Thermodynamic relations: Maxwell relations, cp and cv

Gibbs equations, the four Maxwell relations, the Clapeyron equation, general relations for cp − cv and γ, the speed of sound and the Joule–Thomson coefficient.

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Why it matters

Entropy, internal energy and enthalpy cannot be measured directly, yet property tables for steam, refrigerants and propellants are full of them. Thermodynamic relations convert them into quantities we can measure — p, v, T and specific heats. They also explain why cp > cv, how the boiling point shifts with pressure (Clapeyron) and when a gas cools on throttling (Joule–Thomson), which matters in cryogenic propellant systems.

Key ideas

Exact differentials. If z = z(x, y) is a property, then dz = M dx + N dy with M = (∂z/∂x)_y, N = (∂z/∂y)_x, and the cross-derivatives are equal: (∂M/∂y)_x = (∂N/∂x)_y. Every Maxwell relation is this test applied to a property.

The four fundamental relations (simple compressible system, per unit mass):

  • du = T ds − p dv
  • dh = T ds + v dp (h = u + pv)
  • da = −s dT − p dv (Helmholtz function a = u − Ts)
  • dg = −s dT + v dp (Gibbs function g = h − Ts)

They involve properties only, so they hold for any process between equilibrium states.

Maxwell relations (from the exactness of the four):

  • (∂T/∂v)_s = −(∂p/∂s)_v
  • (∂T/∂p)_s = (∂v/∂s)_p
  • (∂s/∂v)_T = (∂p/∂T)_v
  • (∂s/∂p)_T = −(∂v/∂T)_p

The last two are the useful ones: they turn the change of entropy with v or p at constant T into p–v–T derivatives obtainable from an equation of state.

Clapeyron equation. Applying (∂s/∂v)_T = (∂p/∂T)_v across a phase change (where p depends on T only) gives (dp/dT)_sat = h_fg/(T·v_fg). It lets you find h_fg from p–v–T data alone, and estimates how T_sat changes with pressure. With v_fg ≈ v_g = RT/p it reduces to the Clausius–Clapeyron form.

General relations for specific heats.

  • cv = T·(∂s/∂T)_v, cp = T·(∂s/∂T)_p
  • cp − cv = T·v·β²/κ, where β = (1/v)(∂v/∂T)_p is the volume expansivity and κ = −(1/v)(∂v/∂p)_T the isothermal compressibility.

Consequences: since κ > 0 always, cp ≥ cv; cp = cv when β = 0 (water at 4 °C); for nearly incompressible solids and liquids the difference is small; for an ideal gas β = 1/T and κ = 1/p, which gives cp − cv = R.

Physical reason cp > cv for a gas: at constant pressure part of the heat is spent as boundary work as the gas expands.

Ratio of specific heats. γ = cp/cv = κ/κ_s (isothermal over isentropic compressibility). It sets the speed of sound, a = √(γ·R·T) for an ideal gas, and the isentropic exponents used in nozzles and compressors.

Joule–Thomson coefficient. μ_JT = (∂T/∂p)_h. Positive μ_JT means the gas cools on throttling. For an ideal gas h = h(T), so μ_JT = 0. Real gases cool on throttling below their inversion temperature — the basis of gas liquefaction.

Formulas

du = T·ds − p·dv; dh = T·ds + v·dp (∂s/∂v)_T = (∂p/∂T)_v; (∂s/∂p)_T = −(∂v/∂T)_p (dp/dT)_sat = h_fg / (T·v_fg) cp − cv = T·v·β² / κ cp − cv = R; γ = cp/cv (ideal gas) a = √(γ·R·T) (ideal gas speed of sound) μ_JT = (∂T/∂p)_h

  • T: K; p: Pa (kPa in table work); v: m³/kg; s: J/(kg·K)
  • h_fg: latent heat, kJ/kg; v_fg = v_g − v_f, m³/kg
  • β: volume expansivity, 1/K; κ: isothermal compressibility, 1/Pa
  • cp, cv: J/(kg·K) on a mass basis (with R the specific gas constant), or J/(mol·K) on a molar basis (with R̄ = 8.314)

Worked examples

Example 1 (standard). Use the Clapeyron equation to find the slope of the saturation curve of water at 100 °C, and estimate p_sat at 105 °C. Data at 100 °C: h_fg = 2256.4 kJ/kg, v_fg = 1.6710 m³/kg, p_sat = 101.325 kPa.

