---
title: "Phase Changes of a Pure Substance"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 25
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/25-phase-changes-of-a-pure-substance
---

# Chapter 25 — Phase Changes of a Pure Substance

Water is the one substance everybody has seen in all three states, and it is the working fluid of most of the world’s power stations, the [refrigerant](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-vapourcycle) of the first ice machines, and the engine of the weather. Ice melts at a temperature that does not budge while the heat pours in; a kettle boils at $100{}^{\circ}\mathrm{C}$ at the seaside and at $70{}^{\circ}\mathrm{C}$ on Everest; a sealed bottle of very cold water freezes solid at a tap. This chapter draws the map of the [phases](#def-b1-phase-changes-phase) of a [pure substance](#def-b1-phase-changes-phase) — the $(T, P)$ and $(v, P)$ diagrams — and gives the three tools that go with it: the lever rule (how much of each [phase](#def-b1-phase-changes-phase)), the [latent heat](#def-b1-phase-changes-phase) (what a change costs, in energy and in entropy) and the [Clausius–Clapeyron relation](#thm-b1-phase-changes-clapeyron) (how the boundary lines slope). It ends where the map fails: [metastable states](#def-b1-phase-changes-metastable), which are where the weather and the pressure cooker’s surprises come from.

## 25.1 Phases and the $(T, P)$ diagram

**Definition 25.1 (Pure substance, phase, phase change).**

A *pure substance* is a body of a single chemical species. A *phase* is a part of it that is homogeneous in its physical properties: solid, liquid, gas (*vapour* when it coexists with its liquid). A *phase change* (melting/solidification, vaporization/ condensation, sublimation/deposition) is the passage from one phase to another; at fixed pressure it occurs at a fixed temperature, with heat exchanged but no temperature change: the *latent heat*.

**Proposition 25.2 (Equilibrium diagram).**

In the $(T, P)$ plane each [phase](#def-b1-phase-changes-phase) occupies a region; two [phases](#def-b1-phase-changes-phase) coexist only along a curve (sublimation, fusion, vaporization), all three only at one point, the *[triple point](#prop-b1-phase-changes-PTdiagram)*. The vaporization curve $P = P_s(T)$ — the *[saturation vapour pressure](#prop-b1-phase-changes-PTdiagram)* — ends at the *[critical point](#prop-b1-phase-changes-PTdiagram)* $(T_c, P_c)$, beyond which liquid and gas are no longer distinct. For water: [triple point](#prop-b1-phase-changes-PTdiagram) $273.16\,\mathrm{K}$, $611\,\mathrm{Pa}$; [critical point](#prop-b1-phase-changes-PTdiagram) $647\,\mathrm{K}$, $221\,\mathrm{bar}$. The fusion curve of water leans to the left (ice is less dense than water); for almost every other substance it leans to the right.

**Proof.** *Admitted at this level.* ∎

![The phase diagram of water (not to scale). Heating at atmospheric pressure follows the dashed isobar: ice melts where it crosses the fusion curve, water boils where it crosses the vaporization curve. The fusion curve of water leans left: pressure lowers its melting point.](https://one-course.com/images/onecourse/chapters/physics-3/b1-phase-changes/fig-564579931f11.svg)

*The [phase diagram](#prop-b1-phase-changes-PTdiagram) of water (not to scale). Heating at atmospheric pressure follows the dashed isobar: ice melts where it crosses the fusion curve, water boils where it crosses the vaporization curve. The fusion curve of water leans left: pressure lowers its melting point.*

**Remark 25.3 (Boiling, evaporation and the pressure cooker).**

A liquid *evaporates* at any temperature from its surface, as long as the partial pressure of its vapour above it is below $P_s(T)$; it *boils* when $P_s(T)$ reaches the ambient pressure, so that bubbles of vapour can grow inside it. Water boils at $100{}^{\circ}\mathrm{C}$ because $P_s(100{}^{\circ}\mathrm{C}) = 1.013\,\mathrm{bar}$; at $0.33\,\mathrm{bar}$ on Everest it boils near $70{}^{\circ}\mathrm{C}$, in a pressure cooker at $1.8\,\mathrm{bar}$ near $117{}^{\circ}\mathrm{C}$. The temperature of a boiling liquid is pinned by the pressure — a fact used to define the Celsius scale for two centuries.

## 25.2 The $(v, P)$ diagram and the lever rule

**Proposition 25.4 (Andrews isotherms).**

In the Clapeyron plane $(v, P)$ of the specific volume, an isotherm $T < T_c$ has three parts: a steep branch (liquid, almost incompressible), a horizontal *plateau* at $P = P_s(T)$ where liquid and vapour coexist, and a hyperbola-like branch (vapour). The plateau ends are the specific volumes $v_l(T)$ and $v_v(T)$ of the saturated liquid and vapour; their locus is the *[saturation curve](#prop-b1-phase-changes-andrews)* (the dome), whose summit is the [critical point](#prop-b1-phase-changes-PTdiagram), where the isotherm $T_c$ has a horizontal inflection. Above $T_c$ no plateau exists.

**Proof.** *Admitted at this level.* ∎

**Theorem 25.5 (Lever rule).**

A mass $m$ of specific volume $v$ on the plateau at $T$ splits into a vapour mass fraction $x = m_v/m$ (the *quality*) and a liquid fraction $1 - x$ with

$$
v = (1 - x)v_l + x\,v_v , \qquad\text{i.e.}\qquad
x = \frac{v - v_l}{v_v - v_l} = \frac{LM}{LV} ,
$$

the ratio of the segments cut on the plateau by the state point $M$. The same rule holds for every extensive quantity per unit mass ($u$, $h$, $s$): $h = (1 - x)h_l + xh_v$.

**Proof.** Volumes add: $V = m_lv_l + m_vv_v$; divide by $m$. Energy, [enthalpy](https://one-course.com/books/physics/3/en/chapter/22-the-first-law-of-thermodynamics#def-b1-first-law-enthalpy) and entropy are extensive as well. ∎

![Andrews isotherms in the Clapeyron diagram. Under the dome the isotherm T_1 is a plateau at P_s(T_1); a state M on it is a mixture whose vapour fraction x is the ratio LM/LV. The critical isotherm touches the dome at its summit with a horizontal inflection; above T_c there is no plateau and no distinction between liquid and gas.](https://one-course.com/images/onecourse/chapters/physics-3/b1-phase-changes/fig-7473443ad076.svg)

*[Andrews isotherms](#prop-b1-phase-changes-andrews) in the Clapeyron diagram. Under the dome the isotherm $T_1$ is a plateau at $P_s(T_1)$; a state $M$ on it is a mixture whose vapour fraction $x$ is the ratio $LM/LV$. The critical isotherm touches the dome at its summit with a horizontal inflection; above $T_c$ there is no plateau and no distinction between liquid and gas.*

**Example 25.6 (A kettle and a boiler).**

Water at $100{}^{\circ}\mathrm{C}$, $1\,\mathrm{atm}$: $v_l = 1.04 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$, $v_v = 1.67\,\mathrm{m}^{3}/\mathrm{kg}$ — a ratio of $1600$. A closed $1.0\,\mathrm{L}$ vessel holding $0.50\,\mathrm{kg}$ of water at that temperature has $v =
2.0 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$ and $x = (2.0 - 1.04) \times 10^{-3}/1.67 = 5.8 \times 10^{-4}$: $0.29\,\mathrm{g}$ of vapour occupying half the vessel. Doubling the vapour mass in a sealed vessel barely moves the state point; the pressure is fixed by the temperature alone.

