7.1 Pure Component Vapor Pressure and Phase Equilibria

7.1 Pure Component Vapor Pressure and Phase Equilibria

Phase equilibria and vapor pressure estimation form the cornerstone of separation design, relief system sizing, and process vessel specification on the NCEES PE Chemical Exam. When a pure liquid boils inside a reboiler, flashes across an expansion valve, or evaporates from a storage tank, the system establishes a dynamic balance governed by thermodynamic potential gradients. Understanding how temperature and pressure dictate phase transitions enables chemical engineers to accurately compute operating conditions, prevent cavitation, and size distillation columns.

---

1. Fundamental Thermodynamic Criteria of Phase Equilibrium

Consider an isolated, multi-phase, pure-component system containing $\pi$ distinct phases (e.g., solid, liquid, and vapor) labeled $\alpha, \beta, \dots, \pi$. From the Second Law of Thermodynamics, an isolated system reaches equilibrium when its total entropy is maximized. At specified temperature ($T$) and pressure ($P$), this condition corresponds to minimizing the total Gibbs free energy ($dG_{T,P} = 0$).

For a closed, heterogeneous system at uniform $T$ and $P$, transferring an infinitesimal quantity of moles $dn_i$ from phase $\alpha$ to phase $\beta$ produces a change in total Gibbs energy:

$dG = \mu_i^\alpha (-dn_i) + \mu_i^\beta (dn_i) = \left( \mu_i^\beta - \mu_i^\alpha \right) dn_i$

Where $\mu_i$ is the chemical potential of species $i$, defined as the partial molar Gibbs free energy:

$\mu_i \equiv \left( \frac{\partial G}{\partial n_i} \right)_{T, P, n_{j \ne i}}$

At thermodynamic equilibrium, $dG = 0$ for any arbitrary, reversible mass transfer $dn_i$. Therefore, phase coexistence demands three simultaneous conditions:

1. Thermal Equilibrium: Uniform temperature across all phases. $T^\alpha = T^\beta = \dots = T^\pi$
2. Mechanical Equilibrium: Uniform pressure across all phases (in the absence of interfacial surface tension or hydrostatic heads). $P^\alpha = P^\beta = \dots = P^\pi$
3. Chemical / Phase Equilibrium: Uniform chemical potential (or fugacity) of each component across all phases. $\mu_i^\alpha = \mu_i^\beta = \dots = \mu_i^\pi \iff f_i^\alpha = f_i^\beta = \dots = f_i^\pi$

If $\mu_i^\alpha > \mu_i^\beta$, mass spontaneously transfers from phase $\alpha$ to phase $\beta$ until the chemical potentials equalize. For a pure substance ($C = 1$), the chemical potential equals the molar Gibbs free energy of that phase ($\mu = g$).

---

2. The Gibbs Phase Rule

Josiah Willard Gibbs established the Gibbs Phase Rule, which defines the number of independent intensive variables (degrees of freedom, $F$) that must be specified to completely fix the thermodynamic state of an equilibrium system.

Mathematical Formulation

For a non-reacting system:

$F = C - P + 2$

Where:

• $F$ = degrees of freedom (variance) of the system.
• $C$ = number of non-reacting chemical components.
• $P$ = number of phases coexisting at equilibrium.
• $2$ = represents the two intensive state variables: temperature ($T$) and pressure ($P$).

If the system involves $r$ independent reversible chemical reactions and $s$ special stoichiometric constraints (such as equimolar production from an initial charge), the rule generalizes to:

$F = (C - r - s) - P + 2$

Degrees of Freedom for Pure Fluids ($C = 1$)

