Systems, properties, work and heat

Why it matters

Every thermal calculation — an engine cylinder, a compressor, a battery-pack cooling loop — starts by drawing a boundary, deciding what crosses it and naming the state of what is inside. Getting the system, the properties and the sign of work and heat right at this stage is what makes the first and second laws give correct answers later; most wrong answers in thermodynamics come from a badly chosen boundary or a sign slip, not from hard mathematics.

Key ideas

System, surroundings, boundary. A system is the quantity of matter or region of space chosen for study; everything outside 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 (heat, work) can. Example: gas trapped in a piston–cylinder.
• Open system (control volume): mass flows in and out across a control surface. Example: turbine, nozzle, pump, heat exchanger.
• Isolated system: neither mass nor energy crosses. The universe (system + surroundings) is treated as isolated.

Properties and state. A property is any measurable characteristic of the system at a state (p, T, V, m, U, H, S). Properties depend only on the state, not on how it was reached, so their changes are exact differentials.

• Intensive properties do not depend on the amount of matter (p, T, density).
• Extensive properties scale with mass (V, U, H, S). Extensive divided by mass gives a specific property (v = V/m, u = U/m), which is intensive.
• State postulate: the state of a simple compressible substance is fixed by two independent intensive properties. In the two-phase region p and T are not independent, so you need a pair such as (p, x) or (T, v).

Equilibrium and processes. Thermodynamic equilibrium means thermal, mechanical, phase and chemical equilibrium together. A quasi-static (quasi-equilibrium) process passes through a continuous series of equilibrium states, which is why it can be drawn as a line on a p–V diagram and why ∫p dV gives the boundary work. Common idealised processes: isobaric (p = const), isochoric (V = const), isothermal (T = const), adiabatic (Q = 0) and polytropic (pVⁿ = const). 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 what makes a thermometer meaningful. Use kelvin in every thermodynamic formula: T(K) = T(°C) + 273.15.

Work and heat are energy in transit, not properties.

• Heat (Q) crosses the boundary only because of a temperature difference.
• Work (W) is any other energy interaction — moving boundary, rotating shaft, electrical, spring.
• Both are path functions: their values depend on the process path, so we write δQ and δW (inexact differentials) and never "Q₂ − Q₁".
• Sign convention used here (most Indian texts 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 check which one a question uses.

Forms of work.

• Moving boundary (displacement) work for a quasi-static process: the area under the process curve on a p–V diagram. A non-quasi-static process (free expansion into vacuum) does zero boundary work even though the volume changes, because the gas pushes against nothing.
• Shaft work from torque and speed; electrical work from voltage and current; spring work from the spring force.
• Because work is a path function, two processes between the same end states can give different work — which is exactly why an engine cycle produces net work.

Formulas

W_b = ∫ p dV (from V₁ to V₂)

• W_b: boundary work (J or kJ), p: absolute pressure (Pa or kPa), V: volume (m³). Valid for a quasi-static process only. kPa × m³ = kJ.

W = p (V₂ − V₁) — isobaric process.

W = p₁V₁ ln(V₂/V₁) = m R T ln(V₂/V₁) — isothermal process of an ideal gas.

• m: mass (kg), R: specific gas constant (kJ/kg·K; 0.287 for air), T: absolute temperature (K).

W = (p₁V₁ − p₂V₂) / (n − 1) — polytropic process pVⁿ = const, n ≠ 1.

• n: polytropic index (dimensionless). n = 0 isobaric, n = 1 isothermal (ideal gas), n = γ reversible adiabatic of an ideal gas, n → ∞ isochoric.

p₂ / p₁ = (V₁ / V₂)ⁿ — end states of a polytropic process.

W_shaft = T_q · ω = 2π N T_q / 60

• T_q: torque (N·m), ω: angular speed (rad/s), N: speed (rpm). Gives power in W.

W_spring = ½ k (x₂² − x₁²)

• k: spring stiffness (N/m), x: deflection from free length (m).