## 25.3 Energetics of a phase change

**Proposition 25.7 (Latent heat, enthalpy and entropy of transition).**

At the temperature $T$ and pressure $P_s(T)$ of coexistence, the transition of unit mass from [phase](#def-b1-phase-changes-phase) 1 to [phase](#def-b1-phase-changes-phase) 2 is reversible, isothermal and isobaric, and

$$
L_{12}(T) = h_2 - h_1 , \qquad s_2 - s_1 = \frac{L_{12}}{T} , \qquad
u_2 - u_1 = L_{12} - P_s(v_2 - v_1) .
$$

$L_{12} > 0$ when going toward the less ordered [phase](#def-b1-phase-changes-phase) (fusion, vaporization, sublimation), and $L_{21} = -L_{12}$. For water at $1\,\mathrm{atm}$: $L_f = 334\,\mathrm{kJ}/\mathrm{kg}$ at $0{}^{\circ}\mathrm{C}$, $L_v = 2257\,\mathrm{kJ}/\mathrm{kg}$ at $100{}^{\circ}\mathrm{C}$.

**Proof.** At constant pressure the heat received is $\Delta H$ ([Proposition 22.7](https://one-course.com/books/physics/3/en/chapter/22-the-first-law-of-thermodynamics#prop-b1-first-law-qvqp)); reversible and isothermal, it is $T\Delta S$ ([Proposition 23.3](https://one-course.com/books/physics/3/en/chapter/23-the-second-law-entropy#prop-b1-second-law-entropy-identity)); and $U = H - PV$. ∎

**Example 25.8 (Where the heat of vaporization goes).**

For water at $100{}^{\circ}\mathrm{C}$: $P_s(v_v - v_l) = 1.013 \times 10^5 \times
1.67 = 169\,\mathrm{kJ}/\mathrm{kg}$, so $u_v - u_l = 2257 - 169 = 2088\,\mathrm{kJ}/\mathrm{kg}$: $93\%$ of the [latent heat](#def-b1-phase-changes-phase) goes into tearing the molecules apart, $7\%$ into pushing the atmosphere back. The entropy of vaporization, $2257/373 = 6.05\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$, dwarfs that of fusion, $334/273 =
1.22\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$: melting loosens the molecules, boiling frees them.

**Method 25.9 (Calorimetry with phase changes).**

In an insulated vessel at constant pressure, $\sum\Delta H_i = 0$. Guess the final state (which [phases](#def-b1-phase-changes-phase) are present, at what temperature); write each body’s $\Delta H$ as sensible heats $mc\,\Delta T$ plus [latent heats](#def-b1-phase-changes-phase) $\pm mL$ for the changes it undergoes; solve; *check the guess*: a body ending above its boiling point or below its melting point, or a negative mass of a [phase](#def-b1-phase-changes-phase), means the guess was wrong — the final state is then a mixture at the transition temperature, and the unknown becomes the mass transformed ([Example 22.13](https://one-course.com/books/physics/3/en/chapter/22-the-first-law-of-thermodynamics#ex-b1-first-law-ice)).

**Example 25.10 (Steam burns).**

$10\,\mathrm{g}$ of steam at $100{}^{\circ}\mathrm{C}$ condensing on skin gives $0.010 \times 2257 = 22.6\,\mathrm{kJ}$ before it has cooled one degree — as much as $10\,\mathrm{g}$ of boiling water gives in cooling by $540\,\mathrm{K}$. This is why steam scalds so much worse than water at the same temperature, and why a steam radiator needs so little mass flow.

![The same dome in the entropy diagram. An isobar crosses it as a horizontal segment of length L_v/T; the area under the segment is the latent heat. Power-station cycles are drawn in this plane, and the quality at the turbine exit is read off the segment by the lever rule.](https://one-course.com/images/onecourse/chapters/physics-3/b1-phase-changes/fig-16a96e7c78a7.svg)

*The same dome in the [entropy diagram](https://one-course.com/books/physics/3/en/chapter/23-the-second-law-entropy#def-b1-second-law-entropy-TS). An isobar crosses it as a horizontal segment of length $L_v/T$; the area under the segment is the [latent heat](#def-b1-phase-changes-phase). Power-station cycles are drawn in this plane, and the quality at the turbine exit is read off the segment by the lever rule.*

## 25.4 The Clausius–Clapeyron relation

**Theorem 25.11 (Clausius–Clapeyron).**

Along a coexistence curve of [phases](#def-b1-phase-changes-phase) 1 and 2,

$$
\frac{\dd P_s}{\dd T} = \frac{L_{12}(T)}{T\,(v_2 - v_1)} .
$$

**Proof.** Take unit mass round an infinitesimal [Carnot cycle](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-carnotcycle) inside the dome: complete transition $1 \to 2$ along the plateau at $T + \dd T$, pressure $P_s + \dd P_s$ (heat received $L_{12}$ to first order); an infinitesimal adiabatic step down to $T$; the reverse transition along the plateau at $T$, $P_s$; an adiabatic step back. The cycle is reversible, ditherm between $T + \dd T$ and $T$, so its [efficiency](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-efficiency) is $\dd T/T$ ([Theorem 24.5](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#thm-b1-heat-engines-carnot)); its work, the area of the thin rectangle, is $\dd P_s\,(v_2 - v_1)$. Hence $\dd P_s\,(v_2 - v_1)/L_{12} =
\dd T/T$. ∎

![The infinitesimal Carnot cycle that proves the Clausius–Clapeyron relation: two plateaus T apart, joined by adiabatic steps too short to draw; Carnot’s theorem equates the ratio of its area to the heat received with T/T.](https://one-course.com/images/onecourse/chapters/physics-3/b1-phase-changes/fig-b1ef002ae825.svg)

*The infinitesimal [Carnot cycle](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-carnotcycle) that proves the [Clausius–Clapeyron relation](#thm-b1-phase-changes-clapeyron): two plateaus $\dd T$ apart, joined by adiabatic steps too short to draw; Carnot’s theorem equates the ratio of its area to the heat received with $\dd T/T$.*

