Chapter 10: States of Matter: Gaseous and Liquid States
In a Belfast laboratory in the eighteen-sixties, a chemist named Thomas Andrews spent his days watching a sealed glass tube. Inside it was carbon dioxide, squeezed hard enough that part of it lay as a liquid at the bottom with gas above, and a clear curved surface separating the two.
Andrews warmed the tube, a little at a time, and measured. As the temperature rose past 30 °C something happened that no theory of the day predicted. The surface between the liquid and the gas did not boil away. It grew faint, blurred, and then simply ceased to exist. What remained was neither liquid nor gas, but one uniform substance filling the whole tube. Let the tube cool by a degree or two, and the surface quietly reappeared.
Andrews had found what we now call the critical temperature. Above it, no pressure on earth will squeeze that substance into a liquid.
This mattered far beyond one tube. Chemists of the time had been defeated by oxygen, nitrogen and hydrogen, which refused to liquefy however hard they were compressed, and had begun calling them “permanent gases” — as though being a gas were part of what those substances were. Andrews showed there is nothing permanent about it. A substance is a solid, a liquid or a gas because of a contest between the pull its molecules exert on one another and the heat that keeps them apart. Change the terms of that contest and the state changes with it.
Diagram
Fig. 10.0 – Two drawings of the same thick-walled sealed glass tube containing carbon dioxide: on the left, the cooler tube with liquid at the bottom, gas above and a distinct curved surface between them, labelled “below the critical temperature”; on the right, the warmed tube filled with one uniform hazy fluid and no surface at all, labelled “above the critical temperature”
As you work through this chapter, you will collect exactly the ideas you need to reason out each of these questions for yourself.
10.1 Introduction
Everything you can touch, pour or breathe is matter, and almost all of it is found in one of three familiar physical forms. A substance may exist as a solid, as a liquid or as a gas. These are called the three main states of matter, and they are simply three different ways in which the same particles — atoms, molecules or ions — can be arranged and can move.
One Substance, Three States
The most familiar example is water. Cooled below its freezing point it is ice, a hard solid. At room temperature it is a liquid that flows and takes the shape of the vessel holding it. Heated strongly, it escapes as steam or water vapour, a gas. All three are shown side by side in Fig. 10.1.
Diagram
Fig. 10.1 – Three glasses showing the states of water: one with ice cubes, one with liquid water, one with water vapour escaping; label them ice, water and water vapour
The important point is that the substance itself does not change. Every one of those three forms is still made of H₂O molecules. What changes is how closely the molecules are packed and how freely they can move. This is why melting and boiling are called physical changes and not chemical ones.
Heat Converts One State Into Another
The three states of matter are interconvertible, and the agent of change is heat. Supplying heat to a solid converts it into a liquid, and further heating converts that liquid into a gas. Removing heat, that is cooling, drives the changes in the opposite direction. The complete two-way sequence is set out in FC 10.1.
Flowchart
FC 10.1 – Interconversion of the three states of matter: solid → (heat) → liquid → (heat) → gas; and in the reverse direction: gas → (cool) → liquid → (cool) → solid
Why should heat do this? Heating a substance supplies energy to its particles, so they move faster and are able to pull away from one another against the forces holding them together. Cooling takes that energy away, the particles slow down, and the forces of attraction can once again draw them into a closer, more ordered arrangement. The forces themselves are examined in Section 10.2.
Comparing the Three States Point by Point
The key points of differentiation between solids, liquids and gases are collected in Table 10.1. Read the rows as a set of linked consequences rather than as separate facts, because almost every entry follows from the first one: how far apart the particles are.
In a solid the particles are very close together and occupy fixed positions, so a solid keeps both its own shape and its own volume. In a liquid the particles are still close but no longer fixed, so a liquid keeps a definite volume yet flows into the shape of its container. In a gas the particles are far apart and in continuous random motion, so a gas has neither a fixed shape nor a fixed volume and spreads through the whole container.
