Systems, properties, state and thermodynamic equilibrium

Why it matters

Every thermodynamics calculation starts by drawing a boundary and deciding what is inside it. Choose the system badly and you will double-count energy or miss a mass flow; choose it well and a jet engine, a compressor or a pressurised cabin becomes a simple balance. The ideas of property, state and equilibrium are what make it legitimate to describe a gas with just a few numbers such as p, T and v.

Key ideas

System, surroundings, boundary. A system is the quantity of matter or region of space we choose to study. Everything outside it is the surroundings, and the surface separating them is the boundary. The boundary may be real (a cylinder wall) or imaginary (an inlet plane of a duct), fixed or moving (a piston face).

• Closed system (control mass): no mass crosses the boundary; energy can cross as heat or work. Example: gas trapped in a piston–cylinder.
• Open system (control volume): mass crosses the boundary through inlets and exits, carrying energy with it. Example: compressor, nozzle, combustor, turbine.
• Isolated system: neither mass nor energy crosses the boundary. A system plus its surroundings together is often treated as isolated.

Properties. A property is any measurable characteristic of the system at a given state (p, T, V, m, U, H, S). A quantity is a property only if its change depends on the end states alone, not on the path — so heat and work are not properties.

• Intensive properties do not depend on the amount of matter: p, T, density ρ, specific volume v.
• Extensive properties scale with mass: V, U, H, S, m.
• Specific properties are extensive properties per unit mass (v = V/m, u = U/m) and are intensive.
• Test: split the system in two; a property that keeps its value in each half is intensive.

State. The state is the condition of the system fixed by the values of its properties. The state postulate says that a simple compressible system (only p–dV work matters) is fixed by two independent intensive properties. For a single-phase gas, p and T are independent; inside the wet (two-phase) region p and T are not independent, so you need something like quality x as the second property.

Process, path, cycle. A process is a change from one equilibrium state to another; the path is the series of states passed through. A quasi-static (quasi-equilibrium) process proceeds so slowly that the system is infinitesimally close to equilibrium at every instant, so the path can be drawn on a p–v diagram. A cycle returns the system to its initial state, so the net change of every property over a cycle is zero.

Thermodynamic equilibrium requires all of these at once:

• Mechanical — no unbalanced forces (uniform pressure, ignoring gravity).
• Thermal — uniform temperature; no heat flow inside the system or with surroundings at the same T.
• Phase — no net transfer of mass between phases.
• Chemical — no net change in chemical composition.

Properties are strictly defined only at equilibrium states. The continuum assumption (the system contains enough molecules that local averages are meaningful) breaks down for very rarefied gases, for example at very high altitude where the mean free path approaches body dimensions.

Zeroth law. 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 and defines temperature as a property.

Pressure. Absolute pressure is measured from vacuum. Gauge pressure = absolute − atmospheric; vacuum = atmospheric − absolute. Thermodynamic equations always use absolute pressure and absolute temperature.

Formulas

p·V = m·R·T (ideal gas, mass basis) p·V = n·R̄·T (ideal gas, mole basis) R = R̄ / M v = V / m = 1 / ρ p_abs = p_atm + p_gauge T(K) = T(°C) + 273.15

• p: absolute pressure, Pa
• V: volume, m³; v: specific volume, m³/kg; ρ: density, kg/m³
• m: mass, kg; n: amount of substance, mol (or kmol)
• R̄: universal gas constant = 8.314 J/(mol·K) = 8.314 kJ/(kmol·K)
• R: specific gas constant, J/(kg·K); for air R = 287 J/(kg·K) (M = 28.97 kg/kmol)
• M: molar mass, kg/kmol
• T: absolute temperature, K

The ideal-gas equation applies to gases at low pressure and high temperature relative to their critical point (air near ambient is an excellent example); it fails near saturation and at very high pressure.

Worked examples

Example 1 (standard). A rigid tank of 0.5 m³ holds air at 300 kPa (absolute) and 300 K. Find the mass, density and specific volume of the air. Take R = 0.287 kJ/(kg·K).

1. Ideal gas: m = p·V / (R·T)
2. m = (300 kPa × 0.5 m³) / (0.287 kJ/(kg·K) × 300 K) = 150 / 86.1 = 1.742 kg
3. ρ = m / V = 1.742 / 0.5 = 3.484 kg/m³
4. v = 1/ρ = 0.287 m³/kg

Answer: m = 1.742 kg, ρ = 3.48 kg/m³, v = 0.287 m³/kg

Example 2 (GATE level). Rigid tank A (0.2 m³) contains air at 500 kPa and 350 K. Rigid tank B (0.3 m³) contains air at 100 kPa and 300 K. They are joined by a valve, which is opened. The combined system eventually reaches equilibrium with the surroundings at 300 K. Find the final pressure. R = 0.287 kJ/(kg·K).

1. Choose the system: the air in both tanks (closed, total volume fixed = 0.5 m³).
2. m_A = p_A·V_A / (R·T_A) = (500 × 0.2) / (0.287 × 350) = 100 / 100.45 = 0.9955 kg
3. m_B = p_B·V_B / (R·T_B) = (100 × 0.3) / (0.287 × 300) = 30 / 86.1 = 0.3484 kg
4. Total mass m = 1.3440 kg (mass is conserved because the system is closed).
5. Final equilibrium: uniform p and T = 300 K. p₂ = m·R·T₂ / V = 1.3440 × 0.287 × 300 / 0.5 = 231.4 kPa

Note that you cannot average pressures by volume here because the initial temperatures differ; always go through mass.

Common mistakes

• Using gauge pressure or °C in p·V = m·R·T. Both must be absolute.
• Mixing R̄ (8.314 kJ/(kmol·K)) with mass-based equations, or R for air (0.287 kJ/(kg·K)) with moles.
• Calling heat or work a property. They are path functions — energy in transit across a boundary.
• Treating p and T as independent inside the two-phase dome; they are tied together by saturation.
• Calling specific volume extensive. Any "per unit mass" quantity is intensive.
• Assuming a process drawn as a line on a p–v diagram when it is not quasi-static (e.g. free expansion — only end states can be plotted).

For GATE AE

Expect one-mark conceptual questions: classify a property as intensive/extensive or point/path function, identify closed/open/isolated systems, or state what equilibrium requires. Numerical questions use the ideal-gas equation with unit traps (kPa vs Pa, gauge vs absolute, °C vs K, molar vs specific gas constant), often in tank-filling or tank-connection settings. Practise mass-based bookkeeping for connected tanks and getting R for a gas from its molar mass.

Quick check

1. Is specific enthalpy intensive or extensive?
2. How many independent intensive properties fix the state of a simple compressible system?
3. A tyre gauge reads 200 kPa with atmospheric pressure 101 kPa. What is the absolute pressure?
4. Name the four kinds of equilibrium that make up thermodynamic equilibrium.
5. Why is work not a property?

Answers: 1. Intensive. 2. Two. 3. 301 kPa. 4. Mechanical, thermal, phase and chemical. 5. Its value depends on the path between two states, not only on the end states.