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 left out. For instance, when the total system is in uniform motion, it has kinetic energy. This overall kinetic energy is never considered 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 that constitute 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. Except for ideal gases, the averaged potential energy of the atoms and/or molecules in the field of the others (see intermolecular forces) is also a 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 (forming of steam), then 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 physicist studying the thermodynamics of nuclear reactions will include it in the internal energy.

First law of thermodynamics

Classical (phenomenological) thermodynamics is not concerned with the nature of the internal energy, it simply assumes that it exists and may be changed by certain processes. It is assumed 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 it attained that state. Further, the internal energy, that 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 heat leaves the system. The internal energy of the system changes by dU as a consequence of the heat flow, and it is postulated that

$dU=DQ\,$

The symbol DQ indicates simply a small amount of heat, and not a differential of Q. Note that Q is not a function. The symbol dU indicates a differential of the differentiable function U, for instance, when U is seen as a function of V,

$dU\equiv \lim _{\Delta V\rightarrow 0}{\frac {U(V+\Delta V)-U(V)}{\Delta V}}\Delta V=\left({\frac {\partial U}{\partial V}}\right)dV$

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

$dU=-DW\,$

The minus sign is by convention.

As an example of work, we consider as a system a volume V containing gas of pressure p. Work pdV is performed on the system by reversibly (quasi-statically) compressing the gas (dV < 0). The sign convention gives 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. This change in internal energy is caused by an alignment of the microscopic magnetic moments that constitute a magnetizable material.

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 the state function U. Equation (1) postulates in fact the existence of a quantity (U) that keeps track of the work done on/by the system and the heat that flowed 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. The internal energy has the SI dimension joule.

Note that 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).

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

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

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

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

The quantity Q is the partition function. The internal energy is minus its logarithmic derivative

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

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 this problem.
• 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,

$\mathrm {Tr} ({\hat {O}})=\sum _{i}\langle E_{i}\;|\;{\hat {O}}\;|E_{i}\rangle \quad {\hbox{and}}\quad \exp(-\beta {\hat {H}})|E_{i}\rangle =\exp(-\beta E_{i})|E_{i}\rangle$

the thermodynamic average becomes

$U=\langle \langle {\hat {H}}\rangle \rangle ={\frac {1}{Q}}\sum _{i}\;E_{i}\;\exp(-\beta E_{i})$

The partition function normalizes the Boltzmann weights,

$Q=\sum _{i}\exp(-\beta E_{i})\,$

that is,

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