Thermodynamic property relations and Maxwell relations

Fundamental property relations, Maxwell relations and working equations for real fluids, Clapeyron and Gibbs-Helmholtz equations.

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Why it matters

Entropy, internal energy and enthalpy cannot be measured directly, but P, V, T and heat capacities can. Property relations convert the quantities you need (ΔH, ΔS, ΔU of a real fluid, the Joule-Thomson effect, the slope of a vapour-pressure curve) into derivatives of P-V-T data or an equation of state. Every steam table and every simulator property package is built this way.

Key ideas

Fundamental property relations. Combining the first and second laws for a closed system of fixed composition doing only PV work gives dU = T dS − P dV. Defining H = U + PV, A = U − TS (Helmholtz energy) and G = H − TS (Gibbs energy) gives three more forms. Each potential has natural variables: U(S, V), H(S, P), A(T, V), G(T, P). Although derived along a reversible path, these relations connect state functions and so hold for any change between equilibrium states.

Meaning of A and G. At constant T, −ΔA is the maximum total work a closed system can deliver. At constant T and P, −ΔG is the maximum non-PV work (for example electrical work in a cell), and a spontaneous process has ΔG < 0. That is why G at constant T and P becomes the equilibrium criterion for phases and reactions.

Maxwell relations. Because dU, dH, dA and dG are exact differentials, their mixed second derivatives are equal. This gives four Maxwell relations. The two most used come from A and G, because they turn entropy derivatives into measurable P-V-T derivatives:

  • (∂S/∂V)_T = (∂P/∂T)_V
  • (∂S/∂P)_T = −(∂V/∂T)_P Rather than memorising the signs, write dG = −S dT + V dP or dA = −S dT − P dV and equate cross derivatives; it takes seconds and never goes wrong.

Working equations for real fluids. With Maxwell relations, dS and dH (in T, P) and dU and dS (in T, V) are written using Cp, Cv and the equation of state. These give:

  • the energy equation (∂U/∂V)_T = T(∂P/∂T)_V − P, which is zero for an ideal gas and a/V² for a van der Waals gas;
  • (∂H/∂P)_T = V − T(∂V/∂T)_P, zero for an ideal gas;
  • the Joule-Thomson coefficient μ_JT = (∂T/∂P)_H, whose sign tells whether throttling cools or heats a gas;
  • Cp − Cv = T(∂P/∂T)_V(∂V/∂T)_P, which reduces to R for an ideal gas.

Clapeyron equation. At a phase transition, equality of G of the two phases along the coexistence curve gives dP_sat/dT = ΔH/(T ΔV). Neglecting the liquid volume and treating the vapour as ideal gives the Clausius-Clapeyron equation, from which ln P_sat varies linearly with 1/T over moderate ranges.

Gibbs-Helmholtz equation. [∂(G/RT)/∂T]_P = −H/(RT²). It links the temperature dependence of G to H and is the basis of the van't Hoff equation in reaction equilibrium.

Formulas

dU = T dS − P dV dH = T dS + V dP dA = −S dT − P dV dG = −S dT + V dP (∂T/∂V)_S = −(∂P/∂S)_V, (∂T/∂P)_S = (∂V/∂S)_P (∂S/∂V)_T = (∂P/∂T)_V, (∂S/∂P)_T = −(∂V/∂T)_P dH = Cp dT + [V − T(∂V/∂T)_P] dP dS = (Cp/T) dT − (∂V/∂T)_P dP dU = Cv dT + [T(∂P/∂T)_V − P] dV dS = (Cv/T) dT + (∂P/∂T)_V dV μ_JT = (∂T/∂P)_H = [T(∂V/∂T)_P − V] / Cp dP_sat/dT = ΔH_vap / (T·ΔV_vap) (Clapeyron) ln(P₂/P₁) = −(ΔH_vap/R)·(1/T₂ − 1/T₁) (Clausius-Clapeyron, ideal vapour, constant ΔH_vap) [∂(G/RT)/∂T]_P = −H/(R·T²) (Gibbs-Helmholtz)

Symbols: U, H, A, G molar internal energy, enthalpy, Helmholtz and Gibbs energy (J/mol); S molar entropy (J/mol·K); T (K); P (Pa); V molar volume (m³/mol); Cp, Cv molar heat capacities (J/mol·K); μ_JT Joule-Thomson coefficient (K/Pa); ΔH_vap molar latent heat (J/mol); R = 8.314 J/(mol·K). All apply to a single phase of constant composition with only PV work.

