Systems, properties, work and heat

Why it matters

Every thermodynamic calculation, from an engine cylinder to a car's air-conditioning loop, starts by drawing a boundary and deciding what crosses it. Getting the system, the sign of work and heat, and the type of work right is where most wrong answers in exams and in design begin. Displacement work on a P–V diagram is also the starting point for indicated work, mean effective pressure and cycle analysis of IC engines.

Key ideas

System, boundary and surroundings. A system is the quantity of matter or region of space we choose to study; everything else is the surroundings, separated by a boundary that may be real or imaginary, fixed or moving.

• Closed system (control mass): no mass crosses the boundary; energy can (gas trapped in a cylinder with valves closed).
• Open system (control volume): mass and energy cross the boundary (radiator, compressor, nozzle, an engine cylinder during intake).
• Isolated system: neither mass nor energy crosses (system plus its surroundings taken together).

Properties. 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).

• Intensive properties do not depend on mass (P, T, density). Extensive properties do (V, U, H, S). An extensive property divided by mass is a specific property (v = V/m), which behaves as intensive.
• A test: ∮ d(property) = 0 over any cycle. Its differential is exact.

State, process, cycle. The state is fixed by the values of the properties. For a simple compressible pure substance, two independent intensive properties fix the state (the state postulate). A process is a change of state; the path is the series of states passed through. A cycle returns the system to its initial state.

Quasi-static process. A process carried out slowly enough 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, and only for it does displacement work equal ∫P dV. Fast or unresisted processes (free expansion into vacuum, a burst diaphragm) are not quasi-static.

Zeroth law and temperature. If two bodies are each in thermal equilibrium with a third, they are in thermal equilibrium with each other. This makes the thermometer possible. Thermodynamic temperature in kelvin: T(K) = t(°C) + 273.15. A temperature difference of 1 K equals 1 °C.

Work. Work is energy crossing the boundary that could, in principle, be used solely to raise a weight. It is a boundary phenomenon and a path function: the work between two states depends on the path, so we write δW (inexact differential), never dW, and never speak of the "work of a system".

• Displacement (boundary) work: W = ∫P dV, the area under the process curve on the P–V diagram.
• Other modes: shaft work, electrical work, spring work, paddle-wheel (stirring) work. Stirring work can only be done on a system; it is never quasi-static.
• Free expansion: a gas expanding into a vacuum does zero work, because no force resists the boundary motion, even though its volume increases.

Heat. Heat is energy crossing the boundary because of a temperature difference. It is also a path function (δQ). An adiabatic process has Q = 0; it is not the same as an isothermal process.

Sign convention used in this course and in GATE: heat added to the system is positive; work done by the system is positive. Then the first law for a closed system reads Q − W = ΔU. (Some texts and the IUPAC convention take work done on the system as positive; always check which one a question uses.)

Connection to later topics. Path-dependent W and Q combine into the path-independent ΔU (first law); δQ/T for a reversible path becomes entropy (second law).

Formulas

W = ∫ P dV (from V₁ to V₂)

• W: displacement work, J; P: system pressure, Pa; V: volume, m³. Quasi-static processes only.

W = P (V₂ − V₁) — constant-pressure (isobaric) process.

W = 0 — constant-volume (isochoric) process, and free expansion.

W = P₁V₁ ln(V₂/V₁) = m R T ln(P₁/P₂) — isothermal process of an ideal gas (PV = constant).

• m: mass, kg; R: specific gas constant, J/kg·K (air: 287 J/kg·K); T: temperature, K.

W = (P₁V₁ − P₂V₂)/(n − 1) = m R (T₁ − T₂)/(n − 1) — polytropic process PVⁿ = constant, n ≠ 1. With n = γ it is the reversible adiabatic process of an ideal gas.

W = (P₁ + P₂)(V₂ − V₁)/2 — pressure varying linearly with volume (e.g. a piston restrained by a linear spring).

W_shaft = 2π N T / 60 (power, W) — N: speed, rpm; T: torque, N·m.

W_el = V I t — V: voltage, V; I: current, A; t: time, s.

W_spring = ½ k (x₂² − x₁²) — k: spring stiffness, N/m; x: deflection from free length, m.

