General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: states of matter and phase diagrams (202-SN1-RE)

This chapter bridges microscopic structure and macroscopic properties: the stronger the intermolecular forces, the more you must heat to separate the molecules. Everything that follows, heating curves, vapour pressures, phase diagrams, is only a consequence of this competition between thermal motion and cohesion.

This sheet gives the two formulas of the heat balance and the rule that decides between them, the reading of a phase diagram with the anomaly of water, and the counting of atoms in a cubic unit cell, which is the most regular calculation question of the chapter. It is written for CEGEP students in General Chemistry 202-SN1-RE (formerly 202-NYA). Once you have read the sheet, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

During a phase change, the temperature does NOT move: all the energy goes into breaking interactions, not into more motion. This single idea is what separates $q=mc\Delta T$ from $q=n\Delta H$ , and ignoring it spoils an entire heating calculation.

This chapter is part of 202-SN1-RE, General Chemistry

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Intermolecular forces
Hydrogen bonding
Heating curve
Enthalpy of fusion
Enthalpy of vaporization
Saturated vapour pressure
Clausius-Clapeyron equation
Phase diagram
Triple point
Critical point
Cubic unit cell
Packing efficiency

The essentials

The two formulas of the heat balance, and when to use them

• WITHIN a phase, the temperature rises: $q=mc\Delta T$ , with $c$ specific to each phase ( $2.09$ for ice, $4.18$ for liquid water, $2.01$ for steam, in J/(g·°C)).
• DURING a phase change, the temperature does not move: $q=n\Delta H$ , with $\Delta H_{fus}=6.01$ kJ/mol and $\Delta H_{vap}=40.7$ kJ/mol for water.
• A complete balance is therefore cut into as many segments as the heating curve has, and each segment has its own formula.
• Vaporization costs about SEVEN times as much as melting: it dominates every balance, and it is the longest plateau of the curve.
• Increasing order of intermolecular forces: London, present everywhere and growing with molar mass; dipole-dipole; hydrogen bonding.

On the two PLATEAUS, the temperature does not move even though we keep heating: this is where

q=n\Delta H

replaces

q=mc\Delta T

. The vaporization plateau alone takes up three quarters of the balance.

The energy of a plateau does not depend on the temperature reached but on the AMOUNT OF SUBSTANCE: that is why the enthalpies of phase change are in kilojoules per mole, whereas specific heats are in joules per gram.

Vapour pressure and boiling

• Vapour pressure is the pressure exerted by the vapour in equilibrium with its liquid. It depends ONLY on temperature, never on the amount of liquid or the volume of the container.
• Boiling occurs when the vapour pressure EQUALS the external pressure: the boiling point therefore depends on altitude.
• A more VOLATILE liquid has a higher vapour pressure at a given temperature, hence a lower boiling point.
• Clausius-Clapeyron: $\ln\dfrac{P_{2}}{P_{1}}=-\dfrac{\Delta H_{vap}}{R}\left(\dfrac{1}{T_{2}}-\dfrac{1}{T_{1}}\right)$ , with $R=8.314$ J/(mol·K) and $T$ IN KELVINS.
• Surface tension and viscosity both grow with the intermolecular forces; capillary rise is $h=\dfrac{2\gamma\cos\theta}{\rho g r}$ .

Plotted against

1/T

in kelvins,

\ln P

for water lines up: the slope is

-\Delta H_{vap}/R

, and two points are enough to relate pressure and temperature.

In Montreal, water boils at $100$ °C; at altitude, the external pressure being lower, it boils lower and cooking takes longer. The boiling point is therefore not a constant of the substance, contrary to what the tables suggest.

Phase diagrams and crystalline solids

• Three regions separated by three curves: melting, vaporization, sublimation. Their intersection is the TRIPLE POINT, where the three phases coexist.
• The vaporization curve stops at the CRITICAL POINT, beyond which liquid and gas are no longer distinct: this is the supercritical region.
• ANOMALY OF WATER: its melting curve has a NEGATIVE slope, because ice is LESS dense than liquid water. Almost every other substance has a positive slope.
• Four types of solids: molecular (low melting point, insulator), ionic (high melting point, conducts only when molten or dissolved), covalent network (very high melting point, very hard), metallic (conductor, ductile).
• Cubic unit cells: simple ( $Z=1$ , $a=2r$ , $52.4\%$ ); body-centred ( $Z=2$ , $a\sqrt{3}=4r$ , $68.0\%$ ); face-centred ( $Z=4$ , $a\sqrt{2}=4r$ , $74.0\%$ ). Density: $\rho=\dfrac{ZM}{N_{A}a^{3}}$ .

