Thermodynamics · Chapter 1 of 10 · Beginner
Basic Concepts of Thermodynamics
Before any energy is counted, you draw a boundary and decide what is inside it. That single choice, plus a clear language of properties, temperature, and pressure, is the foundation everything else is built on.
Readiness check
This opening chapter needs basic physics and unit sense. Tick only what you can do closed-notes.
- Recall mass, force, pressure, and temperature with their SI units.
- Convert between kPa and Pa, and between °C and kelvin.
- Compute a force from a pressure and an area.
- Distinguish a quantity you measure at an instant from one that builds up over a process.
- Read a simple diagram of a piston-cylinder or a tank.
The core idea
Thermodynamics begins by choosing a system, the matter inside a boundary, and describing its condition with a few properties. Energy crosses the boundary only as heat or work.
system + surroundings = everythingT(K) = T(°C) + 273.15Pabs = Patm + PgaugeA closed system holds a fixed amount of mass; an open system (a control volume) lets mass flow through it. The condition of the system at an instant is its state, fixed by properties such as temperature, pressure, and specific volume. A property does not depend on how the system got there, which is what separates it from heat and work, the two ways energy crosses the boundary. Get the boundary and the language right and the laws in later chapters become bookkeeping.
The skills, taught in order
This chapter fixes the vocabulary used in every later one. Five skills define the system, describe its state, and set up temperature and pressure with consistent units.
1.1 System, surroundings, and boundary
A system is the matter or region you choose to study; everything else is the surroundings, and the surface between them is the boundary. A closed system (control mass) keeps the same mass and only energy crosses the boundary. An open system (control volume) is a region, such as a turbine, that mass flows through. An isolated system exchanges neither mass nor energy. Choosing the boundary well is half of solving the problem.
1.2 Property, state, and equilibrium
A property is any characteristic of the system: pressure, temperature, volume, density. The state is the full set of property values at one instant. Properties are either intensive (independent of size, such as T and P) or extensive (proportional to mass, such as volume and total energy). Dividing an extensive property by mass gives a specific property, written lower-case, such as specific volume v = V/m. A system in equilibrium has no unbalanced driving potential, so its properties are uniform and unchanging.
1.3 Process, path, and cycle
A process is any change of state; the series of states it passes through is the path. Some processes hold one property fixed: isothermal (constant T), isobaric (constant P), isochoric (constant v), or adiabatic (no heat transfer). A cycle returns the system to its starting state, so every property comes back to its initial value. This is why cycles are the backbone of engines and refrigerators.
1.4 Temperature and the zeroth law
Temperature measures the tendency of a system to give up thermal energy. The zeroth law says that if two bodies are each in thermal equilibrium with a third, they are in equilibrium with each other, which is what lets a thermometer work. Always carry absolute temperature in kelvin for ratios and gas laws: T(K) = T(°C) + 273.15.
1.5 Pressure, absolute and gauge
Pressure is normal force per unit area, P = F/A, in pascals (1 Pa = 1 N/m²). A gauge reads the difference from local atmospheric pressure, so absolute pressure is Pabs = Patm + Pgauge; below atmospheric the difference is a vacuum reading. In a static fluid column the pressure changes with depth by ΔP = ρgh, the principle behind every manometer.
| Quantity | Symbol | SI unit |
|---|---|---|
| Pressure | P | Pa = N/m² |
| Temperature | T | K (or °C) |
| Specific volume | v | m³/kg |
| Density | ρ | kg/m³ |
| Energy | E | J (kJ) |
Engineering connection: every energy balance in this course starts by naming the system and its state in these units, then tracking heat and work across the boundary.
Worked example 1: absolute pressure from a manometer
A gas tank is connected to an open mercury manometer. The mercury on the open side stands 50 cm higher than on the tank side. With atmospheric pressure 101.3 kPa and mercury density 13 600 kg/m³, find the gauge and absolute pressure of the gas.
- ProblemFind the gauge and absolute pressure of the gas behind the manometer in Figure 1.
- Given / findh = 0.50 m, ρ = 13 600 kg/m³, g = 9.81 m/s², Patm = 101.3 kPa. Find Pgauge and Pabs.
- AssumptionsMercury is static and incompressible; the gas column weight is negligible.
- ModelThe column height gives a gauge pressure ΔP = ρgh; absolute pressure adds the atmosphere.
- EquationsPgauge = ρghPabs = Patm + Pgauge
- SolvePgauge = 13 600 × 9.81 × 0.50 = 66 708 Pa = 66.7 kPa. Then Pabs = 101.3 + 66.7 = 168.0 kPa.