The same reasoning explains compressibility. Compressing a substance means pushing its particles closer, which is only possible if there is empty space between them to begin with. A gas has large intermolecular space and is therefore easily compressed, a liquid has moderate space and compresses only slightly, and a solid has almost none and is practically non-compressible. For the same reason a small rise in temperature changes the volume of a gas markedly, a liquid moderately, and a solid very little.
Table 10.1: Distinguishing points between Solid, Liquid and Gas
| Sr. No. | Points | Solid | Liquid | Gas |
|---|---|---|---|---|
| 1 | Microscopic view — mean atomic/molecular separation | Mean separation ≈ 3-5 A⁰ | Mean separation ≈ 3-10 A⁰ | Mean separation > 5 A⁰ |
| 2 | Arrangement of particles (atoms/molecules) | Particles are tightly held, and have regular arrangement of atoms/molecules | Particles are loosely packed, irregular arrangement of particles | Particles are more loosely packed, highly irregular arrangement |
| 3 | Movement of particles | Particles cannot move freely as they occupy fixed positions | Particles can move a small distance within the liquid | Particles are in continuous random motion |
| 4 | Shape and volume | Has definite shape and volume | Takes the shape of the container and has definite volume | Takes the shape and the volume of its container |
| 5 | Intermolecular space | Very small intermolecular space | Moderate intermolecular space | Large intermolecular space |
| 6 | Effect of a small change in temperature | Volume change is small | Moderate effect on volume change | Volume change significantly high |
| 7 | Compression or expansion | Practically non-compressible | Small compressibility | Compressible |
Diagram
Table 10.1, row 1 – Three particle pictures for the microscopic view column: solid as a close-packed regular array of spheres, liquid as closely spaced but disordered spheres in a beaker, gas as widely separated spheres moving in random directions inside a balloon
10.2 Intermolecular Forces
Molecules are not indifferent to one another. Bring two molecules close together and they begin to interact: partly by pulling towards each other, and partly, when they get very close, by pushing apart. These forces acting between neighbouring molecules are called intermolecular forces. They include both the attractive forces and the repulsive forces present between neighbouring molecules.
The attractive part of the interaction is strongly distance-dependent. As the distance between two molecules increases, the attractive force between them decreases. This single fact explains a great deal, because the three states of matter differ mainly in how far apart their particles sit.
Since molecules are closest in a solid and furthest apart in a gas, intermolecular forces are strong in solids, less strong in liquids and very weak in gases. In other words, the three physical states of matter can be determined by the strength of the intermolecular forces acting within the substance.
What These Forces Control
Many of the physical properties you measure in the laboratory are really measurements of intermolecular attraction in disguise. Melting point, boiling point, vapour pressure, viscosity, evaporation, surface tension and solubility can all be studied with respect to the strength of the attractive forces between the molecules. A substance whose molecules cling tightly to one another will melt higher, boil higher, evaporate less readily and flow more sluggishly than one whose molecules barely notice each other.
Melting and vaporization illustrate the point neatly. During the melting process the intermolecular forces are only partially overcome — the particles gain enough freedom to slide past one another, but they remain in contact. During the vapourization process the forces are overcome completely, because the molecules must separate widely and move away as an independent gas.
10.2.1 Types of Intermolecular Forces
Not all intermolecular attractions arise in the same way. Some need permanently polar molecules, some need ions, and one kind operates even between atoms that carry no permanent charge separation at all. The types of intermolecular forces are:
- Dipole-dipole interactions
- Ion-dipole interactions
- Dipole-Induced dipole interaction
- London Dispersion Forces
- Hydrogen bonding
i. Dipole–Dipole Interactions
Polar molecules experience dipole-dipole forces, which arise from electrostatic interactions between the dipoles on neighbouring molecules.
What exactly is a polar molecule, or a polar covalent molecule? You met these in Chapter 5. A polar covalent molecule is also described as a “dipole”, meaning that the molecule has two ‘poles’. A covalent bond becomes polar because of the electronegativity difference between the two bonding atoms: the more electronegative atom draws the shared electron pair closer to itself. Hence polarity is observed in compounds containing dissimilar atoms.