• Single Phase ($P = 1$: Superheated Vapor, Subcooled Liquid, or Supercritical Fluid): $F = 1 - 1 + 2 = 2$ Two intensive properties must be specified (e.g., both $T$ and $P$) to locate the thermodynamic state and determine properties such as density, enthalpy, and entropy.
• Two Coexisting Phases ($P = 2$: Saturated Liquid + Saturated Vapor): $F = 1 - 2 + 2 = 1$ Only one intensive variable can be independently specified. Specifying temperature $T$ uniquely fixes the vapor pressure $P^{sat}(T)$. Conversely, specifying system pressure $P$ uniquely fixes the saturation boiling temperature $T^{sat}(P)$. A chemical process engineer cannot independently control both temperature and pressure in an equilibrium reboiler containing pure liquid and vapor.
• Three Coexisting Phases ($P = 3$: Solid + Liquid + Vapor at the Triple Point): $F = 1 - 3 + 2 = 0$ The system is invariant. The triple point exists at a single, unique temperature and pressure (for water, $T_{tp} = 0.01^\circ\text{C} = 273.16\text{ K}$ and $P_{tp} = 0.6117\text{ kPa} = 4.587\text{ mmHg}$). Neither $T$ nor $P$ can be altered without causing at least one phase to completely disappear.
• Critical Point: At the critical point ($T_c, P_c$), the meniscus between liquid and vapor vanishes, and all intensive properties of both phases merge into a single supercritical fluid. This boundary point is invariant ($F = 0$) for a pure component.

Comparison Table: Gibbs Phase Rule Applications in Process Engineering

---

3. The Clapeyron and Clausius-Clapeyron Equations

Along any phase coexistence boundary line on a $P-T$ diagram, the molar Gibbs free energy of the two coexisting phases must remain equal ($g^\alpha = g^\beta$). Moving an infinitesimal distance $dT$ and $dP$ along this equilibrium curve requires:

$dg^\alpha = dg^\beta$

Applying the fundamental property relation $dg = -s dT + v dP$:

$-s^\alpha dT + v^\alpha dP = -s^\beta dT + v^\beta dP$

Rearranging terms:

$\left( v^\beta - v^\alpha \right) dP = \left( s^\beta - s^\alpha \right) dT$

Because the phase transition is reversible at constant $T$ and $P$, the molar entropy of phase change is $\Delta s = s^\beta - s^\alpha = \Delta h / T$. Substituting this yields the Clapeyron Equation:

$\frac{dP^{sat}}{dT} = \frac{\Delta h_{vap}}{T \left( v^V - v^L \right)} = \frac{\Delta H_{vap}}{T \Delta v_{vap}}$

The Clapeyron equation is thermodynamically exact for any first-order phase transition (solid-liquid, liquid-vapor, solid-vapor) across all pressures up to the critical point.

The Clausius-Clapeyron Approximation

For vapor-liquid equilibria at low to moderate pressures (typically $P < 5\text{ bar}$ or $T < 0.75 T_c$), Rudolf Clausius introduced two simplifying assumptions:

1. Negligible Liquid Molar Volume: The molar volume of the liquid phase is negligible compared to the molar volume of the vapor phase ($v^V \gg v^L \implies \Delta v_{vap} = v^V - v^L \approx v^V$).
2. Ideal Gas Vapor Behavior: The vapor phase behaves as an ideal gas ($v^V = R T / P^{sat}$).
3. Constant Latent Heat: The enthalpy of vaporization ($\Delta H_{vap}$) remains constant over the temperature span of interest.

Substituting $v^V = R T / P^{sat}$ into the Clapeyron equation:

$\frac{dP^{sat}}{dT} = \frac{\Delta H_{vap}}{T \left( \frac{R T}{P^{sat}} \right)} = \frac{P^{sat} \Delta H_{vap}}{R T^2}$

Dividing through by $P^{sat}$ yields the differential Clausius-Clapeyron Equation:

$\frac{1}{P^{sat}} \frac{dP^{sat}}{dT} = \frac{d\ln P^{sat}}{dT} = \frac{\Delta H_{vap}}{R T^2}$

Separating variables and integrating between state 1 ($T_1, P_1^{sat}$) and state 2 ($T_2, P_2^{sat}$):

$\int_{P_1^{sat}}^{P_2^{sat}} d\ln P^{sat} = \frac{\Delta H_{vap}}{R} \int_{T_1}^{T_2} \frac{dT}{T^2}$

$\ln\left( \frac{P_2^{sat}}{P_1^{sat}} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \frac{\Delta H_{vap}}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$

Linear Form and Graphical Interpretation

The indefinite integral produces a linear relationship:

$\ln P^{sat} = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T} \right) + C_{const}$

• A plot of $\ln P^{sat}$ versus $1/T$ (with temperature in absolute units, $\text{K}$) yields a straight line with a negative slope equal to $-\Delta H_{vap} / R$.
• The y-intercept represents the integration constant $C_{const}$.
• This equation provides a rapid two-point estimation method on the PE exam when only the normal boiling point ($T_b$ at $1.0\text{ atm}$) and latent heat of vaporization are given.

