Part II, Chapter 4

Phase Equilibria

Clausius-Clapeyron equation, phase diagrams, and the Gibbs phase rule

Historical Context

The systematic study of phase transitions began with the work of Benoit Paul Emile Clapeyron (1834), who derived the equation relating pressure and temperature along a phase boundary. Rudolf Clausius later refined this into the form we use today. J. Willard Gibbs introduced the phase rule in 1876, providing a powerful tool for determining the degrees of freedom in multi-phase, multi-component systems.

Thomas Andrews' experiments on CO2 (1869) revealed the critical point, above which the distinction between liquid and gas disappears. Johannes van der Waals provided the first theoretical explanation with his equation of state, earning the 1910 Nobel Prize.

Key contributors: Clapeyron (1834), Clausius (1850), Andrews (1869), Gibbs (1876), van der Waals (1873).

Derivation 1: Clausius-Clapeyron Equation

Along a phase coexistence curve, the chemical potentials of the two phases must be equal:$\mu^\alpha(T, P) = \mu^\beta(T, P)$. For an infinitesimal change along the curve:

$$d\mu^\alpha = d\mu^\beta \implies -S_m^\alpha\,dT + V_m^\alpha\,dP = -S_m^\beta\,dT + V_m^\beta\,dP$$

Rearranging gives the Clapeyron equation:

Clapeyron Equation

$$\frac{dP}{dT} = \frac{\Delta S_m}{\Delta V_m} = \frac{\Delta H_m}{T \, \Delta V_m}$$

For liquid-vapor equilibrium, assuming the vapor is an ideal gas ($V_m^{\text{gas}} \gg V_m^{\text{liquid}}$) and $V_m^{\text{gas}} = RT/P$:

Clausius-Clapeyron Equation

$$\frac{d \ln P}{dT} = \frac{\Delta H_{\text{vap}}}{RT^2} \quad \Longrightarrow \quad \ln\frac{P_2}{P_1} = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

A plot of $\ln P$ vs $1/T$ gives a straight line with slope $-\Delta H_{\text{vap}}/R$. This is the basis for measuring latent heats from vapor pressure data.

Derivation 2: Gibbs Phase Rule

The phase rule determines the number of degrees of freedom F (independently variable intensive properties) for a system in equilibrium.

Consider C components in P phases. The intensive variables are T, P, and the mole fractions in each phase. The total number of variables:

$$\text{Variables} = 2 + P(C - 1) \quad \text{(T, P, plus compositions)}$$

The equilibrium conditions require equal chemical potentials across all phases for each component:

$$\mu_i^{(1)} = \mu_i^{(2)} = \cdots = \mu_i^{(P)} \quad \text{for each } i = 1, \ldots, C$$

This gives $C(P - 1)$ constraint equations. The degrees of freedom are:

$$F = [2 + P(C-1)] - C(P-1) = C - P + 2$$

Triple Point

C=1, P=3, F=0

Fixed T and P

Coexistence Curve

C=1, P=2, F=1

One variable (e.g., T) fixes P

Single Phase

C=1, P=1, F=2

T and P independently variable

Derivation 3: Critical Point from van der Waals

The van der Waals equation improves on the ideal gas law by accounting for molecular volume and intermolecular attractions:

$$\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$$

At the critical point, the isotherm has an inflection point:

$$\left(\frac{\partial P}{\partial V_m}\right)_T = 0, \qquad \left(\frac{\partial^2 P}{\partial V_m^2}\right)_T = 0$$

Solving these simultaneously yields the critical constants:

$$V_{m,c} = 3b, \qquad P_c = \frac{a}{27b^2}, \qquad T_c = \frac{8a}{27Rb}$$

Critical compressibility factor: $Z_c = P_c V_{m,c} / (RT_c) = 3/8 = 0.375$

Derivation 4: The Lever Rule

In a two-phase region of a binary phase diagram, the overall composition z determines the relative amounts of each phase through the lever rule.