Take the HCl molecule shown in Fig. 10.2 (a). Chlorine is more electronegative than hydrogen, so the bonding electrons spend more time near chlorine.
Diagram
Fig. 10.2 (a) – Polar molecule: an HCl molecule drawn as an elongated cloud with H at one end and Cl at the other, greater charge density shaded towards chlorine, δ⊕ marked over H and δ⊖ over Cl, with the dipole-moment arrow µ pointing from H towards Cl; caption note “unequal sharing of electrons between the bonding atoms”
One end (pole) of the molecule therefore carries a partial positive charge on the hydrogen atom, while at the other end the chlorine atom carries a partial negative charge. These fractional charges are denoted by the Greek letter ‘δ’ (delta). As a result of this polarisation, the molecule possesses a dipole moment.
Dipole moment (\(\mu\)) is the product of the magnitude of the charge (\(Q\)) and the distance between the centres of positive and negative charge (\(r\)):
\[\mu = Q \times r\]
It is designated by the Greek letter \(\mu\) (mu) and its unit is debye (D).
Dipole moment is a vector quantity, which means it has both size and direction. It is depicted by a small arrow with its tail at the positive centre and its head pointing towards the negative centre.
For example, the dipole moment of HF may be represented as an arrow drawn above the Lewis structure of H–F. The crossed arrow above the Lewis structure represents an electron density shift, that is, a shift of electron density from hydrogen towards fluorine.
Thus polar molecules have permanent dipole moments — permanent because the electronegativity difference that creates them never goes away. When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of the other polar molecule. Many such molecules have dipoles, and their mutual interaction is termed dipole-dipole interaction.
These forces are generally weak, with energies of the order of 3-4 kJ mol⁻¹, and they are significant only when the molecules are in close contact, that is, in a solid or a liquid state. In a gas the molecules are too far apart, and moving too fast, for such a weak attraction to hold them together.
For example, \(\mathrm{C_4H_9Cl}\) (butyl chloride), \(\mathrm{CH_3\!-\!O\!-\!CH_3}\) (dimethyl ether) and ICl (iodine chloride, B.P. 27 °C) are dipolar liquids. In liquid ICl the molecules do not lie at random: they line up so that each positive end faces a neighbouring negative end, as shown in Fig. 10.2 (b).
Diagram
Fig. 10.2 (b) – Dipole-dipole interaction: four or five I–Cl molecules arranged head-to-tail and side-by-side, each I marked δ⊕ and each Cl marked δ⊖, showing positive ends facing negative ends of neighbours
Dipole moment and boiling point
In brief, the more polar the substance, the greater the strength of its dipole-dipole interactions. Table 10.2 lists several substances that have similar molecular masses but different dipole moments, so that the effect of polarity alone can be seen. From Table 10.2 it is clear that the higher the dipole moment, the stronger are the intermolecular forces, generally leading to higher boiling points.
Table 10.2: Effect of dipole moments on boiling point (b.p.)
| Substance | Molar Mass (amu) | Dipole Moment (D) | b.p. (K) |
|---|---|---|---|
| CH₃ - CH₂ - CH₃ | 44.10 | 0.1 | 231 |
| CH₃ - O - CH₃ | 46.07 | 1.3 | 248 |
| CH₃ - Cl | 50.49 | 1.9 | 249 |
| CH₃ - CN | 41.05 | 3.9 | 355 |
Notice how carefully this comparison has been set up. All four molar masses lie between about 41 and 51 amu, so the number and mass of the electrons cannot be the deciding factor. The dipole moment climbs from 0.1 D to 3.9 D, and the boiling point climbs alongside it, from 231 K to 355 K. The conclusion is hard to avoid: the more polar the molecule, the more energy is required to separate one molecule from its neighbours.
ii. Ion–Dipole Interactions
An ion-dipole force is the result of electrostatic interactions between an ion (a cation or an anion) and the partial charges on a polar molecule. Here one partner carries a full charge and the other only a fractional one, which is why this interaction is stronger than a plain dipole-dipole attraction.