  1. (dp/dT)_sat = h_fg/(T·v_fg) = 2256.4 / (373.15 × 1.6710) = 3.619 kPa/K
  2. Linear estimate: p_sat(105 °C) ≈ 101.325 + 5 × 3.619 = 119.4 kPa (the table gives about 120.9 kPa; the slope itself increases with T)

Answer: 3.62 kPa/K; p_sat(105 °C) ≈ 119 kPa

Example 2 (GATE level). For liquid water at 300 K: v = 0.001003 m³/kg, β = 2.76 × 10⁻⁴ K⁻¹, κ = 4.52 × 10⁻¹⁰ Pa⁻¹, cp = 4.18 kJ/(kg·K). Find cp − cv and cv. Then show that the same formula gives R for an ideal gas.

  1. cp − cv = T·v·β²/κ = 300 × 0.001003 × (2.76 × 10⁻⁴)² / (4.52 × 10⁻¹⁰)
  2. = 300 × 0.001003 × 7.618 × 10⁻⁸ / 4.52 × 10⁻¹⁰ = 50.7 J/(kg·K)
  3. cv = 4180 − 50.7 = 4129 J/(kg·K) — about 1.2% lower than cp, which is why liquids are given a single c.
  4. Ideal gas: v = RT/p, so β = 1/T and κ = 1/p. Then T·v·β²/κ = T·(RT/p)·(1/T²)·p = R ✓

Answer: cp − cv ≈ 50.7 J/(kg·K), cv ≈ 4.13 kJ/(kg·K)

Common mistakes

  • Mixing a molar cp (J/(mol·K)) with a specific gas constant (J/(kg·K)), or the reverse.
  • Writing the Maxwell relations with the wrong sign; derive them from the Gibbs equations rather than memorising.
  • Using cp − cv = R for liquids or real gases near saturation.
  • Using the Clapeyron equation with T in °C or v_fg in place of v_g carelessly at high pressures.
  • Assuming every gas cools when throttled; ideal gases do not change temperature, and real gases above the inversion temperature heat up.

For GATE AE

Expect one-mark questions identifying a correct Maxwell relation or Gibbs equation, and short numericals on cp − cv = R, γ and the speed of sound — directly useful in gas dynamics. Clapeyron-equation estimates and the Joule–Thomson coefficient of an ideal gas are also tested. Practise deriving each Maxwell relation from its Gibbs equation in under a minute.

Quick check

  1. Write dh in terms of T, s, v and p.
  2. For an ideal gas with molar cp = 29.1 J/(mol·K), find γ.
  3. What is the Joule–Thomson coefficient of an ideal gas?
  4. Why is cp ≥ cv for every substance?
  5. Find the speed of sound in air at 300 K (γ = 1.4, R = 287 J/(kg·K)).

Answers: 1. dh = T ds + v dp. 2. 1.40. 3. Zero. 4. cp − cv = T v β²/κ and κ is always positive. 5. About 347 m/s.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What are Maxwell relations in thermodynamics?Concept

    Maxwell relations are a set of equations in thermodynamics derived from the second law of thermodynamics. They relate different partial derivatives of thermodynamic potentials (such as internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy) to each other. These relations are useful for converting difficult-to-measure quantities into more easily measurable ones.