If the liquid has composition $x$ and the vapor has composition $y$, with overall composition $z$, then mass balance requires:

$$z = f^L x + f^V y \quad \text{where } f^L + f^V = 1$$

Solving:

$$\frac{f^L}{f^V} = \frac{y - z}{z - x} \quad \text{(inverse lever rule)}$$

The fraction of liquid is proportional to the "distance" from the overall composition to the vapor curve, and vice versa -- like a lever balanced at the fulcrum z.

Derivation 5: Ehrenfest Classification of Phase Transitions

Paul Ehrenfest classified phase transitions by the order of the derivative of the Gibbs energy that first shows a discontinuity:

First-Order Transitions

The first derivatives of G (entropy and volume) are discontinuous:

$$\Delta S = -\Delta\left(\frac{\partial G}{\partial T}\right)_P \neq 0$$

Involve latent heat. Examples: melting, boiling, sublimation.

Second-Order Transitions

First derivatives are continuous, but second derivatives (heat capacity, compressibility) diverge:

$$C_P = -T\left(\frac{\partial^2 G}{\partial T^2}\right)_P \to \infty$$

No latent heat. Examples: superconducting transition, ferromagnetic Curie point.

Applications

Distillation

Phase diagrams guide the design of distillation columns. The presence of azeotropes limits separation by simple distillation and requires techniques like extractive or pressure-swing distillation.

Supercritical Fluids

Above the critical point, fluids have unique solvent properties. Supercritical CO2 is used for decaffeination, dry cleaning, and extraction of natural products.

Metallurgy

Binary phase diagrams (e.g., iron-carbon) are essential for understanding steel microstructure, heat treatment, and alloy design.

Climate Science

The Clausius-Clapeyron equation predicts that atmospheric water vapor content increases by about 7% per degree of warming, amplifying climate change effects.

Simulation: Phase Diagram & Phase Rule

Explore the phase diagram of water, Clausius-Clapeyron vapor pressure curves for various substances, and a visualization of the Gibbs phase rule showing degrees of freedom.

Phase Diagram, Clausius-Clapeyron, and Gibbs Phase Rule

Python

script.py152 lines

import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt

# ============================================================
# Phase Diagram and Clausius-Clapeyron Equation
# ============================================================

R = 8.314  # J/(mol K)

# Water properties (approximate)
T_triple = 273.16      # K
P_triple = 611.73      # Pa
T_critical = 647.096   # K
P_critical = 22.064e6  # Pa

# Latent heats (approximate, J/mol)
dH_vap = 40700.0   # vaporization at 373 K
dH_fus = 6010.0    # fusion at 273 K
dH_sub = 46710.0   # sublimation

# Molar volumes (approximate, m^3/mol)
dV_vap = 0.03      # liquid -> gas change

fig, axes = plt.subplots(1, 3, figsize=(18, 6))

# ---- Panel 1: Phase diagram of water (P-T) ----
ax1 = axes[0]

# Solid-liquid line (negative slope for water!)
T_sl = np.linspace(250, 273.16, 100)
# Clausius-Clapeyron: dP/dT = dH_fus / (T * dV_fus)
# For water, dV_fus < 0 (ice is less dense)
dV_fus = -1.6e-6  # m^3/mol (water shrinks on melting)
P_sl = P_triple + (dH_fus / dV_fus) * np.log(T_sl / T_triple)
P_sl = np.clip(P_sl, 0, 1e9)

# Liquid-vapor line (Clausius-Clapeyron)
T_lv = np.linspace(273.16, 647, 300)
P_lv = P_triple * np.exp((dH_vap / R) * (1.0/T_triple - 1.0/T_lv))
P_lv = np.minimum(P_lv, P_critical)

