Thermodynamics · Chapter 4 of 10 · Intermediate
Energy Analysis of Closed Systems
For a fixed mass, the first law becomes ΔU = Q − W. The work is the area under the process on a P-V diagram, and the specific heats turn temperature changes into energy.
Readiness check
This chapter applies the first law to a fixed mass. Tick only what you can do closed-notes.
- State the energy balance Q − W = ΔE.
- Use the ideal-gas law PV = mRT.
- Recognise the area under a curve as an integral.
- Recall Q = mcΔT for a constant specific heat.
- Keep a consistent sign convention for heat and work.
The core idea
For a closed system the first law is ΔU = Q − W. The work done by an expanding gas is the area under its path on a P-V diagram, and specific heats convert a temperature change into energy.
ΔU = Q − WWb = ∫ P dVcp − cv = RA closed system exchanges no mass, so its energy can change only by heat and boundary work. The boundary work is W = ∫P dV, which is exactly the area under the process curve, so the same end states give different work along different paths. For an ideal gas, internal energy depends only on temperature: ΔU = mcvΔT, while enthalpy change is ΔH = mcpΔT, and the two specific heats differ by the gas constant R.
The skills, taught in order
Five skills set the closed-system first law, the boundary-work integral for common paths, and the two specific heats.
4.1 The closed-system first law
With no mass crossing the boundary and negligible change in bulk kinetic and potential energy, the energy balance reduces to ΔU = Q − W. Heat into the system and work done by the system are positive. This single equation, with the right work term, solves most closed-system problems.
4.2 Boundary work
A moving piston does boundary work W = ∫P dV, the area under the process on a P-V diagram. For a constant-pressure path it is simply W = P(V₂ − V₁); for an ideal gas this equals mR(T₂ − T₁). For an isothermal ideal gas, W = mRT ln(V₂/V₁), and for a polytropic path PVn = constant, W = (P₂V₂ − P₁V₁)/(1 − n).
| Process | Held constant | Boundary work W |
|---|---|---|
| Isobaric | P | P(V₂ − V₁) |
| Isochoric | V | 0 |
| Isothermal (ideal gas) | T | mRT ln(V₂/V₁) |
| Polytropic | PVn | (P₂V₂ − P₁V₁)/(1 − n) |
4.3 Specific heats
The specific heat is the energy to raise one kilogram by one kelvin. At constant volume no work is done, so cv relates to internal energy; at constant pressure the gas also does work, so cp is larger. For any ideal gas cp − cv = R, and their ratio k = cp/cv appears throughout the cycle chapters.
| Gas | R (kJ/kg·K) | cp (kJ/kg·K) | cv (kJ/kg·K) | k |
|---|---|---|---|---|
| Air | 0.287 | 1.005 | 0.718 | 1.400 |
| Nitrogen | 0.297 | 1.042 | 0.745 | 1.400 |
| Carbon dioxide | 0.189 | 0.842 | 0.653 | 1.289 |
| Helium | 2.077 | 5.193 | 3.116 | 1.667 |
Representative values near room temperature (Borgnakke and Sonntag, Table A.5). Note cp − cv = R in every row.
4.4 Internal energy and enthalpy of an ideal gas
For an ideal gas, ΔU = mcvΔT and ΔH = mcpΔT, regardless of how pressure and volume change. At constant volume Q = ΔU = mcvΔT, and at constant pressure Q = ΔH = mcpΔT, which is why cp is the one to use for constant-pressure heating.
Engineering connection: these closed-system tools become the four strokes of the Otto and Diesel cycles in Chapter 9.
Worked example 1: constant-pressure heating of air
A piston-cylinder holds 0.5 kg of air at a constant 200 kPa. It is heated so the temperature rises from 25 °C to 250 °C. Using R = 0.287, cv = 0.718, and cp = 1.005 kJ/kg·K, find the boundary work, the change in internal energy, and the heat added.
- ProblemFind W, ΔU, and Q for the constant-pressure heating in Figure 1.
- Given / findm = 0.5 kg, P = 200 kPa, ΔT = 250 − 25 = 225 K, R = 0.287, cv = 0.718, cp = 1.005 kJ/kg·K. Find W, ΔU, Q.
- AssumptionsAir is an ideal gas; the process is at constant pressure; bulk kinetic and potential energy do not change.
- ModelConstant-pressure work for an ideal gas is W = mRΔT; ΔU = mcvΔT; the first law gives Q = ΔU + W (which equals mcpΔT).
- EquationsW = mRΔTΔU = mcvΔTQ = ΔU + W = mcpΔT
- SolveW = 0.5 × 0.287 × 225 = 32.3 kJ. ΔU = 0.5 × 0.718 × 225 = 80.8 kJ. Q = 80.8 + 32.3 = 113.1 kJ, matching mcpΔT = 0.5 × 1.005 × 225 = 113.1 kJ.