- CheckUnits: (kg/m³)(m/s²)(m) = kg/(m·s²) = Pa, as required. The gas is above atmospheric, which matches the mercury standing higher on the open side.
- ConclusionThe manometer reads gauge pressure directly; adding the local atmosphere converts it to the absolute pressure that the gas laws and steam tables require.
Worked example 2: density and mass from the state
Air fills a 240 m³ classroom at 100 kPa and 25 °C. Treating air as an ideal gas with R = 0.287 kJ/kg·K, find its density, its specific volume, and the total mass of air in the room.
- ProblemFind the density, specific volume, and total mass of the air in Figure 2.
- Given / findP = 100 kPa, T = 25 °C, V = 240 m³, R = 0.287 kJ/kg·K. Find ρ, v, and m.
- AssumptionsAir behaves as an ideal gas at room conditions; properties are uniform.
- ModelUse Pv = RT in the form ρ = P/(RT), with T in kelvin.
- EquationsT = 25 + 273.15 = 298.15 Kρ = P/(RT)m = ρV
- Solveρ = 100/(0.287 × 298.15) = 100/85.57 = 1.169 kg/m³. Then v = 1/ρ = 0.856 m³/kg, and m = 1.169 × 240 = 280 kg.
- CheckThe density matches the tabulated value for air at 25 °C and 100 kPa (1.169 kg/m³). Nearly 300 kg of air in an ordinary room is a useful reminder that gases are not massless.
- ConclusionTwo properties fix the state of an ideal gas; everything else, including mass, follows. Keeping T in kelvin is essential.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Confusing heat and temperature | Treating temperature as a stored amount of heat | "Is this a state I can point to, or a transfer across the boundary?" | Temperature is a property; heat is energy crossing the boundary. |
| Using gauge pressure in a gas law | Densities and masses come out wrong | "Is this pressure measured from zero or from the atmosphere?" | Convert to absolute pressure before using Pv = RT or a table. |
| Leaving temperature in °C | Ratios and gas laws are off | "Did I add 273.15?" | Use kelvin for any ratio, gas law, or efficiency. |
| Vague system boundary | Unsure what crosses in and out | "What exactly is inside my boundary?" | Draw the boundary first and label every crossing. |
Practice ladder
A pressure gauge on a tank reads 250 kPa where the atmosphere is 98 kPa. What is the absolute pressure inside?
Show answer
Pabs = Patm + Pgauge = 98 + 250 = 348 kPa. Gauges read the difference from atmosphere, so the absolute value is always higher for a pressurised tank.
The same classroom air (Worked Example 2) is warmed to 35 °C at the same 100 kPa. Does the mass of air in the room go up or down, and by roughly what fraction?
Show answer
Density falls as temperature rises: ρ scales with 1/T. The ratio is 298.15/308.15 = 0.968, so density and mass drop about 3.2 percent. Warm air is less dense, so some air leaves the room.
Classify each as intensive or extensive: pressure, total internal energy, density, volume, specific volume. Then state which become specific properties when divided by mass.
Show answer
Intensive: pressure, density, specific volume. Extensive: total internal energy, volume. Dividing an extensive property by mass gives a specific (intensive) property, such as v = V/m or u = U/m.
Pick a real device (a bicycle pump, a pressure cooker, a car tyre). Draw its system boundary, state whether it is open or closed, and name the two properties you would measure to fix its state.
What good work looks like
A clear boundary, an open or closed classification with a reason, and two independent intensive properties (usually P and T) named as the state.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take one everyday device, draw its system boundary, classify it, and list the properties that fix its state. Note one quantity that crosses the boundary as heat and one as work.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What is the difference between a closed and an open system?
A closed system keeps a fixed mass and exchanges only energy; an open system (control volume) also lets mass flow across its boundary.
2. What makes a quantity a property?
Its value depends only on the current state, not on the path taken to reach it. Heat and work are not properties.
3. State the zeroth law.
If two bodies are each in thermal equilibrium with a third, they are in equilibrium with each other, which is the basis of temperature measurement.
4. Relate absolute and gauge pressure.
Pabs = Patm + Pgauge; a vacuum reading is a negative gauge pressure.
5. Why use kelvin?
Ratios, gas laws, and efficiencies need an absolute scale; only kelvin starts at absolute zero.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 2 (Some Concepts and Definitions) |
| Cross-reference | Çengel and Boles, Ch. 1 · Moran, Ch. 1 |
| Core topics | 1.1 System and boundary · 1.2 Property and state · 1.3 Process and cycle · 1.4 Temperature and zeroth law · 1.5 Pressure |
| Engineering connection | The system boundary and the language of state set up every energy balance that follows. |
| Read next | Chapter 2: Energy, Heat, and Work. |