# Solid-vapor line (sublimation)
T_sv = np.linspace(200, 273.16, 100)
P_sv = P_triple * np.exp((dH_sub / R) * (1.0/T_triple - 1.0/T_sv))

ax1.semilogy(T_lv, P_lv / 1e3, 'r-', lw=2.5, label='Liquid-Vapor')
ax1.semilogy(T_sl, P_sl / 1e3, 'b-', lw=2.5, label='Solid-Liquid')
ax1.semilogy(T_sv, P_sv / 1e3, 'c-', lw=2.5, label='Solid-Vapor')
ax1.plot(T_triple, P_triple/1e3, 'wo', markersize=10, zorder=5)
ax1.annotate('Triple Point', (T_triple, P_triple/1e3),
             xytext=(290, 0.1), fontsize=10, color='yellow',
             arrowprops=dict(arrowstyle='->', color='yellow'))
ax1.plot(T_critical, P_critical/1e3, 'r*', markersize=15, zorder=5)
ax1.annotate('Critical Point', (T_critical, P_critical/1e3),
             xytext=(550, P_critical/1e3*1.5), fontsize=10, color='yellow',
             arrowprops=dict(arrowstyle='->', color='yellow'))

# Label regions
ax1.text(230, 1e3, 'SOLID', fontsize=14, color='#88ccff', fontweight='bold')
ax1.text(350, 10, 'LIQUID', fontsize=14, color='#ff8888', fontweight='bold')
ax1.text(400, 0.1, 'VAPOR', fontsize=14, color='#88ff88', fontweight='bold')

ax1.set_xlabel('Temperature (K)', fontsize=12, color='white')
ax1.set_ylabel('Pressure (kPa)', fontsize=12, color='white')
ax1.set_title('Phase Diagram of Water', fontsize=14, fontweight='bold', color='white')
ax1.legend(fontsize=9, facecolor='black', edgecolor='gray', labelcolor='white')
ax1.set_facecolor('#1a1a2e')
ax1.tick_params(colors='white')
ax1.grid(True, alpha=0.3)
ax1.set_xlim(200, 700)

# ---- Panel 2: Clausius-Clapeyron vapor pressure ----
ax2 = axes[1]
substances = {
    'Water': (40700, 373.15, 1e5),
    'Ethanol': (38600, 351.4, 1e5),
    'Benzene': (30720, 353.2, 1e5),
    'Acetone': (31300, 329.2, 1e5),
}
colors_sub = ['#ff6b6b', '#ffa502', '#2ed573', '#1e90ff']

T_range = np.linspace(250, 450, 300)
for (name, (dH, Tb, Pb)), clr in zip(substances.items(), colors_sub):
    P_vap = Pb * np.exp((dH / R) * (1.0/Tb - 1.0/T_range))
    ax2.plot(1000.0/T_range, np.log10(P_vap/1e3), color=clr, lw=2, label=name)

ax2.set_xlabel('1000/T (1/K)', fontsize=12, color='white')
ax2.set_ylabel('log10(P / kPa)', fontsize=12, color='white')
ax2.set_title('Clausius-Clapeyron Vapor Pressure', fontsize=14, fontweight='bold', color='white')
ax2.legend(fontsize=10, facecolor='black', edgecolor='gray', labelcolor='white')
ax2.set_facecolor('#1a1a2e')
ax2.tick_params(colors='white')
ax2.grid(True, alpha=0.3)

# ---- Panel 3: Gibbs Phase Rule ----
ax3 = axes[2]
# F = C - P + 2 visualization for C = 1, 2, 3
C_vals = [1, 2, 3]
colors_c = ['#ff6b6b', '#2ed573', '#1e90ff']
P_phases = np.arange(1, 6)

for C, clr in zip(C_vals, colors_c):
    F = C - P_phases + 2
    F_clipped = np.maximum(F, 0)
    valid = F >= 0
    ax3.plot(P_phases[valid], F_clipped[valid], 'o-', color=clr, lw=2.5, markersize=10,
             label='C = ' + str(C) + ' components')

ax3.axhline(y=0, color='white', alpha=0.3, linestyle='--')
ax3.set_xlabel('Number of phases P', fontsize=12, color='white')
ax3.set_ylabel('Degrees of freedom F', fontsize=12, color='white')
ax3.set_title('Gibbs Phase Rule: F = C - P + 2', fontsize=14, fontweight='bold', color='white')
ax3.legend(fontsize=10, facecolor='black', edgecolor='gray', labelcolor='white')
ax3.set_facecolor('#1a1a2e')
ax3.tick_params(colors='white')
ax3.grid(True, alpha=0.3)
ax3.set_xticks([1, 2, 3, 4, 5])
ax3.set_yticks([0, 1, 2, 3, 4])