The triple point opens the vaporization curve, the critical point closes it: beyond, liquid and gas can no longer be told apart. The three curves separate three regions.

The count $Z$ comes from the fractions: a corner atom is shared among eight cells, a face atom between two, and a body-centre atom belongs entirely to its cell. These three fractions give the three values of $Z$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using the specific heat during a phase change

the whole question, 6 marks

What not to write

“Ice goes from $0$ °C to $0$ °C as it melts: $q=mc\Delta T=50\times 4.18\times 0=0$ J.”

What to write

“During a plateau, $\Delta T=0$ but the energy is not zero: $q=n\Delta H_{fus}=2.77\times 6.01=16.7$ kJ.”

On the plateau, the temperature does not move and yet the ice receives

16.7

kJ: with

\Delta T=0

, the formula

mc\Delta T

would wipe out that entire segment.

Why: The energy of the plateau goes into breaking the interactions between molecules, not into increasing their motion. That is exactly why the temperature stays constant while we keep heating.

2. Making vapour pressure depend on the amount of liquid

3 marks on the reasoning question

What not to write

“There is twice as much water in the second container, so the vapour pressure there is twice as great.”

What to write

“Vapour pressure depends ONLY on temperature: it is the same in both containers as long as some liquid remains.”

With twice as much liquid, the two pressure gauges read the same pressure: vapour pressure depends only on temperature, as long as some liquid remains.

Why: It is a dynamic equilibrium: as many molecules evaporate as condense, and this rate depends only on thermal motion. The amount of liquid changes only the DURATION for which the equilibrium can be maintained.

3. Taking the boiling point for a constant of the substance

3 marks, and the pressure-cooker question missed

What not to write

“Water boils at $100$ °C, everywhere and always.”

What to write

“Water boils when its vapour pressure equals the EXTERNAL pressure. At $101.3$ kPa, that happens at $100$ °C; at altitude, the pressure being lower, water boils lower.”

A liquid boils where its curve CROSSES the line of the external pressure. Lower that line, as at altitude, and both boiling points go down.

Why: The table gives the NORMAL boiling point, that is, under one atmosphere. A pressure cooker raises the pressure so that water boils higher and cooks faster.

4. Leaving the temperatures in degrees Celsius in Clausius-Clapeyron

the whole question, 5 marks

What not to write

“ $\ln\dfrac{P_{2}}{P_{1}}=-\dfrac{\Delta H_{vap}}{R}\left(\dfrac{1}{80}-\dfrac{1}{100}\right)$ .”

What to write

“Temperatures go in KELVINS: $\dfrac{1}{353.15}-\dfrac{1}{373.15}$ . With degrees Celsius, the difference of the reciprocals is about sixteen times too large.”

Why: The equation comes from an integration in which $T$ is an absolute temperature. The reciprocals of $353$ and $373$ differ by $1.5\times 10^{-4}$ , those of $80$ and $100$ by $2.5\times 10^{-3}$ : the ratio exceeds $16$ .

5. Giving the melting curve of water the slope of other substances

4 marks on the reasoning question

What not to write

“Compressing ice makes it even more solid, like any other substance.”

What to write

“The melting curve of water has a NEGATIVE slope: compressing ice makes it MELT, because the liquid is denser than the solid.”

The solid-liquid boundary of water LEANS TO THE LEFT: compressing ice makes it melt. For carbon dioxide, as for almost every substance, it leans to the right.

Why: Pressure always favours the DENSER phase. For almost every substance, that is the solid; for water, whose ice floats, it is the liquid. That is also why ice melts under the blade of a skate.

6. Confusing the triple point and the critical point

2 marks, but the question comes up on almost every exam

What not to write

“At the critical point, the three phases coexist.”

