Work and heat transfer in quasi-static processes

Why it matters

Work and heat are the only two ways energy crosses the boundary of a closed system, so every engine, compressor and refrigeration calculation depends on getting them right. Compressor power, the expansion work in a piston engine and the energy absorbed by a pressurised tank all come from the same ∫p dV idea. Getting the sign and the path wrong is the single largest source of lost marks in thermodynamics.

Key ideas

Work is energy transfer across a boundary that could, in principle, be used to raise a weight. Heat is energy transfer across a boundary caused only by a temperature difference. Both are:

• boundary phenomena — they exist only while crossing the boundary; a system contains energy, not heat or work;
• path functions — their amounts depend on the process, so we write δW and δQ (inexact differentials), never dW or ΔW;
• transient — they vanish once the process stops.

Sign convention used here (engineering): Q is positive when heat is added to the system; W is positive when work is done by the system. Some books (and chemistry texts) take work done on the system as positive; always check which one a question uses.

Quasi-static process. The process is carried out so slowly that the system passes through a continuous series of equilibrium states. Only then is the pressure uniform and defined at every instant, so the boundary (displacement) work can be written as W = ∫ p dV and drawn as the area under the path on a p–V diagram. A quasi-static process is not automatically reversible: friction at the piston can still dissipate work. Quasi-static and free of dissipation (friction, resistance, unrestrained expansion, heat transfer across finite ΔT) is what makes it reversible.

Path dependence. Between the same two states, a different path encloses a different area, so the work differs. Over a cycle, the net work equals the area enclosed by the loop — clockwise on p–V means net work output.

Free (unrestrained) expansion. A gas rushing into an evacuated space does no boundary work (W = 0) even though its volume increases, because nothing resists it. The process is not quasi-static, so ∫p dV does not apply.

Other work modes.

• Shaft work: W = T·θ (torque × angle); power = T·ω.
• Electrical work: W = V·I·t.
• Spring work: W = ½·k·(x₂² − x₁²).
• Paddle-wheel (stirring) work: work done on the system, always negative in our convention; it cannot be done quasi-statically in reverse.

Heat. An adiabatic process has Q = 0 (perfect insulation, or a process too fast for heat to flow). Adiabatic does not mean isothermal: temperature can change through work. Heat transfer mechanisms — conduction, convection, radiation — are treated in a later topic.

Formulas

W = ∫ p dV (quasi-static boundary work) W = p·(V₂ − V₁) (constant pressure, isobaric) W = 0 (constant volume, isochoric) W = p₁·V₁·ln(V₂/V₁) (isothermal ideal gas, pV = constant) W = (p₁·V₁ − p₂·V₂) / (n − 1) (polytropic pVⁿ = constant, n ≠ 1) W = ½·(p₁ + p₂)·(V₂ − V₁) (pressure linear in V, e.g. linear spring) W_spring = ½·k·(x₂² − x₁²)

• p: absolute pressure, Pa (kPa × m³ gives kJ)
• V: volume, m³
• n: polytropic index (dimensionless)
• k: spring constant, N/m; x: spring deflection, m
• W: work, J; positive when done by the system

Worked examples

Example 1 (standard). Air in a cylinder, 0.1 m³ at 100 kPa, is compressed quasi-statically following pV¹·³ = constant to 0.02 m³. Find the final pressure and the work.

1. p₂ = p₁·(V₁/V₂)ⁿ = 100 × (0.1/0.02)¹·³ = 100 × 5¹·³ = 810.3 kPa
2. W = (p₁·V₁ − p₂·V₂) / (n − 1)
3. p₁V₁ = 100 × 0.1 = 10 kJ; p₂V₂ = 810.3 × 0.02 = 16.21 kJ
4. W = (10 − 16.21) / 0.3 = −20.69 kJ

Answer: p₂ = 810.3 kPa, W = −20.7 kJ (negative: work is done on the air). For comparison, an isothermal compression between the same volumes needs only 100 × 0.1 × ln(0.2) = −16.1 kJ.

Example 2 (GATE level). A gas in a piston–cylinder (piston area 0.05 m²) is at 150 kPa and 0.01 m³. The piston just touches a linear spring of stiffness 100 kN/m. Heat is added until the volume is 0.025 m³. Find the final pressure, the total boundary work and the share stored in the spring. The process is quasi-static.

1. Piston travel: x = ΔV / A = 0.015 / 0.05 = 0.3 m
2. Extra spring force = k·x = 100 kN/m × 0.3 m = 30 kN, so extra pressure Δp = 30 / 0.05 = 600 kPa
3. p₂ = 150 + 600 = 750 kPa; p varies linearly with V
4. W = ½·(p₁ + p₂)·(V₂ − V₁) = ½ × (150 + 750) × 0.015 = 6.75 kJ
5. Spring share: ½·k·x² = ½ × 100 × 0.3² = 4.5 kJ; the remaining 2.25 kJ (= 150 × 0.015) pushes back the atmosphere and the piston weight.

Answer: p₂ = 750 kPa, W = 6.75 kJ, of which 4.5 kJ goes into the spring

Common mistakes

• Using ∫p dV for a non-quasi-static process such as free expansion (the work there is zero).
• Treating "quasi-static" as identical to "reversible".
• Plugging in gauge pressure; work needs absolute pressure.
• Sign slips: compression work is negative in the W-by-system convention. Check against the physical picture.
• Forgetting kPa·m³ = kJ and reporting 200 J instead of 200 kJ.
• Using W = p₁V₁ ln(V₂/V₁) for a non-isothermal process, or the polytropic formula with n = 1 (it divides by zero — use the log form).
• Assuming an adiabatic process is isothermal.

For GATE AE

Expect numericals on boundary work for isobaric, isothermal and polytropic paths, spring-loaded pistons with linear p–V, and comparisons of work along different paths between the same states (area reasoning). One-mark questions test point vs path functions, the sign convention and free expansion. Practise drawing every process on a p–V diagram before calculating, and keep kPa·m³ = kJ in mind.

Quick check

1. Gas expands from 1 m³ to 3 m³ at a constant 100 kPa. Find the work.
2. What is the boundary work in a free expansion into vacuum?
3. Why is δW an inexact differential?
4. On a p–V diagram, which way must a cycle run to give net work output?
5. In the polytropic work formula, what happens when n = 1?

Answers: 1. 200 kJ. 2. Zero. 3. Work depends on the path, so it is not the change of a property. 4. Clockwise. 5. The formula becomes 0/0; use W = p₁V₁ ln(V₂/V₁) instead.