Worked examples

Example 1 (standard): isothermal entropy change of an ideal gas from a Maxwell relation. Given: 1 mol of ideal gas is compressed isothermally from 1 bar to 10 bar. Find ΔS and ΔH.

  1. (∂S/∂P)_T = −(∂V/∂T)_P. For V = RT/P, (∂V/∂T)_P = R/P.
  2. ΔS = −∫R/P dP = −R·ln(P₂/P₁) = −8.314 × ln 10 = −19.14 J/mol·K.
  3. (∂H/∂P)_T = V − T(∂V/∂T)_P = RT/P − T·R/P = 0, so ΔH = 0. Answer: ΔS = −19.1 J/(mol·K), ΔH = 0

Example 2 (GATE level): isothermal expansion of a van der Waals gas. Given: 1 mol of CO₂ obeying the van der Waals equation (a = 0.3655 Pa·m⁶/mol², b = 4.282 × 10⁻⁵ m³/mol) expands reversibly and isothermally at 300 K from V₁ = 5.0 × 10⁻⁴ to V₂ = 2.0 × 10⁻³ m³/mol. Find ΔU, ΔS, Q and W.

  1. From P = RT/(V − b) − a/V²: (∂P/∂T)_V = R/(V − b).
  2. (∂U/∂V)_T = T(∂P/∂T)_V − P = RT/(V − b) − [RT/(V − b) − a/V²] = a/V².
  3. ΔU = ∫a/V² dV = a(1/V₁ − 1/V₂) = 0.3655 × (2000 − 500) = 548.3 J/mol. (An ideal gas would give zero.)
  4. (∂S/∂V)_T = (∂P/∂T)_V = R/(V − b), so ΔS = R·ln[(V₂ − b)/(V₁ − b)] = 8.314 × ln(1.95718 × 10⁻³ / 4.5718 × 10⁻⁴) = 8.314 × 1.4542 = 12.09 J/mol·K.
  5. Reversible isothermal: Q = T·ΔS = 300 × 12.09 = 3627 J/mol.
  6. W = Q − ΔU = 3627 − 548 = 3079 J/mol (done by the gas). Check by direct integration: W = RT·ln[(V₂ − b)/(V₁ − b)] + a(1/V₂ − 1/V₁) = 3627 − 548 = 3079 J/mol. Answer: ΔU = 548 J/mol, ΔS = 12.09 J/(mol·K), Q = 3627 J/mol, W = 3079 J/mol

Common mistakes

  • Dropping the minus sign in (∂S/∂P)_T = −(∂V/∂T)_P.
  • Assuming ΔU = 0 or ΔH = 0 for any isothermal process; that is true only for an ideal gas.
  • Using Clausius-Clapeyron near the critical point, where the vapour is far from ideal and the liquid volume is not negligible.
  • Applying dU = T dS − P dV to a reacting or open system without the chemical potential terms (added in the partial molar properties topic).
  • Confusing A (constant T, V) with G (constant T, P) as the spontaneity criterion.

For GATE CH

  • MCQs asking which expression is a valid Maxwell relation or which derivative equals a given quantity.
  • Derivation-style questions: (∂U/∂V)_T or (∂H/∂P)_T for an ideal gas, van der Waals gas or a gas obeying V = RT/P + b.
  • NAT problems on Clausius-Clapeyron (vapour pressure at another temperature, or latent heat from two vapour-pressure points) and Joule-Thomson coefficients.
  • Practise writing dG and dA, reading off the Maxwell relations, and differentiating a given EOS cleanly.

Quick check

  1. What are the natural variables of G?
  2. Write the Maxwell relation that gives (∂S/∂V)_T.
  3. For a van der Waals gas, what is (∂U/∂V)_T?
  4. What is (∂H/∂P)_T for an ideal gas?
  5. Under what assumptions does the Clapeyron equation reduce to the Clausius-Clapeyron equation?