---

4. Empirical Vapor Pressure Formulations: The Antoine Equation

While the Clausius-Clapeyron equation provides fundamental insight, real vapor pressure curves plotted as $\ln P^{sat}$ vs $1/T$ exhibit distinct curvature over wide temperature ranges. This curvature arises because $\Delta H_{vap}$ decreases monotonically with increasing temperature (reaching zero at the critical point, $T_c$), and real vapor deviations ($Z < 1$) become pronounced at higher pressures.

To capture this curvature empirically without complex multi-parameter equations of state, Louis Charles Antoine (1888) introduced an effective temperature offset parameter ($C$).

Antoine Equation Formulations

On the NCEES PE Chemical Exam and in engineering reference literature, the Antoine equation appears in two common algebraic forms:

Form 1: Base-10 Logarithm (Common in Perry's Chemical Engineers' Handbook)

$\log_{10} P^{sat} = A - \frac{B}{T + C}$

• Typical units: $P^{sat}$ in $\text{mmHg}$ (or $\text{bar}$), and $T$ in $^\circ\text{C}$.
• Solved explicitly for temperature (bubble/dew point calculation): $T = \frac{B}{A - \log_{10} P^{sat}} - C$

Form 2: Natural Logarithm (Common in NCEES Reference Handbook & NIST)

$\ln P^{sat} = A - \frac{B}{T + C}$

• Typical units: $P^{sat}$ in $\text{kPa}$ (or $\text{bar}$), and $T$ in $\text{K}$ (or $^\circ\text{C}$).
• Solved explicitly for temperature: $T = \frac{B}{A - \ln P^{sat}} - C$

Extended Vapor Pressure Models (DIPPR 101 & Wagner)

For high-accuracy applications spanning from the triple point to the critical point, process simulators (Aspen, HYSYS) employ the Wagner Equation:

$\ln P_r^{sat} = \frac{A \tau + B \tau^{1.5} + C \tau^3 + D \tau^6}{1 - \tau}$

Where reduced vapor pressure is $P_r^{sat} = P^{sat}/P_c$, dimensionless temperature departure is $\tau = 1 - T_r = 1 - (T/T_c)$, and $A, B, C, D$ are substance-specific constants. The Wagner equation correctly models the infinite slope of $\Delta H_{vap}$ as $T \to T_c$ and avoids unphysical behavior outside the Antoine temperature range.

---

5. Cox Charts and Reference Fluid Methods

Before digital computers, chemical engineers utilized graphical Cox Charts to linearize vapor pressure curves across vast temperature spans (from sub-ambient to supercritical conditions).

Operating Principle

Edwin R. Cox (1923) observed that if $\log_{10} P^{sat}$ is plotted against a special, non-linear temperature scale calibrated so that the vapor pressure of a reference substance (usually water) forms a perfectly straight line, the vapor pressure curves of almost all other hydrocarbons and organic liquids also become straight lines:

$\log_{10} P^{sat} = A - \frac{B}{T^*}$

Where $T^*$ is the transformed reference temperature function.

Key Practical Properties of Cox Charts

1. Infinite Convergence Point: For homologous series (e.g., straight-chain paraffinic alkanes: methane through n-decane), the straight lines converge toward a single common focal point at high temperature and pressure.
2. Extrapolation Accuracy: Because the lines are straight on Cox coordinates, drawing a line through just two reliable data points (such as the normal boiling point and room-temperature vapor pressure) enables accurate extrapolation toward elevated pressures without significant error.

---

6. Summary Comparison: Vapor Pressure Estimation Models

---

7. Comprehensive Worked Numerical Example: Vacuum Distillation Reboiler Analysis

Problem Statement

A chemical plant purifies n-hexane ($\text{C}6\text{H}{14}$, $MW = 86.18\text{ g/mol}$) in a continuous vacuum distillation column. The reboiler operates at an absolute pressure of $P_{reb} = 250.0\text{ mmHg}$ ($0.3289\text{ atm} = 33.33\text{ kPa}$).

