Internal energy

In thermodynamics, a system is any object, any quantity of matter, any region, etc. selected for study and mentally set apart from everything else which is then called its surroundings. The imaginary envelope enclosing the system and separating it from its surroundings is called the boundary of the system. In this article the boundaries will be referred to as the walls of the system.

The internal energy of a system is simply its energy. The adjective "internal" refers to the fact that some energy contributions are not considered. For instance, when the total system is in uniform motion, it has kinetic energy. This overall kinetic energy is never seen as part of the internal energy; one could call it external energy. Or, if the system is at constant non-zero height above the surface the Earth, it has constant potential energy in the gravitational field of the Earth. Gravitational energy is only taken into account when it plays a role in the phenomenon of interest, for instance in a colloidal suspension, where the gravitation influences the up- downward motion of the small particles comprising the colloid. In all other cases, gravitational energy is assumed not to contribute to the internal energy; one may call it again external energy.

On the other hand, a contribution to the internal energy that is always included is the kinetic energy of the atoms or molecules constituting the system. In an atomic gas, it is the energy associated with translations of the atoms, while in a molecular gas molecular rotations contribute to the internal energy as well. In a solid, the internal energy acquires contributions from vibrations, among other effects. Except for ideal gases, the averaged potential energy of molecules in the field of the other molecules (see intermolecular forces) is also an important component of the internal energy.

In general, the energies that are not changing in the processes of interest are left out of the definition of internal energy. For instance, when a system consists of a vessel filled with water and the process of interest is evaporation (formation of steam), the kinetic energy of the water molecules and the interaction between them are included in the internal energy. As long as no chemical bonds are broken, the energies contained in these bonds are not included. If the temperatures are not too high, say below 200 to 300 °C, the intramolecular vibrational energies are ignored as well. Chemists and engineers never include relativistic contributions, of the type E = mc², or nuclear contributions (say the fusion energy of protons with oxygen-nuclei). However, a plasma physicist studying the thermodynamics of fusion reactions will include nuclear energy in the internal energy of a plasma.

First law of thermodynamics

Classical (phenomenological) thermodynamics is not concerned with the nature of the internal energy, it simply postulates that it exists and may be changed by certain processes. Further it is postulated that internal energy, usually denoted by either U or E, is a state function, that is, its value depends upon the state of the system and not upon the nature or history of the past processes by which the system attained its state. In addition, the internal energy, which henceforth will be written as U, is assumed to be a differentiable function of the independent variables that uniquely specify the state of the system. An example of such a state variable is the volume V of the system.

When the system has thermally conducting walls, an amount of heat DQ can go through the wall in either direction: if DQ > 0, heat enters the system and if DQ < 0 the system loses heat to its surroundings. The symbol DQ indicates simply a small amount of heat, and not a differential of Q. Note that, because it is not a function, Q does not have a differential. The internal energy of the system changes by dU as a consequence of the heat flow, and it is postulated that

$dU=DQ\,$

By convention DQ is the heat absorbed by the system, i.e., the heat it receives from its surroundings. The symbol dU indicates a differential of the differentiable function U.

Most thermodynamic systems are such that work can be performed on them or by them. When a small amount of work DW is performed by the system, the internal energy decreases,

$dU=-DW\,$

By convention DW is the work by the system on its surroundings, which gives the minus sign in this equation.

As an example of work, we consider as a system a volume V containing gas of pressure p. A small amount of work pdV is performed on the system by reversibly (quasi-statically) compressing the gas (dV < 0). The sign convention of DW is such that DW and dV have the same sign

$DW=pdV\,\quad dV<0,\quad DW<0$

Due to the fact that the work is performed reversibly, the small amount of work DW is proportional to the differential dV. If dV > 0 (expansion), work DW > 0 is performed by the system. Hence the change in internal energy obtains indeed a minus sign:

$dU=-DW=-pdV\,$

Note that other forms of work than pdV are possible. For instance, DW = −HdM, the product of an external magnetic field H with a small change in molar magnetization dM, is a change in internal energy caused by an alignment of the microscopic magnetic moments that constitute a magnetizable material.

An important form of doing work is the reversible addition of substance,

$DW=-\mu dn,\,$

here μ (a function of thermodynamic parameters as T, p, etc.) is the chemical potential of the pure substance added to the system. The infinitesimal quantity dn is the amount (expressed in moles) of substance added. The chemical potential μ is the amount of energy that the system gains when reversibly, adiabatically (DQ = 0), and isochorically (dV = 0) a mole of substance is added to it.

