5.3 Fundamental Property Relations and Maxwell Equations

5.3 Fundamental Property Relations and Maxwell Equations

Chemical engineers cannot directly measure abstract thermodynamic properties such as internal energy ($U$), enthalpy ($H$), entropy ($S$), Helmholtz energy ($A$), or Gibbs free energy ($G$) using field instrumentation. Plant transmitters measure only temperature ($T$), pressure ($P$), volume ($V$), and compositions ($x_i$). Fundamental property relations and Maxwell equations provide the mathematical bridge that converts differential changes in unmeasurable thermodynamic potentials into rigorous expressions involving only measurable $P$-$V$-$T$ coordinates and heat capacities. On the NCEES PE Chemical Exam, mastery of these relations is essential for deriving enthalpy departure functions, evaluating Joule-Thomson cooling, and calculating real fluid property changes.

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1. The Four Fundamental Property Relations

For a closed, homogeneous system of constant composition undergoing an internally reversible process, combining the First Law ($dU = \delta Q_{rev} - \delta W_{rev}$) with the Second Law ($\delta Q_{rev} = T dS$ and $\delta W_{rev} = P dV$) yields the fundamental property relation for internal energy:

$dU = T dS - P dV$

Although derived assuming a reversible process, because $U$, $T$, $S$, $P$, and $V$ are all state functions, $dU = T dS - P dV$ is universally valid for any process (reversible or irreversible) between equilibrium states in a homogeneous fluid of constant composition.

By applying Legendre transforms to $U$, three additional thermodynamic potentials are systematically generated:

1. Enthalpy ($H \equiv U + PV$): $dH = d(U + PV) = dU + P dV + V dP = (T dS - P dV) + P dV + V dP$ $dH = T dS + V dP$

2. Helmholtz Free Energy ($A \equiv U - TS$): $dA = d(U - TS) = dU - T dS - S dT = (T dS - P dV) - T dS - S dT$ $dA = -S dT - P dV$

3. Gibbs Free Energy ($G \equiv H - TS = U + PV - TS$): $dG = d(H - TS) = dH - T dS - S dT = (T dS + V dP) - T dS - S dT$ $dG = -S dT + V dP$

Natural (Canonical) Variables

Each thermodynamic potential possesses a set of natural variables that appear as the differentials on the right-hand side of its fundamental relation:

• $U = U(S, V) \implies \left(\frac{\partial U}{\partial S}\right)_V = T, \quad \left(\frac{\partial U}{\partial V}\right)_S = -P$
• $H = H(S, P) \implies \left(\frac{\partial H}{\partial S}\right)_P = T, \quad \left(\frac{\partial H}{\partial P}\right)_S = V$
• $A = A(T, V) \implies \left(\frac{\partial A}{\partial T}\right)_V = -S, \quad \left(\frac{\partial A}{\partial V}\right)_T = -P$
• $G = G(T, P) \implies \left(\frac{\partial G}{\partial T}\right)_P = -S, \quad \left(\frac{\partial G}{\partial P}\right)_T = V$

[!NOTE] The Chemical Engineer's Preferred Potential: Gibbs Free Energy ($G$)
In chemical process design, reactions and phase changes almost universally occur under controlled temperature and pressure conditions ($T, P$). Because the natural variables of Gibbs free energy are precisely $T$ and $P$, $dG = -S dT + V dP$ forms the computational foundation for phase equilibria (VLE, LLE), fugacity, and chemical reaction equilibrium constants ($K_{eq}$).

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2. Exact Differentials and the Four Maxwell Relations

Mathematically, if a continuous state function $z(x, y)$ has continuous second partial derivatives, its differential $dz = M dx + N dy$ is an exact differential. By Schwarz's theorem (Euler's reciprocity relation), the order of mixed second partial differentiation is commutative:

$\frac{\partial^2 z}{\partial y \, \partial x} = \frac{\partial^2 z}{\partial x \, \partial y} \implies \left( \frac{\partial M}{\partial y} \right)_x = \left( \frac{\partial N}{\partial x} \right)_y$

Applying this reciprocity condition to each of the four fundamental property relations yields the four classical Maxwell relations:

Maxwell Relation 1 (from $dU = T dS - P dV$):

$\left( \frac{\partial T}{\partial V} \right)_S = -\left( \frac{\partial P}{\partial S} \right)_V$

Maxwell Relation 2 (from $dH = T dS + V dP$):

$\left( \frac{\partial T}{\partial P} \right)_S = \left( \frac{\partial V}{\partial S} \right)_P$

Maxwell Relation 3 (from $dA = -S dT - P dV$):

$\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V$

Maxwell Relation 4 (from $dG = -S dT + V dP$):

$\left( \frac{\partial S}{\partial P} \right)_T = -\left( \frac{\partial V}{\partial T} \right)_P$

[!TIP] Mnemonic Strategy for Maxwell Relations: Notice the symmetry across the equations: the cross-derivatives relate $(T, S)$ to $(P, V)$.

