6.3 Residual Properties and Enthalpy/Entropy Departure Functions

6.3 Residual Properties and Enthalpy/Entropy Departure Functions

Thermodynamic property evaluations on the NCEES PE Chemical Exam extend far beyond ideal gas heat capacities and steam tables. In high-pressure chemical reactors, multi-stage gas compressors, Joule-Thomson refrigeration loops, and supercritical fluid processes, enthalpy changes ($\Delta H$) and entropy changes ($\Delta S$) deviate significantly from ideal gas behavior. Because enthalpy and entropy cannot be measured directly with physical gauges, chemical engineers evaluate real fluid behavior using residual properties (also termed departure functions).

Residual properties link measurable $PVT$ volumetric equations of state directly to unmeasurable caloric properties ($H, S, G, U$) through exact mathematical derivations grounded in the Maxwell relations.

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1. The Theoretical Foundation of Residual Properties

A residual property ($M^R$) is defined as the difference between the actual molar thermodynamic property of a fluid ($M$) and the molar property the fluid would exhibit as an ideal gas ($M^{ig}$) at the identical temperature, pressure, and composition:

$M^R(T, P) \equiv M(T, P) - M^{ig}(T, P)$

Where $M$ can represent molar volume ($v$), enthalpy ($h$), entropy ($s$), internal energy ($u$), Gibbs free energy ($g$), or Helmholtz free energy ($a$).

Why the Hypothetical Ideal Gas Reference?

All real gases approach ideal behavior as pressure approaches zero ($P \to 0$). However, $M^{ig}(T, P)$ is evaluated at the actual process pressure $P$, representing a hypothetical state where molecules possess their real thermal heat capacities ($C_p^{ig}(T)$) but zero intermolecular forces and zero physical molecular volume.

The Volume Residual ($v^R$)

The residual molar volume derives directly from the compressibility factor $Z$:

$v^R = v - v^{ig} = \frac{Z R T}{P} - \frac{R T}{P} = \frac{R T}{P}(Z - 1)$

Notice that if a fluid behaves ideally, $Z = 1$, and the residual volume is identically zero ($v^R = 0$).

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2. Fundamental Derivation of Departure Functions from Maxwell Relations

From the fundamental property relation for a single-phase fluid of constant composition:

$dG = V dP - S dT$

For an ideal gas at the same $T$ and $P$:

$dG^{ig} = V^{ig} dP - S^{ig} dT$

Subtracting the ideal gas differential from the real fluid differential defines the residual Gibbs free energy differential ($dG^R$):

$dG^R = (V - V^{ig}) dP - (S - S^{ig}) dT = V^R dP - S^R dT$

1. Residual Gibbs Free Energy & Fugacity Coefficient

At constant temperature ($dT = 0$):

$\left( \frac{\partial G^R}{\partial P} \right)_T = V^R = \frac{R T}{P}(Z - 1)$

Integrating from the zero-pressure ideal gas reference state ($P = 0$, where $G^R = 0$) to process pressure $P$:

$\frac{G^R}{R T} = \int_0^P \frac{Z - 1}{P} dP$

In chemical thermodynamics, the fugacity coefficient ($\phi$) of a pure substance is defined by:

$G - G^{ig} = R T \ln\left( \frac{f}{P} \right) = R T \ln \phi$

Therefore, the fugacity coefficient is the direct exponential of the residual Gibbs free energy:

$\ln \phi = \frac{G^R}{R T} = \int_0^P \frac{Z - 1}{P} dP$

2. Residual Enthalpy Departure ($H^R$)

Applying the Gibbs-Helmholtz relation ($H = -T^2 [\partial(G/T)/\partial T]_P$):

$\frac{H^R}{R T} = -T \left[ \frac{\partial (G^R / R T)}{\partial T} \right]_P = -T \int_0^P \left( \frac{\partial Z}{\partial T} \right)_P \frac{dP}{P}$

$H^R = -R T^2 \int_0^P \left( \frac{\partial Z}{\partial T} \right)_P \frac{dP}{P} = \int_0^P \left[ v - T \left( \frac{\partial v}{\partial T} \right)_P \right] dP$

3. Residual Entropy Departure ($S^R$)

From the definition $G^R = H^R - T S^R$, solving for residual entropy yields:

$\frac{S^R}{R} = \frac{H^R}{R T} - \frac{G^R}{R T} = -T \int_0^P \left( \frac{\partial Z}{\partial T} \right)_P \frac{dP}{P} - \int_0^P (Z - 1) \frac{dP}{P}$