What to write

“It is at the TRIPLE POINT that the three phases coexist. The CRITICAL POINT ends the vaporization curve: beyond it, liquid and gas are no longer distinct.”

Why: The two points are at opposite ends of the same curve: the triple point at its start, the critical point at its end. Confusing them makes it impossible to read the supercritical region.

7. Comparing boiling points without taking molar mass into account

3 marks on ranking questions

What not to write

“Dibromine has only London forces and hydrogen sulfide is polar, so the second boils higher.”

What to write

“London forces grow strongly with molar mass: $Br_{2}$ ( $160$ g/mol) boils at $59$ °C, $H_{2}S$ ( $34$ g/mol) at $-60$ °C. Here mass wins over polarity.”

Dibromine, with no polarity at all, boils

119

°C higher than polar hydrogen sulfide: its London forces, strong because the molecule is heavy, win.

Why: London forces are weak only per unit of contact: on a large, highly polarizable molecule, their sum far exceeds a dipole-dipole interaction between small molecules.

8. Counting the atoms of a unit cell without their fractions

the whole question, 5 marks, and a density three times too large

What not to write

“The face-centred cubic unit cell has $8$ corners and $6$ faces, so $14$ atoms.”

What to write

“A corner counts for $\tfrac{1}{8}$ and a face for $\tfrac{1}{2}$ : $Z=8\times\tfrac{1}{8}+6\times\tfrac{1}{2}=1+3=4$ atoms per unit cell.”

A corner atom (orange) is shared among eight cells, a face atom between two, and the one at the centre belongs only to its own cell:

Z

1

2

and

4

, never

14

Why: A corner atom is shared among the eight cells that meet there, a face atom between the two cells that share the face. Only a body-centre atom belongs entirely to its cell.

9. Forgetting to convert the unit cell edge in the density calculation

3 marks, and a result with no connection to reality

What not to write

“ $a=361.5$ pm, so $\rho=\dfrac{4\times 63.55}{6.022\times 10^{23}\times 361.5^{3}}$ .”

What to write

“ $a=361.5$ pm $=3.615\times 10^{-8}$ cm, so $a^{3}=4.724\times 10^{-23}$ cm $^{3}$ and $\rho=8.94$ g/cm $^{3}$ .”

Why: Density is expected in grams per cubic centimetre, so the unit cell edge must be in centimetres. One picometre is $10^{-10}$ cm, and the error is cubed.

10. Believing that a solid must always melt before becoming a gas

3 marks, and the phase diagram reading missed

What not to write

“Dry ice melts first, then evaporates.”

What to write

“Under one atmosphere, carbon dioxide SUBLIMES directly: its triple-point pressure is $5.1$ atm, so the liquid state does not exist at ambient pressure.”

Under one atmosphere, the horizontal line passes below the triple point of carbon dioxide: it goes from solid to gas without ever touching the liquid region.

Why: Passing through the liquid is possible only if the pressure is ABOVE that of the triple point. It is the reading of a phase diagram that says so, and it is exactly what freeze-drying exploits.

Which method to choose

Which formula for each segment of a heating process

What the temperature does on the segment considered

If the temperature CHANGES and the substance stays in the same phase → $q=mc\Delta T$ , with the $c$ of THIS phase

Example: ice from $-20$ to $0$ °C: $c=2.09$ J/(g·°C)
If the temperature stays CONSTANT and the substance changes state → $q=n\Delta H$ , with the enthalpy of the phase change involved

Example: melting at $0$ °C: $\Delta H_{fus}=6.01$ kJ/mol
If the question asks for the TOTAL energy → add up all the segments, watching the units: joules for some, kilojoules for others

Example: five segments from ice to steam
If the question gives an energy and asks for the final temperature → use up the segments one by one until the energy runs out, and note where you stop

if the energy runs out during a plateau, the final temperature is that of the plateau
If the question mixes two substances at different temperatures → equate the heat given up and the heat received, checking whether a phase change occurs

ice that does not melt completely leaves the mixture at exactly $0$ °C

Always sketch the heating curve before calculating, even freehand: it shows the number of segments, and a forgotten segment is the most frequent error of the chapter.

