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—dS is its differential—and is size-extensive, i.e., S is linear in the size of the system.

Since there are three forms of contact with the surroundings, the system has three independent variables. Choose the variables S, V, and n, all three size-extensive. The differential of the internal energy is

$dU=TdS-pdV+\mu dn.\,\qquad \qquad \qquad \qquad \qquad \qquad \qquad (1)$

The explicit expression for U central to this section is,

$U=TS-pV+\mu n.\,\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2)$

Before equation (2) is proved, we will first derive that it has the following consequence

$\mu n=G\,\quad {\hbox{with}}\quad G\equiv U-TS+pV.\,\qquad \qquad \qquad \qquad \qquad (3)$.

Indeed, from (2) and (1), respectively,

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

so that

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

And from the second equation (3), equation (1), and the last equation,

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

The quantity G = U + pV − TS = nμ is known as the Gibbs free energy, and it follows that the chemical potential, μ = G/n, is the Gibbs free energy per mole.

In order to prove (2) two identical systems with same values for the three size-extensive variables n, V, S (and hence also same values for U, T, and p) are considered. It is clear that for the "super system" consisting of the two identical system holds

$U_{\mathrm {super} }=U(2S,2V,2n)=2U(S,V,n)\,$

The same kind of equation holds when we separate the original system into two equal parts

$U({\tfrac {1}{2}}S,{\tfrac {1}{2}}V,{\tfrac {1}{2}}n)={\tfrac {1}{2}}U(S,V,n)\,$

Clearly, for arbitrary real positive 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, n and V. By Euler's theorem,

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

Since U is a (state) function equation (1) can be written as

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

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,

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

the partition function becomes,

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

and the thermodynamic average becomes

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

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.