# The Third Law and Low Temperatures

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/the-third-law-and-low-temperatures

## The Unreachable Floor

The third law comes in two related statements. **Nernst**: as $T \to 0$, the entropy change of any isothermal process tends to zero. **Planck**, stronger: the entropy of a perfect crystal tends to zero itself,

$$\lim_{T\to 0} S = 0$$

The statistical reading makes this natural. With $S = k_B\ln W$, a perfect crystal at zero temperature has exactly one accessible arrangement, so $W = 1$ and $S = 0$. Unlike energy, entropy therefore has a genuine absolute zero, which is why tables list absolute $S^\circ$ rather than differences.

Real systems can retain **residual entropy** if they freeze into a disordered state. Carbon monoxide, whose CO and OC orientations differ by very little energy, keeps roughly $k_B\ln 2$ per molecule, the crystal is not perfect, so Planck's statement does not apply to it.

A striking consequence: heat capacities must vanish as $T \to 0$. Since $S(T) = \int_0^T (C/T)\,dT'$ must converge, $C$ has to fall faster than $T$. Experiment agrees, metals show $C = \gamma T + AT^3$, the linear term from electrons and the cubic from phonons.

The third law also implies the **unattainability of absolute zero**. Cooling methods remove entropy in finite steps, and as $S \to 0$ for every accessible state, each step removes less. Reaching exactly zero would take infinitely many steps. Absolute zero is a limit, not a destination.

Practical cooling uses staged techniques: liquid nitrogen to 77 K, liquid helium to 4.2 K, pumped helium-3 to about 0.3 K, dilution refrigeration to a few millikelvin, and **adiabatic demagnetisation** below that, magnetise a paramagnetic salt isothermally to order the spins, then demagnetise it adiabatically so the spins re-disorder at the expense of the lattice's thermal energy.

> **Common pitfall:** reading unattainability as a practical difficulty to be engineered around. It is a statement about the structure of thermodynamics: each cooling step removes a smaller entropy increment than the last, so no finite sequence of steps reaches zero however good the apparatus.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. The entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.

**Answer:** True

**Why:** True, this is Planck’s statement of the third law. With exactly one accessible arrangement, $S = k_B\ln 1 = 0$, which gives entropy a genuine absolute zero that energy does not have.

Page: https://tryals.app/practice/physics-ii/the-third-law-and-low-temperatures/the-entropy-of-a-perfect-crystal-approaches-zero-as-the-temperature

### 2. Which statements help explain why cooling becomes progressively harder near absolute zero?

A. Heat capacities must vanish as the temperature approaches zero
B. Metals show a linear electronic and a cubic phonon contribution
C. Absolute zero can be reached with sufficiently good apparatus
D. Some crystals retain residual entropy at low temperature

**Answer:** A. Heat capacities must vanish as the temperature approaches zero; B. Metals show a linear electronic and a cubic phonon contribution; D. Some crystals retain residual entropy at low temperature

**Why:** Heat capacities must vanish for the entropy integral to converge, metals show $\gamma T + AT^3$, and imperfect crystals such as CO keep residual entropy. Unattainability is a consequence of the law itself, not of imperfect engineering.

Page: https://tryals.app/practice/physics-ii/the-third-law-and-low-temperatures/which-statements-help-explain-why-cooling-becomes-progressively

### 3. A crystal of carbon monoxide retains a residual entropy of $k_B \ln 2$ per molecule. For one mole, compute this in J/K, using $R = 8.314$ J/mol/K, to one decimal place.

**Answer:** 5.8 (within ±0.15)

**Why:** $S = R\ln 2 = 8.314 \times 0.693 = 5.8$ J/K per mole. The two nearly equivalent molecular orientations freeze in at random, so the crystal is not perfect and Planck’s statement does not apply.

Page: https://tryals.app/practice/physics-ii/the-third-law-and-low-temperatures/a-crystal-of-carbon-monoxide-retains-a-residual-entropy-of-kb-ln-2

### 4. Arrange these cooling techniques in order of the lowest temperature they reach, warmest first.

**Answer:**

1. Liquid nitrogen at 77 K
2. Liquid helium-4 at 4.2 K
3. Pumped helium-3 at about 0.3 K
4. Dilution refrigeration at a few millikelvin
5. Adiabatic demagnetisation below a millikelvin

**Why:** Each technique precools the next: nitrogen, then helium-4, then pumped helium-3, then a dilution refrigerator, and finally adiabatic demagnetisation. No single method spans the whole range.

Page: https://tryals.app/practice/physics-ii/the-third-law-and-low-temperatures/arrange-these-cooling-techniques-in-order-of-the-lowest-temperature

### 5. Match each statement of the third law or its consequences to what it says.

**Answer:**

- Nernst statement → Isothermal entropy changes vanish as T approaches zero
- Planck statement → The entropy of a perfect crystal itself tends to zero
- Unattainability → No finite sequence of steps reaches absolute zero
- Vanishing heat capacities → Required for the entropy integral to converge

**Why:** Nernst constrains entropy *changes*; Planck fixes the absolute value itself. Unattainability and vanishing heat capacities both follow, the latter because $S = \int (C/T)dT$ must converge at the lower limit.

Page: https://tryals.app/practice/physics-ii/the-third-law-and-low-temperatures/match-each-statement-of-the-third-law-or-its-consequences-to-what-it

### 6. Thermodynamics sets an arbitrary reference point for internal energy, but assigns an absolute value of zero to entropy. What fundamental distinction explains why absolute zero is fixed for entropy alone?

A. A single ground state leaves no residual microstate ambiguity
B. Internal energy cannot be measured during thermodynamic changes
C. Energy balances depend strictly on reference frame velocity choices
D. Residual disorder prevents absolute ground state energy vanishing

**Answer:** A. A single ground state leaves no residual microstate ambiguity

**Why:** Energy zero-points are arbitrary conventions because only energy differences govern work and heat. In contrast, statistical mechanics ties entropy to microstate count W; when a unique ground state yields W = 1, ln W becomes strictly zero.

Page: https://tryals.app/practice/physics-ii/the-third-law-and-low-temperatures/thermodynamics-sets-an-arbitrary-reference-point-for-internal-energy