The NCEES PE Chemical Reference Handbook supplies the following Antoine constants for n-hexane: $\log_{10}\left(P^{sat}[\text{mmHg}]\right) = A - \frac{B}{T[^\circ\text{C}] + C}$ Where:

• $A = 6.87776$
• $B = 1171.53$
• $C = 224.366$
• Valid range: $-13^\circ\text{C}$ to $95^\circ\text{C}$

Calculate:

1. The degrees of freedom ($F$) for the boiling liquid-vapor mixture inside the reboiler.
2. The operating temperature ($T_{reb}$) inside the reboiler in $^\circ\text{C}$ and $\text{K}$.
3. The normal boiling point ($T_b$ at $P = 760.0\text{ mmHg}$) of n-hexane.
4. The latent heat of vaporization ($\Delta H_{vap}$) at the reboiler operating temperature using the combined Clausius-Clapeyron and Antoine derivative relationship: $\frac{d\ln P^{sat}}{dT} = \frac{\Delta H_{vap}}{R T^2}$
5. The percentage reduction in reboiler operating temperature achieved by pulling vacuum ($250\text{ mmHg}$) compared to atmospheric operation ($760\text{ mmHg}$).

---

Step 1: Degrees of Freedom Analysis

Apply the Gibbs Phase Rule for a non-reacting system: $F = C - P + 2$

• Number of components: $C = 1$ (pure n-hexane).
• Number of phases: $P = 2$ (saturated liquid + saturated vapor). $F = 1 - 2 + 2 = \mathbf{1}$

There is 1 degree of freedom. Because the plant operator fixes the operating pressure at $P_{reb} = 250.0\text{ mmHg}$, the system variance is exhausted ($1 - 1 = 0$). The boiling temperature is completely constrained by thermodynamics and cannot be adjusted independently.

---

Step 2: Reboiler Operating Temperature ($T_{reb}$)

Rearrange the Antoine equation to solve explicitly for $T$: $\log_{10}(P^{sat}) = A - \frac{B}{T + C} \implies T = \frac{B}{A - \log_{10}(P^{sat})} - C$

Evaluate $\log_{10}(250.0)$: $\log_{10}(250.0) = 2.397940$

Substitute constants: $T_{reb} = \frac{1171.53}{6.87776 - 2.397940} - 224.366$ $T_{reb} = \frac{1171.53}{4.479820} - 224.366 = 261.5127 - 224.366 = \mathbf{37.15^\circ\text{C}}$

Convert to absolute temperature: $T_{reb} = 37.147 + 273.15 = \mathbf{310.30\text{ K}}$

---

Step 3: Normal Boiling Point Determination

At atmospheric pressure, $P^{sat} = 760.0\text{ mmHg}$: $\log_{10}(760.0) = 2.880814$

$T_b = \frac{1171.53}{6.87776 - 2.880814} - 224.366 = \frac{1171.53}{3.996946} - 224.366$ $T_b = 293.106 - 224.366 = \mathbf{68.74^\circ\text{C}} \quad (341.89\text{ K})$

(Literature experimental normal boiling point of n-hexane is $68.7^\circ\text{C}$; calculation matches exactly).

---

Step 4: Latent Heat of Vaporization ($\Delta H_{vap}$) at Reboiler Conditions

From the Clausius-Clapeyron equation: $\frac{d\ln P^{sat}}{dT} = \frac{\Delta H_{vap}}{R T^2} \implies \Delta H_{vap} = R T^2 \left( \frac{d\ln P^{sat}}{dT} \right)$

Relate natural log to base-10 log: $\ln P^{sat} = \ln(10) \cdot \log_{10} P^{sat} = 2.302585 \cdot \left( A - \frac{B}{T + C} \right)$

Differentiating with respect to temperature ($T$ in $^\circ\text{C}$ or $\text{K}$, since $d(T[^\circ\text{C}]) = d(T[\text{K}])$): $\frac{d\ln P^{sat}}{dT} = 2.302585 \cdot \frac{B}{(T[^\circ\text{C}] + C)^2}$

