Systems, properties, work and heat

Why it matters

Thermodynamics is the accounting system for energy in every engine, compressor, furnace, heat-treatment cell and refrigeration unit. Before you can apply the laws, you must define exactly what you are analysing (the system), describe its state with properties, and recognise the two ways energy crosses a boundary: work and heat. Most wrong answers in thermodynamics come from a badly drawn boundary or a sign error, not from hard mathematics.

Key ideas

System, surroundings and boundary. A system is the quantity of matter or region of space chosen for study; everything else is the surroundings; the boundary separates them and may be real or imaginary, fixed or moving.

• Closed system (control mass): no mass crosses the boundary; energy can (gas in a piston–cylinder).
• Open system (control volume): mass and energy cross the boundary (turbine, compressor, nozzle, heat exchanger).
• Isolated system: neither mass nor energy crosses (an idealisation; the universe = system + surroundings is treated as isolated).

Properties and state. A property is any measurable characteristic that depends only on the state, not on how the state was reached (p, V, T, U, H, S). The change in a property between two states is path-independent; mathematically its differential is exact (dV, dU). Intensive properties are independent of the amount of matter (p, T, ρ, specific volume v); extensive properties are proportional to it (V, m, U). An extensive property divided by mass is a specific property (v = V/m, u = U/m). A simple compressible substance needs two independent intensive properties to fix its state (state postulate).

Equilibrium, processes and cycles. A system is in thermodynamic equilibrium when it is in thermal, mechanical, phase and chemical equilibrium. A quasi-static (quasi-equilibrium) process proceeds so slowly that the system passes through a continuous series of equilibrium states; only such a process can be drawn as a line on a p–V diagram. Common processes: isobaric (constant p), isochoric (constant V), isothermal (constant T), adiabatic (no heat), polytropic (pVⁿ = constant). A cycle returns the system to its initial state, so every property change over a cycle is zero.

Zeroth law and temperature. If bodies A and B are each in thermal equilibrium with C, they are in thermal equilibrium with each other. This is the basis of thermometry. Absolute temperature T (K) = t (°C) + 273.15. A temperature difference has the same value in K and °C.

Work. Work is energy transfer across a boundary that could, in principle, be used to raise a weight. Boundary (displacement) work of a quasi-static process is ∫p dV, the area under the process curve on a p–V diagram. Different paths between the same end states give different areas, so work is a path function (inexact differential δW). Other forms: shaft work, electrical work, spring work, stretching work. Free expansion into a vacuum involves no work, because there is no resisting pressure, even though the volume changes.

Heat. Heat is energy transfer across a boundary caused only by a temperature difference. It is also a path function. An adiabatic process has Q = 0 — not the same as constant temperature.

Sign convention used here (the usual one in Indian textbooks and GATE): heat added to the system is positive; work done by the system is positive. Some books use work done on the system as positive; always state your convention.

Specific heats and ideal gas. Specific heat c is the energy needed to raise unit mass by one kelvin. For an ideal gas pV = mRT, with R = R̄/M (R̄ = 8.314 kJ/kmol·K; for air R ≈ 0.287 kJ/kg·K), c_p − c_v = R and γ = c_p/c_v (≈ 1.4 for air).

Formulas

W = ∫ p·dV

• Boundary work (J) of a quasi-static process; p in Pa, V in m³.

W = p·(V₂ − V₁)

• Constant pressure.

W = 0

• Constant volume.

W = p₁·V₁·ln(V₂/V₁) = m·R·T·ln(p₁/p₂)

• Isothermal ideal gas.

W = (p₁·V₁ − p₂·V₂)/(n − 1)

• Polytropic pVⁿ = C, n ≠ 1 (adiabatic reversible: n = γ).

p₂/p₁ = (V₁/V₂)ⁿ, T₂/T₁ = (V₁/V₂)^(n−1) = (p₂/p₁)^((n−1)/n)

• Polytropic relations, ideal gas.

p·V = m·R·T

• Ideal gas; R specific gas constant (J/kg·K); T in K.

Q = m·c·ΔT

• Sensible heat (J); c specific heat (J/kg·K).

W_spring = ½·k·(x₂² − x₁²)

• Spring work (J); k stiffness (N/m).

Worked examples

Example 1 (standard): polytropic compression. Given: air at 100 kPa occupies 0.1 m³ and is compressed quasi-statically to 0.02 m³ following pV^1.3 = constant. Find the final pressure and the work.

1. p₂ = p₁·(V₁/V₂)^1.3 = 100 × 5^1.3 = 810.3 kPa
2. W = (p₁V₁ − p₂V₂)/(n − 1) = (100 × 0.1 − 810.3 × 0.02)/(1.3 − 1)
3. W = (10 − 16.207)/0.3 = −20.69 kJ (kPa × m³ = kJ) Answer: p₂ ≈ 810 kPa; W ≈ −20.7 kJ, i.e. about 20.7 kJ of work done on the air (For comparison, an isothermal compression over the same volume ratio needs only 100 × 0.1 × ln 5 = 16.1 kJ.)

Example 2 (GATE level): gas against a linear spring. Given: a gas in a piston–cylinder restrained by a linear spring expands so that its pressure rises linearly with volume from 150 kPa at 0.05 m³ to 300 kPa at 0.10 m³. Find the work done by the gas and the share stored in the spring if the atmosphere outside the piston is at 100 kPa (piston area constant, piston weight negligible).

1. Linear p–V path, so the area is a trapezium: W = ½·(p₁ + p₂)·(V₂ − V₁) = ½ × (150 + 300) × 0.05 = 11.25 kJ.
2. Work done pushing back the atmosphere: p_atm·ΔV = 100 × 0.05 = 5 kJ.
3. The rest goes into the spring: 11.25 − 5 = 6.25 kJ. Answer: W_gas = 11.25 kJ; spring energy stored = 6.25 kJ

Example 3 (quick): sensible heat. Heating 5 kg of water (c = 4.18 kJ/kg·K) from 20 °C to 80 °C: Q = 5 × 4.18 × 60 = 1254 kJ.

Common mistakes

• Treating heat or work as properties ("the heat in the gas"); they exist only while crossing a boundary.
• Using °C instead of K in pV = mRT or in temperature ratios.
• Mixing kPa with m³ and expecting J (kPa·m³ = kJ).
• Mixing sign conventions, especially "work done on the system".
• Assuming adiabatic means isothermal.
• Calculating ∫p dV for a free expansion (it is zero) or for a non-quasi-static process.

For GATE PI

Expect short numericals on boundary work for isobaric, isothermal, polytropic and spring-loaded processes, identification of intensive versus extensive properties and of path versus point functions, ideal-gas state calculations, and the area-under-curve interpretation of work on p–V diagrams. Practise the polytropic work formula with the correct sign and the unit bookkeeping kPa × m³ = kJ.

Quick check

1. Is specific volume intensive or extensive?
2. What is the boundary work in a rigid tank?
3. Air expands isothermally at 300 K, 1 kg, from 1 bar to 0.5 bar. Work done? (R = 0.287 kJ/kg·K)
4. Why is work called a path function?

Answers: 1. Intensive 2. Zero 3. 0.287 × 300 × ln 2 ≈ 59.7 kJ (by the gas) 4. Its value depends on the process path between two states (the area under the p–V curve), not just on the end states