Ideal gas mixtures and thermodynamic relations

Why it matters

Air, exhaust gas, fuel–air charge and refrigerant blends are all mixtures, and you need their gas constant, specific heats and partial pressures before any engine, combustion or psychrometric calculation. The general thermodynamic relations (Maxwell relations, Tds equations, Clapeyron equation) are what let property tables be built from measurable p, v, T data and explain results such as why throttling cools a real gas.

Key ideas

Ideal-gas mixture. A mixture of non-reacting ideal gases is itself an ideal gas. Each component behaves as if it alone occupied the whole volume at the mixture temperature, because the molecules do not interact.

Describing composition.

• Mass fraction x_i = m_i/m (used for per-kg properties: R, c_p, c_v, u, h, s).
• Mole fraction y_i = n_i/n (used for per-kmol properties and for partial pressures).
• Conversion: n_i = m_i/M_i, and the apparent molar mass M = m/n = Σ y_i M_i.

Dalton's law of additive pressures. The mixture pressure equals the sum of the partial pressures, each component's partial pressure being the pressure it would exert alone in the full volume V at the mixture temperature T. For ideal gases p_i = y_i p.

Amagat's law of additive volumes. The mixture volume equals the sum of the component volumes, each being the volume the component would occupy alone at the mixture p and T. For ideal gases V_i = y_i V, so volume fraction = mole fraction = pressure fraction. (Gas analysers such as the Orsat apparatus report volume fractions — that is, mole fractions.)

Properties of the mixture. Extensive properties add: U = Σ m_i u_i, H = Σ m_i h_i. So per-kg properties are mass-weighted (c_p = Σ x_i c_p,i) and per-kmol properties are mole-weighted. Entropy of each component must be evaluated at its own partial pressure, which is why mixing two different gases generates entropy even at equal T and p.

General thermodynamic relations. For a simple compressible substance, two independent properties fix the state, so the properties are linked by exact differentials:

• Gibbs (Tds) equations: T ds = du + p dv and T ds = dh − v dp. They connect property changes and hold for any process, reversible or not, because they involve only properties.
• Helmholtz function a = u − Ts and Gibbs function g = h − Ts. During phase change at constant T and p, g_f = g_g.
• Maxwell relations follow from the equality of mixed second derivatives; they let you replace an unmeasurable entropy derivative with a measurable p–v–T one.
• Clapeyron equation: the slope of the saturation curve is fixed by the latent heat and the change in specific volume. It lets you get h_fg from p–T–v data alone.
• c_p − c_v for any substance depends on the volume expansivity β and isothermal compressibility κ; it is R for an ideal gas and almost zero for liquids and solids, which is why one "c" is used for them.
• Joule–Thomson coefficient μ_JT = (∂T/∂p)_h. Throttling cools the gas when μ_JT > 0 (below the inversion temperature). For an ideal gas μ_JT = 0.

Formulas

m = Σ m_i, n = Σ n_i, n_i = m_i / M_i x_i = m_i / m, y_i = n_i / n — mass and mole fractions (dimensionless). M = m / n = Σ y_i M_i — apparent molar mass (kg/kmol). R = R̄ / M = Σ x_i R_i — mixture gas constant (kJ/kg·K), R̄ = 8.314 kJ/kmol·K. p = Σ p_i, p_i = y_i p — Dalton's law (kPa). V = Σ V_i, V_i = y_i V — Amagat's law (m³). pV = m R T = n R̄ T — mixture equation of state (p in kPa, V in m³, T in K). c_p = Σ x_i c_p,i, c_v = Σ x_i c_v,i — mixture specific heats (kJ/kg·K). ΔS_mix = −R̄ Σ n_i ln y_i — entropy generated by adiabatic mixing of different ideal gases at the same T and 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 equation. h_fg in kJ/kg, T in K, v_fg in m³/kg, dp/dT in kPa/K. ln(p₂/p₁) = (h_fg / R)(1/T₁ − 1/T₂) — Clausius–Clapeyron, for low pressures where v_g ≫ v_f and the vapour is ideal. c_p − c_v = T v β² / κ, with β = (1/v)(∂v/∂T)_p, κ = −(1/v)(∂v/∂p)_T. For an ideal gas c_p − c_v = R. μ_JT = (∂T/∂p)_h

Worked examples

Example 1 (standard, mixture properties). A rigid tank holds 3 kg of N₂ (M = 28) and 2 kg of CO₂ (M = 44) at 300 K and 200 kPa. Find the mole fractions, apparent molar mass, gas constant, tank volume and partial pressures. Also find c_p of the mixture, taking c_p,N₂ = 1.040 and c_p,CO₂ = 0.846 kJ/kg·K.