$S^R = \int_0^P \left[ \frac{R}{P} - \left( \frac{\partial v}{\partial T} \right)_P \right] dP$

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3. Volume-Explicit vs. Pressure-Explicit Integral Forms

While the derivations above integrate with respect to pressure, cubic equations of state (vdW, SRK, PR) express pressure as an explicit function of molar volume: $P = f(T, v)$. Converting the departure integrals from $(T, P)$ coordinates to $(T, v)$ coordinates via integration by parts yields the general volume-explicit departure formulas:

$\frac{H^R}{R T} = Z - 1 + \frac{1}{R T} \int_\infty^v \left[ T \left( \frac{\partial P}{\partial T} \right)_v - P \right] dv$

$\frac{S^R}{R} = \ln Z + \frac{1}{R} \int_\infty^v \left[ \left( \frac{\partial P}{\partial T} \right)_v - \frac{R}{v} \right] dv$

$\ln \phi = Z - 1 - \ln Z - \frac{1}{R T} \int_\infty^v \left( P - \frac{R T}{v} \right) dv$

These universal integral equations allow analytical evaluation of departure functions for any analytical equation of state.

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4. Analytical Residual Expressions for Virial and Cubic Equations of State

1. Truncated Virial Equation ($Z = 1 + B P / R T$)

For low to moderate pressures where the two-term virial equation is valid, analytical differentiation yields:

$\ln \phi = \frac{B P}{R T} = Z - 1$

$\frac{H^R}{R T} = \frac{P}{R} \left( \frac{B}{T} - \frac{dB}{dT} \right)$

$\frac{S^R}{R} = -\frac{P}{R} \left( \frac{dB}{dT} \right)$

2. Soave-Redlich-Kwong (SRK)

Integrating the SRK equation of state yields:

$\ln \phi = Z - 1 - \ln(Z - B) - \frac{A}{B} \ln\left( 1 + \frac{B}{Z} \right)$

$\frac{H^R}{R T} = Z - 1 - \frac{A}{B} \left[ 1 + \frac{m \sqrt{T_r}}{\sqrt{\alpha}} \right] \ln\left( 1 + \frac{B}{Z} \right)$

$\frac{S^R}{R} = \ln(Z - B) - \frac{A}{B} \left[ \frac{m \sqrt{T_r}}{\sqrt{\alpha}} \right] \ln\left( 1 + \frac{B}{Z} \right)$

3. Peng-Robinson (PR)

Integrating the Peng-Robinson equation of state yields the industry standard departure formulations:

$\ln \phi = Z - 1 - \ln(Z - B) - \frac{A}{2\sqrt{2} B} \ln\left[ \frac{Z + (1 + \sqrt{2})B}{Z + (1 - \sqrt{2})B} \right]$

$\frac{H^R}{R T} = Z - 1 - \frac{A}{2\sqrt{2} B} \left[ 1 + \frac{\kappa \sqrt{T_r}}{\sqrt{\alpha}} \right] \ln\left[ \frac{Z + (1 + \sqrt{2})B}{Z + (1 - \sqrt{2})B} \right]$

$\frac{S^R}{R} = \ln(Z - B) - \frac{A}{2\sqrt{2} B} \left[ \frac{\kappa \sqrt{T_r}}{\sqrt{\alpha}} \right] \ln\left[ \frac{Z + (1 + \sqrt{2})B}{Z + (1 - \sqrt{2})B} \right]$

Where $A = a P / (R T)^2$ and $B = b P / R T$.

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5. Generalized Three-Parameter Lee-Kesler Departure Tables

When cubic equations are not solved analytically on the PE exam, candidates utilize the generalized Lee-Kesler departure tables in the NCEES Reference Handbook. These express dimensionless residual enthalpy and entropy as two-term acentric factor expansions:

$\frac{H^R}{R T_c} = \left( \frac{H^R}{R T_c} \right)^{(0)} + \omega \left( \frac{H^R}{R T_c} \right)^{(1)}$

$\frac{S^R}{R} = \left( \frac{S^R}{R} \right)^{(0)} + \omega \left( \frac{S^R}{R} \right)^{(1)}$

$\log_{10}(\phi) = (\log_{10}\phi)^{(0)} + \omega (\log_{10}\phi)^{(1)}$

Both the simple fluid terms (superscript $0$) and correction terms (superscript $1$) are tabulated against reduced temperature ($T_r$) and reduced pressure ($P_r$).

