Entropy (classical thermodynamics)

In thermodynamics, entropy is a measure of how much of the energy of a system is potentially available to do work and how much of it is potentially manifest as heat. In classical thermodynamics, the entropy is defined only for a system in thermodynamic equilibrium. A thermodynamic system is any physical object or region of space that can be described by its thermodynamic quantities such as temperature, pressure, volume and density. In simple terms, the second law of thermodynamics states that for a system, the intensive thermodynamic quantities such as temperature, pressure, and chemical potential tend to become more uniform as time goes by, unless there is an outside influence which works to maintain the differences.

There are two related definitions of entropy. The first definition is the thermodynamic definition. It was developed by Rudolf Clausius and essentially describes how to measure the entropy of an isolated system in thermodynamic equilibrium. It makes no reference to the microscopic nature of matter. The second definition is the statistical definition developed by Ludwig Boltzmann in the 1870s. This definition describes the entropy as a measure of the number of possible microscopic configurations of the individual atoms and molecules of the system (microstates) which would give rise to the observed macroscopic state (macrostate) of the system. Boltzmann then went on to show that this definition of entropy was equal to the thermodynamic entropy to within a constant number which has since been known as Boltzmann's constant.

This article is concerned with the thermodynamic definition of entropy. Although thermodynamic entropy is a self-contained subject, it should be understood in parallel with the statistical definition. When the thermodynamic definition becomes most difficult to understand, the statistical definition brings a simple explanation, and where the link between the statistical theory and experiment becomes extended, the thermodynamic theory delivers a straightforward answer.

Clausius defined the change in entropy $dS$ of a thermodynamic system, during a reversible process, as

$dS = \frac{\delta Q}{T} \!$

where

δQ is a small amount of heat introduced reversibly to the system,

T is a constant absolute temperature

Note that the small amount δQ of energy transferred by heating is denoted by δQ rather than dQ, because Q is not a state function while the entropy is.

Clausius gave the quantity S the name "entropy", from the Greek word τρoπή, "transformation". Since this definition involves only differences in entropy, the entropy itself is only defined up to an arbitrary additive constant.

When a process is irreversible, the above definition must be replaced by the statement that the entropy change is equal to the amount of energy required to return the system to its original state by a reversible transformation at a constant temperature, divided by that temperature. This is explained in more detail below.

Introduction

In a thermodynamic system, modeled as a universe consisting of surroundings and systems and made up of quantities of matter, pressure differences, density differences, and temperature differences all tend to equalize over time. In the example of ice melting, the difference in temperature between a warm room, which is the surroundings, and a cold glass of water and ice, which is the system, is equalized as heat from the warm surroundings are transferred to the cooler ice and water mixture.

A thermodynamic model system

Over time the temperature of the glass and its contents and the temperature of the room achieve balance. The entropy of the room has decreased. However, as calculated in the example, the entropy of the system of ice and water has increased more than the entropy of the surrounding room has decreased. In an isolated system such as the room and ice water taken together, the dispersal of energy from warmer to cooler regions always results in a net increase in entropy. Thus, when the universe of the room and ice water system has reached a temperature equilibrium, the entropy change from the initial state is at a maximum. The entropy of the thermodynamic system is a measure of how far the equalization has progressed.

A special case of entropy increase, the entropy of mixing, occurs when two or more different substances are mixed. If the substances are at the same temperature and pressure, there will be no net exchange of heat or work - the entropy increase will be entirely due to the mixing of the different substances.

From a macroscopic perspective, in classical thermodynamics the entropy is interpreted simply as a state function of a thermodynamic system: that is, a property depending only on the current state of the system, independent of how that state came to be achieved. The state function has the important property that, when multiplied by a reference temperature, it can be understood as a measure of the amount of energy in a physical system that cannot be used to do thermodynamic work; i.e., work mediated by thermal energy. More precisely, in any process where the system gives up energy ΔE, and its entropy falls by ΔS, a quantity at least T_R ΔS of that energy must be given up to the system's surroundings as unusable heat (T_R is the temperature of the system's external surroundings). Otherwise the process will not go forward.

Definition

According to the Clausius equality, for a reversible process

$\oint \frac{\delta Q}{T}=0$

That means the line integral $\int_L \frac{\delta Q}{T}$ is path independent.

So we can define a state function S called entropy, which satisfied

$dS = \frac{\delta Q}{T} \!$

With this we can only obtain the difference of entropy by integrating the above formula. To obtain the absolute value, we need Third Law of Thermodynamics, which states that S=0 at absolute zero for perfect crystals.

Entropy change in irreversible transformations

See also: exergy

We now consider irreversible transformations. For any irreversible process, since entropy is a state function, we can always connect the initial and terminal status with an imaginary reversible process and integrating on that path to calculate the difference in entropy.

Now reverse the reversible process and combine it with the said irreversible process. Applying Clausius inequality on this loop,

$-\Delta S+\int\frac{\delta Q}{T}=\oint\frac{\delta Q}{T}< 0$

Thus,

$\Delta S \ge \int \frac{\delta Q}{T} \,\!$

where the equality holds if the transformation is reversible.

Notice that if $\delta Q=0$, then ΔS ≥ 0. This is the Second Law of Thermodynamics.

Suppose a system is thermally and mechanically isolated from the environment. For example, consider an insulating rigid box divided by a movable partition into two volumes, each filled with gas. If the pressure of one gas is higher, it will expand by moving the partition, thus performing work on the other gas. Also, if the gases are at different temperatures, heat can flow from one gas to the other provided the partition is an imperfect insulator. Our above result indicates that the entropy of the system as a whole will increase during these process (it could in principle remain constant, but this is unlikely.) Typically, there exists a maximum amount of entropy the system may possess under the circumstances. This entropy corresponds to a state of stable equilibrium, since a transformation to any other equilibrium state would cause the entropy to decrease, which is forbidden. Once the system reaches this maximum-entropy state, no part of the system can perform work on any other part. It is in this sense that entropy is a measure of the energy in a system that cannot be used to do work.

