Third law of thermodynamics
{{ safesubst:#invoke:Unsubst$N=Use dmy dates date=__DATE__ $B= }} {{#invoke:Sidebar collapsible  bodyclass = plainlist  titlestyle = paddingbottom:0.3em;borderbottom:1px solid #aaa;  title = Thermodynamics  imagestyle = display:block;margin:0.3em 0 0.4em;  image =  caption = The classical Carnot heat engine  listtitlestyle = background:#ddf;textalign:center;  expanded = Laws
 list1name = branches  list1title = Branches  list1 = Template:Startflatlist
 list2name = laws  list2title = Laws  list2 = Template:Startflatlist
 list3name = systems  list3title = Systems  list3 =
State 

Processes 
Cycles 
 list4name = sysprop  list4title = System properties
 list4 =
Functions of state 

Process functions 
 list5name = material  list5title = Material properties  list5 =
 list6name = equations  list6title = Equations  list6 = Template:Startflatlist
 Maxwell relations
 Onsager reciprocal relations
 Bridgman's equations
 Table of thermodynamic equations
 list7name = potentials  list7title = Potentials  list7 = Template:Startflatlist
 list8name = hist/cult  list8title = Template:Hlist  list8 =
History 

Philosophy 
Theories 

Key publications 
Timelines 
Template:Hlist 
 list9name = scientists  list9title = Scientists  list9 = Template:Startflatlist
 Bernoulli
 Carnot
 Clapeyron
 Clausius
 Carathéodory
 Duhem
 Gibbs
 von Helmholtz
 Joule
 Maxwell
 von Mayer
 Onsager
 Rankine
 Smeaton
 Stahl
 Thompson
 Thomson
 Waterston
 below = Book:Thermodynamics
}} The third law of thermodynamics is sometimes stated as follows, regarding the properties of systems in equilibrium at absolute zero temperature:
The entropy of a perfect crystal, at absolute zero (zero kelvins), is exactly equal to zero.
At zero kelvin the system must be in a state with the minimum possible energy, and this statement of the third law holds true if the perfect crystal has only one minimum energy state. Entropy is related to the number of possible microstates, and for a system containing a certain collection of particles, quantum mechanics indicates that there is only one unique state (called the ground state) with minimum energy.^{[1]} If the system does not have a welldefined order (if its order is glassy, for example), then in practice there will remain some finite entropy as the system is brought to very low temperatures as the system becomes locked into a configuration with nonminimal energy. The constant value is called the residual entropy of the system.^{[2]}
The Nernst–Simon statement of the third law of thermodynamics is in regard to thermodynamic processes, and whether it is possible to achieve absolute zero in practice:
The entropy change associated with any condensed system undergoing a reversible isothermal process approaches zero as temperature approaches 0 K, where condensed system refers to liquids and solids.
A simpler formulation of the Nernst–Simon statement might be:
It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its absolutezero value in a finite number of operations.
Physically, the Nernst–Simon statement implies that it is impossible for any procedure to bring a system to the absolute zero of temperature in a finite number of steps.^{[3]}
Contents
History
The 3rd law was developed by the chemist Walther Nernst during the years 1906–12, and is therefore often referred to as Nernst's theorem or Nernst's postulate. The third law of thermodynamics states that the entropy of a system at absolute zero is a welldefined constant. This is because a system at zero temperature exists in its ground state, so that its entropy is determined only by the degeneracy of the ground state.
In 1912 Nernst stated the law thus: "It is impossible for any procedure to lead to the isotherm T = 0 in a finite number of steps."^{[4]}
An alternative version of the third law of thermodynamics as stated by Gilbert N. Lewis and Merle Randall in 1923:
 If the entropy of each element in some (perfect) crystalline state be taken as zero at the absolute zero of temperature, every substance has a finite positive entropy; but at the absolute zero of temperature the entropy may become zero, and does so become in the case of perfect crystalline substances.
This version states not only ΔS will reach zero at 0 K, but S itself will also reach zero as long as the crystal has a ground state with only one configuration. Some crystals form defects which causes a residual entropy. This residual entropy disappears when the kinetic barriers to transitioning to one ground state are overcome.^{[5]}
With the development of statistical mechanics, the third law of thermodynamics (like the other laws) changed from a fundamental law (justified by experiments) to a derived law (derived from even more basic laws). The basic law from which it is primarily derived is the statisticalmechanics definition of entropy for a large system:
where S is entropy, k_{B} is the Boltzmann constant, and is the number of microstates consistent with the macroscopic configuration. The counting of states is from the reference state of absolute zero, which corresponds to the entropy of S_{0}.
Explanation
In simple terms, the third law states that the entropy of a perfect crystal of a pure substance approaches zero as the absolute temperature approaches zero. The alignment of a perfect crystal leaves no ambiguity as to the position of the components of the system and the orientation of each part of the crystal is identical. As the energy of the crystal is reduced, the unique vibrations of each atom are reduced to nothing. At that point no part of the crystal is unique, hence it is in essence one thing. This law provides an absolute reference point for the determination of entropy at any other temperature. Any increase in the entropy of a system, determined relative to this zero point, is the absolute entropy of that system. Mathematically, the absolute entropy of any system at zero temperature is the natural log of the number of ground states times Boltzmann's constant k_{B}=1.38x10^{−23,} JK^{−1}.