**Corollary 25.12 (Saturation vapour pressure of a liquid).**

If the vapour is a [perfect gas](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-model) and $v_v \gg v_l$, so that $v_v - v_l \approx
RT/(MP_s)$, and if $L_v$ varies little,

$$
\frac{\dd P_s}{P_s} = \frac{ML_v}{R}\,\frac{\dd T}{T^2} , \qquad
\ln\frac{P_s(T)}{P_s(T_0)} = \frac{ML_v}{R}\Bigl(\frac{1}{T_0} - \frac{1}{T}\Bigr) :
$$

the vapour pressure rises roughly exponentially with temperature. For water near $100{}^{\circ}\mathrm{C}$, $ML_v/R = 0.018 \times 2.257 \times 10^6/
8.314 = 4890\,\mathrm{K}$ and $\dd P_s/\dd T = 2.257 \times 10^6/(373 \times 1.67) =
3.6\,\mathrm{kPa}/\mathrm{K}$: each [kelvin](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-temperature) of extra boiling temperature costs $36\,\mathrm{mbar}$ of extra pressure.

**Proof.** Substitute $v_v - v_l \approx RT/(MP_s)$ in the theorem and separate variables; integrate with $L_v$ constant. ∎

![The saturation vapour pressure of water. The curve is close to an exponential in -1/T; the boiling temperature is where it crosses the ambient pressure.](https://one-course.com/images/onecourse/chapters/physics-3/b1-phase-changes/fig-a77641d43b27.svg)

*The [saturation vapour pressure](#prop-b1-phase-changes-PTdiagram) of water. The curve is close to an exponential in $-1/T$; the boiling temperature is where it crosses the ambient pressure.*

**Example 25.13 (Ice under a skate).**

For ice–water, $L_f = 334\,\mathrm{kJ}/\mathrm{kg}$ and $v_l - v_s = (1.000 - 1.091)
\times 10^{-3} = -9.1 \times 10^{-5}\,\mathrm{m}^{3}/\mathrm{kg}$: $\dd P/\dd T = 3.34 \times 10^5/(273 \times
(-9.1 \times 10^{-5})) = -1.3 \times 10^{7}\,\mathrm{Pa}/\mathrm{K} = -130\,\mathrm{bar}/\mathrm{K}$. A $70\,\mathrm{kg}$ skater on one blade of $30\,\mathrm{cm}$ by $3\,\mathrm{mm}$ ($9\,\mathrm{cm}^{2}$) exerts about $8\,\mathrm{bar}$, lowering the melting point by $0.06\,\mathrm{K}$ — pressure melting is not why skates glide (a thin surface layer of ice is liquid-like on its own, and friction heats); but the negative slope is real, and it is why a wire loaded with weights cuts through a block of ice that refreezes behind it.

## 25.5 Metastable states

**Definition 25.14 (Supercooling, superheating).**

A [phase](#def-b1-phase-changes-phase) can persist beyond its equilibrium boundary: liquid water cooled below $0{}^{\circ}\mathrm{C}$ without freezing (*supercooling*), heated above its boiling point without boiling (*superheating*), vapour compressed beyond $P_s$ without condensing (supersaturation). Such *metastable states* require the absence of *nucleation* sites — dust, scratches, dissolved gas, ice crystals — on which the new [phase](#def-b1-phase-changes-phase) can start; a disturbance tips them suddenly and irreversibly to equilibrium.

**Example 25.15 (The supercooled bottle).**

Still water at $-8{}^{\circ}\mathrm{C}$ in a clean bottle, knocked: ice crystals shoot through it and the temperature jumps to $0{}^{\circ}\mathrm{C}$. The process is adiabatic and fast, so the sensible heat $c\,\Delta T$ freezes a fraction $x = c\,\Delta T/L_f = 4.18 \times 8/334 =
0.10$ of the water; the entropy created, $c\ln(273/265) - xL_f/273 =
124 - 122 = +2\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$, is positive as it must be. Clouds are full of supercooled droplets down to $-40{}^{\circ}\mathrm{C}$; aircraft icing and the seeding of clouds both exploit that.

**Remark 25.16 (Boiling chips and bubble chambers).**

A bubble of radius $r$ in a liquid at pressure $P$ holds vapour at $P + 2\sigma/r$ (surface tension $\sigma$, Year 2 volume); it can only grow if $P_s(T) > P + 2\sigma/r$, so a clean liquid at its boiling point has no bubble small enough to start and superheats by several [kelvin](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-temperature) before erupting — the “bumping” that boiling chips prevent by offering cavities full of trapped gas. The bubble chamber of particle physics ran on the same instability in reverse: a superheated liquid boils first along the track of an ionizing particle.

**Example 25.17 (Humidity and dew).**

Air holds water vapour at a partial pressure $P_v \leq P_s(T)$; the *[relative humidity](#ex-b1-phase-changes-dew)* is $P_v/P_s(T)$, the *[dew point](#ex-b1-phase-changes-dew)* the temperature at which $P_s$ falls to $P_v$. Air at $25{}^{\circ}\mathrm{C}$ ($P_s = 3.17\,\mathrm{kPa}$) and $60\%$ humidity has $P_v = 1.9\,\mathrm{kPa}$, $\rho_v = P_vM/(RT) = 14\,\mathrm{g}/\mathrm{m}^{3}$, and dews on any surface below about $17{}^{\circ}\mathrm{C}$ — the grass at dawn, the cold bottle, the bathroom mirror. Lifted, the same air cools adiabatically at $9.8\,\mathrm{K}/\mathrm{km}$ ([Problem 25.1](#pb-b1-phase-changes-1)) and reaches its [dew point](#ex-b1-phase-changes-dew) a kilometre or so up: that is where the flat bottoms of the cumulus are.

## 25.6 Exercises

**Exercise 25.1 ★.**

Using the diagram of the course, give the [phase](#def-b1-phase-changes-phase) of water (a) at $20{}^{\circ}\mathrm{C}$ under $0.5\,\mathrm{kPa}$; (b) at $-5{}^{\circ}\mathrm{C}$ under $1\,\mathrm{bar}$; (c) at $400{}^{\circ}\mathrm{C}$ under $200\,\mathrm{bar}$; (d) at $0.01{}^{\circ}\mathrm{C}$ under $611\,\mathrm{Pa}$. Why does a puddle dry at $20{}^{\circ}\mathrm{C}$ although water “boils at 100”?