Q = m c ΔT — sensible heat, no phase change. c: specific heat, J/kg·K; ΔT in K (same as °C).

T(K) = t(°C) + 273.15

Worked examples

Example 1 (standard). A gas in a cylinder expands quasi-statically according to PV¹·³ = constant from P₁ = 1 MPa, V₁ = 0.05 m³ to V₂ = 0.2 m³. Find the final pressure and the work done.

1. Final pressure: P₂ = P₁ (V₁/V₂)ⁿ = 1000 kPa × (0.05/0.2)¹·³ = 1000 × 0.25¹·³ = 164.9 kPa.
2. Work: W = (P₁V₁ − P₂V₂)/(n − 1).
3. P₁V₁ = 1000 kPa × 0.05 m³ = 50 kJ; P₂V₂ = 164.94 kPa × 0.2 m³ = 32.99 kJ.
4. W = (50 − 32.99)/0.3 = 56.7 kJ, positive, so work is done by the gas.

Check: an isothermal expansion over the same volume ratio would give P₁V₁ ln 4 = 50 × 1.386 = 69.3 kJ. The polytropic work is smaller because pressure falls faster (n > 1), so the area under the curve is smaller.

Example 2 (GATE level). A vertical cylinder of cross-section A = 0.05 m² contains gas at 100 kPa and 0.01 m³ under a weightless, frictionless piston. Atmospheric pressure is 100 kPa. A linear spring of stiffness k = 50 kN/m just touches the top of the piston. The gas is heated until its volume is 0.03 m³. During the process a paddle wheel inside the cylinder also does 3 kJ of work on the gas. Find (a) the final pressure, (b) the boundary work done by the gas, (c) the net work of the system.

1. Piston rise: x = ΔV/A = (0.03 − 0.01)/0.05 = 0.4 m.
2. Force balance on the piston: P = P_atm + k x / A = 100 + (50 × 0.4)/0.05 = 100 + 400 = 500 kPa final.
3. P varies linearly with V (spring is linear), so W_b = (P₁ + P₂)(V₂ − V₁)/2 = (100 + 500)/2 × 0.02 = 6 kJ.
4. Split check: against atmosphere P_atm ΔV = 100 × 0.02 = 2 kJ; into the spring ½ k x² = 0.5 × 50 × 0.4² = 4 kJ; total 6 kJ. Checks.
5. Paddle-wheel work is done on the gas, so it is negative in our convention: W_paddle = −3 kJ.
6. Net work: W = 6 − 3 = 3 kJ (done by the system).

Common mistakes

• Using W = ∫P dV for a non-quasi-static process such as free expansion; the answer there is zero work.
• Mixing sign conventions: adding paddle-wheel or compressor work as positive when the question takes work by the system as positive.
• Treating heat and work as properties ("the heat in the gas"). Only their difference, Q − W = ΔU, is path-independent.
• Using gauge pressure in ∫P dV when the atmosphere pushes on the other side; work against the atmosphere is part of the boundary work.
• Forgetting kPa × m³ = kJ, or using °C instead of K in PV = mRT.
• Assuming adiabatic means isothermal. In an adiabatic compression the temperature rises.

For GATE ME

This topic appears as short conceptual MCQs (intensive vs extensive properties, path vs point functions, quasi-static processes, free expansion) and as short numericals on boundary work: polytropic, isothermal and spring-loaded pistons, often combined with a paddle wheel or electrical heater to test sign conventions. Practise reading work as area on P–V diagrams, including cycles where the enclosed area is the net work and its direction (clockwise = net work output) gives the sign.

Quick check

1. Is specific volume intensive or extensive?
2. What work is done when a gas expands freely into an evacuated vessel?
3. 2 kg of gas is heated at a constant 300 kPa from 0.4 m³ to 0.6 m³. What is the work done?
4. Why is work written δW and not dW?

Answers: 1. Intensive (it is an extensive property per unit mass). 2. Zero, because nothing resists the boundary motion. 3. W = 300 × 0.2 = 60 kJ, done by the gas. 4. Work is a path function, so its differential is inexact.