Ideal gas mixtures and thermodynamic relations

Why it matters

Intake charge, exhaust gas and atmospheric air are all mixtures of gases, so engine and emission calculations need mixture properties built from the properties of the components. Thermodynamic relations (Maxwell relations, the Tds equations and the Clapeyron equation) let us compute properties that cannot be measured directly, such as entropy and latent heat, from P, v and T data. They are how steam and refrigerant tables are actually built.

Key ideas

Describing a mixture.

• Mass fraction: mf_i = m_i/m. Mole fraction: y_i = n_i/n, where n = m/M is the amount in kmol.
• Apparent molar mass: M = m/n = Σ y_i M_i. Mixture gas constant: R = R_u/M, with R_u = 8.314 kJ/kmol·K.

Dalton's law of additive pressures. In an ideal-gas mixture each component behaves as if it alone occupied the whole volume at the mixture temperature. Its partial pressure is P_i = y_i P, and the total pressure is the sum of partial pressures.

Amagat's law of additive volumes. Each component, at the mixture's T and P, would occupy a partial volume V_i = y_i V. For ideal gases Dalton and Amagat give identical results; for real gases they are approximations, and a compressibility chart with Kay's pseudo-critical properties is used instead.

Mixture properties. Extensive properties add: U = Σ m_i u_i, H = Σ m_i h_i. Per unit mass, specific heats are mass-weighted: c_p = Σ mf_i c_p,i. Per kmol they are mole-weighted. Entropy of each component is evaluated at its own partial pressure, not at the mixture pressure.

Entropy of mixing. When different ideal gases at the same T and P mix, each expands from its own volume into the whole volume, so entropy rises even though no heat or work crosses the boundary. Mixing is irreversible. Mixing two samples of the same gas produces no entropy change.

Thermodynamic relations. For a simple compressible substance, combining the first and second laws gives the Gibbs (Tds) equations, which relate property changes and hold for any process because they contain only properties:

• T ds = du + P dv and T ds = dh − v dP.

From the four energy functions u, h, a = u − Ts (Helmholtz) and g = h − Ts (Gibbs) and the reciprocity of exact differentials come the Maxwell relations, which convert unmeasurable entropy derivatives into measurable P–v–T derivatives.

Clapeyron equation. During phase change P and T are linked, and the slope of the saturation curve gives the latent heat: (dP/dT)_sat = h_fg/(T v_fg). For liquid–vapour equilibrium at low pressure, treating the vapour as an ideal gas and neglecting v_f gives the Clausius–Clapeyron form, ln(P₂/P₁) = (h_fg/R)(1/T₁ − 1/T₂).

General specific-heat relation. c_p − c_v = T v β²/κ_T, where β is the volume expansivity and κ_T the isothermal compressibility. Since κ_T > 0, c_p ≥ c_v always; for an ideal gas this reduces to c_p − c_v = R; for liquids and solids c_p ≈ c_v.

Joule–Thomson coefficient. μ_JT = (∂T/∂P)_h describes the temperature change in throttling. μ_JT > 0 means the fluid cools on throttling (the basis of the refrigeration expansion valve); μ_JT = 0 for an ideal gas; above the inversion temperature μ_JT < 0 and the gas warms.

Formulas

y_i = n_i / n, n_i = m_i / M_i — y: mole fraction; n: kmol; m: kg; M: kg/kmol.

M = Σ y_i M_i, R = R_u / M — R_u = 8.314 kJ/kmol·K; R in kJ/kg·K.

P_i = y_i P (Dalton), V_i = y_i V (Amagat) — ideal-gas mixtures.

c_p = Σ mf_i c_p,i, c_v = Σ mf_i c_v,i — per unit mass, kJ/kg·K.

ΔS_mix = −R_u Σ n_i ln y_i — ideal gases initially at the same T and P, mixed at that T and total P; kJ/K.

T ds = du + P dv, T ds = dh − v dP — Gibbs equations.

(∂T/∂v)_s = −(∂P/∂s)_v, (∂T/∂P)_s = (∂v/∂s)_P, (∂P/∂T)_v = (∂s/∂v)_T, (∂v/∂T)_P = −(∂s/∂P)_T — Maxwell relations.

