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Thermodynamics · Lesson 1 of 7

Thermal Equilibrium and Thermodynamic States

“Use the zeroth law to understand temperature and describe equilibrium with measurable state variables.”

Learning Objectives

• Distinguish a macroscopic thermodynamic description from the motion of a body as a whole. • Explain how insulating and conducting boundaries affect thermal equilibrium. • Apply the zeroth law to temperature measurement. • Use the ideal-gas equation and distinguish intensive from extensive variables. • Explain why rapid expansion cannot be represented by a sequence of equilibrium states.

What does thermodynamics describe?

Rubbing your palms makes them warmer, while a steam engine uses energy supplied through heating to move a piston. Thermodynamics studies thermal states and the conversion and transfer of energy through heat and work. Its starting point is the behaviour of a whole sample: a litre of gas, a metal block or water in a vessel. It does not require the position and velocity of every molecule. Pressure P, volume V, absolute temperature T, amount of substance n, and composition are enough to describe many equilibrium situations.

Mechanics asks how forces change a body’s motion. Thermodynamics asks how its internal state changes. A moving bullet has kinetic energy because its centre of mass moves; that fact alone does not make it hotter. On impact, some of that organised kinetic energy becomes internal energy of the bullet and wood, and their temperatures may rise. The two descriptions can cooperate, but centre-of-mass speed and temperature are different quantities.

A microscopic description would track the huge number of molecules and their motions. Kinetic theory instead uses statistical information about molecular motion. Thermodynamics works with directly measurable bulk variables. Molecules continue moving even when the bulk variables are steady: thermodynamic equilibrium does not mean every molecule is motionless or mechanically balanced.

Systems, surroundings and thermal contact

Choose the matter you want to study and call it the system; everything outside it is its surroundings. A boundary separates them. Here the usual system is a fixed amount of gas in a cylinder. A boundary may conduct energy as heat, insulate against heat transfer, or move so that work can cross it. Specifying the boundary is essential: “insulated” tells us about heat transfer, not automatically about work.

An adiabatic wall is a thermal insulator: it prevents heat transfer across it. A diathermic wall conducts energy as heat when there is a temperature difference. Two gas containers separated by a fixed insulating wall can remain at different temperatures. Replacing that wall by a conducting one allows energy to pass from the hotter sample to the colder one until thermal equilibrium is reached. At equilibrium there is no net heat transfer between the samples.

Thermal boundaries: fixed volumesABHot initiallyCold initiallyInsulating partitionNo heat crossesABHot initiallyCold initiallyConducting partitionHeat: hot → coldZeroth law: compare A and B through CC: thermometerA: TA = TCB: TB = TCTherefore TA = TB
Thermal boundaries and the zeroth law— The top panels compare fixed insulating and conducting partitions. The lower panel shows a third system C acting as the thermal comparator. Schematic, not to scale.

Thermal equilibrium means equal temperatures; it does not require equal pressures, volumes, masses or internal energies. Constraints decide which other variables can change. If both vessels are rigid, their volumes stay fixed while their pressures can change during heating and cooling. A movable conducting partition can also permit volume changes. A movable insulating partition can transmit work, so insulation alone does not guarantee that the states on either side remain unchanged.

The zeroth law gives temperature its meaning

Place A and B separately in thermal contact with a third system C, while keeping A and B thermally isolated from one another. Wait until each is in thermal equilibrium with C. Now thermally isolate C and connect A and B by a conducting boundary. No further net heat transfer occurs. This observation is the zeroth law: systems separately in thermal equilibrium with the same third system are in thermal equilibrium with one another.

Equality of temperatureLaTeX
Temperature labels thermal-equilibrium states. Equality of temperature, rather than equality of heat or internal energy, is the criterion.

A thermometer plays the role of C. A calibrated property, such as the length of a liquid column or the pressure of a fixed-volume gas, changes with temperature. Allowing the thermometer to reach thermal equilibrium with a body lets us assign the same temperature to both. A small thermometer limits disturbance of the body. Calibration supplies numerical values; the zeroth law explains why comparing those values is physically meaningful. Fowler named the law in 1931, after the first and second laws were already established, so it received the earlier number “zeroth”.

A thermometer is a comparator

Problem
A thermometer gives the same stable reading when placed first in A and then in B. Must A and B have equal internal energies?

  1. 1.The stable reading means that the thermometer reaches thermal equilibrium separately with A and with B.
  2. 2.The zeroth law therefore gives TA = TB. If A and B are brought into thermal contact without other changes, no net heat flows because of a temperature difference.
  3. 3.Equal internal energies do not follow. A large bucket of water and a small cup can have the same temperature while storing different total internal energies.

Equilibrium states and an equation of state

An equilibrium state has macroscopic variables that remain constant under the stated constraints. Full thermodynamic equilibrium also excludes unbalanced internal gradients that drive change. A steady flow maintained by external heating and cooling is not automatically equilibrium. Mechanical equilibrium of a body means zero net force and torque; thermal equilibrium is about the absence of a temperature-driven energy transfer. These are related conditions, not interchangeable definitions.