- CheckThe two routes to Q agree exactly, because cp = cv + R. The work is the smaller share; most of the heat raises internal energy.
- ConclusionAt constant pressure the heat added equals the enthalpy change, and it divides cleanly into work out and internal-energy rise. This is why cp governs constant-pressure heating.
Worked example 2: constant-volume heating in a rigid tank
A rigid tank holds 2 kg of air at 100 kPa and 27 °C. It is heated until the temperature reaches 227 °C. With cv = 0.718 kJ/kg·K, find the boundary work, the heat added, and the final pressure.
- ProblemFind W, Q, and P₂ for the constant-volume heating in Figure 2.
- Given / findm = 2 kg, P1 = 100 kPa, T1 = 300 K, T2 = 500 K, cv = 0.718 kJ/kg·K. Find W, Q, P2.
- AssumptionsAir is an ideal gas; the tank is rigid, so the volume is fixed.
- ModelNo volume change means W = 0, so the first law gives Q = ΔU = mcvΔT; pressure follows P/T = constant.
- EquationsW = 0Q = ΔU = mcvΔTP2 = P1(T2/T1)
- SolveW = 0. Q = 2 × 0.718 × (500 − 300) = 2 × 0.718 × 200 = 287 kJ. P2 = 100 × (500/300) = 167 kPa.
- CheckWith no work, every kilojoule of heat becomes internal energy, so the same temperature rise needs less heat than the constant-pressure case used cp for. Pressure scales with absolute temperature, as it must in a rigid tank.
- ConclusionAt constant volume the heat equals the internal-energy change and cv is the right specific heat. The path, not just the end states, decides the work.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Swapping cp and cv | Heat off by the factor k | "Is the volume or the pressure held constant?" | Use cv at constant volume, cp at constant pressure. |
| Work where there is none | Nonzero W for a rigid tank | "Did the volume change?" | No volume change means zero boundary work. |
| Constant-pressure work on the wrong path | W = PΔV used when P varies | "Is pressure really constant along the path?" | Integrate the actual P-V path; only constant pressure gives PΔV. |
| ΔU from pressure or volume alone | Internal energy tied to P or V for an ideal gas | "Did the temperature change?" | For an ideal gas, ΔU depends only on ΔT. |
Practice ladder
Air expands at a constant 150 kPa from 0.02 m³ to 0.05 m³. Find the boundary work.
Show answer
W = P(V₂ − V₁) = 150 × (0.05 − 0.02) = 150 × 0.03 = 4.5 kJ. The work is positive because the gas expands against the piston.
The rigid tank (Worked Example 2) is heated to 600 K instead of 500 K. What are the new heat input and final pressure?
Show answer
Q = mcvΔT = 2 × 0.718 × (600 − 300) = 430.8 kJ. P₂ = 100 × (600/300) = 200 kPa. Both scale with the larger temperature change.
0.3 kg of air is compressed isothermally at 300 K from 0.10 m³ to 0.04 m³. Find the work and the heat (R = 0.287 kJ/kg·K).
Show answer
W = mRT ln(V₂/V₁) = 0.3 × 0.287 × 300 × ln(0.04/0.10) = 25.83 × (−0.9163) = −23.7 kJ. Isothermal means ΔU = 0, so Q = W = −23.7 kJ: work is done on the gas and an equal heat is rejected.
Explain why heating a gas at constant pressure always needs more heat than heating it the same amount at constant volume, using work and the two specific heats.
What good work looks like
At constant pressure the gas also does boundary work, so the heat must cover both ΔU and W; that extra work is exactly the gap cp − cv = R per kilogram per kelvin.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse a real closed-system process (a sealed syringe, a pressure cooker coming up to pressure). Sketch its P-V path, find the boundary work, and apply ΔU = Q − W with the correct specific heat.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the closed-system first law.
ΔU = Q − W, with heat in and work out taken positive.
2. What is boundary work geometrically?
The area under the process path on a P-V diagram, W = ∫P dV.
3. Relate the two specific heats.
cp − cv = R for an ideal gas, and k = cp/cv.
4. Give ΔU and ΔH for an ideal gas.
ΔU = mcvΔT and ΔH = mcpΔT, depending only on temperature change.
5. Why is rigid-tank work zero?
The volume cannot change, so ∫P dV = 0 and all heat goes to internal energy.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 5 (The First Law of Thermodynamics) |
| Cross-reference | Çengel and Boles, Ch. 4 · Moran, Ch. 3 |
| Core topics | 4.1 Closed-system first law · 4.2 Boundary work · 4.3 Specific heats · 4.4 Ideal-gas ΔU and ΔH |
| Engineering connection | Boundary work and specific heats become the strokes of the gas power cycles. |
| Read next | Chapter 5: Mass and Energy Analysis of Control Volumes. |