W_elec = V I Δt

• V: voltage (V), I: current (A), Δt: time (s).

pv = RT — ideal-gas equation of state (p in kPa, v in m³/kg, R in kJ/kg·K, T in K).

Worked examples

Example 1 (standard). Air in a piston–cylinder expands at a constant pressure of 200 kPa from 0.05 m³ to 0.15 m³. Find the boundary work.

1. Process is isobaric, so W = p (V₂ − V₁).
2. W = 200 kPa × (0.15 − 0.05) m³ = 200 × 0.10 = 20 kPa·m³.
3. 1 kPa·m³ = 1 kJ, and the volume increased, so the work is done by the gas (positive).

W = +20 kJ

Example 2 (GATE level). A gas at 1 MPa occupies 0.05 m³. It expands quasi-statically to 0.2 m³ following pV¹·³ = constant. Find the final pressure and the work, and compare with an isothermal expansion between the same volumes.

1. Final pressure: p₂ = p₁ (V₁/V₂)ⁿ = 1000 kPa × (0.05/0.2)¹·³ = 1000 × 0.25¹·³ = 1000 × 0.16494 = 164.9 kPa.
2. Polytropic work: W = (p₁V₁ − p₂V₂)/(n − 1) = (1000 × 0.05 − 164.9 × 0.2)/(1.3 − 1) = (50 − 32.99)/0.3 = 17.01/0.3 = 56.7 kJ.
3. Isothermal work for comparison: W = p₁V₁ ln(V₂/V₁) = 50 × ln 4 = 50 × 1.3863 = 69.3 kJ.
4. The isothermal curve lies above the polytropic one (pressure falls more slowly), so it encloses more area and gives more work.

p₂ ≈ 165 kPa, W ≈ 56.7 kJ (polytropic) against 69.3 kJ (isothermal)

Example 3 (spring-loaded piston). A gas pushes a piston restrained by a linear spring, so its pressure rises linearly from 100 kPa to 400 kPa while the volume grows from 0.10 m³ to 0.25 m³. Find the work.

1. A linear p–V path gives a trapezium: W = ½ (p₁ + p₂)(V₂ − V₁).
2. W = 0.5 × (100 + 400) × (0.25 − 0.10) = 250 × 0.15 = 37.5 kJ.

W = 37.5 kJ

Common mistakes

• Using gauge pressure or °C in p dV and pV = mRT. Use absolute pressure and kelvin.
• Treating heat or work as a property, writing Q₂ − Q₁, or assuming the work between two states is fixed regardless of path.
• Applying ∫p dV to a free (unresisted) expansion. It is not quasi-static, so the boundary work is zero.
• Mixing sign conventions within one problem, especially when a question states "work done on the gas".
• Dividing by (n − 1) when n = 1. The isothermal case needs the logarithmic formula.
• Forgetting that kPa × m³ = kJ and then multiplying by 1000 a second time.
• Calling temperature an extensive property, or calling specific volume extensive. A specific property is always intensive.

For GATE ME

This topic shows up as one-mark concept questions (intensive or extensive, point or path function, which process is quasi-static, what the zeroth law establishes) and as the first step of longer numericals: boundary work for isobaric, isothermal, polytropic or spring-loaded processes, often read off a p–V diagram as an area. Practise computing work from the polytropic formula and from p–V geometry, deciding the sign from the direction of volume change, and checking units (kPa·m³ = kJ).

Quick check

1. Is specific enthalpy an intensive or an extensive property?
2. What is the boundary work when a gas expands freely into an evacuated rigid chamber?
3. For pVⁿ = constant, which value of n gives a constant-volume process?
4. Air at 150 kPa expands at constant pressure from 0.02 m³ to 0.06 m³. How much work does it do?
5. Why are heat and work called path functions?

Answers: 1. Intensive. 2. Zero, because the expansion is not quasi-static and meets no resisting pressure. 3. n → ∞. 4. 150 × 0.04 = 6 kJ. 5. Their values depend on the process followed between two states, not only on the end states.