**Solution of Exercise 25.1.**

(a) Vapour: $0.5\,\mathrm{kPa}$ is below $P_s(20{}^{\circ}\mathrm{C}) = 2.3\,\mathrm{kPa}$. (b) Solid. (c) Above $T_c = 374{}^{\circ}\mathrm{C}$: supercritical fluid, no liquid–gas distinction. (d) The [triple point](#prop-b1-phase-changes-PTdiagram): all three. The puddle evaporates from its surface as long as the vapour pressure in the air is below $P_s(20{}^{\circ}\mathrm{C})$; boiling is only the case where bubbles can form inside the liquid.

**Exercise 25.2 ★.**

$100\,\mathrm{g}$ of steam at $100{}^{\circ}\mathrm{C}$ are bubbled into $1.0\,\mathrm{kg}$ of water at $20{}^{\circ}\mathrm{C}$ in an insulated vessel. Final temperature ($L_v = 2257\,\mathrm{kJ}/\mathrm{kg}$, $c = 4.18\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$). Compare with adding $100\,\mathrm{g}$ of water at $100{}^{\circ}\mathrm{C}$.

**Solution of Exercise 25.2.**

$0.1 \times 2257 + 0.1 \times 4.18\,(100 - T_f) = 1.0 \times 4.18\,(T_f - 20)$: $351.1 = 4.60\,T_f$, $T_f = 76{}^{\circ}\mathrm{C}$. With hot water instead: $100 - T_f = 10(T_f - 20)$, $T_f = 27{}^{\circ}\mathrm{C}$ — the [latent heat](#def-b1-phase-changes-phase) is worth $540\,\mathrm{K}$ of sensible heat.

**Exercise 25.3 ★.**

A rigid $2.0\,\mathrm{L}$ vessel contains $1.0\,\mathrm{kg}$ of water at $150{}^{\circ}\mathrm{C}$ in liquid–vapour equilibrium ($P_s =
4.76\,\mathrm{bar}$, $v_l = 1.09 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$, $v_v = 0.393\,\mathrm{m}^{3}/\mathrm{kg}$). Quality, mass of vapour, volume occupied by the liquid.

**Solution of Exercise 25.3.**

$v = 2.0 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$; $x = (2.0 - 1.09) \times 10^{-3}/0.392 = 2.3 \times 10^{-3}$: $2.3\,\mathrm{g}$ of vapour filling $0.91\,\mathrm{L}$; the $0.998\,\mathrm{kg}$ of liquid occupy $0.998 \times 1.09 = 1.09\,\mathrm{L}$.

**Exercise 25.4 ★.**

Entropy of vaporization of water at $100{}^{\circ}\mathrm{C}$, per kilogram and per [mole](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-scales); entropy of fusion per [mole](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-scales) at $0{}^{\circ}\mathrm{C}$. Many ordinary liquids have a molar entropy of vaporization near $88\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ (Trouton’s rule): is water ordinary?

**Solution of Exercise 25.4.**

$2257/373 = 6.05\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$, i.e. $\times 0.018 = 109\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$; fusion $334/273 \times 0.018 = 22\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$. Water’s $109$ is well above Trouton’s $88$: hydrogen bonds make liquid water more ordered than an ordinary liquid, so vaporizing it gains more disorder.

**Exercise 25.5 ★★.**

Boiling temperature of water on the summit of Everest, $P =
0.33\,\mathrm{bar}$, from [Corollary 25.12](#cor-b1-phase-changes-vapourpressure) with $ML_v/R = 4890\,\mathrm{K}$; and at $2000\,\mathrm{m}$ ($P = 0.80\,\mathrm{bar}$). What becomes of a $3\,\mathrm{min}$ egg?

**Solution of Exercise 25.5.**

$1/T = 1/373.15 - \ln(P/P_0)/4890$. Everest: $\ln(0.33/1.013) = -1.12$, $1/T = 2.680 \times 10^{-3} + 2.29 \times 10^{-4}$, $T = 344\,\mathrm{K} =
71{}^{\circ}\mathrm{C}$. At $0.80\,\mathrm{bar}$: $\ln = -0.236$, $T = 366\,\mathrm{K} =
93{}^{\circ}\mathrm{C}$. At $71{}^{\circ}\mathrm{C}$ the white sets slowly and the yolk hardly at all: the $3\,\mathrm{min}$ egg becomes a $10\,\mathrm{min}$ egg or more.

**Exercise 25.6 ★★.**

Temperature inside a pressure cooker whose valve opens at $1.8\,\mathrm{bar}$ absolute; the same if its valve were set at $2.5\,\mathrm{bar}$. Steam at $1.8\,\mathrm{bar}$ has $v_v = 0.98\,\mathrm{m}^{3}/\mathrm{kg}$: how many grams of steam does a $6\,\mathrm{L}$ cooker contain above $4\,\mathrm{L}$ of water?

**Solution of Exercise 25.6.**

$\ln(1.8/1.013) = 0.575$: $1/T = 2.680 \times 10^{-3} - 1.18 \times 10^{-4}$, $T = 390\,\mathrm{K} = 117{}^{\circ}\mathrm{C}$; at $2.5\,\mathrm{bar}$: $\ln = 0.903$, $T = 401\,\mathrm{K} = 128{}^{\circ}\mathrm{C}$. Steam: $2\,\mathrm{L}/0.98 =
2.0\,\mathrm{g}$ — the cooker holds almost nothing but water and a little steam.

**Exercise 25.7 ★★.**

Slope of the fusion curve of water at $0{}^{\circ}\mathrm{C}$ ($L_f =
334\,\mathrm{kJ}/\mathrm{kg}$, $v_s = 1.091 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$, $v_l = 1.000 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$). At what pressure does ice melt at $-1{}^{\circ}\mathrm{C}$? Ice at the bottom of a $3\,\mathrm{km}$ ice sheet ($\rho = 917\,\mathrm{kg}/\mathrm{m}^{3}$): melting temperature there, and a consequence for glacier flow.

**Solution of Exercise 25.7.**

$\dd P/\dd T = 3.34 \times 10^5/[273 \times (-9.1 \times 10^{-5})] = -1.3 \times 10^{7}\,\mathrm{Pa}/\mathrm{K}
= -130\,\mathrm{bar}/\mathrm{K}$: ice melts at $-1{}^{\circ}\mathrm{C}$ under about $130\,\mathrm{bar}$. Under $3\,\mathrm{km}$ of ice, $P = 917 \times 9.8 \times 3000 =
270\,\mathrm{bar}$: melting at $-2{}^{\circ}\mathrm{C}$; a glacier bed can be at its melting point while the surface is far colder, and the film of meltwater lets the ice sheet slide.

**Exercise 25.8 ★★.**

One kilogram of water boils away at $100{}^{\circ}\mathrm{C}$ under $1\,\mathrm{atm}$ ($v_v = 1.673\,\mathrm{m}^{3}/\mathrm{kg}$, $v_l = 1.04 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$). Work received, heat received, $\Delta U$, $\Delta H$, $\Delta S$ of the water; entropy created if the heat comes from a flame at $1500\,\mathrm{K}$.