When a small amount of heat DQ flows in or out the system and simultaneously a small amount of work DW is done by or on the system, the first law of thermodynamics states that the internal energy changes as follows

$dU=DQ-DW\,\qquad \qquad \qquad (1)$

Note that the sum of two small quantities, both not necessarily differentials, gives a differential of U. The first law, equation (1), postulates the existence of a state function, U, that accumulates the work done on/by the system and the heat that flows in/out the system.

Internal energy is an extensive property—that is, its magnitude depends on the amount of substance in a given state. Often one considers the molar energy, energy per amount of substance (amount expressed in moles); this is an intensive property. Also the specific energy (energy per kilogram) is an intensive property. The internal energy has the SI dimension joule.

Note that thus far only a change in internal energy was defined. An absolute value can be obtained by defining a zero (reference) point with U₀ = 0 and integration

$\int _{0}^{1}dU=U_{1}-U_{0}=U_{1}$

The reference point could be the zero of absolute temperature (zero kelvin).

Explicit expression

Consider a one-component thermodynamical system that allows heat exchange DQ, work −pdV, and matter exchange μdn. The second law of thermodynamics states that there exists a variable, entropy (commonly denoted by S) that is given by

$dS={\frac {DQ}{T}}.$

This relation holds when the heat exchange occurs reversibly. By the second law, the entropy S is a state function. It is size-extensive, i.e., S is linear in the size of the system, and has dimension J/K (joule per degree kelvin).

Since there are three forms of contact with the surroundings, the system has exactly three independent variables. Choose the variables S, V, and n, which all three are size-extensive. By results shown earlier, cf. equation (1), and use of DQ = TdS, the differential of the internal energy is

$dU=TdS-pdV+\mu dn.\,\qquad \qquad \qquad \qquad \qquad \qquad \qquad (2)$

It will be shown that an explicit expression for the internal energy U of the system under consideration is,

$U=TS-pV+\mu n.\,\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (3)$

In order to prove (3), we consider first two identical systems with the same values for the three size-extensive independent variables S, V, n (and hence also the same values for U, T, and p). Clearly, the "super system" consisting of the two identical system has twice the entropy, volume, and amount of substance, and the sum of the energies of the two systems is 2U, or

$U(2S,2V,2n)=2U(S,V,n).\,$

The same kind of equation holds when we separate the original system into two equal parts, then the energy, entropy, volume and amount of substance are halved for each of the two parts. Clearly then, for an arbitrary real positive number λ,

$U(\lambda S,\lambda V,\lambda n)=\lambda U(S,V,n).\,$

That is, the internal energy U is a homogeneous function of order 1 of the size-extensive variables, S, V and n. By Euler's theorem,

$dU=\left({\frac {\partial U}{\partial S}}\right)_{V,n}S+\left({\frac {\partial U}{\partial V}}\right)_{S,n}V+\left({\frac {\partial U}{\partial n}}\right)_{S,V}n.\qquad \qquad \qquad \qquad (4)$

Since U is a function of the three variables, equation (2) can be written as

${\begin{aligned}dU&=\left({\frac {\partial U}{\partial S}}\right)_{V,n}dS+\left({\frac {\partial U}{\partial V}}\right)_{S,n}dV+\left({\frac {\partial U}{\partial n}}\right)_{S,V}dn\\&=TdS-pdV+\mu dn\\\end{aligned}}$

From which may be deduced (since S, V, and n are independent) that

$T=\left({\frac {\partial U}{\partial S}}\right)_{V,n},\quad p=-\left({\frac {\partial U}{\partial V}}\right)_{S,n},\quad \mu =\left({\frac {\partial U}{\partial n}}\right)_{S,V}$

Inserting this into equation (4), the result (3) follows.

Gibbs free energy

At first sight it may look as if equations (2) and (3) are contradictory, for, when taking the differential dU of (3), the term SdT −Vdp + ndμ arises, which is absent in equation (2). For the consistency of equations (2) and (3) it is necessary that SdT − Vdp + ndμ = 0. It is now shown that the latter expression yields an equation for the Gibbs free energy.