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6. The Three-Step Thermodynamic Process Path

Because enthalpy ($H$) and entropy ($S$) are state functions, their changes between State 1 ($T_1, P_1$) and State 2 ($T_2, P_2$) depend only on the initial and final states, not the physical path.

Direct integration of real fluid properties between $(T_1, P_1)$ and $(T_2, P_2)$ is difficult because heat capacity varies with both temperature and pressure ($C_p(T, P)$). Chemical engineers replace the real process path with a three-step hypothetical path:

Real State 1 (T_1, P_1) --------------------------> Real State 2 (T_2, P_2)
      |                                                   ^
      | Step 1: Depressurization                          | Step 3: Pressurization
      | ΔH_1 = -H^R(T_1, P_1)                             | ΔH_3 = +H^R(T_2, P_2)
      v                                                   |
Ideal Gas (T_1, P -> 0) ------> Step 2 ---------> Ideal Gas (T_2, P -> 0)
                          Ideal Sensible Heating
                          ΔH_2 = ∫ C_p^ig(T) dT

Governing Equations for Real Changes

1. Enthalpy Change ($\Delta H$):

$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3$

$\Delta H = -H^R(T_1, P_1) + \int_{T_1}^{T_2} C_p^{ig}(T) dT + H^R(T_2, P_2)$

2. Entropy Change ($\Delta S$):

$\Delta S = -S^R(T_1, P_1) + \left[ \int_{T_1}^{T_2} \frac{C_p^{ig}(T)}{T} dT - R \ln\left( \frac{P_2}{P_1} \right) \right] + S^R(T_2, P_2)$

[!IMPORTANT] Sign Convention Rule:

• In Step 1, the fluid transitions from the real state to the ideal gas state, which is the reverse of departure; hence, we subtract the initial residual property: $-H^R(T_1, P_1)$ and $-S^R(T_1, P_1)$.
• In Step 3, the fluid transitions from the ideal gas state to the final real state; hence, we add the final residual property: $+H^R(T_2, P_2)$ and $+S^R(T_2, P_2)$.

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7. Summary Table: Departure Function Formulations Across Models

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8. Comprehensive Worked Numerical Example: High-Pressure Carbon Dioxide Gas Heater Sizing

Problem Statement

A supercritical carbon dioxide ($\text{CO}_2$) extraction unit operates an in-line electric process preheater. Pure $\text{CO}_2$ gas enters the heater at State 1 ($T_1 = 310.0\text{ K}$, $P_1 = 8.00\text{ MPa}$) at a molar flow rate of $\dot{n} = 500.0\text{ mol/s}$.

The gas is heated isobarically ($P_2 = P_1 = 8.00\text{ MPa} = 80.0\text{ bar}$) to State 2 ($T_2 = 420.0\text{ K}$) before entering the extraction column.

Thermodynamic Data for $\text{CO}_2$ ($MW = 44.01\text{ g/mol}$, $T_c = 304.2\text{ K}$, $P_c = 7.38\text{ MPa}$):

• At State 1 ($310.0\text{ K}, 8.00\text{ MPa}$): Tabulated molar enthalpy departure is $H^R_1 = -6,850.0\text{ J/mol}$.
• At State 2 ($420.0\text{ K}, 8.00\text{ MPa}$): Tabulated molar enthalpy departure is $H^R_2 = -1,420.0\text{ J/mol}$.
• Ideal gas heat capacity for $\text{CO}_2$ across this range is given by the mean value: $\bar{C}_p^{ig} = 41.50\text{ J/(mol}\cdot\text{K)}$.

Calculate:

1. The reduced temperatures ($T_{r1}, T_{r2}$) and reduced pressure ($P_r$) to confirm real fluid regime.
2. The ideal gas enthalpy change ($\Delta H^{ig}$) from $T_1$ to $T_2$.
3. The actual molar enthalpy change ($\Delta H$) of the $\text{CO}_2$ stream.
4. The required electrical heating duty ($\dot{Q}$) in kilowatts ($\text{kW}$).
5. The heating duty that would have been calculated if the engineer had mistakenly assumed ideal gas behavior, and the engineering consequences of that assumption.

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Step 1: Reduced Coordinates

$T_{r1} = \frac{T_1}{T_c} = \frac{310.0\text{ K}}{304.2\text{ K}} = \mathbf{1.019}$

$T_{r2} = \frac{T_2}{T_c} = \frac{420.0\text{ K}}{304.2\text{ K}} = \mathbf{1.381}$

$P_r = \frac{P}{P_c} = \frac{8.00\text{ MPa}}{7.38\text{ MPa}} = \mathbf{1.084}$

Because $T_{r1} \approx 1.02$ and $P_r = 1.08$, State 1 is right in the dense supercritical transition region just above the critical point, where intermolecular attractive forces are extremely strong and real gas deviations are severe.