Substitute $T_{reb} = 37.147^\circ\text{C}$, $B = 1171.53$, and $C = 224.366$: $T[^\circ\text{C}] + C = 37.147 + 224.366 = 261.513$ $(T + C)^2 = (261.513)^2 = 68,389.05$ $\frac{d\ln P^{sat}}{dT} = \frac{2.302585 \times 1171.53}{68,389.05} = \frac{2697.547}{68,389.05} = \mathbf{0.039444\text{ K}^{-1}}$

Now calculate $\Delta H_{vap}$ at $T = 310.30\text{ K}$ using $R = 8.31447\text{ J/(mol}\cdot\text{K)}$: $T^2 = (310.30\text{ K})^2 = 96,286.09\text{ K}^2$ $\Delta H_{vap} = \left( 8.31447\text{ J/(mol}\cdot\text{K)} \right) \times \left( 96,286.09\text{ K}^2 \right) \times \left( 0.039444\text{ K}^{-1} \right)$ $\Delta H_{vap} = 8.31447 \times 3797.91 = \mathbf{31,577\text{ J/mol}} = \mathbf{31.58\text{ kJ/mol}}$

In specific mass units: $\Delta h_{vap} = \frac{31,577\text{ J/mol}}{0.08618\text{ kg/mol}} = 366,408\text{ J/kg} = \mathbf{366.4\text{ kJ/kg}}$

(Literature experimental value for n-hexane at $37^\circ\text{C}$ is $31.6\text{ kJ/mol}$; exact agreement demonstrates the thermodynamic consistency between the Antoine derivative and latent heat).

---

Step 5: Vacuum Temperature Reduction

$\Delta T = T_b - T_{reb} = 68.74^\circ\text{C} - 37.15^\circ\text{C} = \mathbf{31.59^\circ\text{C}}$ $\text{Percentage Reduction (Absolute Scale)} = \frac{341.89\text{ K} - 310.30\text{ K}}{341.89\text{ K}} \times 100\% = \mathbf{9.24\%}$

Operating the reboiler under a vacuum of $250\text{ mmHg}$ lowers the boiling point by $31.6^\circ\text{C}$. In industrial practice, this thermal reduction permits the use of low-pressure utility waste steam ($100^\circ\text{C}$) as the reboiler heating medium instead of expensive high-pressure steam, and prevents heat-sensitive chemicals from thermal degradation and fouling.

---

8. Critical PE Exam Traps & Pitfalls

Trap 1: The Antoine Equation Unit Catastrophe
Antoine constants are purely empirical curve-fit parameters tied to exact units. You must verify three items before pressing buttons on your calculator:

1. Base-10 vs. Natural Logarithm: Does the formula specify $\log_{10}(P)$ or $\ln(P)$? Entering a natural logarithm into a base-10 equation produces an error factor of $\ln(10) \approx 2.303$.
2. Temperature Scale: Does the parameter $C$ expect Celsius ($^\circ\text{C}$) or Kelvin ($\text{K}$)? If $C \approx 200-250$, the equation almost certainly expects $^\circ\text{C}$. If $C \approx -30$ to $-50$, it expects $\text{K}$.
3. Pressure Units: Does the formula evaluate pressure in $\text{mmHg}$, $\text{bar}$, $\text{kPa}$, or $\text{psia}$? Look at the table header carefully.

Trap 2: Extrapolating the Clausius-Clapeyron Equation Near the Critical Point
The Clausius-Clapeyron equation assumes $\Delta v_{vap} \approx v^V$ and ideal gas behavior ($Z = 1$). Near the critical point ($T_r > 0.85$), saturated liquid volume expands dramatically, vapor compressibility drops to $Z_c \approx 0.27$, and $\Delta H_{vap}$ plummets toward zero. Using Clausius-Clapeyron to estimate vapor pressures above $0.8 T_c$ results in errors exceeding $30-50%$.

Trap 3: Misapplying the Phase Rule to Partially Miscible or Immiscible Systems
When two liquids are completely immiscible (e.g., liquid water + liquid toluene in a steam stripping vessel), they form two distinct liquid phases. Candidates often count the liquid mixture as a single phase ($P = 1$). For water + toluene + vapor: $C = 2, P = 3 \implies F = 2 - 3 + 2 = 1$. Fixing total system pressure completely fixes the three-phase boiling temperature, meaning the mixture boils at a temperature lower than the normal boiling point of either pure component.