# Annotate special points
ax3.annotate('Triple point\n(C=1, P=3, F=0)', (3, 0),
             xytext=(3.5, 1), fontsize=9, color='yellow',
             arrowprops=dict(arrowstyle='->', color='yellow'))

for ax in axes:
    ax.spines['bottom'].set_color('gray')
    ax.spines['left'].set_color('gray')
    ax.spines['top'].set_color('gray')
    ax.spines['right'].set_color('gray')

fig.patch.set_facecolor('#0a0a1a')
plt.tight_layout()
plt.savefig('output.png', dpi=100, bbox_inches='tight', facecolor='#0a0a1a')
plt.close()

print("Phase Equilibria Analysis")
print("=" * 50)
print("Water triple point: T = 273.16 K, P = 611.73 Pa")
print("Water critical point: T = 647.1 K, P = 22.06 MPa")
print()
print("Clausius-Clapeyron slopes at triple point:")
dPdT_lv = dH_vap * P_triple / (R * T_triple**2)
print("  dP/dT (liquid-vapor) = " + str(round(dPdT_lv, 2)) + " Pa/K")
dPdT_sl = dH_fus / (T_triple * dV_fus)
print("  dP/dT (solid-liquid) = " + str(round(dPdT_sl/1e6, 2)) + " MPa/K (negative for water!)")
print()
print("Gibbs Phase Rule F = C - P + 2:")
print("  Single component: triple point has F=0 (invariant)")
print("  Binary system: can have up to 4 phases in equilibrium (F=0)")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Simulation: Binary Phase Diagrams

Compare an ideal binary system (benzene-toluene following Raoult's law) with a non-ideal system showing a minimum boiling azeotrope (ethanol-water).

Binary Phase Diagrams: Ideal vs Azeotropic

Python

script.py138 lines

import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt

# ============================================================
# Binary Phase Diagram: Ideal and Non-Ideal Solutions
# ============================================================

R = 8.314  # J/(mol K)

# Ideal solution: Raoult's law
# P_total = x_A * P_A* + x_B * P_B*
# y_A = x_A * P_A* / P_total

fig, axes = plt.subplots(1, 2, figsize=(14, 6))

# ---- Ideal binary T-x diagram (boiling point diagram) ----
ax1 = axes[0]

# Antoine equation parameters (simplified) for benzene and toluene
# ln(P*/kPa) = A - B/(T + C)
# Using Clausius-Clapeyron instead for simplicity
T_b_A = 353.2  # benzene boiling point (K)
T_b_B = 383.8  # toluene boiling point (K)
dH_A = 30720.0
dH_B = 33180.0
P_atm = 101.325  # kPa

x_A = np.linspace(0, 1, 500)
x_B = 1 - x_A

# Find bubble and dew temperatures at P = 1 atm
T_bubble = np.zeros_like(x_A)
T_dew = np.zeros_like(x_A)

for i, xa in enumerate(x_A):
    # Bubble point: solve P_atm = xa * PA*(T) + (1-xa) * PB*(T)
    for T_try in np.linspace(T_b_A - 5, T_b_B + 5, 2000):
        PA_star = P_atm * np.exp((dH_A/R)*(1/T_b_A - 1/T_try))
        PB_star = P_atm * np.exp((dH_B/R)*(1/T_b_B - 1/T_try))
        P_calc = xa * PA_star + (1 - xa) * PB_star
        if P_calc >= P_atm:
            T_bubble[i] = T_try
            break
    else:
        T_bubble[i] = T_b_B

