# Phase Transitions and Critical Phenomena

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/phase-transitions-and-critical-phenomena

## Coexistence and Its End

At a phase transition two phases coexist, and coexistence has a precise condition: equal temperature, equal pressure, and **equal chemical potential**. The last is what actually selects the transition point, the stable phase at given $T$ and $P$ is whichever has the lower $\mu$.

Ehrenfest classified transitions by which derivative of $G$ first jumps. In a **first-order** transition the first derivatives jump, so volume and entropy change discontinuously and **latent heat** $L = T\Delta S$ is absorbed. Melting and boiling are the everyday examples: heat goes in and the temperature does not move. In a **continuous** (second-order) transition the first derivatives are smooth and the second ones diverge, no latent heat, but a diverging heat capacity. The ferromagnetic transition at the Curie point is the standard case.

Along a first-order coexistence line the **Clausius-Clapeyron equation** fixes the slope:

$$\frac{dP}{dT} = \frac{L}{T\Delta V}$$

The sign of $\Delta V$ therefore sets the direction the line leans. Water is the famous anomaly: ice is less dense than liquid water, so $\Delta V < 0$ on melting and the solid-liquid line slopes **backwards**: which is why pressure melts ice.

For a real gas the van der Waals isotherms below $T_c$ contain an unphysical region where $(\partial P/\partial V)_T > 0$, which would be mechanically unstable. The **Maxwell construction** replaces that loop with a horizontal tie-line placed so the two enclosed areas are equal, and that line is the coexistence region.

Raising the temperature shrinks the coexistence region until, at the **critical point**, the two phases become identical and the distinction disappears. Near it, properties follow power laws with **critical exponents** that are strikingly universal, wildly different substances share the same exponents, because near the critical point the microscopic details stop mattering.

> **Common pitfall:** assuming heat always raises temperature. During a first-order transition every joule goes into the latent heat of rearrangement, and the temperature is pinned until the change is complete, which is exactly what makes an ice bath a reliable fixed point.

## Practice questions

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

### 1. During a first-order phase transition, adding heat does not raise the temperature.

**Answer:** True

**Why:** True, every joule goes into the latent heat until the transition completes. This is exactly why an ice-water bath holds 0 °C so reliably that it serves as a calibration fixed point.

Page: https://tryals.app/practice/physics-ii/phase-transitions-and-critical-phenomena/during-a-first-order-phase-transition-adding-heat-does-not-raise-the

### 2. Why does the solid-liquid coexistence line of water slope backwards?

A. Ice is less dense than liquid water, making the volume change negative on melting
B. The melting of ice involves an anomalous decrease in entropy as bonds break
C. Latent heat is released rather than absorbed when ice transitions to water
D. Water possesses an exceptionally large latent heat of fusion across all pressures

**Answer:** A. Ice is less dense than liquid water, making the volume change negative on melting

**Why:** In $dP/dT = L/(T\Delta V)$, melting ice has $\Delta V < 0$ because ice is the less dense phase, so the slope is negative. Almost every other substance has a denser solid and a forward-sloping line.

Page: https://tryals.app/practice/physics-ii/phase-transitions-and-critical-phenomena/why-does-the-solid-liquid-coexistence-line-of-water-slope-backwards

### 3. Which statements about the critical point are correct?

A. Latent heat is at its maximum there
B. The distinction between liquid and gas disappears there
C. It terminates the liquid-vapour coexistence line
D. Different substances share the same critical exponents

**Answer:** B. The distinction between liquid and gas disappears there; C. It terminates the liquid-vapour coexistence line; D. Different substances share the same critical exponents

**Why:** At the critical point the two phases become indistinguishable, so the coexistence line ends and the latent heat falls to **zero**, not a maximum. The universality of critical exponents is one of the deepest results in the subject.

Page: https://tryals.app/practice/physics-ii/phase-transitions-and-critical-phenomena/which-statements-about-the-critical-point-are-correct

### 4. A substance has a latent heat of vaporisation of 40 kJ/mol and boils at 350 K. Compute its entropy of vaporisation in J/mol/K, to the nearest whole number.

**Answer:** 114 (within ±2)

**Why:** $\Delta S = L/T = 40{,}000/350 = 114$ J/mol/K. Most liquids cluster near 88 (Trouton’s rule); a markedly higher value signals strong association in the liquid, such as hydrogen bonding.

Page: https://tryals.app/practice/physics-ii/phase-transitions-and-critical-phenomena/a-substance-has-a-latent-heat-of-vaporisation-of-40-kj-mol-and-boils

### 5. The latent heat of a transition is zero at the critical point.

**Answer:** True

**Why:** True, at the critical point the phases become identical, so $\Delta S$ and $\Delta V$ both vanish and $L = T\Delta S = 0$. The transition changes continuously from first-order to nothing at all as the critical point is approached.

Page: https://tryals.app/practice/physics-ii/phase-transitions-and-critical-phenomena/the-latent-heat-of-a-transition-is-zero-at-the-critical-point

### 6. Chemical potential equivalence determines phase coexistence, whereas free energy derivatives classify the transition order. Why does equal chemical potential not preclude a discontinuous jump in volume?

A. Equilibrium requires only that entropy remains strictly continuous
B. Volume depends on the slope of chemical potential with pressure
C. Discontinuities arise because chemical potentials balance dynamically
D. Equal chemical potential guarantees identical densities in both phases

**Answer:** B. Volume depends on the slope of chemical potential with pressure

**Why:** Chemical potential continuity means $\mu_1 = \mu_2$, but its first derivative $(\partial \mu / \partial P)_T = v$ can differ across phases. Mistaking equal potentials for equal densities ignores derivative jumps. In second-order transitions, even higher derivatives remain coupled to distinct response functions.

Page: https://tryals.app/practice/physics-ii/phase-transitions-and-critical-phenomena/chemical-potential-equivalence-determines-phase-coexistence-whereas