Reading a phase diagram

What the question asks about the diagram

If the question asks for the state of a substance at a given $(P,T)$ → locate the region in which the point falls

Example: $CO_{2}$ at $1$ atm and $-50$ °C: solid region
If the question asks what happens when heating at constant pressure → follow a HORIZONTAL line and note each curve crossed

Example: under $1$ atm, $CO_{2}$ goes directly from solid to gas
If the question asks what happens when compressing at constant temperature → follow a VERTICAL line upward

Example: compressing ice at $-1$ °C makes it melt
If the question asks where the three phases coexist → TRIPLE point, at the intersection of the three curves

Example: $0.0098$ °C and $0.61$ kPa for water
If the question asks from when liquid and gas can no longer be told apart → CRITICAL point, at the end of the vaporization curve

beyond it, the supercritical fluid dissolves like a liquid and diffuses like a gas

The slope of the melting curve is the only piece of information on the diagram that changes sign from one substance to another. Check which way it leans before drawing any conclusion about the effect of compression.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a complete heating balance

When to use it: The question takes a given mass from one temperature to another, crossing one or more phase changes.

1 Sketch the heating curve freehand and NUMBER the segments: each rise and each plateau counts as one.
2 Convert the mass to moles once and for all, since the plateaus use amounts of substance.
3 Calculate each segment separately, with its formula and its unit written out: joules for the $mc\Delta T$ terms, kilojoules for the $n\Delta H$ terms.
4 Convert all the results to the SAME unit before adding.
5 Add up and comment on the breakdown: which segment dominates the balance, and why.

Concluding sentence

“The heating breaks down into five segments: $2.09$ kJ for the ice, $16.7$ kJ for the melting, $20.9$ kJ for the liquid water, $113$ kJ for the vaporization and $2.01$ kJ for the steam, for a total of $155$ kJ, of which $73\%$ is for vaporization alone.”

The trap: Mixing joules and kilojoules at the moment of adding. Specific heats are in joules per gram and enthalpies in kilojoules per mole: this is the only place in the calculation where a factor of a thousand can sneak in.

Marking: Typically 1 mark per segment correctly identified and calculated, 1 mark for the consistent conversion of units, 1 mark for the total and its comment.

Calculating a density from a unit cell

When to use it: The question gives the type of unit cell and the parameter $a$ , or the reverse, and asks for a density or an atomic radius.

1 Determine $Z$ by counting the fractions: corners for an eighth, faces for a half, the centre for a whole.
2 Convert the unit cell parameter to CENTIMETRES before any calculation, writing the conversion on its own line.
3 Apply $\rho=\dfrac{ZM}{N_{A}a^{3}}$ , the unit of $\rho$ then being the gram per cubic centimetre.
4 Check the order of magnitude: a few grams per cubic centimetre for a common metal, up to about twenty for the densest.

Concluding sentence

“The unit cell being face-centred, $Z=4$ ; with $a=3.615\times 10^{-8}$ cm, the relation $\rho=\dfrac{ZM}{N_{A}a^{3}}$ gives $8.94$ g/cm $^{3}$ , in agreement with the value for copper.”

The trap: Using the relation $a=2r$ for a face-centred unit cell. Each type of unit cell has its own relation between $a$ and $r$ , and it is the FACE diagonal that carries the atoms in contact in a face-centred unit cell.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The number of segments compared with the curve

Count the terms of the balance and compare them with the number of portions of the heating curve. A missing segment shows immediately.

Four terms for a passage from ice to steam: a plateau has been forgotten.
The dominant vaporization

In any balance that goes up to steam, the vaporization plateau must account for the largest share of the total, generally more than half.

A vaporization plateau at $12\%$ of the balance: the enthalpy of fusion was used in its place.
Kelvins in Clausius-Clapeyron

Check that both temperatures exceed $200$ and that the difference of the reciprocals is of the order of $10^{-4}$ .

A difference of $2.5\times 10^{-3}$ : the temperatures were left in degrees Celsius.
The order of magnitude of a density

A common metal lies between $2$ and $12$ g/cm $^{3}$ . A result of $10^{20}$ or $10^{-20}$ signals a forgotten conversion of $a$ .

