Solution
= Solution
\b[<Entropy> at <absolute zero>.] The <Third law of thermodynamics> concerns the low-temperature limit of equilibrium states. In the zero-<entropy> formulation, the <entropy> of a perfect crystal with a nondegenerate <ground state> tends to zero as $T\to0$. The statistical interpretation is $S=k_B\ln\Omega$: a unique accessible <ground state> has $\Omega=1$. The entropy-difference formulation of the <Third law of thermodynamics> states that <entropy> differences associated with regular changes of mechanical parameters tend to zero as <absolute zero> is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate <ground state> can carry <residual entropy>.
For a phase of fixed composition and <pressure>,
$$
S(T,p)-S(0,p)=\int_0^T\frac{C_p(t,p)}{t}\,dt.
$$
A finite limiting <entropy> constrains this <integral>. Ordinary equilibrium <heat capacities> therefore vanish as <temperature> approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic <phonons> and $C\propto T$ for a normal degenerate <Fermi liquid>. A constant positive <heat capacity> extrapolated to zero would give a logarithmically divergent <entropy> difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.
\b[<Gibbs free energy>.] For fixed particle numbers,
$$
dG=-S\,dT+V\,dp,
\qquad
\boxed{S=-\left(\frac{\partial G}{\partial T}\right)_p.}
$$
Thus, when $S\to0$, the <Gibbs free energy> approaches its zero-temperature value with a horizontal tangent:
$$
\boxed{G(T,p)\to H(0,p),\qquad
\left(\frac{\partial G}{\partial T}\right)_p\to0,
\qquad G-H=-TS\to0.}
$$
The equality $G(0,p)=H(0,p)$ uses the <enthalpy> $H=U+pV$; neither limiting value needs to be zero. The <low-temperature Gibbs free-energy expansion> makes the approach quantitative. For $C_p=aT^r$ with $r>0$ and $S(0)=0$,
$$
S=\frac{a}{r}T^r,\qquad
G(T,p)=G(0,p)-\frac{a}{r(r+1)}T^{r+1}.
$$
The second <derivative> is $\partial_T^2G=-C_p/T\leq0$, although it need not tend to zero. For a reaction between such equilibrium phases, $\partial_T\Delta G=-\Delta S\to0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine <residual entropy> $S_0$ is retained instead, $G=G(0,p)-S_0T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.
\b[Why <helium> remains a <liquid>.] <Helium> atoms attract each other weakly and are exceptionally light. Their <zero-point energy> is large compared with the binding that would localize them in a <solid>. Both <helium-4> and <helium-3> can therefore remain <liquid> down to the lowest <temperatures> at low <pressure>; substantial applied <pressure> is needed to solidify them. This is not a violation of the third law. A quantum <liquid> can have a unique low-entropy <ground state>, and <zero-point motion> does not itself imply thermal <entropy>. The third law does not require every substance to crystallize at atmospheric <pressure>.
A <helium-4> atom has integer <spin> zero and is a <boson>. Below the <lambda point> near $2.17\,\mathrm K$ at saturated <vapor pressure>, its <liquid> is <superfluid>. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the <condensate fraction> need not equal the <superfluid fraction>. The <two-fluid model of superfluidity> separates an entropy-carrying <normal component of a superfluid> from the <superfluid> component, whose <entropy> is zero. At sufficiently low <temperature> the dominant thermal excitations are acoustic <phonons>: their linear dispersion in three dimensions gives $U-U_0\propto VT^4$, $C_V\propto VT^3$ and $S\propto VT^3$. Higher-energy <rotons> are exponentially suppressed. Thus the <normal component of a superfluid> and <entropy> disappear as $T\to0$, while <superfluid> flow and quantum motion can persist.
A <helium-3> atom has <spin> $1/2$ and is a <fermion>. Well below its <Fermi temperature>, but above its <superfluid> transition, <liquid> <helium-3> is a degenerate interacting <Fermi liquid>. Only <quasiparticles> close to the <Fermi surface> are thermally excited, giving $C_V\propto T$ and $S\propto T$. At <temperatures> of order a millikelvin, <helium-3> also becomes <superfluid> through formation of <Cooper pairs> of <fermions>. This is not <Bose-Einstein condensation> of individual <helium-3> atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic <quasiparticle> contribution to <heat capacity> is exponentially suppressed at <temperatures> well below the <pairing gap>. Gapless collective modes can still give thermal terms following a <power law>. The <entropy> nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.<The Extraordinary Phases of Liquid 3He by David Morris Lee>[1996 Nobel lecture, discussion of <helium> isotopes and paired <superfluid> phases]{ref}
<Helium-3> also illustrates the importance of <nuclear spin> <entropy>. Over an intermediate low-temperature range, localized <spins> in the <solid> can retain approximately $k_B\ln2$ of <entropy> per atom, while the degenerate <liquid> has much less. Reversible conversion of <liquid> to the higher-entropy <solid> then absorbs <heat>; adiabatic compression along the <melting> curve can cool the sample. This is <Pomeranchuk cooling>. At still lower <temperature>, interactions between the nuclear <spins> remove this nearly free-spin <entropy>, so a constant $k_B\ln2$ cannot be extrapolated to <absolute zero>.<The Pomeranchuk Effect by Robert Coleman Richardson>[1996 Nobel lecture, <entropy> comparison and compression cooling]{ref}
\b[Liquefaction and cooling methods.] A gas must be below its <liquid-gas critical point> to form a distinct <liquid> phase by compression. Compression above the <critical temperature> alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression <heat>, cools the compressed gas in a <heat exchanger> and uses expansion to lower its <temperature> further.
A throttling valve operates approximately at constant <specific enthalpy> when kinetic-energy changes and <heat> leakage are negligible. The <Joule-Thomson coefficient> is
$$
\mu_{\rm JT}=\left(\frac{\partial T}{\partial p}\right)_H
=\frac{T(\partial V/\partial T)_p-V}{C_p}
$$
using consistent molar quantities, or consistent total quantities for a fixed amount. A <pressure> drop cools the gas if $\mu_{\rm JT}>0$. An <ideal gas> has zero <Joule-Thomson coefficient>; real intermolecular interactions are essential. In particular, room-temperature <helium> warms on throttling, so it must first be brought into a region of positive <Joule-Thomson coefficient>. Expansion through an engine or turbine can instead remove <work> and produce approximately isentropic cooling.
A regenerative liquefier sends the cold returning gas through a <counterflow heat exchanger> to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces <liquid>. Modern <helium> liquefaction commonly combines work-producing expansion with a final throttle. https://trc.nist.gov/cryogenics/Papers/Review/2020-Review_of_Refrigeration_Methods.pdf[Review of Refrigeration Methods, <helium> liquefiers and Claude cycles]{ref}
Once <liquid> exists, pumping away vapour lowers its <saturation pressure> and boiling <temperature>. The escaping vapour carries <latent heat> away, cooling the remaining <liquid>. At still lower <temperatures>, a <dilution refrigerator> exploits the different <entropy> carried by <helium-3> in concentrated and dilute <helium> mixtures; <adiabatic demagnetization> and <Pomeranchuk cooling> provide other routes.
\b[Approaching zero.] The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach <absolute zero> in finitely many stages. As the <entropy> available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero <entropy> cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. \b[Quantum liquids and refrigeration methods can approach $0\,\mathrm K$ while remaining consistent with vanishing equilibrium <entropy>; they do not provide a finite-operation route to <absolute zero>.]