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Step 2: Ideal Gas Enthalpy Change (Step 2)

$\Delta H^{ig} = \int_{T_1}^{T_2} C_p^{ig}(T) dT = \bar{C}_p^{ig} (T_2 - T_1)$ $\Delta H^{ig} = (41.50\text{ J/(mol}\cdot\text{K)}) \times (420.0 - 310.0)\text{ K} = 41.50 \times 110.0 = \mathbf{4,565.0\text{ J/mol}}$

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Step 3: Actual Molar Enthalpy Change

Apply the three-step departure formulation: $\Delta H = -H^R_1 + \Delta H^{ig} + H^R_2$ $\Delta H = -(-6,850.0\text{ J/mol}) + 4,565.0\text{ J/mol} + (-1,420.0\text{ J/mol})$ $\Delta H = +6,850.0 + 4,565.0 - 1,420.0 = \mathbf{9,995.0\text{ J/mol}}$

Thermodynamic Insight: In heating the fluid from $310\text{ K}$ to $420\text{ K}$, we must supply $4,565\text{ J/mol}$ to increase molecular kinetic energy (sensible heat), PLUS we must supply $(6,850 - 1,420) = 5,430\text{ J/mol}$ of energy to pull the molecules apart against their strong mutual attractive forces. The real fluid requires more than double the energy of an ideal gas!

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Step 4: Required Electric Heater Duty

For an open steady-state heating unit with no shaft work ($\dot{W}_s = 0$): $\dot{Q} = \dot{n} \Delta H$ $\dot{Q} = (500.0\text{ mol/s}) \times (9,995.0\text{ J/mol}) = 4,997,500\text{ W} = \mathbf{4,997.5\text{ kW}} \approx \mathbf{4,998\text{ kW}} \text{ (5.00 MW)}$

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Step 5: Comparison with Ideal Gas Assumption

If ideal gas behavior had been assumed: $\dot{Q}_{ideal} = \dot{n} \Delta H^{ig} = (500.0\text{ mol/s}) \times (4,565.0\text{ J/mol}) = 2,282,500\text{ W} = \mathbf{2,282.5\text{ kW}}$

$\text{Underprediction} = \frac{\dot{Q}_{ideal} - \dot{Q}_{real}}{\dot{Q}_{real}} \times 100\% = \frac{2,282.5 - 4,997.5}{4,997.5} \times 100\% = \mathbf{-54.33\%}$

An engineer who designs the heater using the ideal gas law would undersize the heating element by $54.3%$. In the actual plant, the heater would fail to reach the target extraction temperature of $420\text{ K}$, stalling extraction and potentially overloading the electrical supply.

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9. Critical PE Exam Traps & Pitfalls

Trap 1: The Initial Departure Sign Error
The most common error in residual enthalpy calculations is writing $\Delta H = H^R_1 + \Delta H^{ig} + H^R_2$. The initial state departure must be subtracted ($-H^R_1$) because the first step depressurizes the real gas to the ideal gas state. Since $H^R$ is typically negative for attractive gases, subtracting a negative number yields a positive contribution: $-(-6,850) = +6,850\text{ J/mol}$.

Trap 2: Mixing Pressure Units in the Residual Entropy Formulation
In the entropy departure equation, the ideal gas pressure change is $-R \ln(P_2 / P_1)$. This pressure ratio is strictly part of the ideal gas step. Do not insert $P_2 / P_1$ into the residual entropy functions $S^R$. Furthermore, $P_1$ and $P_2$ must be in the same absolute units so the ratio is dimensionless.

Trap 3: Misinterpreting Reference Table Headers
Handbooks and exam reference manuals often tabulate departure values with a leading negative sign, such as $-(H - H^{ig}) / (R T_c)$ or $-(H^R) / (R T_c)$ to report positive tabulated numbers. If the table heading already has a minus sign, the tabulated value is positive, but the actual residual property $H^R$ is negative! Always verify whether the table header includes the negative sign before plugging numbers into your equation.

Trap 4: Confusing Fugacity with Fugacity Coefficient
Remember that the fugacity coefficient $\phi = f / P$ is dimensionless, whereas fugacity $f = \phi P$ has units of pressure ($\text{bar}$, $\text{kPa}$, or $\text{psia}$). At low pressures, $\lim_{P \to 0} \phi = 1.00$ and $\lim_{P \to 0} f = P$.