A $\rho=7\times 10^{-24}$ g/cm $^{3}$ : the unit cell parameter was left in picometres.
The consistency between melting slope and densities

Check that the slope given to the melting curve matches the relative densities: pressure always favours the denser phase.

A positive slope given to water even though ice floats: the two statements are incompatible.

The typical problem, taken apart

From ice at $-20$ °C to steam at $120$ °C

A $50.0$ g sample of ice, initially at $-20.0$ °C, is heated until it becomes water vapour at $120.0$ °C, under a constant pressure of $101.3$ kPa.

Calculate the total energy required and determine the share of each step.

Data: $c(\text{ice})=2.09$ , $c(\text{water})=4.18$ , $c(\text{steam})=2.01$ J/(g·°C); $\Delta H_{fus}=6.01$ kJ/mol; $\Delta H_{vap}=40.7$ kJ/mol; $M=18.02$ g/mol.

Fifty grams of ice start at

-20.0

°C and end up as steam at

120.0

°C: two phase changes to cross, under constant pressure.

Step 1

The heating curve has FIVE segments: rise of the ice, melting plateau, rise of the liquid water, vaporization plateau, rise of the steam.

Why

This breakdown is done before any calculation and is sketched freehand. Each segment will have its own formula, and it is the omission of a segment, not a calculation error, that costs the marks here.

Step 2

$n=\dfrac{50.0}{18.02}=2.77$ mol, an amount that will serve for both plateaus.

Why

The plateaus use molar enthalpies whereas the rises use specific heats: converting once and for all avoids doing the division twice.

Step 3

Segment 1, ice from $-20.0$ to $0$ °C: $q_{1}=50.0\times 2.09\times 20.0=2090$ J $=2.09$ kJ.

Why

The specific heat used is that of ICE, not that of liquid water: the three phases have three different values, and the difference reaches a factor of two.

Step 4

Segment 2, melting at $0$ °C: $q_{2}=2.77\times 6.01=16.7$ kJ, at constant temperature.

Why

$\Delta T$ is zero on this segment, and yet the energy is not: this is the central trap of the chapter, and it is defused by changing formula.

Step 5

Segment 3, liquid water from $0$ to $100$ °C: $q_{3}=50.0\times 4.18\times 100=20\,900$ J $=20.9$ kJ.

Why

The temperature interval here is five times larger than in segment 1, and the specific heat twice as large: the segment costs ten times as much.

Step 6

Segment 4, vaporization at $100$ °C: $q_{4}=2.77\times 40.7=113$ kJ.

Why

The enthalpy of vaporization is nearly seven times that of fusion, because the molecules must be completely separated and not merely released from the order of the crystal.

Step 7

Segment 5, steam from $100$ to $120$ °C: $q_{5}=50.0\times 2.01\times 20.0=2010$ J $=2.01$ kJ.

Why

We are back to a specific heat, that of steam this time. The result is comparable to segment 1, which is consistent with similar intervals and specific heats.

Step 8

Total: $2.09+16.7+20.9+113+2.01=155$ kJ, of which $113$ kJ for vaporization alone, or $73\%$ .

Why

Commenting on the breakdown is the part of the question that distinguishes an answer that understood from one that merely calculated: vaporization dominates, and by far.

Step 9

Check: the ratio of the two plateaus is $\dfrac{113}{16.7}=6.8$ , to be compared with the ratio of the enthalpies $\dfrac{40.7}{6.01}=6.8$ .

Why

This ratio does not depend on the mass chosen: it therefore checks both plateaus at once, without redoing any multiplication.

The conclusion, written out

“ $155$ kJ must be supplied to take $50.0$ g of ice from $-20.0$ °C to $120.0$ °C under atmospheric pressure, of which $113$ kJ, nearly three quarters, for vaporization at $100$ °C alone.”

The classic mistake on this problem: Treating the passage from $0$ °C to $0$ °C as a segment of zero energy and announcing $138$ kJ. The $17$ kJ gap corresponds exactly to the melting plateau, and the total obtained remains plausible enough not to raise an alarm.