**Solution of Exercise 25.8.**

$W = -P(v_v - v_l) = -1.013 \times 10^5 \times 1.672 = -169\,\mathrm{kJ}$ (the water pushes the atmosphere back); $Q = L_v = 2257\,\mathrm{kJ}$; $\Delta U = 2257 - 169
= 2088\,\mathrm{kJ}$; $\Delta H = 2257\,\mathrm{kJ}$; $\Delta S = 2257/373 = 6.05\,\mathrm{kJ}/\mathrm{K}$. From a flame at $1500\,\mathrm{K}$: $S_{\mathrm{exch}} = 2257/1500 = 1.50\,\mathrm{kJ}/\mathrm{K}$, created $4.5\,\mathrm{kJ}/\mathrm{K}$ — most of the flame’s quality is wasted heating water.

**Exercise 25.9 ★★.**

Saturation pressures: $1.23\,\mathrm{kPa}$ ($10{}^{\circ}\mathrm{C}$), $1.70\,\mathrm{kPa}$ ($15{}^{\circ}\mathrm{C}$), $2.34\,\mathrm{kPa}$ ($20{}^{\circ}\mathrm{C}$), $3.17\,\mathrm{kPa}$ ($25{}^{\circ}\mathrm{C}$), $4.24\,\mathrm{kPa}$ ($30{}^{\circ}\mathrm{C}$). A room at $22{}^{\circ}\mathrm{C}$ has a [relative humidity](#ex-b1-phase-changes-dew) of $70\%$: vapour pressure, [dew point](#ex-b1-phase-changes-dew) (interpolate), mass of water vapour in $50\,\mathrm{m}^{3}$. A window pane at $12{}^{\circ}\mathrm{C}$: does it fog? Heating the room to $25{}^{\circ}\mathrm{C}$ without adding water: new humidity.

**Solution of Exercise 25.9.**

$P_s(22{}^{\circ}\mathrm{C}) \approx 2.34 + 0.4 \times 0.83 = 2.67\,\mathrm{kPa}$, $P_v = 0.70 \times 2.67 = 1.87\,\mathrm{kPa}$; [dew point](#ex-b1-phase-changes-dew) where $P_s = 1.87$: $15 +
5 \times 0.17/0.64 = 16{}^{\circ}\mathrm{C}$. Mass: $\rho_v = 1870 \times 0.018/
(8.314 \times 295) = 13.7\,\mathrm{g}/\mathrm{m}^{3}$, $0.69\,\mathrm{kg}$ in the room. Pane at $12{}^{\circ}\mathrm{C}$: $P_s \approx 1.42\,\mathrm{kPa} < 1.87$, it fogs. At $25{}^{\circ}\mathrm{C}$ the humidity falls to $1.87/3.17 = 59\%$.

**Exercise 25.10 ★★★.**

*The [triple point](#prop-b1-phase-changes-PTdiagram).* (a) Show from the state-function character of $h$ that, at the [triple point](#prop-b1-phase-changes-PTdiagram), $L_s = L_f + L_v$ (sublimation = fusion + vaporization). (b) Show with Clausius–Clapeyron that the sublimation curve is steeper than the vaporization curve at the [triple point](#prop-b1-phase-changes-PTdiagram) (neglect $v_s$, $v_l$ before $v_v$). (c) Water: $L_f = 334\,\mathrm{kJ}/\mathrm{kg}$, $L_v = 2500\,\mathrm{kJ}/\mathrm{kg}$, $T = 273.16\,\mathrm{K}$, $P = 611\,\mathrm{Pa}$, vapour a [perfect gas](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-model): both slopes in $\mathrm{Pa}/\mathrm{K}$. (d) Why does snow disappear on a dry cold day without ever melting?

**Solution of Exercise 25.10.**

(a) $h$ is a state function: going solid $\to$ liquid $\to$ vapour at the [triple point](#prop-b1-phase-changes-PTdiagram) or solid $\to$ vapour directly gives the same $\Delta h$: $L_s = L_f + L_v$. (b) With $v_s, v_l \ll v_v$ both slopes are $L/(Tv_v)$ at the same $T$ and $v_v$: their ratio is $L_s/L_v > 1$. (c) $v_v = RT/(MP) =
8.314 \times 273.16/(0.018 \times 611) = 206\,\mathrm{m}^{3}/\mathrm{kg}$; vaporization $2.50 \times 10^6/(273.16 \times 206) = 44\,\mathrm{Pa}/\mathrm{K}$, sublimation $2.83 \times 10^6/
(273.16 \times 206) = 50\,\mathrm{Pa}/\mathrm{K}$. (d) Below $0{}^{\circ}\mathrm{C}$ at $1\,\mathrm{bar}$ no liquid exists; if the air’s vapour pressure is below $P_s$ over ice ($260\,\mathrm{Pa}$ at $-10{}^{\circ}\mathrm{C}$) the snow sublimates — the dry cold wind eats it.

**Exercise 25.11 ★★★.**

*Superheated water.* A bubble of radius $r$ in water at pressure $P_0$ contains vapour at $P_0 + 2\sigma/r$ with $\sigma = 0.059\,\mathrm{N}/\mathrm{m}$ (admitted). (a) Minimum radius of a bubble that can grow in water at $105{}^{\circ}\mathrm{C}$ under $1.013\,\mathrm{bar}$, using $\dd P_s/\dd T = 3.6\,\mathrm{kPa}/\mathrm{K}$. (b) Same at $101{}^{\circ}\mathrm{C}$. (c) A cup of water superheated to $105{}^{\circ}\mathrm{C}$ in a microwave oven is disturbed by a spoon: fraction of the water that flashes to steam, and the volume of steam produced per litre of water — comment on the danger. (d) Why do boiling chips prevent this?

**Solution of Exercise 25.11.**

(a) $P_s(105{}^{\circ}\mathrm{C}) \approx 1.013 + 5 \times 0.036 = 1.19\,\mathrm{bar}$: $\Delta P = 0.18\,\mathrm{bar}$, $r^\ast = 2 \times 0.059/(1.8 \times 10^4) = 6.6\,\text{µ}\mathrm{m}$. (b) $\Delta P = 3.6\,\mathrm{kPa}$, $r^\ast = 33\,\text{µ}\mathrm{m}$. (c) $x = c\,\Delta T/L_v =
4.18 \times 5/2257 = 0.93\%$: $9.3\,\mathrm{g}$ per litre, i.e. $9.3 \times 10^{-3}
\times 1.67 = 16\,\mathrm{L}$ of steam produced in a fraction of a second inside a cup — the water is blown out, boiling. (d) Chips hold air in their pores: bubbles of tens of micrometres already exist, so boiling starts as soon as $P_s$ reaches $P_0$.

