State Variables, Equilibrium, and the Zeroth Law
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You should be comfortable with basic calculus (partial derivatives) and have a general intuition for what temperature and pressure represent physically, even if you have not yet defined them rigorously.
Why this matters
Before we can calculate anything, we must agree on what we are measuring and when the system is ready to be measured.
Thermodynamics is a theory of equilibrium. It does not describe how a gas rushes into a vacuum or how a cup of coffee cools down minute by minute; it describes the stable states where those processes have finished. To navigate the landscape of these stable states, we need a precise vocabulary: state variables that depend only on the current condition, not the history, and a logical foundation (the Zeroth Law) that justifies the very concept of temperature. This lecture establishes the ground rules for the entire unit.
The Idea: Equilibrium and the State
Imagine a sealed, insulated box containing a gas. If you shake it, the molecules move chaotically. But if you leave it alone for a long time, the macroscopic properties—pressure, volume, and temperature—stop changing. This is thermodynamic equilibrium. The key insight is that the state of the system is fully described by a small set of macroscopic variables, regardless of how it got there. If you compress the gas slowly and then let it expand back to its original volume and pressure, it returns to the exact same state. This path-independence is what defines a "state function." We do not need to track the position of every molecule (which would be impossible); we only need to know the collective averages. The Zeroth Law provides the logical bridge that allows us to assign a single number, temperature, to these states, ensuring that if system A is in equilibrium with B, and B with C, then A and C are in equilibrium with each other. This transitivity is the foundation of thermometry.
State Variables vs. Path Functions
We distinguish between properties that define the state and those that describe the process. State variables, such as pressure \( P \), volume \( V \), and temperature \( T \), are point functions. Their change depends only on the initial and final states. In contrast, work \( W \) and heat \( Q \) are path functions. The amount of work done by a gas expanding depends on the specific path taken in the \( P\text{-}V \) diagram, not just the endpoints.
Mathematically, a state function \( X \) has an exact differential \( \mathrm{d}X \). For a simple compressible system, the state is often fixed by two independent variables. If we choose \( P \) and \( V \), then \( T \) is determined by the equation of state. This is the principle of state postulates: for a simple system, the number of independent intensive variables required to fix the state is two.
Here, \( U \) is internal energy. Because \( U \) is a state function, this differential is exact. This means the order of partial differentiation does not matter, a property we will exploit extensively in later lectures to derive Maxwell relations.
The Zeroth Law and Thermometry
The Zeroth Law is named "Zeroth" because it was formulated after the First and Second Laws but is logically prior to them. It states: If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
This law justifies the use of thermometers. A thermometer is a third system. If it reads the same value for two different systems, those systems are at the same temperature. Without this transitivity, temperature would not be a well-defined property of the state. It allows us to define an empirical temperature scale. While we will later define temperature statistically via entropy, the Zeroth Law ensures that this statistical definition corresponds to the operational measurement we perform in the lab.
Thermal equilibrium implies that no net heat flows between systems. This is a static condition. It is distinct from mechanical equilibrium (no net force) and chemical equilibrium (no net reaction). A system in full thermodynamic equilibrium satisfies all three simultaneously.
Equations of State
An equation of state (EOS) relates the state variables. For an ideal gas, this is the familiar \( PV = nRT \). However, real gases deviate from this behaviour. The EOS is a constraint that reduces the degrees of freedom. If we know \( P \) and \( V \), the EOS tells us \( T \). If we know \( P \) and \( T \), it tells us \( V \).
For a general substance, the EOS can be complex. It might be expressed as \( f(P, V, T) = 0 \). The specific form depends on the intermolecular forces. In this unit, we will often assume ideal gas behaviour for simplicity, but we must always remember that the EOS is an empirical input, not a consequence of the First Law alone. The First Law tells us energy is conserved; the EOS tells us how the matter behaves under compression and heating.
This relation is valid only in the limit of low density and high temperature. At high pressures, we must use more complex forms like the van der Waals equation, which accounts for molecular volume and attraction. The choice of EOS determines the specific values of work and heat for a given process.
A common misconception is that "equilibrium" means "no motion." Molecules are still moving rapidly in thermal equilibrium. Equilibrium is a statistical steady state where macroscopic fluxes (like heat flow or mass diffusion) are zero, but the microscopic motion continues unabated. Confusing static equilibrium with dynamic equilibrium leads to errors in calculating entropy changes, as entropy is a measure of the number of accessible microstates, which remains high even in equilibrium. C
Worked example
Problem. A cylinder contains 2.0 mol of an ideal gas at \(1.0 \times 10^5\ \mathrm{Pa}\) and \(0.040\ \mathrm{m^3}\). It is compressed slowly and isothermally to \(0.020\ \mathrm{m^3}\). Find the temperature, the final pressure, and the work done on the gas.
- Temperature from the equation of state: \(T = \dfrac{P_1V_1}{nR} = \dfrac{1.0 \times 10^5 \times 0.040}{2.0 \times 8.314} = 241\ \mathrm{K}\).
- Isothermal, so \(P_1V_1 = P_2V_2\): \(P_2 = 1.0 \times 10^5 \times \dfrac{0.040}{0.020} = 2.0 \times 10^5\ \mathrm{Pa}\).
- Work done by the gas: \(W_{\mathrm{by}} = \displaystyle\int_{V_1}^{V_2} P\,\mathrm{d}V = nRT\ln\frac{V_2}{V_1} = P_1V_1\ln\frac{V_2}{V_1}\).
- Substitute: \(W_{\mathrm{by}} = 4000\ \mathrm{J} \times \ln(0.5) = 4000 \times (-0.693) = -2.77 \times 10^{3}\ \mathrm{J}\).
- Work done on the gas: \(W_{\mathrm{on}} = -W_{\mathrm{by}} = +2.77 \times 10^{3}\ \mathrm{J}\). Since \(\Delta U = 0\) for an isothermal ideal gas, the same 2.77 kJ leaves as heat.
Units check. \(\mathrm{Pa} \times \mathrm{m^3} = \mathrm{N\,m^{-2}} \times \mathrm{m^3} = \mathrm{J}\); the logarithm is dimensionless.
Three quick tests
- Is heat a state function?
Answer
No, heat is a path function. It depends on the process, not just the initial and final states. - If system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, what can we say about A and C?
Answer
They are in thermal equilibrium with each other (Zeroth Law). - For a simple compressible system, how many independent intensive variables are needed to fix the state?
Answer
Two (e.g., P and V, or T and P).
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.