# Dew point: solve 1/P_atm = ya/PA*(T) + (1-ya)/PB*(T) where ya = xa
    for T_try in np.linspace(T_b_A - 5, T_b_B + 5, 2000):
        PA_star = P_atm * np.exp((dH_A/R)*(1/T_b_A - 1/T_try))
        PB_star = P_atm * np.exp((dH_B/R)*(1/T_b_B - 1/T_try))
        if PA_star > 0 and PB_star > 0:
            P_inv = xa / PA_star + (1 - xa) / PB_star
            if 1.0 / P_inv <= P_atm:
                T_dew[i] = T_try
                break
    else:
        T_dew[i] = T_b_B

ax1.plot(x_A, T_bubble - 273.15, 'r-', lw=2.5, label='Bubble line (liquid)')
ax1.plot(x_A, T_dew - 273.15, 'b-', lw=2.5, label='Dew line (vapor)')
ax1.fill_between(x_A, T_bubble - 273.15, T_dew - 273.15, alpha=0.15, color='orange',
                  label='Two-phase region')
ax1.set_xlabel('x_benzene (liquid) / y_benzene (vapor)', fontsize=11, color='white')
ax1.set_ylabel('Temperature (C)', fontsize=11, color='white')
ax1.set_title('Benzene-Toluene T-x-y Diagram', fontsize=14, fontweight='bold', color='white')
ax1.legend(fontsize=9, facecolor='black', edgecolor='gray', labelcolor='white')
ax1.set_facecolor('#1a1a2e')
ax1.tick_params(colors='white')
ax1.grid(True, alpha=0.3)

# ---- Azeotropic system (non-ideal) ----
ax2 = axes[1]

# Simple model with positive deviation (ethanol-water like)
T_b_eth = 351.4  # K ethanol
T_b_wat = 373.15  # K water
T_azeotrope = 351.1  # K (minimum boiling azeotrope)
x_azeo = 0.90  # mole fraction ethanol at azeotrope

# Construct approximate curves
x = np.linspace(0, 1, 500)
# Bubble line: interpolate with dip at azeotrope
T_bub_az = T_b_wat - (T_b_wat - T_azeotrope) * np.exp(-((x - x_azeo)**2) / 0.08)
T_bub_az = np.minimum(T_bub_az, T_b_wat * (1 - x) + T_b_eth * x + 5 * x * (1 - x))
T_bub_az[0] = T_b_wat
T_bub_az[-1] = T_b_eth

# Smooth
from scipy.ndimage import gaussian_filter1d
T_bub_az = gaussian_filter1d(T_bub_az, sigma=10)

# Dew line: slightly above bubble
T_dew_az = T_bub_az + 3.0 * x * (1 - x) * (1 + 2 * np.exp(-((x - x_azeo)**2)/0.01))
T_dew_az = np.maximum(T_dew_az, T_bub_az)

# At azeotrope, bubble = dew
idx_az = np.argmin(np.abs(x - x_azeo))
T_dew_az[idx_az-5:idx_az+5] = T_bub_az[idx_az-5:idx_az+5]

ax2.plot(x, T_bub_az - 273.15, 'r-', lw=2.5, label='Bubble line')
ax2.plot(x, T_dew_az - 273.15, 'b-', lw=2.5, label='Dew line')
ax2.fill_between(x, T_bub_az - 273.15, T_dew_az - 273.15, alpha=0.15, color='orange')
ax2.plot(x_azeo, T_azeotrope - 273.15, 'y*', markersize=20, zorder=5)
ax2.annotate('Azeotrope', (x_azeo, T_azeotrope - 273.15),
             xytext=(0.6, T_azeotrope - 273.15 - 3), fontsize=11, color='yellow',
             arrowprops=dict(arrowstyle='->', color='yellow'))

