Properties

Thermodynamic properties describe the state of a substance. Learn systems and boundaries, pressure, temperature and specific volume, the ideal-gas law, phases and saturation, quality and how to read steam tables.

Before calculating energy flows, you must describe the substance: how hot, how pressurised, how much space it takes up, whether it's liquid, vapour or both. These are properties, and two independent properties usually fix the whole state.

Air in an engine behaves like an ideal gas; water in a power plant changes phase and needs property tables. Knowing which model to use is the first skill of thermodynamics.

An analogy

A property is like a GPS coordinate for the substance: give two coordinates (say pressure and temperature) and every other property — density, energy, enthalpy — can be looked up.

Systems, boundaries and states

A closed system holds a fixed mass (gas in a sealed cylinder); an open system (control volume) has mass flowing through (a turbine, pump, nozzle). The state is described by properties. Intensive properties (p, T, v) don't depend on size; extensive ones (V, U, mass) do.

Always use absolute values

Gas laws need absolute temperature (kelvin) and absolute pressure. A tyre gauge reads 220 kPa gauge, which is about 321 kPa absolute. Forgetting this is the most common thermodynamics mistake.

Phases and saturation

At a given pressure, water boils at one temperature (100 °C at 101.3 kPa, 180 °C at 1 MPa). Between saturated liquid and saturated vapour is the wet region, described by quality x. Beyond the critical point (22.06 MPa, 374 °C) liquid and vapour are indistinguishable — supercritical power plants operate there.

Reading property tables

Steam tables list v, u, h, s for saturated states (by T or p) and for superheated vapour. Interpolate linearly between rows. Modern practice uses software (CoolProp, REFPROP, EES) but the logic is identical.

Real-gas behaviour

At high pressure or near saturation, gases deviate from pv = RT. The compressibility factor Z = pv/RT from generalized charts (reduced p and T) or equations of state (van der Waals, Peng–Robinson) corrects for it — essential for natural-gas pipelines, CO₂ and refrigerants.

Formulas

Worked example: Mass of air in a tank

Given: V = 0.5 m³; p = 800 kPa (abs), T = 27 °C; R = 0.287 kJ/kg·K

Find: Mass, and pressure if heated to 60 °C.

Answer: 4.65 kg; 888 kPa at 60 °C

Worked example: Wet steam enthalpy

Given: p = 100 kPa, x = 0.9; h_f = 417.5, h_fg = 2257.5 kJ/kg

Find: Specific enthalpy.

Answer: ≈ 2449 kJ/kg

Common mistakes

Applications

Key takeaways

Going further