Third law of thermodynamics

Suppose a system consisting of a crystal lattice with volume V of N identical atoms at T = 0 K, and an incoming photon of wavelength λ and energy ε.

Initially, there is only one accessible microstate:

$${\displaystyle S_{0}=k_{\text{B}}\ln \Omega =k_{\text{B}}\ln {1}=0.}$$

Let's assume the crystal lattice absorbs the incoming photon. There is a unique atom in the lattice that interacts and absorbs this photon. So after absorption, there is N possible microstates accessible by the system, each of the microstates corresponding to one excited atom, and the other atoms remaining at ground state.

The entropy, energy, and temperature of the closed system rises and can be calculated. The entropy change is:

$${\displaystyle \Delta S=S-S_{0}=k_{\text{B}}\ln {\Omega }}$$

From the second law of thermodynamics:

$${\displaystyle \Delta S=S-S_{0}={\frac {\delta Q}{T}}}$$

Hence:

$${\displaystyle \Delta S=S-S_{0}=k_{\text{B}}\ln(\Omega )={\frac {\delta Q}{T}}}$$

Calculating entropy change:

$${\displaystyle S-0=k_{\text{B}}\ln {N}=1.38\times 10^{-23}\times \ln {\left(3\times 10^{22}\right)}=70\times 10^{-23}\,\mathrm {J\,K^{-1}} }$$

We assume N = 3 × 10²² and λ = 1 cm. The energy change of the system as a result of absorbing the single photon whose energy is ε:

$${\displaystyle \delta Q=\varepsilon ={\frac {hc}{\lambda }}={\frac {6.62\times 10^{-34}\,\mathrm {J\cdot s} \times 3\times 10^{8}\,\mathrm {m\,s^{-1}} }{0.01\,\mathrm {m} }}=2\times 10^{-23}\,\mathrm {J} }$$

The temperature of the closed system rises by:

$${\displaystyle T={\frac {\varepsilon }{\Delta S}}={\frac {2\times 10^{-23}\,\mathrm {J} }{70\times 10^{-23}\,\mathrm {J\,K^{-1}} }}=0.02857\,\mathrm {K} }$$

This can be interpreted as the average temperature of the system over the range from ${\displaystyle 0<S<70\times 10^{-23}\,\mathrm {J\,K^{-1}} }$.[7] 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 half-integer, for which time-reversal 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. Ground-state 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 nearly-degenerate 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. However, ferromagnetic materials do not, in fact, have zero entropy at zero temperature, because the spins of the unpaired electrons are all aligned and this gives a ground-state spin degeneracy. Materials that remain paramagnetic at 0 K, by contrast, may have many nearly-degenerate ground states (for example, in a spin glass), or may retain dynamic disorder (a quantum spin liquid).

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 closed system to zero temperature in a finite number of finite operations.[8]

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 X₂ to X₁. One can think of a multistage nuclear demagnetization setup where a magnetic field is switched on and off in a controlled way.[9] 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.

A non-quantitative 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.[10] A modern, quantitative analysis follows.

Supposed that the heat capacity of a sample in the low temperature region has the form of a power law C(T,X) = C₀T^α asymptotically as T → 0, and we wish to find which values of α are compatible with the third law. We have

${\displaystyle \int _{T_{0}}^{T}{\frac {C(T',X)}{T'}}dT'={\frac {C_{0}}{\alpha }}(T^{\alpha }-T_{0}^{\alpha }).}$

(11)

By the discussion of third law (above), this integral must be bounded as T₀ → 0, which is only possible if α > 0. So the heat capacity must go to zero at absolute zero

${\displaystyle \lim _{T\to 0}C(T,X)=0.}$

(12)

if it has the form of a power law. The same argument shows that it cannot be bounded below by a positive constant, even if we drop the power-law assumption.

On the other hand, 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. But clearly a constant heat capacity does not satisfy Eq. (12). That is, a gas with a constant heat capacity all the way to absolute zero violates the third law of thermodynamics. We can verify this more fundamentally by substituting C_V in Eq. (14), which yields

${\displaystyle S(T,V)=S(T_{0},V)+{\frac {3}{2}}R\ln {\frac {T}{T_{0}}}.}$

(13)

In the limit T₀ → 0 this expression diverges, again contradicting the third law of thermodynamics.

The conflict is resolved 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

${\displaystyle C_{V}={\frac {\pi ^{2}}{2}}R{\frac {T}{T_{\text{F}}}}}$

(14)

with the Fermi temperature T_F given by

${\displaystyle T_{\text{F}}={\frac {1}{8\pi ^{2}}}{\frac {N_{\text{A}}^{2}h^{2}}{MR}}\left({\frac {3\pi ^{2}N_{\text{A}}}{V_{\text{m}}}}\right)^{2/3}.}$

(15)

Here N_A is the Avogadro constant, V_m the molar volume, and M the molar mass.

For Bose gases

${\displaystyle C_{V}=1.93..R\left({\frac {T}{T_{\text{B}}}}\right)^{3/2}}$

(16)

with T_B given by

${\displaystyle T_{\text{B}}={\frac {1}{11.9..}}{\frac {N_{\text{A}}^{2}h^{2}}{MR}}\left({\frac {N_{\text{A}}}{V_{\text{m}}}}\right)^{2/3}.}$

(17)

The specific heats given by Eq. (14) and (16) both satisfy Eq. (12). Indeed, they are power laws with α=1 and α=3/2 respectively.

Even within a purely classical setting, the density of a classical ideal gas at fixed particle number becomes arbitrarily high as T goes to zero, so the interparticle spacing goes to zero. The assumption of non-interacting particles presumably breaks down when they are sufficiently close together, so the value of C_V gets modified away from its ideal constant value.

The only liquids near absolute zero are ³He and ⁴He. Their heat of evaporation has a limiting value given by

${\displaystyle L=L_{0}+C_{p}T}$

(18)

with L₀ and C_p constant. If we consider a container, partly filled with liquid and partly gas, the entropy of the liquid–gas mixture is

${\displaystyle S(T,x)=S_{l}(T)+x\left({\frac {L_{0}}{T}}+C_{p}\right)}$

(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.

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.

The thermal expansion coefficient is defined as

${\displaystyle \alpha _{V}={\frac {1}{V_{m}}}\left({\frac {\partial V_{m}}{\partial T}}\right)_{p}.}$

(20)

With the Maxwell relation

${\displaystyle \left({\frac {\partial V_{m}}{\partial T}}\right)_{p}=-\left({\frac {\partial S_{m}}{\partial p}}\right)_{T}}$

(21)

and Eq. (8) with X = p it is shown that

${\displaystyle \lim _{T\to 0}\alpha _{V}=0.}$

(22)

So the thermal expansion coefficient of all materials must go to zero at zero kelvin.