ax2.set_xlabel('x_ethanol', fontsize=11, color='white')
ax2.set_ylabel('Temperature (C)', fontsize=11, color='white')
ax2.set_title('Ethanol-Water Azeotrope (approx.)', fontsize=14, fontweight='bold', color='white')
ax2.legend(fontsize=9, facecolor='black', edgecolor='gray', labelcolor='white')
ax2.set_facecolor('#1a1a2e')
ax2.tick_params(colors='white')
ax2.grid(True, alpha=0.3)

for ax in axes:
    ax.spines['bottom'].set_color('gray')
    ax.spines['left'].set_color('gray')
    ax.spines['top'].set_color('gray')
    ax.spines['right'].set_color('gray')

fig.patch.set_facecolor('#0a0a1a')
plt.tight_layout()
plt.savefig('output.png', dpi=100, bbox_inches='tight', facecolor='#0a0a1a')
plt.close()

print("Binary Phase Diagram Analysis")
print("=" * 50)
print("Benzene-Toluene (ideal solution):")
print("  T_b(benzene) = 80.1 C, T_b(toluene) = 110.6 C")
print("  Obeys Raoult law: no azeotrope")
print()
print("Ethanol-Water (non-ideal, positive deviation):")
print("  Minimum boiling azeotrope at x_eth = 0.90, T = 78.0 C")
print("  Cannot be separated by simple distillation beyond azeotrope")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Chapter Summary

• Clausius-Clapeyron: $d\ln P/dT = \Delta H_{\text{vap}}/(RT^2)$ relates vapor pressure to temperature.

• Gibbs Phase Rule: $F = C - P + 2$ determines degrees of freedom in multi-phase systems.

• Critical Point: Where liquid and vapor become indistinguishable; an inflection in the P-V isotherm.

• Lever Rule: Determines relative phase amounts in two-phase regions.

• Ehrenfest Classification: First-order transitions have latent heat; second-order have divergent heat capacity.

Practice Problems

Problem 1:Water boils at 100 degrees C at 1 atm. Using the Clausius-Clapeyron equation with $\Delta H_{\text{vap}} = 40.7\,\text{kJ/mol}$, find the boiling point at an altitude where $P = 0.70\,\text{atm}$.

Solution:

1. Clausius-Clapeyron: $\ln\frac{P_2}{P_1} = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$.

2. Given: $P_1 = 1\,\text{atm}$, $T_1 = 373.15\,\text{K}$, $P_2 = 0.70\,\text{atm}$.

3. $\ln(0.70) = -0.3567 = -\frac{40700}{8.314}\left(\frac{1}{T_2} - \frac{1}{373.15}\right)$.

4. $\frac{1}{T_2} - \frac{1}{373.15} = \frac{-0.3567 \times 8.314}{-40700} = 7.287 \times 10^{-5}\,\text{K}^{-1}$.

5. $\frac{1}{T_2} = \frac{1}{373.15} + 7.287 \times 10^{-5} = 2.680 \times 10^{-3} + 7.287 \times 10^{-5} = 2.753 \times 10^{-3}$.

6. $T_2 = 363.2\,\text{K} = 90.1°\text{C}$. Water boils about 10 degrees C lower at this altitude (approximately 3000 m), which is why cooking at high altitude takes longer.

Problem 2:Apply the Gibbs phase rule to a ternary system (3 components) with 2 phases in equilibrium. How many degrees of freedom are there? What if all three phases coexist?

Solution:

1. Gibbs phase rule: $F = C - P + 2$ where $C$ = components, $P$ = phases.

2. For $C = 3, P = 2$: $F = 3 - 2 + 2 = 3$. Three degrees of freedom.

3. This means we can independently vary temperature, pressure, and one composition variable while maintaining two-phase equilibrium.

4. For $C = 3, P = 3$: $F = 3 - 3 + 2 = 2$. Two degrees of freedom (e.g., T and P).

5. For $C = 3, P = 4$: $F = 3 - 4 + 2 = 1$. A univariant line in $(T, P)$ space.

6. Maximum phases: $F \geq 0$ requires $P \leq C + 2 = 5$. Five-phase coexistence is a single invariant point ($F = 0$).

Problem 3:In a binary liquid-vapor phase diagram, the overall composition is $z = 0.40$, the liquid phase has $x = 0.25$, and the vapor phase has $y = 0.60$. Find the fraction of each phase using the lever rule.