• Relations holding temperature constant ($dA$ and $dG$) feature entropy derivatives on the left: $(\partial S / \partial V)_T$ and $(\partial S / \partial P)_T$.
• Watch the negative signs: $V$ and $P$ carry negative signs when paired with $S$ and $T$ as dictated by the signs in $dA = -S dT - P dV$ and $dG = -S dT + V dP$.

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3. Measurable $P$-$V$-$T$ Coefficients and Bridgman Transformations

To apply Maxwell relations to practical fluid mechanics and heat transfer, chemical engineers express volumetric derivatives using two standardized physical properties:

1. Isobaric Thermal Expansivity ($\beta$): The fractional change in volume with temperature at constant pressure: $\beta \equiv \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P$

2. Isothermal Compressibility ($\kappa_T$): The fractional reduction in volume with pressure at constant temperature (defined with a negative sign so that $\kappa_T > 0$ for stable phases): $\kappa_T \equiv -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T$

From the cyclic triple product rule of partial differentiation:

$\left( \frac{\partial P}{\partial T} \right)_V \left( \frac{\partial T}{\partial V} \right)_P \left( \frac{\partial V}{\partial P} \right)_T = -1 \implies \left( \frac{\partial P}{\partial T} \right)_V = -\frac{(\partial V / \partial T)_P}{(\partial V / \partial P)_T} = \frac{\beta V}{\kappa_T V} = \frac{\beta}{\kappa_T}$

Derivation of Internal Energy Variation with Volume: $(\partial U / \partial V)_T$

How does internal energy change during isothermal expansion? Start with $dU = T dS - P dV$. Divide by $dV$ holding $T$ constant:

$\left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial S}{\partial V} \right)_T - P$

Substitute Maxwell Relation 3 [$(\partial S / \partial V)_T = (\partial P / \partial T)_V$]:

$\left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial P}{\partial T} \right)_V - P = T \left( \frac{\beta}{\kappa_T} \right) - P$

For an ideal gas ($P = RT/V$): $\left( \frac{\partial P}{\partial T} \right)_V = \frac{R}{V} = \frac{P}{T} \implies \left( \frac{\partial U}{\partial V} \right)_T = T \left(\frac{P}{T}\right) - P = 0$ This rigorously proves Joule's Law: the internal energy of an ideal gas is strictly a function of temperature alone ($U = U(T)$).

Derivation of Enthalpy Variation with Pressure: $(\partial H / \partial P)_T$

How does enthalpy change during isothermal compression? Start with $dH = T dS + V dP$. Divide by $dP$ holding $T$ constant:

$\left( \frac{\partial H}{\partial P} \right)_T = T \left( \frac{\partial S}{\partial P} \right)_T + V$

Substitute Maxwell Relation 4 [$(\partial S / \partial P)_T = -(\partial V / \partial T)_P$]:

$\left( \frac{\partial H}{\partial P} \right)_T = V - T \left( \frac{\partial V}{\partial T} \right)_P = V (1 - \beta T)$

For an ideal gas ($V = RT/P \implies (\partial V / \partial T)_P = R/P = V/T$): $\left( \frac{\partial H}{\partial P} \right)_T = V - T \left(\frac{V}{T}\right) = 0$ This proves that ideal gas enthalpy is strictly independent of pressure ($H = H(T)$). For real fluids at high pressure, $(\partial H / \partial P)_T \ne 0$; integrating this expression from zero pressure to system pressure yields the enthalpy departure function:

$H - H^{ig} = \int_0^P \left[ V - T \left(\frac{\partial V}{\partial T}\right)_P \right] dP$

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4. Joule-Thomson Expansion and the Inversion Curve

When a high-pressure fluid stream is continuously throttled through a flow restriction (porous plug, cracked orifice, or throttling valve) with no heat exchange ($\dot{Q} = 0$) and no shaft work ($\dot{W}_s = 0$), the process is isenthalpic ($h_1 = h_2$).