Learn by heart

• Within a phase: $q=mc\Delta T$ . During a plateau: $q=n\Delta H$ , at CONSTANT temperature.
• Three different specific heats for water: $2.09$ (ice), $4.18$ (liquid), $2.01$ (steam), in J/(g·°C).
• $\Delta H_{vap}$ is about SEVEN times $\Delta H_{fus}$ : the vaporization plateau dominates every balance.
• Vapour pressure depends ONLY on temperature, never on the amount of liquid.
• Boiling: vapour pressure EQUALS external pressure, so the boiling point depends on altitude.
• Clausius-Clapeyron requires KELVINS, and the difference of the reciprocals is of the order of $10^{-4}$ .
• Water: NEGATIVE melting slope, ice being less dense. Every other substance: positive slope.
• Unit cell: corner for $\tfrac{1}{8}$ , face for $\tfrac{1}{2}$ , centre for $1$ . Hence $Z=1$ , $2$ and $4$ .

Frequently asked questions

Why does the temperature not rise during a phase change?

Because all the energy supplied goes into breaking the interactions between molecules, not into increasing their motion. Yet it is that motion which temperature measures. As long as bonds remain to be broken, the thermometer stays still while we keep heating, and that is why the calculation changes formula.

Does water always boil at one hundred degrees?

No. It boils when its vapour pressure equals the external pressure. Under one atmosphere, that happens at one hundred degrees. At altitude, where the pressure is lower, it boils lower, which lengthens cooking times. In a pressure cooker, the pressure is higher and water boils above one hundred degrees.

Why does ice melt under pressure?

Because ice is less dense than liquid water, which is an anomaly among pure substances. Pressure always favours the denser phase. The solid-liquid boundary of water therefore leans to the left on the phase diagram, and compressing ice just below zero turns it to liquid.

What is the difference between the triple point and the critical point?

The triple point is the one combination of pressure and temperature at which the three phases coexist; it lies at the intersection of the three curves of the diagram. The critical point ends the vaporization curve: beyond it, liquid and gas are no longer distinct and form a supercritical fluid.

How many atoms are in a face-centred cubic unit cell?

Four. The eight corner atoms count for only one eighth each, since they are shared among eight cells, which makes one atom. The six face atoms count for one half each, which makes three. The total is therefore four atoms per unit cell, against two for a body-centred cell and one for a simple one.

Why does dry ice not melt?

Because the triple point of carbon dioxide lies above five atmospheres. Under ambient pressure, the liquid state simply does not exist for this substance: when heated, the solid goes directly to the gas state. This is called sublimation, and it is the principle of freeze-drying.

Practise it

Corrected exercises: States of matter and phase diagrams, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn1-states-of-matter. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: states of matter and phase diagrams (202-SN1-RE)

The essentials

The two formulas of the heat balance, and when to use them

Vapour pressure and boiling

Phase diagrams and crystalline solids

The mistakes that cost marks

1. Using the specific heat during a phase change

2. Making vapour pressure depend on the amount of liquid

3. Taking the boiling point for a constant of the substance

4. Leaving the temperatures in degrees Celsius in Clausius-Clapeyron

5. Giving the melting curve of water the slope of other substances

6. Confusing the triple point and the critical point

7. Comparing boiling points without taking molar mass into account

8. Counting the atoms of a unit cell without their fractions

9. Forgetting to convert the unit cell edge in the density calculation

10. Believing that a solid must always melt before becoming a gas

Which method to choose

Which formula for each segment of a heating process

Reading a phase diagram

How the answer is expected to be written

Writing up a complete heating balance

Calculating a density from a unit cell

Check before you hand in

The typical problem, taken apart

From ice at −20-20−20 °C to steam at 120120120 °C

Learn by heart

Frequently asked questions

Why does the temperature not rise during a phase change?

Does water always boil at one hundred degrees?

Why does ice melt under pressure?

What is the difference between the triple point and the critical point?

How many atoms are in a face-centred cubic unit cell?

Why does dry ice not melt?

Corrected exercises: States of matter and phase diagrams, 202-SN1-RE

See also

Looking for a 202-SN1-RE chemistry tutor in Montreal?

From ice at $-20$ °C to steam at $120$ °C