  2. 2.Explain the significance of specific heat capacities, cp and cv, in thermodynamics.Concept

    Specific heat capacities, cp and cv, are measures of the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. cp is the specific heat capacity at constant pressure, while cv is at constant volume. They are crucial for understanding how substances absorb and transfer heat under different conditions, and they play a key role in thermodynamic processes and calculations.

  3. 3.How are Maxwell relations derived from thermodynamic potentials?Concept

    Maxwell relations are derived using the properties of exact differentials and the symmetry of second derivatives. By starting with the fundamental thermodynamic equations for internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy, and applying the condition that mixed partial derivatives are equal, we can derive the four Maxwell relations.

  4. 4.Why is cp generally greater than cv for gases?Application

    cp is generally greater than cv for gases because, at constant pressure, the system does work on the surroundings as it expands. This requires additional energy compared to the constant volume process, where no work is done. Therefore, more heat is needed to raise the temperature of the gas at constant pressure than at constant volume.

  5. 5.What happens to the values of cp and cv as a gas approaches ideal behavior?Application

    In the ideal-gas limit u and h depend on temperature only, so cv = du/dT and cp = dh/dT become functions of T alone (independent of pressure), and their difference becomes cp − cv = R. On a molar basis that is the universal constant R̄ = 8.314 J/(mol·K); on a mass basis it is the specific gas constant R = R̄/M, e.g. 0.287 kJ/(kg·K) for air. The general relation cp − cv = T·v·β²/κ reduces to R because β = 1/T and κ = 1/p for an ideal gas.

  6. 6.How can Maxwell relations be used to determine changes in entropy?Application

    Maxwell relations can be used to express changes in entropy in terms of measurable quantities like temperature, volume, and pressure. For example, one of the Maxwell relations can be rearranged to express the change in entropy as a function of changes in volume and temperature, which can be measured experimentally.

  7. 7.Calculate the change in internal energy for 2 moles of an ideal gas when its temperature increases by 10 K at constant volume. Assume cv = 20.8 J/mol·K.Numerical

    The change in internal energy (ΔU) at constant volume is given by ΔU = n·cv·ΔT. Here, n = 2 moles, cv = 20.8 J/mol·K, and ΔT = 10 K. Therefore, ΔU = 2 moles × 20.8 J/mol·K × 10 K = 416 J.

  8. 8.If the specific heat capacity at constant pressure (cp) for a gas is 29 J/mol·K, and the gas constant (R) is 8.314 J/mol·K, what is the specific heat capacity at constant volume (cv)?Numerical

    For an ideal gas, the relation cp - cv = R holds. Given cp = 29 J/mol·K and R = 8.314 J/mol·K, we can find cv by rearranging the equation: cv = cp - R = 29 J/mol·K - 8.314 J/mol·K = 20.686 J/mol·K.

  9. 9.Explain how the ratio of cp to cv (γ) affects the speed of sound in a gas.Application

    Sound waves are nearly isentropic, so the speed of sound is a = √((∂p/∂ρ)_s), which for an ideal gas is a = √(γ·R·T) with R the specific gas constant (equivalently √(γ·R̄·T/M)). A larger γ means the gas is stiffer under isentropic compression, so the pressure rise per unit density change is larger and sound travels faster. For air at 300 K (γ = 1.4, R = 287 J/(kg·K)) a ≈ 347 m/s; a monatomic gas with γ ≈ 1.67 and the same R would be about 9% faster.

  10. 10.What is the physical interpretation of the difference between cp and cv for a real gas?Application

    The general relation is cp − cv = T·v·β²/κ, where β is the volume expansivity and κ the isothermal compressibility. Physically, heating at constant pressure makes the substance expand, so extra energy goes into boundary work against the surroundings and, for a real gas, into work against intermolecular attraction as the molecules move apart. Because β and κ depend on the state, cp − cv for a real gas varies with temperature and pressure; it reduces to R only in the ideal-gas limit and is small for liquids and solids.

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