**Exercise 25.12 ★★★.**

*A [refrigerant](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-vapourcycle)’s cycle.* A [heat pump](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-machine) runs R-134a round [Definition 24.16](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-vapourcycle): evaporation at $-10{}^{\circ}\mathrm{C}$ ($2.0\,\mathrm{bar}$; $h_l = 187\,\mathrm{kJ}/\mathrm{kg}$, $h_v =
392\,\mathrm{kJ}/\mathrm{kg}$), condensation at $40{}^{\circ}\mathrm{C}$ ($10.2\,\mathrm{bar}$; $h_l = 256\,\mathrm{kJ}/\mathrm{kg}$, $h_v = 419\,\mathrm{kJ}/\mathrm{kg}$). The fluid leaves the condenser as saturated liquid and the evaporator as saturated vapour; the compressor delivers it at $h = 430\,\mathrm{kJ}/\mathrm{kg}$. Admit that in the valve $h$ is conserved and that the compressor’s work per kilogram is $\Delta h$ (steady-flow balances, Year 2 volume); in the exchangers, isobaric, $q = \Delta h$ ([Proposition 22.7](https://one-course.com/books/physics/3/en/chapter/22-the-first-law-of-thermodynamics#prop-b1-first-law-qvqp)). (a) Quality at the evaporator inlet. (b) Heat taken per kilogram in the evaporator, heat given in the condenser, work of compression; check the first law. (c) $e_p$ and $e_f$; compare with the Carnot values between the two saturation temperatures. (d) Mass flow rate for $6\,\mathrm{kW}$ of heating.

**Solution of Exercise 25.12.**

(a) After the valve $h = 256\,\mathrm{kJ}/\mathrm{kg}$ at $-10{}^{\circ}\mathrm{C}$: $x =
(256 - 187)/(392 - 187) = 0.34$. (b) Evaporator $q_c = 392 - 256 =
136\,\mathrm{kJ}/\mathrm{kg}$; compressor $w = 430 - 392 = 38\,\mathrm{kJ}/\mathrm{kg}$; condenser $q_h = 256 - 430 = -174\,\mathrm{kJ}/\mathrm{kg}$; $136 + 38 - 174 = 0$. (c) $e_p =
174/38 = 4.6$, $e_f = 136/38 = 3.6$; Carnot $313/50 = 6.3$ and $263/50 =
5.3$: about $0.7$ of the ideal. (d) $\dot m = 6000/174000 = 34\,\mathrm{g}/\mathrm{s}$, $2.1\,\mathrm{kg}$ a minute.

![A pressure cooker: at 1.8\, bar water boils at 117 C, and the weighted valve is what sets the pressure.](https://one-course.com/images/onecourse/chapters/physics-3/b1-phase-changes/img-3a8cb16dd599.jpg)

*A pressure cooker: at $1.8\,\mathrm{bar}$ water boils at $117{}^{\circ}\mathrm{C}$, and the weighted valve is what sets the pressure.*

## 25.7 Problem: Water in four scenes

**Problem 25.1.**

Weekend problem — the pressure cooker, the cloud, the sealed vessel and the supercooled bottle: one substance, four uses of its phase diagram

Data for water: $L_v = 2257\,\mathrm{kJ}/\mathrm{kg}$ at $100{}^{\circ}\mathrm{C}$ and about $2.2\,\mathrm{MJ}/\mathrm{kg}$ at $117{}^{\circ}\mathrm{C}$, $2.45\,\mathrm{MJ}/\mathrm{kg}$ near $20{}^{\circ}\mathrm{C}$; $L_f = 334\,\mathrm{kJ}/\mathrm{kg}$; $c = 4.18\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$ (liquid); $M = 18\,\mathrm{g}/\mathrm{mol}$; $ML_v/R = 4890\,\mathrm{K}$ near $100{}^{\circ}\mathrm{C}$. Saturation pressures: $0.61\,\mathrm{kPa}$ ($0{}^{\circ}\mathrm{C}$), $1.23\,\mathrm{kPa}$ ($10{}^{\circ}\mathrm{C}$), $1.70\,\mathrm{kPa}$ ($15{}^{\circ}\mathrm{C}$), $2.34\,\mathrm{kPa}$ ($20{}^{\circ}\mathrm{C}$), $3.17\,\mathrm{kPa}$ ($25{}^{\circ}\mathrm{C}$), $4.24\,\mathrm{kPa}$ ($30{}^{\circ}\mathrm{C}$), $101.3\,\mathrm{kPa}$ ($100{}^{\circ}\mathrm{C}$). Saturated specific volumes: $v_l = 1.0 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$ (cold water), $1.9 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$ at $360{}^{\circ}\mathrm{C}$; $v_v = 0.0217\,\mathrm{m}^{3}/\mathrm{kg}$ at $300{}^{\circ}\mathrm{C}$, $0.0108$ at $340{}^{\circ}\mathrm{C}$, $0.0088$ at $350{}^{\circ}\mathrm{C}$; [critical point](#prop-b1-phase-changes-PTdiagram) $374{}^{\circ}\mathrm{C}$, $221\,\mathrm{bar}$, $v_c = 3.1 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$. Air: $M = 29\,\mathrm{g}/\mathrm{mol}$, $c_p =
1.0\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$, $\gamma = 1.4$; $g = 9.8\,\mathrm{m}/\mathrm{s}^{2}$; $R =
8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$.

**Part I — The pressure cooker.**

1. Why can water in an open pan at sea level not be heated above $100{}^{\circ}\mathrm{C}$ , however strong the flame?
2. Starting from the [Clausius–Clapeyron relation](#thm-b1-phase-changes-clapeyron) , derive $\ln[P_s(T)/P_s(T_0)] = (ML_v/R)(1/T_0 - 1/T)$ , stating the approximations.
3. Temperature in a cooker whose valve opens at $1.8\,\mathrm{bar}$ absolute.
4. The valve is a weight resting on an orifice of $20\,\mathrm{mm}^{2}$ ; atmospheric pressure $1.0\,\mathrm{bar}$ : mass of the weight.
5. $2.0\,\mathrm{L}$ of water are brought from $20{}^{\circ}\mathrm{C}$ to the working temperature by a $2.0\,\mathrm{kW}$ plate (neglect the pot): heat and time.
6. Once there, the plate is left at $2.0\,\mathrm{kW}$ : mass of steam leaving the valve per minute. What should the cook do?
7. Cooking reactions roughly double in rate for every $10\,\mathrm{K}$ : by what factor does the cooker shorten a $30\,\mathrm{min}$ sea-level recipe? And how long would the recipe take on Everest, where water boils at $70{}^{\circ}\mathrm{C}$ ?