For a fixed amount and composition of a simple ideal gas, P, V and T are linked. Choosing any two independent variables fixes the third. Pressure is measured in pascals, volume in cubic metres, T in kelvin, and n in moles. The constant R is about 8.314 J mol⁻¹ K⁻¹. Some books use the symbol μ instead of n for the number of moles; both describe the same quantity here.

Ideal-gas equation of stateLaTeX
This model applies to an ideal gas; real gases can deviate, especially at high density or near liquefaction. Use absolute pressure and kelvin temperature.
Only two independent variables

Problem
A fixed sample has P = 1.00×10⁵ Pa, V = 2.00×10⁻³ m³ and T = 300 K. Find n. Then find its pressure at 330 K if the vessel remains rigid.

  1. 1.First n = PV/(RT) = (1.00×10⁵)(2.00×10⁻³)/(8.314×300) ≈ 0.0802 mol.
  2. 2.For the same n and V, P/T = nR/V is constant, so P₂ = P₁(T₂/T₁) = 1.10×10⁵ Pa.
  3. 3.The pressure rises because the state changes; the volume cannot expand. Using 30/27 or Celsius ratios would be incorrect.

At fixed n, the equation PV = nRT defines a surface in P–V–T space. An equilibrium state corresponds to a point on it. An isotherm is the curve cut from that surface at a fixed T; on a P–V plot it satisfies P = nRT/V. During a sudden free expansion, pressure and temperature need not be uniform enough for one pair of values to describe the whole gas. The initial and final equilibrium states lie on the surface; the transient expansion is not a path of equilibrium points on it. An exploding gas mixture is another case where a single bulk P and T may fail during the change.

Intensive and extensive variables

Imagine dividing a uniform equilibrium sample into two equal parts without changing its local state. Each part has half the volume, mass, mole number and internal energy: these are extensive quantities. Each has the original pressure, temperature and density: these are intensive quantities. Density is intensive because both mass and volume halve together, leaving their ratio unchanged.

VariableTypeWhen a uniform sample is halved
P, TIntensiveRemain the same
ρ = m/VIntensiveRemains the same
V, m, n, UExtensiveEach becomes half its original value

Scaling checks complement unit checks. In PV = nRT, P and T are intensive, while V and n scale with sample size, so both sides scale equally. Heat and work for corresponding processes also scale with amount, but that does not make them state variables: they describe energy transferred during a path. Internal energy U does belong to the state.

Thermodynamics also uses quantities such as entropy and enthalpy. They are macroscopic state functions, but they are not additional independent coordinates that every simple gas requires. Entropy has a precise thermodynamic definition beyond this chapter; “disorder” is only a limited intuition. Enthalpy is defined as H = U + PV, rather than a universal quantity of heat contained inside matter. The present chapter can be developed using P, V, T and U without adding a separate theory of these quantities.

Quick Quiz

Quiz

Quick check

A and B are separately in thermal equilibrium with C. Which conclusion follows?

Explanation

The zeroth law equates temperatures; it does not equate other properties.

Quick check

Two rigid gas vessels have different temperatures and are separated by an insulating fixed wall. What does the insulation guarantee?

Explanation

An adiabatic wall blocks heat transfer. Rigidity additionally prevents boundary-volume work.

Quick check

Which pair consists only of extensive state variables?

Explanation

V and U halve when a uniform sample is divided into equal parts.

Quick check

A fixed ideal-gas sample at 300 K is heated to 360 K at fixed volume. Its pressure becomes

Explanation

At fixed n and V, P is proportional to absolute T, so P₂/P₁ = 360/300 = 1.20.

Quick check

Why is a rapid free expansion not generally a curve of equilibrium states?

Explanation

Transient gradients prevent a unique equilibrium description. The final settled state can obey the equation of state.

Practice Problems

Practice Problems
  1. Explain the different questions answered by mechanics and thermodynamics for a bullet before and after it strikes wood. Distinguish bulk kinetic energy from internal energy.
  2. Two rigid gas vessels start at 280 K and 340 K. Describe what happens when an insulating partition is replaced by a conducting one. State which equality is required at final thermal equilibrium and which geometrical quantity cannot change.
  3. Describe how a third system C can establish that A and B have equal temperatures without initially bringing A and B into direct thermal contact.
  4. A fixed amount of ideal gas has pressure 1.20×10⁵ Pa at 300 K in a rigid container. Find its pressure at 350 K. Explain why Celsius ratios cannot be used.
  5. A uniform gas sample is divided into four equal portions at the same state. For P, V, T, mass, density and internal energy, state the value of each portion relative to the original and classify each variable.
  6. A partition is suddenly removed between gas and vacuum. Explain why the initial and final equilibrium states can satisfy PV = nRT even though the intermediate motion is not a well-defined equilibrium path.

Key Takeaways

Key Takeaways

• Thermodynamics describes bulk internal states using macroscopic variables. • Thermal equilibrium requires equal temperatures; the zeroth law makes temperature comparison possible. • Boundary conditions determine whether heat, work and volume changes are possible. • For fixed n in an ideal gas, PV = nRT leaves two independent state variables. • P, T and density are intensive; V, mass, n and U are extensive.