State vs Path Variables

Core Concepts

In this article, we distinguish between State vs Path variables and functions, describing each type and explaining how to convert between them.

Topics Covered in Other Articles

Thermodynamic Variables: State vs Path

In chemistry, we like to use a variety of different variables and properties to describe thermodynamic systems. When we know quantities like pressure, temperature, enthalpy, and Gibbs free energy, we can accurately predict physical and chemical behavior. Physical chemists divide these variables into two categories based on their mathematical and conceptual properties. These two categories are state variables and path variables. To understand much of thermodynamics, it is important to compare and contrast state vs path variables.

State Variables

State variables are quantities that depend on the present state of a system. Any previous states or changes to the system do not affect state variables. Many of the most basic physical and chemical parameters of a system are state variables. Examples include temperature, pressure, volume, moles, and concentration. Many advanced thermodynamic parameters, like entropy, Gibbs free energy, kinetic energy, and internal energy, are also state variables.

When talking about specific substances, state variables can be extensive or intensive. Extensive state variables depend on how much you have that substance, such as mass or volume. Intensive state variables, by contrast, do not change based on the quantity of the substance, such as density or electronegativity. Tungsten, for instance, can occupy a variety of different masses or volumes, but its density and electronegativity remain unchangeable.

State Functions

To illustrate how (extrinsic) state variables change, physical chemists use state functions. These functions only depend on state variables and thermodynamic constants. Using the Ideal Gas Law, we can solve for pressure, generating the following state function for pressure:

    \begin{gather*} {P=\frac{nRT}{V}} \end{gather*}

Since state variables only depend on the present state, we only need the initial and final states to calculate change in a state function. Any intermediate states, or “paths”, between the two states do not factor into the calculation.

    \begin{gather*} {\Delta P=P_{final}-P_{initial}} \end{gather*}

Using this, state variables like entropy, enthalpy, and Gibbs free energy, can be calculated using Hess’s Law, which generates state functions by subtracting the sum of the product values minus those of the reactant values. These functions yield the change in that thermodynamic variable for the reaction.

    \begin{gather*} {\Delta H_{rxn}=\sum \Delta H_{f, products} - \sum \Delta H_{f,reactants}} \\ {\Delta S_{rxn}=\sum S_{products} - \sum S_{reactants}} \\ {\Delta G_{rxn}=\sum \Delta G_{f, products} - \sum \Delta G_{f,reactants}} \end{gather*}

List of Important State Variables

T – Temperature

mol – Moles

m – Mass

\rho – Density

V – Potential Energy

S – Entropy

H – Enthalpy

P – Pressure

V – Volume

\chi – Mole Fraction

MM – Molecular Weight

K – Kinetic Energy

G – Gibbs Free Energy

U – Internal Energy

Path Variables

Path variables, on the other hand, depend instead on the particular sequence that converted one state into another. Importantly, path variables don’t describe a particular thermodynamic state, but rather a process, such as a chemical reaction. In thermodynamics, heat and work are the most important path variables.

Path Functions

The classic context in which chemists use path variables involves the isothermal expansion and compression of gases. Specifically, when you have a piston full of gas, you can compress the gas, thus increasing its pressure. This is because compression reduces the volume while the temperature stays the same (as implied by the term “isothermal”). We know this from our state function for pressure.

    \begin{gather*} {P=\frac{nRT}{V}} \end{gather*}

This act of compression counts as performing work on the gas. You can calculate the work according to the following function:

    \begin{gather*} {w=-\int_{V_{i}}^{V_{f}}PdV} \end{gather*}

Work: One Step vs Two Steps

Our expression for work depends on the initial and final volume of the gas, as well as pressure. If you do the compression in one step, we consider the gas’s pressure as not changing. Chemists call this an “irreversible” compression. With constant pressure, you can integrate to generate the following formula:

    \begin{gather*} {w=-P\left(V_{f}-V_{i}\right)} \end{gather*}

However, if the compression happens in multiple separate steps, we end up with a different amount of work. Specifically, let’s say compress over two steps, reaching some intermediate volume (V_{int}) before reaching the final volume. The gas’s pressure reaches some equilibrium at that intermediate volume (P_{int}), affecting the second compression. Both of these steps are still irreversible since they happen (basically) instantly. 

    \begin{gather*} {w=-P_{i}\left(V_{int}-V_{i}\right)-P_{int}\left(V_{f}-V_{int}\right)} \end{gather*}

Since the P_{int} is higher than P_{i}, after a partial compression, we would expect less work is required to compress over two steps than one. Allowing the gas to reach equilibrium before further compression relieves much of the gas’s stress. If you carry out a compression over infinitely small steps, you reach the minimum work required to compress the gas.

All this is to emphasize that even though the compressed gas ends up in the same final state, the work required to make the compression changes based on the particular path taken.

State vs Path: Converting Between The Two Variables

carnot engine, showing the difference between state vs path variables
Diagram of heat engine, involving two path variables (work and heat) and a state variable (temperature)

In thermodynamics, state functions often involve path variables, and path functions often involve state variables. This may not make much sense at first, since state changes shouldn’t depend on the specifics of the path. In most cases, however, these relationships simply demonstrate that a limited number of paths are possible for a given state change. 

For instance, let’s look at the change in internal energy, a state variable. Due to the First Law of Thermodynamics, we know that change in internal energy depends on heat and work.

    \begin{gather*} {\Delta U =w+q} \end{gather*}

In this formula, we can witness that our work and heat path variables must then sum to change in internal energy. This places a thermodynamic limitation on our possible paths, given a certain change in internal energy.

Another important example relates to what we used before to calculate work:

    \begin{gather*} {w=-\int_{V_{i}}^{V_{f}}PdV} \end{gather*}

We see here that work then depends on two state variables. However, we can still end up with different values for work depending on how exactly the pressure changes. Work is not dependent simply on the initial and final pressures and volumes.

Finally, as you may observe in this article, entropy, a state variable, depends on heat:

    \begin{gather*} {\Delta S =\frac{q_{rev}}{T}} \end{gather*}

However, entropy depends specifically on the heat absorbed or emitted in the reversible path of a given state change. In this way, entropy depends on the heat of one specific path, making reversible heat a quasi-state function.

State vs Path Variables Practice Problems

Problem 1

Activation energy is the amount of energy needed to form a transition state to initiate a reaction. The use of catalysts allows the formation of alternate transition states with different activation energies without changing the product formed by the reaction. Based on this information, is activation energy therefore a path variable or a state variable?

Problem 2

You observe pure-carbon diamond crystals to have a certain length. Is length an intensive state variable or an extensive state variable?

State vs Path Variables Practice Problem Solutions

1: Activation energy is a path variable

2: Length is an extensive state variable