**Part II — The cloud.**

8. Define [relative humidity](#ex-b1-phase-changes-dew) . Air at $25{}^{\circ}\mathrm{C}$ and $60\%$ : partial pressure of vapour, and mass of vapour per cubic metre ( [perfect gas](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-model) ).
9. [Dew point](#ex-b1-phase-changes-dew) of that air (interpolate the table). Explain the dew on the grass at dawn and the drops on a cold bottle.
10. A parcel of dry air rises without exchanging heat. From the reversible adiabatic law and the [hydrostatic relation](https://one-course.com/books/physics/3/en/chapter/21-fluid-statics#thm-b1-fluid-statics-fundamental) $\dd P/\dd z = -\rho g$ , show that $\dd T/\dd z = -g/c_p$ and evaluate it.
11. The [dew point](#ex-b1-phase-changes-dew) of the rising parcel itself falls by about $2\,\mathrm{K}/\mathrm{km}$ as the pressure drops. Height at which the parcel of question 8 becomes saturated — the base of the cloud.
12. Above the base, the vapour that condenses releases its [latent heat](#def-b1-phase-changes-phase) into the parcel. By how much does condensing $1\,\mathrm{g}$ of water per kilogram of air warm that air? What does this do to the cooling rate of the parcel as it keeps rising, and why does it matter for thunderstorms?
13. A cumulus of $1\,\mathrm{km}^{3}$ holds $1\,\mathrm{g}$ of liquid water per cubic metre: mass of water, and the energy released by its condensation, in joules and in $\mathrm{GW}\mathrm{h}$ .
14. A bathroom mirror at $18{}^{\circ}\mathrm{C}$ , air at $25{}^{\circ}\mathrm{C}$ and $80\%$ : does it fog? At what humidity would it stop?

**Part III — The sealed vessel.** A rigid steel vessel of $1.0\,\mathrm{L}$ is filled with a mass $m$ of water and sealed with no air, then heated slowly.

15. $m = 0.50\,\mathrm{kg}$ at $20{}^{\circ}\mathrm{C}$ : pressure inside, specific volume, mass of vapour (use the [perfect gas](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-model) for $v_v$ ), fraction of the volume filled by liquid.
16. Heated: does the liquid level rise or fall? At roughly what temperature does the liquid fill the vessel, and what happens to the pressure after that? (Use $v_l$ at $360{}^{\circ}\mathrm{C}$ .)
17. $m = 0.10\,\mathrm{kg}$ : does the liquid level rise or fall on heating? At roughly what temperature does the last drop evaporate (use the $v_v$ table)?
18. $m = 0.31\,\mathrm{kg}$ : describe what an observer sees at the meniscus as the temperature approaches $374{}^{\circ}\mathrm{C}$ .
19. Sketch the three heating paths in the $(v, P)$ diagram with the saturation dome, and say which side of the [critical point](#prop-b1-phase-changes-PTdiagram) each passes.
20. The same $1.0\,\mathrm{L}$ vessel with $0.50\,\mathrm{kg}$ of water is now sealed *with* the air it contained at $20{}^{\circ}\mathrm{C}$ , $1.0\,\mathrm{bar}$ , and heated to $100{}^{\circ}\mathrm{C}$ : total pressure inside (Dalton), and the force on a lid of $100\,\mathrm{cm}^{2}$ .

**Part IV — The supercooled bottle.**

21. A clean, still bottle of water has cooled to $-8{}^{\circ}\mathrm{C}$ without freezing. What is this state called, why is it possible, and what happens when the bottle is knocked? Why does the temperature then settle at exactly $0{}^{\circ}\mathrm{C}$ ?
22. Fraction of the water that freezes (the process is fast, hence adiabatic).
23. Entropy created per kilogram; sign check.
24. The reverse surprise: a cup of water heated to $105{}^{\circ}\mathrm{C}$ in a microwave oven without boiling, then disturbed. Fraction of the water that flashes to steam, volume of steam produced per litre ( $v_v = 1.67\,\mathrm{m}^{3}/\mathrm{kg}$ ), and why the cup erupts.
25. [Lost work](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#thm-b1-heat-engines-lostwork) : the supercooled litre could have driven a reversible engine with the $0{}^{\circ}\mathrm{C}$ surroundings; taking $T_0 = 273\,\mathrm{K}$ , the work thrown away by letting it freeze on a knock is $T_0S_{\mathrm{created}}$ . Evaluate it per litre and compare with lifting the bottle by one metre. Conclude: what does a [metastable state](#def-b1-phase-changes-metastable) store?

**Solution of Problem 25.1.**

**1.** At $1\,\mathrm{atm}$ water boils at $100{}^{\circ}\mathrm{C}$; while liquid remains, any heat supplied makes vapour at that temperature ([latent heat](#def-b1-phase-changes-phase)) instead of warming the liquid: the temperature is pinned by the pressure.

**2.** $\dd P_s/\dd T = L_v/[T(v_v - v_l)]$ with $v_l \ll v_v = RT/(MP_s)$ (perfect-gas vapour): $\dd P_s/P_s = (ML_v/R)\,\dd T/T^2$; with $L_v$ taken constant, integrate: $\ln(P_s/P_0) = (ML_v/R)(1/T_0 - 1/T)$.

**3.** $1/T = 1/373.15 - \ln(1.8/1.013)/4890 = 2.562 \times 10^{-3}$: $T =
390\,\mathrm{K} = 117{}^{\circ}\mathrm{C}$.

**4.** $m = \Delta P\,A/g = 0.8 \times 10^5 \times 2.0 \times 10^{-5}/9.8 =
0.16\,\mathrm{kg}$.

**5.** $Q = 2.0 \times 4.18 \times 97 = 811\,\mathrm{kJ}$; $t = 811/2.0 =
405\,\mathrm{s} \approx 7\,\mathrm{min}$.

**6.** $2000/(2.2 \times 10^6) = 0.91\,\mathrm{g}/\mathrm{s} = 55\,\mathrm{g}/\mathrm{min}$ — two litres gone in $37\,\mathrm{min}$. Turn the plate down to the least that keeps the valve hissing: the temperature is set by the pressure, not by the flame.

**7.** $+17\,\mathrm{K}$: $2^{1.7} = 3.2$ times faster, $30\,\mathrm{min}$ $\to$ $9\,\mathrm{min}$. Everest, $-30\,\mathrm{K}$: $2^{-3}$, four hours.

**8.** [Relative humidity](#ex-b1-phase-changes-dew) $= P_v/P_s(T)$. $P_v = 0.60 \times 3.17 =
1.90\,\mathrm{kPa}$; $\rho_v = P_vM/(RT) = 1900 \times 0.018/(8.314 \times 298) =
13.8\,\mathrm{g}/\mathrm{m}^{3}$.

