l) \) to \( H_2O(g) \) at \( T = 100^\circ C \) and 1 atmosphere pressure is **Option B**.

To solve the problem of the process \( H_2O(l) \) to \( H_2O(g) \) at \( T = 100^\circ C \) and 1 atmosphere pressure, we can analyze the thermodynamic aspects of the phase change from liquid to gas. ### Step-by-Step Solution: 1. **Identify the Process**: The process described is the phase change of water from liquid to gas (evaporation) at a specific temperature and pressure. 2. **Understand the Conditions**: ...
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