Calculate the change in entropy of the system for the following process: 10.0g of ice at -10.0°C is converted into water vapour 115.0°C at a constant pressure of 1 bar. Calculate this by dividing this process up into individual steps for heating and each phase transition. The heat capacities and enthalpy changes for H₂O are: Cp (H₂O, solid, 1bar) = 37.6 Cp (H₂O, gas, 1 bar) = 33.6 AHfusion (H₂0, 1bar, Tfus) = 6.01 J mol K J mol K J mol K C„(H₂O, liquid, 1bar) = 75.3 J mol K AHvaporization(H20, 1bar, Trap) = 40.7 J mol K

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**Title: Calculating the Change in Entropy for a Phase Transition Process**

**Introduction:**
This section details the steps necessary to calculate the change in entropy as 10.0g of ice at -10.0°C is transformed into water vapor at 115.0°C at a constant pressure of 1 bar. This is achieved by segmenting the process into discrete steps for heating and phase transitions.

**Heat Capacities and Enthalpy Changes for H₂O:**

- **Solid Phase Heat Capacity:**
  \[
  \bar{C}_p(H_2O, \text{solid}, 1 \text{ bar}) = 37.6 \frac{J}{\text{mol K}}
  \]

- **Liquid Phase Heat Capacity:**
  \[
  \bar{C}_p(H_2O, \text{liquid}, 1 \text{ bar}) = 75.3 \frac{J}{\text{mol K}}
  \]

- **Gas Phase Heat Capacity:**
  \[
  \bar{C}_p(H_2O, \text{gas}, 1 \text{ bar}) = 33.6 \frac{J}{\text{mol K}}
  \]

- **Enthalpy of Fusion:**
  \[
  \Delta H_\text{fusion}(H_2O, 1 \text{ bar}, T_\text{fus}) = 6.01 \frac{J}{\text{mol}}
  \]

- **Enthalpy of Vaporization:**
  \[
  \Delta H_\text{vaporization}(H_2O, 1 \text{ bar}, T_\text{vap}) = 40.7 \frac{J}{\text{mol}}
  \]

**Discussion:**
To calculate the total entropy change, consider each step:
1. Heating ice from -10.0°C to 0°C.
2. Melting ice to liquid water at 0°C.
3. Heating water from 0°C to 100°C.
4. Vaporizing water at 100°C.
5. Heating water vapor from 100°C to 115.0°C.

By summing the entropy changes for each step, the overall entropy change can be determined. Each heat capacity value and enthalpy change plays a crucial role in these calculations, highlighting the extensive overlaps between
Transcribed Image Text:**Title: Calculating the Change in Entropy for a Phase Transition Process** **Introduction:** This section details the steps necessary to calculate the change in entropy as 10.0g of ice at -10.0°C is transformed into water vapor at 115.0°C at a constant pressure of 1 bar. This is achieved by segmenting the process into discrete steps for heating and phase transitions. **Heat Capacities and Enthalpy Changes for H₂O:** - **Solid Phase Heat Capacity:** \[ \bar{C}_p(H_2O, \text{solid}, 1 \text{ bar}) = 37.6 \frac{J}{\text{mol K}} \] - **Liquid Phase Heat Capacity:** \[ \bar{C}_p(H_2O, \text{liquid}, 1 \text{ bar}) = 75.3 \frac{J}{\text{mol K}} \] - **Gas Phase Heat Capacity:** \[ \bar{C}_p(H_2O, \text{gas}, 1 \text{ bar}) = 33.6 \frac{J}{\text{mol K}} \] - **Enthalpy of Fusion:** \[ \Delta H_\text{fusion}(H_2O, 1 \text{ bar}, T_\text{fus}) = 6.01 \frac{J}{\text{mol}} \] - **Enthalpy of Vaporization:** \[ \Delta H_\text{vaporization}(H_2O, 1 \text{ bar}, T_\text{vap}) = 40.7 \frac{J}{\text{mol}} \] **Discussion:** To calculate the total entropy change, consider each step: 1. Heating ice from -10.0°C to 0°C. 2. Melting ice to liquid water at 0°C. 3. Heating water from 0°C to 100°C. 4. Vaporizing water at 100°C. 5. Heating water vapor from 100°C to 115.0°C. By summing the entropy changes for each step, the overall entropy change can be determined. Each heat capacity value and enthalpy change plays a crucial role in these calculations, highlighting the extensive overlaps between
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