The rate of temperature change with respect to pressure drop at constant enthalpy is defined as the Joule-Thomson coefficient ($\mu_{JT}$):

$\mu_{JT} \equiv \left( \frac{\partial T}{\partial P} \right)_H$

Applying the cyclic triple product rule to $(T, P, H)$:

$\left( \frac{\partial T}{\partial P} \right)_H \left( \frac{\partial P}{\partial H} \right)_T \left( \frac{\partial H}{\partial T} \right)_P = -1$

$\mu_{JT} = -\frac{(\partial H / \partial P)_T}{(\partial H / \partial T)_P} = -\frac{1}{C_p} \left( \frac{\partial H}{\partial P} \right)_T$

Substituting $(\partial H / \partial P)_T = V - T (\partial V / \partial T)_P$:

$\mu_{JT} = \frac{T \left( \frac{\partial V}{\partial T} \right)_P - V}{C_p} = \frac{V (\beta T - 1)}{C_p}$

Physical Behavior During Throttling ($dP < 0$)

Because pressure always drops during flow through a restriction ($dP < 0$), the sign of $\mu_{JT}$ dictates the temperature response:

$\Delta T \approx \mu_{JT} \Delta P$

1. $\mu_{JT} > 0$ (Cooling Regime):
   • Condition: $\beta T > 1$, or $T (\partial V / \partial T)_P > V$.
   • Result: Fluid cools upon expansion ($dT < 0$ when $dP < 0$).
   • Industrial application: Liquefaction of natural gas (Linde-Hampson cycle), refrigeration expansion valves.
2. $\mu_{JT} < 0$ (Heating Regime):
   • Condition: $\beta T < 1$, or $T (\partial V / \partial T)_P < V$.
   • Result: Fluid warms upon expansion ($dT > 0$ when $dP < 0$).
   • Critical safety hazard: Hydrogen, helium, and neon at ambient temperature possess negative Joule-Thomson coefficients. Throttling high-pressure hydrogen gas can heat it above its autoignition temperature, creating spontaneous fire hazards!
3. $\mu_{JT} = 0$ (Inversion Curve):
   • Condition: $\beta T = 1$, or $T (\partial V / \partial T)_P = V$.
   • The locus of states where $\mu_{JT} = 0$ defines the Joule-Thomson inversion curve on a $T$-$P$ diagram. The maximum temperature on this boundary is the maximum inversion temperature ($T_{inv, max}$). Below $T_{inv}$, gases can be cooled by throttling.

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5. Fundamental Relation for Heat Capacity Difference: $C_p - C_v$

The general thermodynamic relationship connecting constant-pressure and constant-volume heat capacities is derived from entropy differentials:

$C_p - C_v = T \left( \frac{\partial P}{\partial T} \right)_V \left( \frac{\partial V}{\partial T} \right)_P$

Substituting $(\partial P / \partial T)_V = \beta / \kappa_T$ and $(\partial V / \partial T)_P = V \beta$:

$C_p - C_v = \frac{T V \beta^2}{\kappa_T}$

Engineering Consequences of $C_p - C_v$

1. Thermodynamic Stability: For any stable thermodynamic phase, absolute temperature $T > 0$, volume $V > 0$, and isothermal compressibility $\kappa_T > 0$ (compression reduces volume). Because $\beta^2$ is inherently non-negative: $C_p - C_v \ge 0 \implies C_p \ge C_v$ $C_p$ can never be smaller than $C_v$ for a stable substance!
2. Incompressible Liquids: For ideal liquids, thermal expansivity is very small ($\beta \approx 0$). Therefore, $C_p \approx C_v = C$.
3. Ideal Gases: For an ideal gas, $\beta = 1/T$ and $\kappa_T = 1/P$. Substituting gives: $C_p - C_v = \frac{T V (1/T^2)}{1/P} = \frac{P V}{T} = R$ Re-deriving Mayer's classical relation $C_p - C_v = R$.

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6. Summary Table: Thermodynamic Potentials, Maxwell Relations, and Measurable Equivalents

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7. Critical PE Exam Traps & Pitfalls

[!WARNING] PE Exam Trap 1: The Negative Sign in the Gibbs Maxwell Relation
The Maxwell relation derived from $dG$ is $(\partial S / \partial P)_T = -(\partial V / \partial T)_P$. Forgetting the negative sign will cause you to calculate the wrong sign for entropy changes during compression ($dP > 0$), falsely showing that entropy increases during isothermal gas compression!

