Thermodynamics · Chapter 3 of 10 · Intermediate
Properties of Pure Substances
Real working fluids boil, condense, and superheat, so their properties come from tables, not a single formula. Reading those tables, and knowing when a gas is simple enough for Pv = RT, is the practical heart of the subject.
Readiness check
This chapter is about reading property data. Tick only what you can do closed-notes.
- Interpolate linearly between two rows of a table.
- Recall specific volume v = V/m with units.
- Use the ideal-gas law Pv = RT with T in kelvin.
- Describe what happens as water is heated from liquid to vapor.
- Read absolute pressure and temperature from a state.
The core idea
A pure substance moves through liquid, two-phase mixture, and vapor along a saturation dome. Inside the dome, pressure and temperature are linked, so a third number, the quality, is needed to fix the state.
v = vf + x vfgx = mvapor / mtotalPv = RT (ideal gas)As a liquid is heated at constant pressure it reaches the saturation temperature and begins to boil. Through the two-phase region the temperature stays fixed while the fluid turns to vapor; only after the last drop evaporates does the temperature climb again into the superheated region. Inside the dome, T and P are not independent, so we report quality x, the vapor mass fraction, and interpolate properties between the saturated-liquid (f) and saturated-vapor (g) values. Far from the dome, a gas obeys the ideal-gas law closely.
The skills, taught in order
Five skills cover phase change, the structure of the tables, quality, and the ideal-gas model with its limits.
3.1 Phases and phase change
At a fixed pressure, heating a liquid raises its temperature to the saturation temperature, where boiling begins. The temperature then holds constant while latent heat converts liquid to vapor, and rises again only once the substance is fully vapor. The reverse happens on cooling. This constant-temperature plateau during phase change is why steam carries so much energy.
3.2 The property tables
Because no simple equation fits a real fluid across phases, properties are tabulated. Saturated tables list paired values at each saturation temperature or pressure: subscript f for saturated liquid, g for saturated vapor, and fg for the difference. Separate tables cover superheated vapor and compressed liquid. Interpolate linearly between rows when your state falls between tabulated values.
| P (kPa) | Tsat (°C) | hf (kJ/kg) | hfg (kJ/kg) | hg (kJ/kg) |
|---|---|---|---|---|
| 10 | 45.81 | 191.81 | 2392.82 | 2584.63 |
| 100 | 99.62 | 417.44 | 2258.02 | 2675.46 |
| 200 | 120.23 | 504.68 | 2201.96 | 2706.63 |
Saturated water, pressure entry (Borgnakke and Sonntag, Table B.1.2). Latent heat hfg falls as pressure rises.
3.3 Quality of a saturated mixture
Inside the dome, the quality x = mvapor/mtotal runs from 0 (saturated liquid) to 1 (saturated vapor). Any specific property is the f-value plus quality times the fg-difference: v = vf + x vfg, and likewise for u, h, and s. This single weighting is the workhorse of the chapter.
3.4 Superheated vapor and compressed liquid
Right of the dome the vapor is superheated: its temperature exceeds the saturation value at that pressure, and T and P become independent again, so a superheated table is entered with both. Left of the dome the compressed liquid is nearly incompressible, and is often approximated by the saturated-liquid value at the same temperature.
3.5 The ideal-gas law and compressibility
Well away from the dome, a gas obeys Pv = RT, or PV = mRT, with R the specific gas constant. This breaks down near saturation and at high pressure, where the compressibility factor Z = Pv/RT departs from one. For air at ordinary conditions Z is within a percent of unity, so the ideal-gas law is excellent.
Engineering connection: steam tables drive the power cycles of Chapter 10, while the ideal-gas law drives the gas cycles of Chapter 9.
Worked example 1: quality and properties of wet steam
A rigid tank holds 2 kg of a water-steam mixture at 200 kPa with a quality of 0.60. Using the saturated table (vf = 0.001061 m³/kg, vg = 0.88573 m³/kg, hf = 504.68 kJ/kg, hfg = 2201.96 kJ/kg), find the specific volume, the specific enthalpy, and the tank volume.
- ProblemFind v, h, and the total volume for the wet steam in Figure 1.
- Given / findm = 2 kg, P = 200 kPa, x = 0.60, plus the saturated f and g values listed. Find v, h, and V.
- AssumptionsEquilibrium two-phase mixture at the saturation pressure; properties from the table.
- ModelWeight the f-value by quality: v = vf + x vfg and h = hf + x hfg; then V = mv.
- Equationsvfg = vg − vf = 0.88467 m³/kgv = vf + x vfgh = hf + x hfg
- Solvev = 0.001061 + 0.60 × 0.88467 = 0.5319 m³/kg. h = 504.68 + 0.60 × 2201.96 = 1825.9 kJ/kg. V = 2 × 0.5319 = 1.064 m³.