An irreversible process degrades the performance of a thermodynamic system designed to do work and results in entropy generation. Total entropy generation during a reversible process is zero. Thus entropy is a measure of the irreversibility and may be used to compare engineering processes and machines.

Heat engines

Figure 1: Heat engine diagram

Clausius' identification of S as a significant quantity was motivated by the study of reversible and irreversible thermodynamic transformations. A thermodynamic transformation is a change in a system's thermodynamic properties, such as temperature and volume. A transformation is reversible if it is a quasistatic process which means that it is infinitesimally close to thermodynamic equilibrium at all times. Otherwise, the transformation is irreversible. To illustrate this, consider a gas enclosed in a piston chamber, whose volume may be changed by moving the piston. If we move the piston slowly enough, the density of the gas is always homogeneous, so the transformation is reversible. If we move the piston quickly, pressure waves are created, so the gas is not in equilibrium, and the transformation is irreversible.

A heat engine is a thermodynamic system that can undergo a sequence of transformations which ultimately return it to its original state. Such a sequence is called a cyclic process, or simply a cycle. During some transformations, the engine may exchange energy with the environment. The net result of a cycle is (i) mechanical work done by the system (which can be positive or negative, the latter meaning that work is done on the engine), and (ii) heat energy transferred from one part of the environment to another. By the conservation of energy, the net energy lost by the environment is equal to the work done by the engine.

If every transformation in the cycle is reversible, the cycle is reversible, and it can be run in reverse, so that the energy transfers occur in the opposite direction and the amount of work done switches sign.

For the case of a heat engine working between two temperatures $T_H$ and $T_L$. The work produced by the engine on a cycle is given by the First law of thermodynamics

$|W| = |Q_H|-|Q_L|$

with the entropy production in the universe during the same time is given by the reduction of entropy in the hot source and the increase in the cold sink

$\Delta S = -\frac{|Q_H|}{T_H}+\frac{|Q_L|}{T_L}$

Combining these two relations

$|W| = \left(1-\frac{T_L}{T_H}\right)|Q_H| - T_L\,\Delta S$

The first is the maximum possible work for a heat engine, given by a reversible engine, as one operating along a Carnot cycle. Finally

$|W| = |W|_\mathrm{max} - T_L\,\Delta S$

This equation tells us that the production of work is reduced by the generation of entropy, being the term $T_L\,\Delta S$ the lost work by the machine.

Correspondingly, the amount of heat discarded to the cold sink is increased by the entropy generation

$|Q_L| = \frac{T_L}{T_H}|Q_H| + T_L\,\Delta S = |Q_L|_\mathrm{min}+T_L\,\Delta S$

Refrigerators

The same principle can be applied to a refrigerator. In this case the entropy production is

$\Delta S = \frac{|Q_H|}{T_H}-\frac{|Q_L|}{T_L}$

and the work to extract heat $|Q_L|$ from the cold source is

$|W| = \frac{|Q_L|}{T_H/T_L-1}+T_H\,\Delta S$

The first term is the minimum required work, which corresponds to a reversible refrigerator, so we have

$|W| = |W|_\mathrm{min}+T_H\,\Delta S$

i.e., the refrigerator compressor has to perform extra work to compensate for the energy waste due to entropy production.

Entropy measurement

In thermal experiments, it is quite difficult to measure the entropy of a system.^{[citation needed]} The techniques for doing so are based on the thermodynamic definition of the entropy, and require extremely careful calorimetry.

For simplicity, we will examine a mechanical system, whose thermodynamic state may be specified by its volume V and pressure P. In order to measure the entropy of a specific state, we must first measure the heat capacity at constant volume and at constant pressure (denoted C_V and C_P respectively), for a successive set of states intermediate between a reference state and the desired state. The heat capacities are related to the entropy S and the temperature T by

$C_X = T \left(\frac{\partial S}{\partial T}\right)_X \,\!$

where the X subscript refers to either constant volume or constant pressure. This may be integrated numerically to obtain a change in entropy:

$\Delta S = \int \frac{C_X}{T} dT \,\!$

We can thus obtain the entropy of any state (P,V) with respect to a reference state (P₀,V₀). The exact formula depends on our choice of intermediate states. For example, if the reference state has the same pressure as the final state,

$S(P,V) = S(P, V_0) + \int^{T(P,V)}_{T(P,V_0)} \frac{C_P(P,V(T,P))}{T} dT \,\!$

In addition, if the path between the reference and final states lies across any first order phase transition, the latent heat associated with the transition must be taken into account.

The entropy of the reference state must be determined independently. Ideally, one chooses a reference state at an extremely high temperature, at which the system exists as a gas. The entropy in such a state would be that of a classical ideal gas plus contributions from molecular rotations and vibrations, which may be determined spectroscopically. Choosing a low temperature reference state is sometimes problematic since the entropy at low temperatures may behave in unexpected ways. For instance, a calculation of the entropy of ice by the latter method, assuming no entropy at zero temperature, falls short of the value obtained with a high-temperature reference state by 3.41 J/(mol·K). This is due to the "zero-point" entropy of ice mentioned earlier.

See also

• Entropy
• Enthalpy
• Fundamental thermodynamic relation
• Thermodynamic free energy
• History of entropy
• Entropy (statistical views)

Further reading

• Goldstein, Martin, and Inge F., 1993. The Refrigerator and the Universe. Harvard Univ. Press. A gentle introduction at a lower level than this entry.

External links

• Thermodynamic entropy