[!WARNING] PE Exam Trap 2: Blindly Assuming $(\partial H / \partial P)_T = 0$ for Real Fluids
While $(\partial H / \partial P)_T = 0$ is true for an ideal gas, it is never zero for real gases at elevated pressures or for liquids. When calculating high-pressure boiler feed pumps or high-pressure gas compressors, the mechanical enthalpy contribution $\int [V - T(\partial V/\partial T)_P] dP$ is non-zero and must be included.

[!WARNING] PE Exam Trap 3: Reversing the Sign of Joule-Thomson Temperature Change
Remember that throttling expansion involves a pressure drop ($\Delta P = P_2 - P_1 < 0$). If $\mu_{JT} = +0.25\text{ K/bar}$, then $\Delta T = \mu_{JT} \Delta P = (+0.25) \times (-20\text{ bar}) = -5.0\text{ K}$ (a $5.0\text{ K}$ drop in temperature). Candidates frequently forget that $\Delta P$ is negative, erroneously predicting that positive $\mu_{JT}$ produces heating.

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8. Step-by-Step Worked Numerical Example: Real Gas Enthalpy Departure and Joule-Thomson Throttling

Problem Statement

A high-pressure chemical synthesis reactor feed loop compresses pure nitrogen gas ($N_2$, $MW = 28.01\text{ g/mol}$) at a constant temperature of $T = 300.0\text{ K}$ from an initial pressure of $P_1 = 0.100\text{ MPa}$ ($1.00\text{ bar}$) to a reactor feed pressure of $P_2 = 10.00\text{ MPa}$ ($100.0\text{ bar}$).

Over this pressure range at $300.0\text{ K}$, nitrogen behaves according to the truncated volumetric equation of state:

$V = \frac{RT}{P} + b - \frac{a}{RT}$

Where:

• $R = 8.314\text{ J/(mol}\cdot\text{K)}$
• $C_p = 29.12\text{ J/(mol}\cdot\text{K)}$ (constant over this range)
• $a = 0.1370\text{ J}\cdot\text{m}^3/\text{mol}^2$
• $b = 3.870 \times 10^{-5}\text{ m}^3/\text{mol}$

Calculate:

1. Derive the analytical expression for $(\partial H / \partial P)_T$ for this gas.
2. Calculate the numerical value of $(\partial H / \partial P)_T$ in units of $\text{m}^3/\text{mol}$ (or $\text{J/(mol}\cdot\text{Pa)}$) at $300.0\text{ K}$.
3. Calculate the molar enthalpy change ($\Delta H$) of nitrogen during this isothermal compression from $1.00\text{ bar}$ to $100.0\text{ bar}$. Compare this to an ideal gas.
4. Derive the analytical expression for the Joule-Thomson coefficient ($\mu_{JT}$) and evaluate its numerical value at $300.0\text{ K}$ in units of $\text{K/bar}$.
5. If high-pressure nitrogen at $10.00\text{ MPa}$ and $300.0\text{ K}$ is suddenly throttled across an emergency relief valve to atmospheric pressure ($0.100\text{ MPa}$), estimate the resulting stream exit temperature ($T_2$).
6. Calculate the Joule-Thomson inversion temperature ($T_{inv}$) for this equation of state.

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Step-by-Step Solution

Step 1: Analytical Derivation of $(\partial H / \partial P)_T$

From fundamental thermodynamics: $\left( \frac{\partial H}{\partial P} \right)_T = V - T \left( \frac{\partial V}{\partial T} \right)_P$

Differentiate the equation of state with respect to $T$ at constant $P$: $V = \frac{RT}{P} + b - \frac{a}{RT}$ $\left( \frac{\partial V}{\partial T} \right)_P = \frac{R}{P} + 0 - \left( -\frac{a}{R T^2} \right) = \frac{R}{P} + \frac{a}{RT^2}$

Substitute $(\partial V / \partial T)_P$ into the enthalpy derivative: $\left( \frac{\partial H}{\partial P} \right)_T = \left( \frac{RT}{P} + b - \frac{a}{RT} \right) - T \left( \frac{R}{P} + \frac{a}{RT^2} \right)$ $\left( \frac{\partial H}{\partial P} \right)_T = \frac{RT}{P} + b - \frac{a}{RT} - \frac{RT}{P} - \frac{a}{RT}$ $\mathbf{\left( \frac{\partial H}{\partial P} \right)_T = b - \frac{2a}{RT}}$ (Notice that $P$ has completely canceled out; for this equation of state, $(\partial H / \partial P)_T$ depends solely on temperature!)