Answers: 1. T and P. 2. (∂S/∂V)_T = (∂P/∂T)_V. 3. a/V². 4. Zero. 5. Liquid volume negligible compared with vapour volume, vapour ideal, and (for the integrated form) ΔH_vap constant.

Interview questions

All Chemical Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What are thermodynamic property relations?Concept

    Thermodynamic property relations are mathematical equations that relate different thermodynamic properties of a system. These relations help in determining the changes in properties like pressure, volume, temperature, and entropy without direct measurement. They are derived from the fundamental laws of thermodynamics and are essential for simplifying complex calculations in thermodynamic processes.

  2. 2.Explain Maxwell's relations in thermodynamics.Concept

    Maxwell's relations are a set of four equations derived from the second law of thermodynamics and the concept of exact differentials. They relate the partial derivatives of thermodynamic potentials (such as internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy) to measurable properties like temperature, pressure, and volume. These relations are useful for converting difficult-to-measure properties into ones that are easier to measure.

  3. 3.Why are Maxwell's relations important in chemical engineering thermodynamics?Application

    Maxwell's relations are important because they provide a way to calculate changes in entropy, volume, and other properties using measurable quantities like temperature and pressure. This is particularly useful in chemical engineering for designing processes and equipment where direct measurement of certain properties is challenging. They simplify the analysis of thermodynamic systems and help in understanding the behavior of substances under different conditions.

  4. 4.How can you derive Maxwell's relations from the fundamental thermodynamic equations?Concept

    Write the fundamental relation for each potential: dU = T dS − P dV, dH = T dS + V dP, dA = −S dT − P dV, dG = −S dT + V dP. Each is an exact differential, so the mixed second derivatives are equal. For example, from dG = −S dT + V dP, ∂²G/∂P∂T gives −(∂S/∂P)_T = (∂V/∂T)_P, i.e. (∂S/∂P)_T = −(∂V/∂T)_P. Repeating for U, H and A gives the other three.

  5. 5.Explain the significance of the Gibbs-Helmholtz equation in thermodynamics.Concept

    The Gibbs-Helmholtz equation, [∂(G/RT)/∂T]_P = −H/(RT²), relates how G/RT changes with temperature to the enthalpy. Applied to the standard Gibbs energy change of a reaction, where ΔG° = −RT ln K, it gives the van't Hoff equation d ln K/dT = ΔH°/(RT²). That is how we predict the shift of an equilibrium constant with temperature from heat-of-reaction data.

  6. 6.Why is the concept of exact differentials important in deriving thermodynamic property relations?Concept

    Exact differentials are important because they ensure that the derived relations are path-independent and only depend on the initial and final states of the system. This is crucial in thermodynamics, where we often deal with state functions like internal energy, enthalpy, and entropy. Using exact differentials allows us to derive relations like Maxwell's equations, which are fundamental in connecting different thermodynamic properties.

  7. 7.Calculate the change in entropy for an ideal gas when the temperature changes from 300 K to 400 K at constant volume. Assume the molar heat capacity at constant volume, Cv, is 20 J/mol·K.Numerical

    At constant volume, dS = (Cv/T) dT for an ideal gas, so per mole ΔS = Cv·ln(T₂/T₁). ΔS = 20 × ln(400/300) = 20 × 0.2877 = 5.75 J/(mol·K). The entropy rises because heat is added at constant volume.

  8. 8.What is the physical interpretation of the Helmholtz free energy in a thermodynamic system?Concept

    The Helmholtz free energy (A) is a thermodynamic potential that measures the useful work obtainable from a closed system at constant temperature and volume. It is defined as A = U - TS, where U is the internal energy, T is the temperature, and S is the entropy. The Helmholtz free energy is particularly useful in systems where volume is constant, such as in certain chemical reactions and processes.

  9. 9.If the pressure of a system is increased while keeping the temperature constant, how does it affect the Gibbs free energy?Application

    The Gibbs free energy (G) is affected by changes in pressure at constant temperature according to the relation dG = VdP - SdT. At constant temperature (dT = 0), the change in Gibbs free energy is dG = VdP. This means that if the pressure increases, the Gibbs free energy will also increase, assuming the volume is positive. This relationship is important for understanding how pressure changes influence the spontaneity of processes.

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