Ideal gas relations and polytropic processes

Why it matters

Air and combustion gases in compressors, turbines, nozzles and piston engines behave very nearly as ideal gases, so a handful of relations give temperatures, pressures and work for every stage. Real compression and expansion lie between isothermal and adiabatic and are modelled as polytropic processes pVⁿ = C. These relations are the building blocks of every gas-power cycle.

Key ideas

Ideal gas model. p·v = R·T. Molecules occupy negligible volume and exert no forces on each other except during collisions. It is accurate when the pressure is low and the temperature high compared with the critical values (air at ordinary conditions is within about 1%). The compressibility factor Z = p·v/(R·T) measures departure from ideal behaviour; Z = 1 for an ideal gas.

Energy depends on T only. For an ideal gas u = u(T) and h = h(T), so du = cv·dT and dh = cp·dT for every process. With constant specific heats (a "perfect" or calorically perfect gas) these integrate directly.

Specific-heat relations. cp − cv = R, γ = cp/cv, hence cv = R/(γ − 1) and cp = γ·R/(γ − 1). For air: R = 0.287, cp = 1.005, cv = 0.718 kJ/(kg·K), γ = 1.4. Monatomic gases have γ ≈ 1.67, diatomic ≈ 1.4.

The polytropic family pVⁿ = C.

• n = 0 → isobaric (p constant)
• n = 1 → isothermal (T constant, ideal gas)
• n = γ → reversible adiabatic (isentropic)
• n → ∞ → isochoric (V constant)

On a p–v diagram, starting from the same state, the expansion curves get steeper as n increases: isobaric (flat) → isothermal → isentropic → isochoric (vertical).

Linking p, v, T in a polytropic process. Combining pVⁿ = C with pV = mRT gives the temperature–volume and temperature–pressure relations below. For isentropic processes simply put n = γ.

Heat in a polytropic process. The first law and the work expression give q = w·(γ − n)/(γ − 1). Equivalently, the process behaves as if it had a polytropic specific heat cn = cv·(n − γ)/(n − 1), so q = cn·(T₂ − T₁). For 1 < n < γ, cn is negative: during compression the gas heats up yet rejects heat, and during expansion it cools yet receives heat. Real compressors with jacket cooling (n ≈ 1.2–1.35) behave like this.

Work: closed versus steady flow. For a closed system the boundary work is w = ∫p dv. For a reversible steady-flow device (compressor, turbine) with negligible KE and PE changes the shaft work is w = −∫v dp, which for a polytropic process is n times the closed-system work. This is why a compressor's specific work is larger than a piston's for the same pressure ratio.

Formulas

p·v = R·T; p·V = m·R·T cp − cv = R; γ = cp / cv Δu = cv·(T₂ − T₁); Δh = cp·(T₂ − T₁) (any process) p₁·V₁ⁿ = p₂·V₂ⁿ T₂/T₁ = (V₁/V₂)ⁿ⁻¹ = (p₂/p₁)^((n−1)/n) w = R·(T₁ − T₂)/(n − 1) = (p₁v₁ − p₂v₂)/(n − 1) (closed system, n ≠ 1) w = R·T·ln(v₂/v₁) = R·T·ln(p₁/p₂) (isothermal) q = w·(γ − n)/(γ − 1); cn = cv·(n − γ)/(n − 1) w_flow = n·R·(T₁ − T₂)/(n − 1) (reversible steady flow, ΔKE = ΔPE = 0)

• p: absolute pressure, kPa; v: specific volume, m³/kg; T: absolute temperature, K
• R: specific gas constant, kJ/(kg·K); cp, cv, cn: kJ/(kg·K)
• n: polytropic index; γ: ratio of specific heats
• w, q: work done by and heat added to the gas per kg, kJ/kg

Worked examples

Example 1 (standard). Air at 100 kPa and 300 K is compressed isentropically in a closed cylinder with a volume ratio V₁/V₂ = 16 (as in a diesel engine). Find T₂, p₂ and the work per kg. γ = 1.4, cv = 0.718 kJ/(kg·K).

1. T₂ = T₁·(V₁/V₂)^(γ−1) = 300 × 16⁰·⁴ = 909.4 K
2. p₂ = p₁·(V₁/V₂)^γ = 100 × 16¹·⁴ = 4850 kPa
3. Adiabatic, so w = −Δu = −cv·(T₂ − T₁) = −0.718 × 609.4 = −437.6 kJ/kg

Answer: T₂ = 909 K, p₂ = 4.85 MPa, w = −437.6 kJ/kg (work input)

Example 2 (GATE level). Air at 100 kPa and 300 K is compressed in a closed cylinder following pV¹·²⁵ = C to 800 kPa. Find T₂, the work and the heat per kg. R = 0.287, cv = 0.718 kJ/(kg·K), γ = 1.4.

1. T₂ = T₁·(p₂/p₁)^((n−1)/n) = 300 × 8⁰·² = 454.7 K
2. w = R·(T₁ − T₂)/(n − 1) = 0.287 × (300 − 454.7)/0.25 = −177.6 kJ/kg
3. Δu = cv·(T₂ − T₁) = 0.718 × 154.7 = 111.1 kJ/kg
4. q = Δu + w = 111.1 − 177.6 = −66.5 kJ/kg
5. Check: q = w·(γ − n)/(γ − 1) = −177.6 × 0.15/0.4 = −66.6 kJ/kg ✓

Answer: T₂ = 454.7 K, w = −177.6 kJ/kg, q = −66.5 kJ/kg (heat rejected while the air heats up)

Common mistakes

• Using the exponent n in the T–p relation instead of (n − 1)/n.
• Using °C or gauge pressure in any of these ratios.
• Assuming "adiabatic" means isentropic for a real (irreversible) machine — isentropic needs reversible and adiabatic.
• Using w = (p₁v₁ − p₂v₂)/(n − 1) with n = 1; use the logarithmic form.
• Using closed-system work for a steady-flow compressor (it is n times smaller than the flow work).
• Assuming heat must be added whenever the gas temperature rises; check the sign of cn.

For GATE AE

Expect numericals on isentropic temperature and pressure ratios (compressors, intakes, nozzles), work and heat in polytropic processes, and comparisons of processes on p–v diagrams by slope. One-mark questions test which n corresponds to which process, γ for monatomic versus diatomic gases, and cp − cv = R. Practise moving fluently between the T–v and T–p forms of the polytropic relations.

Quick check

1. What process is pVⁿ = C with n = 0?
2. Air (γ = 1.4) is compressed isentropically through a pressure ratio of 10 from 300 K. Find T₂.
3. Find cv for a gas with R = 0.287 kJ/(kg·K) and γ = 1.4.
4. For 1 < n < γ, is heat added or rejected during polytropic compression?
5. If pressure doubles at constant temperature, what happens to the volume of an ideal gas?

Answers: 1. Isobaric. 2. About 579 K. 3. 0.7175 kJ/(kg·K). 4. Rejected. 5. It halves.