1. Moles: n_N₂ = 3/28 = 0.10714 kmol; n_CO₂ = 2/44 = 0.04545 kmol; n = 0.15260 kmol.
2. y_i = n_i/n: y_N₂ = 0.7021, y_CO₂ = 0.2979.
3. M = m/n = 5/0.15260 = 32.77 kg/kmol; R = R̄/M = 8.314/32.77 = 0.2537 kJ/kg·K.
4. V = mRT/p = 5 × 0.2537 × 300/200 = 1.903 m³.
5. p_i = y_i p: p_N₂ = 0.7021 × 200 = 140.4 kPa; p_CO₂ = 59.6 kPa (sum 200 kPa ✓).
6. Mass fractions 0.6 and 0.4: c_p = Σ x_i c_p,i = 0.6 × 1.040 + 0.4 × 0.846 = 0.962 kJ/kg·K.

M ≈ 32.8 kg/kmol, R ≈ 0.254 kJ/kg·K, V ≈ 1.90 m³, p_N₂ ≈ 140.4 kPa, p_CO₂ ≈ 59.6 kPa, c_p ≈ 0.962 kJ/kg·K

Example 2 (GATE level, entropy of mixing). An insulated vessel has two compartments, one with 1 kmol of O₂ and the other with 1 kmol of N₂, both at 300 K and 100 kPa. The partition is removed. Find the entropy generated.

1. Same T, no work, no heat, ideal gases: final T = 300 K and final p = 100 kPa.
2. Each gas expands from 100 kPa to its partial pressure 50 kPa, so y_O₂ = y_N₂ = 0.5.
3. ΔS_mix = −R̄ Σ n_i ln y_i = −8.314 × (1 × ln 0.5 + 1 × ln 0.5) = 8.314 × 2 × 0.6931 = 11.53 kJ/K.
4. The vessel is insulated, so S_gen = ΔS = 11.53 kJ/K. (If both compartments held the same gas, ΔS would be zero — the Gibbs paradox.)

S_gen ≈ 11.5 kJ/K

Example 3 (Clapeyron equation). Saturation pressures of water are 84.61 kPa at 95 °C and 120.90 kPa at 105 °C; at 100 °C, v_fg = 1.6707 m³/kg. Estimate h_fg at 100 °C.

1. Central difference: dp/dT ≈ (120.90 − 84.61)/10 = 3.629 kPa/K.
2. h_fg = T v_fg (dp/dT) = 373.15 × 1.6707 × 3.629 = 2262 kJ/kg.
3. The steam-table value is 2256.4 kJ/kg, so the estimate is within 0.3 %.

h_fg ≈ 2262 kJ/kg

Common mistakes

• Using mole fractions to average per-kg properties (c_p, R) or mass fractions to get partial pressures.
• Taking the partial volume as the whole tank volume. Under Amagat's law each component's volume is at mixture p and T; under Dalton's law each component's pressure is at the full volume.
• Evaluating component entropies at the mixture pressure instead of at their partial pressures, which misses the entropy of mixing.
• Using °C or a v_fg in L/kg in the Clapeyron equation.
• Thinking throttling always cools a gas. It cools only when μ_JT > 0; an ideal gas shows no temperature change.
• Applying c_p − c_v = R to liquids.

For GATE ME

Mixtures appear as computations of R, M, partial pressures and specific heats, often as the first step of a combustion or psychrometry question, and as entropy-of-mixing problems. Thermodynamic relations appear as conceptual one-mark questions (Maxwell relations, Clapeyron slope, Joule–Thomson coefficient, which property is constant during phase change) and occasionally as Clapeyron numericals. Practise converting between mass and mole fractions quickly and recognising which Maxwell relation a given partial derivative needs.

Quick check

1. Air is 21 % O₂ and 79 % N₂ by volume. What is the partial pressure of O₂ at 101.3 kPa?
2. Are per-kg specific heats of a mixture averaged by mass or by mole fraction?
3. Which property is equal for saturated liquid and saturated vapour at the same T?
4. What is μ_JT for an ideal gas?
5. Which Maxwell relation equals (∂s/∂v)_T?

Answers: 1. 0.21 × 101.3 ≈ 21.3 kPa. 2. By mass fraction. 3. The Gibbs function g = h − Ts. 4. Zero. 5. (∂p/∂T)_v.