(dP/dT)_sat = h_fg / (T v_fg) — Clapeyron; T in K, h_fg in kJ/kg, v_fg in m³/kg, P in kPa.

ln(P₂/P₁) = (h_fg/R)(1/T₁ − 1/T₂) — Clausius–Clapeyron, low-pressure liquid–vapour.

c_p − c_v = T v β² / κ_T — β = (1/v)(∂v/∂T)_P in 1/K; κ_T = −(1/v)(∂v/∂P)_T in 1/kPa.

μ_JT = (∂T/∂P)_h — K/kPa.

Worked examples

Example 1 (standard). A rigid vessel contains 2 kg of N₂ (M = 28) and 1 kg of CO₂ (M = 44) at 300 K and 200 kPa. Find the mole fractions, partial pressures, mixture gas constant, vessel volume and the mixture c_p (c_p,N₂ = 1.039, c_p,CO₂ = 0.846 kJ/kg·K).

1. Moles: n_N₂ = 2/28 = 0.07143 kmol; n_CO₂ = 1/44 = 0.02273 kmol; n = 0.09416 kmol.
2. Mole fractions: y_N₂ = 0.07143/0.09416 = 0.759; y_CO₂ = 0.241.
3. Partial pressures: P_i = y_i P → P_N₂ = 151.7 kPa, P_CO₂ = 48.3 kPa.
4. M = m/n = 3/0.09416 = 31.86 kg/kmol; R = R_u/M = 8.314/31.86 = 0.2609 kJ/kg·K.
5. Volume: V = m R T / P = 3 × 0.2609 × 300/200 = 1.174 m³.
6. c_p = Σ mf_i c_p,i = (2 × 1.039 + 1 × 0.846)/3 = 0.975 kJ/kg·K.

If the two gases had first been held separately at 300 K and 200 kPa and then allowed to mix, the entropy generated would be ΔS = −8.314 × (0.07143 ln 0.759 + 0.02273 ln 0.241) = 0.433 kJ/K.

Example 2 (GATE level, Clapeyron equation). Saturation pressures of water are 84.61 kPa at 95 °C and 120.90 kPa at 105 °C. At 100 °C, v_f = 0.001043 m³/kg and v_g = 1.6719 m³/kg. Estimate h_fg at 100 °C and compare with the table value of 2256.5 kJ/kg.

1. Slope by central difference: (dP/dT)_sat ≈ (120.90 − 84.61)/(105 − 95) = 3.629 kPa/K.
2. v_fg = 1.6719 − 0.0010 = 1.6708 m³/kg; T = 373.15 K.
3. h_fg = T v_fg (dP/dT)_sat = 373.15 × 1.6708 × 3.629 = 2263 kJ/kg.
4. Error vs 2256.5 kJ/kg ≈ 0.3%, which comes only from the finite-difference slope. Units: K × m³/kg × kPa/K = kPa·m³/kg = kJ/kg.

Common mistakes

• Averaging specific heats by mole fraction when they are per kg (or by mass fraction when they are per kmol).
• Evaluating each component's entropy at the total pressure instead of its partial pressure, which loses the mixing entropy.
• Using R_u where the specific gas constant R is needed, or forgetting kmol vs kg.
• Using °C in the Clapeyron equation, or forgetting that v_fg, not v_g alone, appears (the difference is small only at low pressure).
• Believing c_p − c_v = R for liquids or real gases; the general relation is c_p − c_v = T v β²/κ_T.
• Assuming every gas cools on throttling; an ideal gas does not, and real gases above their inversion temperature warm up.

For GATE ME

This area produces conceptual MCQs (Maxwell relations, which relation follows from which function, sign of the Joule–Thomson coefficient, c_p − c_v for liquids and ideal gases) and short numericals on mixture molar mass, gas constant, partial pressure and entropy of mixing, and on the Clapeyron equation. Practise deriving one Maxwell relation from dg = v dP − s dT so you can rebuild the rest in the exam.

Quick check

1. A mixture has y_O₂ = 0.21 at a total pressure of 100 kPa. What is the partial pressure of O₂?
2. What is c_p − c_v for an ideal gas, and approximately for a liquid?
3. What is the Joule–Thomson coefficient of an ideal gas?
4. Which Maxwell relation comes from dg = v dP − s dT?

Answers: 1. 21 kPa. 2. R for an ideal gas; approximately zero for a liquid. 3. Zero. 4. (∂v/∂T)_P = −(∂s/∂P)_T.