**9.** $P_s(T_d) = 1.90\,\mathrm{kPa}$: between $15{}^{\circ}\mathrm{C}$ ($1.70\,\mathrm{kPa}$) and $20{}^{\circ}\mathrm{C}$ ($2.34\,\mathrm{kPa}$): $T_d = 15 + 5 \times 0.20/0.64
= 16.6{}^{\circ}\mathrm{C}$. Grass radiating to the night sky, or a bottle from the [refrigerator](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-machine), falls below $T_d$: the vapour in contact with it is then above saturation and condenses.

**10.** $TP^{(1-\gamma)/\gamma} = \text{const}$: $\dd T/T = \frac{\gamma - 1}{\gamma}
\dd P/P$; $\dd P = -\rho g\,\dd z$ with $\rho = PM/(RT)$: $\dd T/T = -\frac{\gamma -
1}{\gamma}\frac{Mg}{RT}\dd z$, so $\dd T/\dd z = -\frac{\gamma - 1}{\gamma}\frac{Mg}{R} =
-g/c_p$ since $c_p = \gamma R/[(\gamma - 1)M]$. Value $9.8/1000 = 9.8\,\mathrm{K}/\mathrm{km}$.

**11.** $T - T_d$ closes at $9.8 - 2 = 7.8\,\mathrm{K}/\mathrm{km}$: $z = (25 - 16.6)/
7.8 = 1.1\,\mathrm{km}$.

**12.** $1 \times 2.45 = 2.45\,\mathrm{kJ}$ per kilogram of air: $\Delta T =
2.45\,\mathrm{K}$. The release offsets part of the adiabatic cooling: a saturated parcel cools at only $5\text{ to }6\,\mathrm{K}/\mathrm{km}$, stays warmer and lighter than its surroundings and keeps rising — the [latent heat](#def-b1-phase-changes-phase) is the fuel of the thunderstorm.

**13.** $10^9 \times 1\,\mathrm{g} = 1000\,\mathrm{t}$; $10^6 \times 2.45 \times 10^6 =
2.5 \times 10^{12}\,\mathrm{J} = 0.68\,\mathrm{GW}\mathrm{h}$ — forty minutes of a large power plant.

**14.** $P_v = 0.80 \times 3.17 = 2.54\,\mathrm{kPa}$; $P_s(18{}^{\circ}\mathrm{C})
\approx 1.70 + 0.6 \times 0.64 = 2.08\,\mathrm{kPa} < 2.54$: it fogs, until the humidity drops below $2.08/3.17 = 66\%$.

**15.** $P = P_s(20{}^{\circ}\mathrm{C}) = 2.3\,\mathrm{kPa}$; $v = 2.0 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$; $v_v = RT/(MP_s) = 8.314 \times 293/(0.018 \times 2340) = 58\,\mathrm{m}^{3}/\mathrm{kg}$; $x =
(2.0 - 1.0) \times 10^{-3}/58 = 1.7 \times 10^{-5}$: $9\,\mathrm{mg}$ of vapour; liquid $0.50 \times 10^{-3} = 0.50\,\mathrm{L}$, half the vessel.

**16.** $v = 2.0 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg} < v_c$: the isochore meets the liquid branch of the dome — the liquid expands faster than it evaporates, the level *rises*, and the vessel is full of liquid when $v_l(T) =
2.0 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg}$, a little above $360{}^{\circ}\mathrm{C}$ (about $190\,\mathrm{bar}$). Beyond, a nearly incompressible liquid heated at constant volume: the pressure climbs by tens of bar per [kelvin](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-temperature) — the vessel bursts. Never seal a vessel full of liquid.

**17.** $v = 1.0 \times 10^{-2}\,\mathrm{m}^{3}/\mathrm{kg} > v_c$: the level *falls*; the last drop goes when $v_v(T) = 0.010$, between $340{}^{\circ}\mathrm{C}$ ($0.0108$) and $350{}^{\circ}\mathrm{C}$ ($0.0088$): about $344{}^{\circ}\mathrm{C}$, $150\,\mathrm{bar}$.

**18.** $v = 3.2 \times 10^{-3}\,\mathrm{m}^{3}/\mathrm{kg} \approx v_c$: the meniscus stays near mid-height, grows flatter and fainter as the two densities converge, the fluid turns milky (critical opalescence), and at $374{}^{\circ}\mathrm{C}$, $221\,\mathrm{bar}$ the meniscus vanishes: one [phase](#def-b1-phase-changes-phase).

**19.** Three vertical lines: $v = 2.0 \times 10^{-3}$ leaves the dome through the liquid branch, left of C; $10^{-2}$ through the vapour branch, right of C; $3.2 \times 10^{-3}$ through the summit.

**20.** Air: $0.5\,\mathrm{L}$ at $1.0\,\mathrm{bar}$, $293\,\mathrm{K}$, same volume at $373\,\mathrm{K}$: $P_{\text{air}} = 1.0 \times 373/293 = 1.27\,\mathrm{bar}$; plus $P_s = 1.01\,\mathrm{bar}$: $2.3\,\mathrm{bar}$ inside, $1.3\,\mathrm{bar}$ above the outside: $1.3 \times 10^5 \times 10^{-2} = 1.3\,\mathrm{kN}$ on the lid — the weight of $130\,\mathrm{kg}$.

**21.** Supercooled, metastable: no [nucleation](#def-b1-phase-changes-metastable) site (clean water, no dust, no motion) for the first crystal. A knock nucleates ice, which grows through the bottle releasing $L_f$; the temperature rises to $0{}^{\circ}\mathrm{C}$ and stops there, the only temperature at which ice and water coexist at $1\,\mathrm{atm}$: the freezing halts as soon as the released heat has brought the mixture to it.

**22.** $x = c\,\Delta T/L_f = 4.18 \times 8/334 = 0.10$.

**23.** $\Delta s = c\ln(273/265) - xL_f/273 = 124.3 - 122.3 =
+2.0\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$: adiabatic, hence all created; positive, as a spontaneous process must be.

**24.** $x = 4.18 \times 5/2257 = 0.93\%$: $9.3\,\mathrm{g}$ of steam per litre, $9.3 \times 10^{-3} \times 1.67 = 16\,\mathrm{L}$ of vapour born in the bulk in a fraction of a second: the cup erupts.

**25.** $T_0S_{\mathrm{created}} = 273 \times 2.0 = 550\,\mathrm{J}$ per litre, against $mgh = 9.8\,\mathrm{J}$: fifty-six times more. A [metastable state](#def-b1-phase-changes-metastable) stores work — what a reversible path could have extracted — and a knock turns it all into entropy.