Entropy Changes in Phase Transitions

1. The Entropy of Phase Transition

Recall the fundamental definition of entropy change from our previous discussions:
$\Delta S=\int \frac{d{Q}_{rev}}{T}$
A phase transition, such as the melting of ice at 0°C (273.15 K) and 1 atm, is a prime example of a reversible process that occurs at a constant temperature. Because the temperature $T$ is constant, we can pull it out of the integral:
$\Delta S=\frac{1}{T}\int d{Q}_{rev}=\frac{Q_{rev}}{T}$
The heat absorbed or released during a phase transition is the latent heat. At constant pressure, this heat transfer is equal to the change in enthalpy, $\Delta H$. Therefore, for a phase transition, the formula becomes exceptionally straightforward.

To understand this core concept, please review the following brief article and video.

To summarize, the entropy change during a phase transition is given by:
$\Delta S_{trans}=\frac{\Delta H_{trans}}{T_{trans}}$
where:

• $\Delta H_{trans}$ is the molar enthalpy of transition (e.g., fusion, vaporization). Its units are typically J/mol or kJ/mol.
• $T_{trans}$ is the absolute temperature (in Kelvin) at which the transition occurs.

If you are working with mass instead of moles, the formula is:
$\Delta S_{trans}=\frac{m\cdot L_{trans}}{T_{trans}}$
where $m$ is the mass and $L_{trans}$ is the specific latent heat (e.g., J/g or J/kg).

2. Physical Interpretation: Disorder, Signs, and Magnitude

The sign of $\Delta S$ tells us whether the system is becoming more or less ordered. A phase transition is a perfect illustration of this.

Let's watch a short video that connects the sign of $\Delta S$ to the physical process.

As the video explains:

• Melting (Solid $\to$ Liquid) & Vaporization (Liquid $\to$ Gas): These are endothermic processes ($\Delta H>0$). The system absorbs heat to break intermolecular bonds, allowing molecules more freedom of movement. This increases the number of possible microstates, so disorder increases and $\Delta S>0$.
• Freezing (Liquid $\to$ Solid) & Condensation (Gas $\to$ Liquid): These are exothermic processes ($\Delta H<0$). The system releases heat as molecules settle into a more structured arrangement. This decreases the number of available microstates, so order increases and $\Delta S<0$.

This aligns perfectly with our intuition about the states of matter: a gas is more disordered than a liquid, which is more disordered than a solid.

3. Combining Concepts: Multi-Step Processes

JEE Advanced problems rarely test a single concept in isolation. A common problem type involves a substance undergoing both a temperature change and a phase change. Since entropy is a state function, we can calculate the total entropy change by summing the $\Delta S$ for a series of reversible steps that connect the initial and final states.

Problem: Calculate the total entropy change when 2 moles of supercooled liquid water at -10°C and 1 atm are converted to ice at -10°C.

Given:

• $C_{p,m}(\text{liquid water})=75.3\text{ J/(mol⋅K)}$
• $C_{p,m}(\text{ice})=37.7\text{ J/(mol⋅K)}$
• $\Delta H_{fus}(\text{at }{0}^{\circ }\text{C})=6010\text{ J/mol}$

Analysis:
The process is happening at a constant temperature of -10°C. However, this is not the equilibrium phase transition temperature. The process as stated (liquid to solid at -10°C) is irreversible and spontaneous.

To calculate $\Delta S$, we must construct a hypothetical reversible path between the initial state (liquid water at -10°C) and the final state (ice at -10°C).

The logical reversible path is:

1. Step 1: Heat the liquid water from -10°C (263.15 K) to 0°C (273.15 K).
2. Step 2: Freeze the liquid water to ice at the equilibrium temperature of 0°C (273.15 K).
3. Step 3: Cool the ice from 0°C (273.15 K) back down to -10°C (263.15 K).

The following image illustrates this exact strategy.

Calculation:

• Step 1: $\Delta S_1$ (Heating liquid)
  This is an entropy change with varying temperature, which we covered in the last lesson.
  $\Delta S_1=n{C}_{p,m}(\text{liquid})\ln \left(\frac{T_{final}}{T_{initial}}\right)=(2\text{ mol})(75.3\frac{\text{J}}{\text{mol⋅K}})\ln \left(\frac{273.15}{263.15}\right)\approx +5.63\text{ J/K}$

• Step 2: $\Delta S_2$ (Phase change at 0°C)
  This is the core of today's lesson. Freezing is exothermic, so $\Delta H$ is negative.
  $\Delta H_{freezing}=-\Delta H_{fus}=-6010\text{ J/mol}$
  $\Delta S_2=\frac{n\cdot \Delta H_{freezing}}{T_{fus}}=\frac{(2\text{ mol})(-6010\frac{\text{J}}{\text{mol}})}{273.15\text{ K}}\approx -44.01\text{ J/K}$

• Step 3: $\Delta S_3$ (Cooling ice)
  Another temperature-dependent change.
  $\Delta S_3=n{C}_{p,m}(\text{ice})\ln \left(\frac{T_{final}}{T_{initial}}\right)=(2\text{ mol})(37.7\frac{\text{J}}{\text{mol⋅K}})\ln \left(\frac{263.15}{273.15}\right)\approx -2.83\text{ J/K}$

• Total Entropy Change:
  $\Delta S_{total}=\Delta S_1+\Delta S_2+\Delta S_3=5.63-44.01-2.83=-41.21\text{ J/K}$

The final entropy change is negative, which is what we expect for a spontaneous freezing process. This type of problem is a classic example of applying state functions to bridge an irreversible process with a calculable, reversible path.

Conclusion

Today we've added a crucial tool to your thermodynamics problem-solving kit. By understanding how to handle phase transitions, you can now calculate entropy changes for a much wider range of physical processes.

Key Takeaways:

• For a phase transition occurring at a constant equilibrium temperature $T_{trans}$, the entropy change is calculated with the simple formula: $\Delta S_{trans}=\Delta H_{trans}/T_{trans}$.
• The sign of $\Delta S$ reflects the change in molecular disorder:
  • Positive for melting/vaporization (more disorder).
  • Negative for freezing/condensation (more order).
• To calculate $\Delta S$ for complex or irreversible processes, you can construct a hypothetical path of simpler, reversible steps (like temperature changes and equilibrium phase transitions) and sum their individual entropy changes.

Preview of the Next Lesson:
We are now entering the final module of the course, which focuses on integrated problems. We will start by connecting thermodynamics to other areas of physics. In the next lesson, we will solve problems where work done during a thermodynamic process is converted to heat in a calorimeter, bridging the concepts of P-V diagrams and calorimetry.