Solution:

1. Lever rule: $\frac{f^L}{f^V} = \frac{y - z}{z - x}$.

2. $\frac{f^L}{f^V} = \frac{0.60 - 0.40}{0.40 - 0.25} = \frac{0.20}{0.15} = \frac{4}{3}$.

3. With $f^L + f^V = 1$: $f^L = \frac{4}{7} \approx 0.571$ and $f^V = \frac{3}{7} \approx 0.429$.

4. Verify: $f^L x + f^V y = \frac{4}{7}(0.25) + \frac{3}{7}(0.60) = \frac{1.00 + 1.80}{7} = \frac{2.80}{7} = 0.40 = z$. Correct.

5. Physical interpretation: the system is 57.1% liquid and 42.9% vapor by moles.

6. The name "lever rule" comes from the analogy: imagine a lever balanced at $z$ with weights proportional to the phase amounts at positions $x$ and $y$.

Problem 4:Find the triple point pressure of a substance given: solid-liquid line slope $dP/dT = 1.3 \times 10^7\,\text{Pa/K}$, liquid-vapor Clausius-Clapeyron with $\Delta H_{\text{vap}} = 25\,\text{kJ/mol}$, and the normal boiling point at $T_b = 350\,\text{K}$, $P_b = 1\,\text{atm}$.

Solution:

1. The triple point lies at the intersection of the solid-liquid and liquid-vapor coexistence curves.

2. Liquid-vapor curve: $\ln P = -\frac{\Delta H_{\text{vap}}}{RT} + C$. Using the boiling point: $C = \ln(101325) + \frac{25000}{8.314 \times 350} = 11.526 + 8.583 = 20.109$.

3. The solid-liquid line is nearly vertical: $P = P_{\text{tp}} + (dP/dT)(T - T_{\text{tp}})$.

4. Since the solid-liquid slope is very steep, $T_{\text{tp}} \approx T_{\text{melt}}$ at low pressures. The triple point temperature is typically close to the normal melting point.

5. Estimate $T_{\text{tp}}$ by extrapolating the vapor pressure curve to low pressure. At the triple point, the vapor pressure curve gives: $P_{\text{tp}} = \exp\left(20.109 - \frac{25000}{8.314 T_{\text{tp}}}\right)$.

6. For a typical substance with $T_{\text{tp}} \approx 300\,\text{K}$: $P_{\text{tp}} = \exp(20.109 - 10.03) = \exp(10.08) \approx 2.4 \times 10^4\,\text{Pa} \approx 0.24\,\text{atm}$.

Problem 5:Classify the ferromagnetic-to-paramagnetic transition at the Curie temperature as first-order or second-order using the Ehrenfest classification. What thermodynamic quantities are continuous or discontinuous?

Solution:

1. At the Curie temperature $T_C$, the spontaneous magnetization continuously vanishes: $M(T) \propto (T_C - T)^\beta$ for $T < T_C$.

2. The Gibbs energy $G(T, P)$ and its first derivatives ($S = -(\partial G/\partial T)_P$ and $V = (\partial G/\partial P)_T$) are continuous at $T_C$.

3. There is no latent heat: $\Delta S = 0$ at the transition. No volume discontinuity either.

4. However, the second derivative $C_P = -T(\partial^2 G/\partial T^2)_P$ diverges at $T_C$: $C_P \propto |T - T_C|^{-\alpha}$.

5. By the Ehrenfest classification, this is a second-order (continuous) phase transition.

6. Modern perspective: the Ehrenfest scheme is too rigid for most transitions. The renormalization group classifies transitions by universality class and critical exponents ($\alpha, \beta, \gamma, \delta$), with the ferromagnetic transition in the 3D Ising universality class.

← Part II Overview Next: Chemical Potential →