1.4 Thermodynamic equilibrium and processes

Thermodynamic Equilibrium

Thermodynamic equilibrium is the state where a system's macroscopic properties stop changing over time, with no net exchange of energy or matter with the surroundings. Understanding this state is essential because thermodynamic laws and equations only apply rigorously when a system is in equilibrium (or very close to it). From here, you can analyze how systems move between equilibrium states through different types of processes.

Concept of Thermodynamic Equilibrium

A system reaches thermodynamic equilibrium when its measurable properties (temperature, pressure, volume, etc.) remain constant over time and there are no net flows of energy or matter between the system and its surroundings.

For full thermodynamic equilibrium, three conditions must be satisfied simultaneously:

• Thermal equilibrium: Temperature is uniform throughout the system. No net heat flow occurs between different parts of the system or between the system and surroundings.
• Mechanical equilibrium: Pressure is uniform throughout the system. No unbalanced forces exist, so there's no bulk movement of matter.
• Chemical equilibrium: Chemical potential is uniform for each species in the system. No net chemical reactions or diffusion occur.

If even one of these isn't met, the system is not in thermodynamic equilibrium. For example, a cup of hot coffee in a cool room has thermal non-equilibrium: heat flows from the coffee to the air until both reach the same temperature.

Once a system is in equilibrium, you can apply equations like the ideal gas law ($PV = nRT$) and the first law of thermodynamics to calculate its properties and predict its behavior.

Types of Thermodynamic Processes

A thermodynamic process describes how a system transitions from one equilibrium state to another. Each type of process holds a specific variable constant, which simplifies the analysis.

• Isothermal process (constant temperature): The system exchanges heat with its surroundings to keep $T$ constant. For an ideal gas, internal energy depends only on temperature, so $\Delta U = 0$ and all heat absorbed equals work done: $Q = W$. A common example is the slow expansion of a gas in thermal contact with a large heat reservoir.
• Isobaric process (constant pressure): Pressure stays fixed while volume and temperature change. Work is straightforward to calculate: $W = P \Delta V$. Heating a gas in a cylinder fitted with a freely moving piston is a typical example.
• Isochoric (isovolumetric) process (constant volume): Because the volume doesn't change, no pressure-volume work is done ($W = 0$). All heat added goes directly into changing the internal energy: $Q = \Delta U$. Heating a gas in a rigid, sealed container illustrates this.
• Adiabatic process (no heat exchange): The system is thermally insulated from its surroundings, so $Q = 0$. Any work done changes the internal energy and therefore the temperature: $\Delta U = -W$. Rapid processes often approximate adiabatic conditions because there isn't enough time for significant heat transfer. The compression stroke in an internal combustion engine is a classic example.

Analysis of Quasi-Static Processes

A quasi-static process proceeds slowly enough that the system stays infinitesimally close to equilibrium at every instant. This matters because it means you can use equilibrium equations (like $PV = nRT$) at each step along the way, and the process can be represented as a continuous path on a P-V diagram.

A reversible process is a quasi-static process that also has no dissipative effects such as friction, viscosity, or unrestrained expansion. Every reversible process is quasi-static, but not every quasi-static process is reversible (a quasi-static process with friction, for instance, is irreversible).

Problem-solving approach for quasi-static processes:

1. Identify the type of process (isothermal, isobaric, isochoric, or adiabatic) based on which variable is held constant or which constraint applies.

2. Apply the appropriate thermodynamic equations:

   • Ideal gas law: $PV = nRT$
   • First law of thermodynamics: $\Delta U = Q - W$
   • Work done by the system: $W = \int P \, dV$

3. Use the constraint of the specific process to simplify. For example, in an isochoric process $dV = 0$, so $W = 0$ and the first law reduces to $\Delta U = Q$.

4. Solve for the unknown quantities using the given information.

Reversibility vs. Irreversibility in Thermodynamics

• A reversible process can be reversed so that both the system and surroundings return exactly to their original states. It requires the process to be quasi-static with no dissipative effects. Reversible processes are an idealization: real processes can approach them but never fully achieve them. The Carnot cycle is the most well-known example of a fully reversible cycle.
• An irreversible process leaves some permanent change in the system, the surroundings, or both when reversed. Sources of irreversibility include friction, viscosity, unrestrained expansion, and heat transfer across a finite temperature difference. All real-world processes are irreversible to some degree.

Why does this distinction matter? Reversible processes set the upper bound on efficiency. A Carnot engine operating between two temperatures achieves the maximum possible efficiency for any heat engine between those temperatures. Real engines, which involve irreversibilities, always fall short of this limit.

The second law of thermodynamics formalizes this: for any irreversible process in an isolated system, the total entropy increases ($\Delta S > 0$). For a reversible process, entropy stays constant ($\Delta S = 0$). Entropy never decreases in an isolated system, which is why irreversible processes can't simply be "undone."