Q8(a). Third law: absolute zero cannot be reached

1. Third law (Nernst heat theorem): the entropy change of any isothermal process between equilibrium states tends to zero as T→0; the entropies of all perfect crystals approach the same value (taken as zero). Hence S(0,X₁)=S(0,X₂) for any two values of a variable X (field, pressure).
2. How low temperatures are reached (adiabatic demagnetisation): a paramagnetic salt such as Gd₂(SO₄)₃.8H₂O at about 1 K is magnetised isothermally (field X₁→ X₂; entropy falls, heat leaves to the helium bath), then isolated and demagnetised adiabatically (X₂→ X₁ at constant S), so its temperature falls.
3. The two curves: S(T,X)=S(0,X)+∫₀^T(C_X)/(T) dT. The field-on curve lies below the field-off curve at every T>0, but by the third law the two meet at T=0 (figure).
4. One adiabatic step from (T₁,X₂) to (T₂,X₁) keeps S constant: S(0,X₂)+∫₀^(T₁)C_(X₂)TdT=S(0,X₁)+∫₀^(T₂)C_(X₁)TdT
5. Try to reach T₂=0: the right-hand integral vanishes, and with S(0,X₁)=S(0,X₂): ∫₀^(T₁)C_(X₂)T dT=0
6. This is impossible: heat capacities are positive at every T₁>0, so the integral is positive. No adiabatic step starting above zero ends at zero.
7. Graphically: because the curves converge, each isothermal–adiabatic pair gives a smaller drop in temperature than the one before; an infinite number of steps would be needed. Hence: T=0 cannot be reached in a finite number of steps

Application. nuclear adiabatic demagnetisation has reached 280 pK in silver, very close to zero but never zero; the third law also gives absolute (third-law) entropies from heat-capacity data, used to calculate Δ S°, Δ G° and equilibrium constants.