- CheckThe result sits between vf and vg, as a mixture must. Because vapor is roughly 800 times less dense than liquid, even 60 percent quality already fills most of the volume.
- ConclusionQuality plus the saturated table fixes every property of a wet mixture. The same weighting returns for internal energy and entropy in later chapters.
Worked example 2: an ideal gas in a rigid tank
A rigid 1.0 m³ tank holds air at 500 kPa and 300 K. Treating air as an ideal gas (R = 0.287 kJ/kg·K), find its mass. The tank is then heated to 350 K; find the new pressure.
- ProblemFind the mass of air in Figure 2, then the pressure after heating at constant volume.
- Given / findV = 1.0 m³, P1 = 500 kPa, T1 = 300 K, R = 0.287 kJ/kg·K, T2 = 350 K. Find m and P2.
- AssumptionsAir is an ideal gas; the tank is rigid, so V and m are constant.
- ModelUse PV = mRT for the mass, then the constant-volume relation P/T = constant for the pressure.
- Equationsm = PV/(RT)P2 = P1 (T2/T1)
- Solvem = (500 × 1.0)/(0.287 × 300) = 500/86.1 = 5.81 kg. P2 = 500 × (350/300) = 583 kPa.
- CheckFor air at 500 kPa the reduced pressure is far below critical, so Z is within a percent of one and the ideal-gas law is justified. Pressure rises with temperature at fixed volume, as expected.
- ConclusionTwo properties fix an ideal-gas state, and a rigid tank ties pressure directly to absolute temperature. Always work in kelvin.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Ideal gas near saturation | Properties disagree with the steam table | "Is this state near the dome?" | Use the table inside and near the dome; reserve Pv = RT for superheated, low-pressure gas. |
| Ignoring quality | One value used for a two-phase state | "Is the state inside the dome?" | Inside the dome, weight by quality: v = vf + x vfg. |
| Setting T and P independent inside the dome | Two saturation values that disagree | "Are T and P linked here?" | Inside the dome they are paired; one fixes the other. |
| Forgetting to interpolate | Snapping to the nearest row | "Does my state fall between two rows?" | Interpolate linearly between the bracketing rows. |
Practice ladder
Steam at 100 kPa has quality 0.80. Using hf = 417.44 kJ/kg and hfg = 2258.02 kJ/kg, find its specific enthalpy.
Show answer
h = hf + x hfg = 417.44 + 0.80 × 2258.02 = 417.44 + 1806.4 = 2223.9 kJ/kg. The value lies between hf and hg, closer to the vapor end because quality is high.
The rigid air tank (Worked Example 2) is instead cooled to 250 K. What is the new pressure, and what fraction of the original is it?
Show answer
P2 = 500 × (250/300) = 417 kPa, which is 250/300 = 0.833 of the original. At constant volume, pressure tracks absolute temperature exactly.
Water at 200 kPa has a specific volume of 0.45 m³/kg. Is it wet, saturated, or superheated, and if wet, what is the quality? Use vf = 0.001061 and vg = 0.88573 m³/kg.
Show answer
Since 0.45 lies between vf and vg, it is a wet mixture. x = (v − vf)/(vg − vf) = (0.45 − 0.001061)/0.88467 = 0.507, about 51 percent vapor.
Explain why a steam burn at 100 °C is far worse than a hot-water burn at the same temperature, using the idea of latent heat from the table.
What good work looks like
A reference to hfg: condensing steam releases roughly 2257 kJ/kg of latent heat on the skin before it even begins to cool, far more than the same mass of hot water can give up.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take one real device that uses steam or a refrigerant (an espresso machine, a radiator). Identify two states it passes through, place each relative to the dome, and read or interpolate one property at each.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What happens to temperature during boiling at fixed pressure?
It stays at the saturation temperature while latent heat converts liquid to vapor, then rises again once the fluid is fully vapor.
2. Define quality.
x = mvapor/mtotal, the vapor mass fraction, running from 0 (saturated liquid) to 1 (saturated vapor).
3. How do you get a property inside the dome?
Weight by quality: property = f-value + x times fg-difference, for example v = vf + x vfg.
4. When does the ideal-gas law fail?
Near saturation and at high pressure, where the compressibility factor Z departs from one.
5. What two properties enter a superheated table?
Pressure and temperature, because outside the dome they are independent again.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 3 (Properties of a Pure Substance) and Appendix B |
| Cross-reference | Çengel and Boles, Ch. 3 · Moran, Ch. 3 |
| Core topics | 3.1 Phase change · 3.2 Property tables · 3.3 Quality · 3.4 Superheated and compressed · 3.5 Ideal gas and compressibility |
| Engineering connection | Steam tables drive the vapor cycles; the ideal-gas law drives the gas cycles. |
| Read next | Chapter 4: Energy Analysis of Closed Systems. |