The entropy of a perfect crystal lattice as defined by Nernst's theorem is zero provided that its ground state is unique, because ln(1) = 0. If the system is composed of onebillion atoms, all alike, and lie within the matrix of a perfect crystal, the number of permutations of onebillion identical things taken onebillion at a time is Ω = 1. Hence:
The difference is zero, hence the initial entropy S_{0} can be any selected value so long as all other such calculations include that as the initial entropy. As a result the initial entropy value of zero is selected S_{0} = 0 is used for convenience.
In way of example, suppose a system consists of 1 cm^{3} of matter with a mass of 1 g and 20 g/gmole. The system consists of 3x10^{22} identical atoms at 0 K. If one atom should absorb a photon of wavelength of 1 cm that atom is then unique and the permutations of one unique atom among the 3x10^{22} is N=3x10^{22}. The entropy, energy, and temperature of the system rises and can be calculated. The entropy change is:
From the second law of thermodynamics:
Hence:
Calculating entropy change:
The energy change of the system as a result of absorbing the single photon whose energy is ε:
The temperature of the system rises by:
This can be interpreted as the average temperature of the system over the range from 0 < S < 70x10^{−23} J/K^{[6]} A single atom was assumed to absorb the photon but the temperature and entropy change characterizes the entire system.
An example of a system which does not have a unique ground state is one whose net spin is a halfinteger, for which timereversal symmetry gives two degenerate ground states. For such systems, the entropy at zero temperature is at least k_{B}*ln(2) (which is negligible on a macroscopic scale). Some crystalline systems exhibit geometrical frustration, where the structure of the crystal lattice prevents the emergence of a unique ground state. Groundstate helium (unless under pressure) remains liquid.
In addition, glasses and solid solutions retain large entropy at 0 K, because they are large collections of nearly degenerate states, in which they become trapped out of equilibrium. Another example of a solid with many nearlydegenerate ground states, trapped out of equilibrium, is ice Ih, which has "proton disorder".
For the entropy at absolute zero to be zero, the magnetic moments of a perfectly ordered crystal must themselves be perfectly ordered; from an entropic perspective, this can be considered to be part of the definition of a "perfect crystal". Only ferromagnetic, antiferromagnetic, and diamagnetic materials can satisfy this condition. Materials that remain paramagnetic at 0 K, by contrast, may have many nearlydegenerate ground states (for example, in a spin glass), or may retain dynamic disorder (a quantum spin liquid).{{ safesubst:#invoke:Unsubstdate=__DATE__ $B= {{#invoke:Category handlermain}}{{#invoke:Category handlermain}}^{[citation needed]} }}
Mathematical formulation
Consider a closed system in internal equilibrium. As the system is in equilibrium there are no irreversible processes so the entropy production is zero. During the heat, supply temperature gradients are generated in the material, but the associated entropy production can be kept low enough if the heat is supplied slowly. The increase in entropy due to the added heat δQ is then given by the second part of the Second law of thermodynamics which states that the entropy change of a system is given by
(1) 
The temperature rise dT due to the heat δQ is determined by the heat capacity C(T,X) according to
(2) 
The parameter X is a symbolic notation for all parameters (such as pressure, magnetic field, liquid/solid fraction, etc.) which are kept constant during the heat supply. E.g. if the volume is constant we get the heat capacity at constant volume C_{V}. In the case of a phase transition from liquid to solid, or from gas to liquid the parameter X can be one of the two components. Combining relations (1) and (2) gives
(3) 
Integration of Eq.(3) from a reference temperature T_{0} to an arbitrary temperature T gives the entropy at temperature T
(4) 
We now come to the mathematical formulation of the third law. There are three steps:
1: in the limit T_{0}→0 the integral in Eq.(4) is finite. So that we may take T_{0}=0 and write
(5) 
2. the value of S(0,X) is independent of X. In mathematical form
(6) 
So Eq.(5) can be further simplified to
(7) 
Equation (6) can also be formulated as
(8) 
In words: at absolute zero all isothermal processes are isentropic. Eq.(8) is the mathematical formulation of the third law.
3: as one is free to choose the zero of the entropy it is convenient to take
(9) 
so that Eq.(7) reduces to the final form
(10) 
The physical meaning of Eq.(9) is deeper than just a convenient selection of the zero of the entropy. It is due to the perfect order at zero kelvin as explained before.
Consequences of the third law
Can absolute zero be obtained?