Step 2: Numerical Evaluation of $(\partial H / \partial P)_T$ at $300.0\text{ K}$

Calculate $RT$: $RT = (8.314\text{ J/(mol}\cdot\text{K)}) \times (300.0\text{ K}) = 2,494.2\text{ J/mol}$

Calculate $2a / (RT)$: $\frac{2a}{RT} = \frac{2 \times 0.1370\text{ J}\cdot\text{m}^3/\text{mol}^2}{2,494.2\text{ J/mol}} = \frac{0.2740}{2,494.2} = 1.09855 \times 10^{-4}\text{ m}^3/\text{mol}$

Now evaluate $(\partial H / \partial P)_T$: $\left( \frac{\partial H}{\partial P} \right)_T = (3.870 \times 10^{-5}) - (10.9855 \times 10^{-5}) = \mathbf{-7.1155 \times 10^{-5}\text{ m}^3/\text{mol} \quad (\text{or J/(mol}\cdot\text{Pa)})}$

Step 3: Enthalpy Change During Isothermal Compression

Because $(\partial H / \partial P)_T$ is independent of pressure, integration is direct: $\Delta H = \int_{P_1}^{P_2} \left( \frac{\partial H}{\partial P} \right)_T dP = \left( b - \frac{2a}{RT} \right) (P_2 - P_1)$

Pressure difference in Pascals: $P_2 - P_1 = 10.00 \times 10^6\text{ Pa} - 0.100 \times 10^6\text{ Pa} = 9.90 \times 10^6\text{ Pa}$

Calculate $\Delta H$: $\Delta H = (-7.1155 \times 10^{-5}\text{ J/(mol}\cdot\text{Pa)}) \times (9.90 \times 10^6\text{ Pa}) = \mathbf{-704.4\text{ J/mol} \quad (-0.7044\text{ kJ/mol})}$

Comparison with Ideal Gas: For an ideal gas, $(\partial H / \partial P)T = 0$, so $\Delta H{ideal} = 0$. The real-gas intermolecular attractions cause the fluid enthalpy to decrease by $704.4\text{ J/mol}$ during compression.

Step 4: Joule-Thomson Coefficient at $300.0\text{ K}$

$\mu_{JT} = -\frac{1}{C_p} \left( \frac{\partial H}{\partial P} \right)_T = \frac{1}{C_p} \left( \frac{2a}{RT} - b \right)$ $\mu_{JT} = -\frac{-7.1155 \times 10^{-5}\text{ J/(mol}\cdot\text{Pa)}}{29.12\text{ J/(mol}\cdot\text{K)}} = +2.4435 \times 10^{-6}\text{ K/Pa}$

Convert to engineering units of $\text{K/bar}$ ($1\text{ bar} = 10^5\text{ Pa}$): $\mu_{JT} = (2.4435 \times 10^{-6}\text{ K/Pa}) \times (10^5\text{ Pa/bar}) = \mathbf{+0.2444\text{ K/bar}}$

Step 5: Throttling Temperature Change Across Relief Valve

The pressure change during throttling is negative: $\Delta P = P_2 - P_1 = 0.100\text{ MPa} - 10.00\text{ MPa} = -9.90\text{ MPa} = -99.0\text{ bar}$ $\Delta T \approx \mu_{JT} \Delta P = (+0.2444\text{ K/bar}) \times (-99.0\text{ bar}) = \mathbf{-24.19\text{ K} \quad (-24.2^\circ\text{C})}$

Exiting temperature: $T_2 = T_1 + \Delta T = 300.0\text{ K} - 24.2\text{ K} = \mathbf{275.8\text{ K} \quad (2.65^\circ\text{C})}$ (Because $\mu_{JT} > 0$, the gas undergoes substantial cooling of $24.2^\circ\text{C}$ upon depressurization.)

Step 6: Joule-Thomson Inversion Temperature ($T_{inv}$)

The inversion temperature occurs when $\mu_{JT} = 0$: $\frac{2a}{R T_{inv}} - b = 0 \implies T_{inv} = \frac{2a}{R b}$ $T_{inv} = \frac{2 \times 0.1370\text{ J}\cdot\text{m}^3/\text{mol}^2}{(8.314\text{ J/(mol}\cdot\text{K)}) \times (3.870 \times 10^{-5}\text{ m}^3/\text{mol})} = \frac{0.2740}{3.2175 \times 10^{-4}} = \mathbf{851.6\text{ K} \quad (578.4^\circ\text{C})}$

(Conclusion: Since the operating temperature $300.0\text{ K}$ is well below the inversion temperature of $851.6\text{ K}$, nitrogen will always cool upon throttling at ambient conditions.)