The Gibbs free energy, often denoted by G in honor of the American physicist J. Willard Gibbs, is defined by

$G\equiv U-TS+pV.$

Taking the differential and using equation (2) gives a form for dG,

$dG=dU-TdS-SdT+pdV+Vdp=-SdT+Vdp+\mu dn.\,$

Returning to the differential of equation (3), when we put it equal to equation (2), it gives,

$TdS+SdT-pdV-Vdp+nd\mu +\mu dn=TdS-pdV+\mu dn,\,$

from which follows that

$-SdT+VdP=nd\mu .\,$

And from the equation for dG,

$dG=-SdT+Vdp+\mu dn=nd\mu +\mu dn=d(n\mu ).\,$

This leads to the the conclusion, drawn from the non-contradictory equations (2) and (3), that G = nμ + G₀. Taking that the zero of Gibbs energy G₀ = 0, it follows that the chemical potential μ is the Gibbs free energy per mole.

Statistical thermodynamics definition

Consider a system of constant temperature T, constant number of molecules N, and constant volume V. In statistical thermodynamics one defines for such a system the density operator

${\hat {\rho }}\;{\stackrel {\mathrm {def} }{=}}\,{\frac {e^{-\beta {\hat {H}}}}{\mathrm {Tr} (e^{-\beta {\hat {H}}})}}={\frac {e^{-\beta {\hat {H}}}}{Q}}\quad {\hbox{with}}\quad Q\equiv \mathrm {Tr} (e^{-\beta {\hat {H}}}),$

where ${\hat {H}}$ is the Hamiltonian (energy operator) of the total system, $\mathrm {Tr} ({\hat {O}})$ is the trace of the operator ${\hat {O}}$, β = 1/(kT), and k is Boltzmann's constant. The quantity Q is the partition function.

The thermodynamic average of ${\hat {H}}$ is equal to the internal energy,

$U=\langle \langle {\hat {H}}\rangle \rangle \equiv \mathrm {Tr} ({\hat {\rho }}\,{\hat {H}})={\frac {1}{Q}}\mathrm {Tr} ({\hat {H}}\,e^{-\beta {\hat {H}}}).$

The internal energy is minus the logarithmic derivative of Q,

${\frac {d\ln Q}{d\beta }}={\frac {1}{Q}}{\frac {dQ}{d\beta }}={\frac {1}{Q}}\mathrm {Tr} {\frac {d(e^{-\beta {\hat {H}}})}{d\beta }}=-{\frac {1}{Q}}\mathrm {Tr} ({\hat {H}}\,e^{-\beta {\hat {H}}})=-U.$

Further

${\frac {d\ln Q}{d\beta }}={\frac {d\ln Q}{dT}}\left({\frac {d\beta }{dT}}\right)^{-1}=-kT^{2}{\frac {d\ln Q}{dT}}.$

Hence, the following well-known statistical-thermodynamics expression is obtained for the internal energy U,

$U=kT^{2}\left({\frac {d\ln Q}{dT}}\right)=kT^{2}\left({\frac {\partial \ln Q}{\partial T}}\right)_{V,N}$

Notes:

• The existence of an energy operator ${\hat {H}}$ was simply assumed. The choice of energy terms to be included in this operator, is in fact equivalent to the choice of contributions adding to the internal energy U. Statistical thermodynamics does not solve the problem of defining internal energy.
• When the trace is evaluated in a basis of eigenstates of ${\hat {H}}$, the physical meaning of the density operator becomes clearer. In fact, Boltzmann weight factors will arise. Thus, upon writing,

${\hat {H}}|E_{i}\rangle =E_{i}\;|E_{i}\rangle ,\qquad e^{-\beta {\hat {H}}}|E_{i}\rangle =e^{-\beta E_{i}}\;|E_{i}\rangle ,\qquad \mathrm {Tr} (e^{-\beta {\hat {H}}})=\sum _{i}\langle E_{i}\;|\;e^{-\beta {\hat {H}}}\;|E_{i}\rangle ,$

the partition function becomes [β = 1 /(kT)]:

$Q=\sum _{i}e^{-E_{i}/(kT)},\qquad {\hbox{(sum over eigenstates, not over energy levels)}},$

and the thermodynamic average becomes in a basis of energy eigenstates

$U=\langle \langle {\hat {H}}\rangle \rangle ={\frac {1}{Q}}\sum _{i}\;E_{i}\;e^{-E_{i}/(kT)}.$

The partition function normalizes the Boltzmann weights, exp[− E_i/(kT)]. Indeed,

$\mathrm {Tr} ({\hat {\rho }})={\frac {1}{Q}}\left[\sum _{i}e^{-E_{i}/(kT)}\right]={\frac {Q}{Q}}=1.$

The sum over normalized weights equals unity, as a proper weight function should.