ΔS

To calculate the change in entropy ($\Delta S$) for the condensation of water from a gas to a liquid, we can use the formula:

$\Delta S = \frac{q_{\text{rev}}}{T}$

where:

• $q_{\text{rev}}$ is the heat released during the phase change.
• $T$ is the temperature in kelvins.

Step 1: Calculate $q_{\text{rev}}$

The heat released ($q_{\text{rev}}$) is given by:

$q_{\text{rev}} = \text{mass} \times \frac{\Delta H_{\text{vap}}}{\text{molar mass}}$

Given:

• Mass of water = $28.0 \, \text{g}$
• $\Delta H_{\text{vap}} = 40.7 \, \text{kJ/mol} = 40,700 \, \text{J/mol}$
• Molar mass of water = $18.015 \, \text{g/mol}$

$q_{\text{rev}} = 28.0 \, \text{g} \times \frac{40,700 \, \text{J/mol}}{18.015 \, \text{g/mol}}$

$q_{\text{rev}} = 28.0 \times 2260.89 = 63,305 \, \text{J}$

Step 2: Calculate $\Delta S$

The temperature at the boiling point of water is $100.00^\circ \, \text{C} = 373.15 \, \text{K}$.

$\Delta S = \frac{q_{\text{rev}}}{T} = \frac{63,305 \, \text{J}}{373.15 \, \text{K}}$

$\Delta S = 169.7 \, \text{J/K}$

Final Answer:

The change in entropy is:

$\Delta S = 169.7 \, \text{J/K}$

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Would you like additional details on any of these calculations? Here are 5 related questions for deeper understanding:

1. How is the latent heat of vaporization related to entropy changes?
2. Why is temperature converted to kelvins for thermodynamic calculations?
3. What is the physical interpretation of entropy change during condensation?
4. How does this calculation differ for processes occurring at temperatures other than $100^\circ \, \text{C}$?
5. Could entropy increase during a phase transition? Under what conditions?

Tip: Always ensure your temperature is in kelvins when working with thermodynamic formulas, as kelvin is the SI unit for absolute temperature.