The third law is equivalent to the statement that
 "It is impossible by any procedure, no matter how idealized, to reduce the temperature of any system to zero temperature in a finite number of finite operations".^{[7]}
The reason that T=0 cannot be reached according to the third law is explained as follows: Suppose that the temperature of a substance can be reduced in an isentropic process by changing the parameter X from X_{2} to X_{1}. One can think of a multistage nuclear demagnetization setup where a magnetic field is switched on and off in a controlled way.^{[8]} If there were an entropy difference at absolute zero, T=0 could be reached in a finite number of steps. However, at T=0 there is no entropy difference so an infinite number of steps would be needed. The process is illustrated in Fig.1.
Specific heat
A nonquantitative description of his third law that Nernst gave at the very beginning was simply that the specific heat can always be made zero by cooling the material down far enough.^{[9]} A modern, quantitative analysis follows. Suppose that the heat capacity of a sample in the low temperature region can be approximated by C(T,X)=C_{0}T^{α}, then
(11) 
The integral is finite for T_{0}→0 if α>0. So the heat capacity of all substances must go to zero at absolute zero
(12) 
The molar specific heat at constant volume of a monatomic classical ideal gas, such as helium at room temperature, is given by C_{V}=(3/2)R with R the molar ideal gas constant. Substitution in Eq.(4) gives
(13) 
In the limit T_{0}→0 this expression diverges. Clearly a constant heat capacity does not satisfy Eq.(12). This means that a gas with a constant heat capacity all the way to absolute zero violates the third law of thermodynamics.
The conflict is solved as follows: At a certain temperature the quantum nature of matter starts to dominate the behavior. Fermi particles follow Fermi–Dirac statistics and Bose particles follow Bose–Einstein statistics. In both cases the heat capacity at low temperatures is no longer temperature independent, even for ideal gases. For Fermi gases
(14) 
with the Fermi temperature T_{F} given by
(15) 
Here N_{A} is Avogadro's number, V_{m} the molar volume, and M the molar mass.
For Bose gases
(16) 
with T_{B} given by
(17) 
The specific heats given by Eq.(14) and (16) both satisfy Eq.(12).
Vapor pressure
The only liquids near absolute zero are ³He and ⁴He. Their heat of evaporation has a limiting value given by
(18) 
with L_{0} and C_{p} constant. If we consider a container, partly filled with liquid and partly gas, the entropy of the liquid–gas mixture is
(19) 
where S_{l}(T) is the entropy of the liquid and x is the gas fraction. Clearly the entropy change during the liquid–gas transition (x from 0 to 1) diverges in the limit of T→0. This violates Eq.(8). Nature solves this paradox as follows: at temperatures below about 50 mK the vapor pressure is so low that the gas density is lower than the best vacuum in the universe. In other words: below 50 mK there is simply no gas above the liquid.
Latent heat of melting
The melting curves of ³He and ⁴He both extend down to absolute zero at finite pressure. At the melting pressure liquid and solid are in equilibrium. The third law demands that the entropies of the solid and liquid are equal at T=0. As a result the latent heat of melting is zero and the slope of the melting curve extrapolates to zero as a result of the Clausius–Clapeyron equation.
Thermal expansion coefficient
The thermal expansion coefficient is defined as
(20) 
With the Maxwell relation
(21) 
and Eq.(8) with X=p it is shown that
(22) 
So the thermal expansion coefficient of all materials must go to zero at zero kelvin.
See also
 Adiabatic process
 Ground state
 Laws of thermodynamics
 Residual entropy
 Thermodynamic entropy
 Timeline of thermodynamics, statistical mechanics, and random processes
References
 ↑ J. Wilks The Third Law of Thermodynamics Oxford University Press (1961).Template:Page needed
 ↑ Kittel and Kroemer, Thermal Physics (2nd ed.), page 49.
 ↑ Wilks, J. (1971). The Third Law of Thermodynamics, Chapter 6 in Thermodynamics, volume 1, ed. W. Jost, of H. Eyring, D. Henderson, W. Jost, Physical Chemistry. An Advanced Treatise, Academic Press, New York, page 477.
 ↑ Bailyn, M. (1994). A Survey of Thermodynamics, American Institute of Physics, New York, ISBN 0–88318–7973, page 342.
 ↑ {{#invoke:Citation/CS1citation CitationClass=journal }}
 ↑ {{#invoke:citation/CS1citation CitationClass=book }}
 ↑ Guggenheim, E.A. (1967). Thermodynamics. An Advanced Treatment for Chemists and Physicists, fifth revised edition, NorthHolland Publishing Company, Amsterdam, page 157.
 ↑ F. Pobell, Matter and Methods at Low Temperatures, (SpringerVerlag, Berlin, 2007)Template:Page needed
 ↑ Einstein and the Quantum, A. Douglas Stone, Princeton University Press, 2013.
Further reading
 Goldstein, Martin & Inge F. (1993) The Refrigerator and the Universe. Cambridge MA: Harvard University Press. ISBN 0674753240. Chpt. 14 is a nontechnical discussion of the Third Law, one including the requisite elementary quantum mechanics.
 {{#invoke:Citation/CS1citation
CitationClass=journal }}
 {{#invoke:Citation/CS1citation
CitationClass=journal }} ·