Ideal gas mixtures

Why it matters

Air itself is a mixture, and so are combustion products in a gas-turbine combustor, rocket exhaust, cabin air with water vapour, and the fuel–air charge in a piston engine. To use the ideal-gas relations on them you need the mixture's molar mass, gas constant and specific heats, and the partial pressures of its components. Mixing is also a classic source of irreversibility.

Key ideas

Describing composition.

• Mass fraction mf_i = m_i/m, Σ mf_i = 1 (gravimetric analysis).
• Mole fraction y_i = N_i/N, Σ y_i = 1 (molar or volumetric analysis).
• They are linked through the molar masses: mf_i = y_i·M_i/M.

Apparent molar mass and gas constant. M = m/N = Σ y_i·M_i, and R = R̄/M. For dry air (about 21% O₂, 79% N₂ by mole, ignoring argon) M ≈ 28.97 kg/kmol and R = 0.287 kJ/(kg·K).

Dalton's law of additive pressures. Each component, if it alone occupied the whole mixture volume at the mixture temperature, would exert its partial pressure p_i; the mixture pressure is their sum: p = Σ p_i.

Amagat's law of additive volumes. Each component, if it alone were at the mixture temperature and pressure, would occupy its partial (component) volume V_i; the mixture volume is their sum: V = Σ V_i.

For ideal gases both laws hold exactly, and y_i = p_i/p = V_i/V = N_i/N. That is why a volumetric analysis (such as an Orsat analysis of flue gas) directly gives mole fractions.

Mixture properties. Each component behaves as if alone at T and its partial pressure (Gibbs–Dalton law). So extensive properties add:

• U = Σ m_i·u_i, H = Σ m_i·h_i, S = Σ m_i·s_i(T, p_i)
• Specific heats are mass-weighted on a mass basis, mole-weighted on a molar basis: cp = Σ mf_i·cp,i, c̄p = Σ y_i·c̄p,i.

Entropy of a component must be evaluated at its partial pressure, not the mixture pressure. When gases mix, each one expands from its own pressure to its (lower) partial pressure, so mixing generates entropy even when nothing else happens. For ideal gases initially at the same T and p, the entropy of mixing is ΔS = −R̄·Σ N_i·ln y_i (always positive). Mixing two samples of the same gas at the same state generates no entropy.

Limitations. Real gas mixtures at high pressure need mixing rules or compressibility charts. Condensable components (water vapour in air) are treated in psychrometrics.

Formulas

y_i = N_i/N; mf_i = m_i/m; mf_i = y_i·M_i/M M = Σ y_i·M_i = m/N; R = R̄/M p = Σ p_i; p_i = y_i·p (Dalton) V = Σ V_i; V_i = y_i·V (Amagat) cp = Σ mf_i·cp,i; cv = Σ mf_i·cv,i (mass basis) Δs_i = cp,i·ln(T₂/T₁) − R_i·ln(p_i,2/p_i,1) (each component, at its partial pressure) ΔS_mix = −R̄·Σ N_i·ln y_i (same initial T and p)

• N: amount, kmol; m: mass, kg; M: molar mass, kg/kmol
• R̄ = 8.314 kJ/(kmol·K); R: kJ/(kg·K)
• p, p_i: kPa; V, V_i: m³; T: K
• Molar specific heats c̄ in kJ/(kmol·K)

Worked examples

Example 1 (standard). A mixture contains 3 kg O₂ (M = 32), 5 kg N₂ (M = 28) and 12 kg CO₂ (M = 44) at 300 kPa. Find the mole fractions, the apparent molar mass, the gas constant and the partial pressures.

1. Moles: N_O₂ = 3/32 = 0.0938, N_N₂ = 5/28 = 0.1786, N_CO₂ = 12/44 = 0.2727; N = 0.5450 kmol
2. y_i = N_i/N: y_O₂ = 0.172, y_N₂ = 0.328, y_CO₂ = 0.500
3. M = m/N = 20/0.5450 = 36.69 kg/kmol; R = 8.314/36.69 = 0.2266 kJ/(kg·K)
4. p_i = y_i·p: O₂ 51.6 kPa, N₂ 98.3 kPa, CO₂ 150.1 kPa (sum 300 kPa ✓)

Answer: y = 0.172 / 0.328 / 0.500; M = 36.7 kg/kmol; R = 0.227 kJ/(kg·K); partial pressures 51.6 / 98.3 / 150.1 kPa

Example 2 (GATE level). An insulated rigid tank is divided by a partition. Side A has 2 kmol N₂ at 350 K, 200 kPa; side B has 1 kmol O₂ at 290 K, 100 kPa. The partition is removed. Take c̄v = 20.8 (N₂) and 21.1 (O₂) kJ/(kmol·K), c̄p = c̄v + 8.314. Find the final temperature, final pressure and entropy generated.

1. Volumes: V_A = 2 × 8.314 × 350/200 = 29.10 m³; V_B = 1 × 8.314 × 290/100 = 24.11 m³; V = 53.21 m³
2. Energy (Q = W = 0, so ΔU = 0): 2 × 20.8 × (T − 350) + 1 × 21.1 × (T − 290) = 0 → T = 329.8 K
3. p = N·R̄·T/V = 3 × 8.314 × 329.8/53.21 = 154.6 kPa
4. Final partial pressures: p_N₂ = (2/3) × 154.6 = 103.1 kPa; p_O₂ = (1/3) × 154.6 = 51.5 kPa
5. N₂: ΔS = 2 × [29.114 × ln(329.8/350) − 8.314 × ln(103.1/200)] = 2 × [−1.730 + 5.512] = 7.564 kJ/K
6. O₂: ΔS = 1 × [29.414 × ln(329.8/290) − 8.314 × ln(51.5/100)] = 3.785 + 5.511 = 9.295 kJ/K
7. Isolated, so S_gen = ΣΔS = 16.86 kJ/K

Answer: T = 329.8 K, p = 154.6 kPa, S_gen = 16.9 kJ/K

Common mistakes

• Averaging molar masses by mass fraction instead of mole fraction (M = Σ y_i·M_i, not Σ mf_i·M_i).
• Evaluating component entropy at the mixture pressure instead of the partial pressure — this hides the entropy of mixing.
• Confusing partial volume (at mixture p and T) with the container volume.
• Mixing mass-basis and molar-basis specific heats.
• Expecting entropy generation when two identical gases at the same state are mixed (there is none).

For GATE AE

Expect calculations of mole and mass fractions, apparent molar mass and gas constant, partial pressures, and cp or γ of a mixture such as combustion products, plus adiabatic mixing in a tank with the final temperature, pressure and entropy generated. One-mark questions test Dalton's and Amagat's laws and the fact that y_i = p_i/p = V_i/V. Practise building a small table of N, y, M and c̄p for each component.

Quick check

1. Partial pressures of 30 kPa and 90 kPa: what is the total pressure?
2. A component has y = 0.25 in a mixture at 120 kPa. Find its partial pressure.
3. 1 mol O₂ and 4 mol N₂ are mixed. Find y_N₂.
4. Is M of a mixture weighted by mass fractions or mole fractions?
5. In a 5 m³ He–Ar mixture at 150 kPa, the partial volume of helium is 4 m³. Find p_He.

Answers: 1. 120 kPa. 2. 30 kPa. 3. 0.8. 4. Mole fractions. 5. 120 kPa.