The strength of this interaction depends on the charge and the size of the ion. It also depends on the magnitude of the dipole moment and the size of the polar molecule.
Ion-dipole forces are particularly important in aqueous solutions of ionic substances such as sodium chloride (NaCl). When the ionic compound sodium chloride is dissolved in water, the ions get separated and each ion is surrounded by water molecules, as drawn in Fig. 10.2 (c). This surrounding of the sodium ions by water molecules is called Hydration of sodium ions.
Diagram
Fig. 10.2 (c) – Na⊕ ion (cation) – H₂O interaction: a central Na⁺ ion ringed by six water molecules, each turned so that its δ- oxygen faces the ion and its δ+ hydrogens point outwards; label “hydrated Na⊕ ion”
Look closely at the orientation of the water molecules in that figure. Every water molecule has turned its negative (oxygen) end towards the cation. That orientation is not accidental — it is what makes the attraction possible.
Cations are smaller in size than the isoelectronic anions, that is, anions having the same number of electrons. Because the same charge is packed into a smaller volume, the charge density on the cation (\(\mathrm{Na^+}\)) is more concentrated than on the anion (\(\mathrm{Cl^-}\)). This makes the interaction between \(\mathrm{Na^+}\) and the negative end of the polar \(\mathrm{H_2O}\) molecule stronger than the corresponding interaction between \(\mathrm{Cl^-}\) and the positive end of the polar \(\mathrm{H_2O}\) molecule.
The charge on the ion matters just as much as its size. The more the charge on the cation, the stronger is the ion-dipole interaction. For example, the \(\mathrm{Mg^{2+}}\) ion has a higher charge and a smaller ionic radius (78 pm) than the \(\mathrm{Na^+}\) ion (98 pm). Hence the \(\mathrm{Mg^{2+}}\) ion is surrounded (hydrated) more strongly by water molecules and exerts a strong ion-dipole interaction.
Thus the strength of interaction increases with an increase in charge on the cation and with a decrease in ionic size or radius. Therefore ion-dipole forces increase in the order:
\[\mathrm{Na^+} < \mathrm{Mg^{2+}} < \mathrm{Al^{3+}}\]
iii. Dipole–Induced Dipole Interaction
The two forces described so far need both partners to be charged or polar. But an attraction can also arise between a polar molecule and a completely non-polar one.
When polar molecules (like \(\mathrm{H_2O}\) and \(\mathrm{NH_3}\)) and nonpolar molecules (like benzene) approach each other, the polar molecules induce a dipole in the non-polar molecules. The electric field of the permanent dipole tugs on the loosely held electron cloud of the non-polar molecule and pushes it slightly to one side. Hence ‘Temporary dipoles’ are formed by the shifting of electron clouds in the nonpolar molecules.
For example, ammonia (\(\mathrm{NH_3}\)) is polar and has a permanent dipole moment, while benzene (\(\mathrm{C_6H_6}\)) is non-polar and has zero dipole moment. The force of attraction developed between the polar and nonpolar molecules is of the type dipole-induced dipole interaction. It can be seen in Fig. 10.2 (d).
Diagram
Fig. 10.2 (d) – Dipole-induced dipole interaction: on the left a polar NH₃ molecule with δ+ on H atoms and δ- on N, plus a non-polar C₆H₆ with no charge marks; arrow leading to the right-hand picture where H₃N (δ+ δ-) is drawn beside C₆H₆ that now carries induced δ+ and δ- ends, joined by a dotted line; label the three parts “polar molecule”, “non-polar molecule”, “non-polar molecule with induced dipole”
The induced dipole lasts only as long as the polar molecule stays nearby. Move the two apart and the borrowed polarity vanishes, which is why this force too is significant only at close range.
iv. London Dispersion Force
The study of intermolecular forces present among nonpolar molecules, or among the individual atoms of a noble gas, is very interesting. Consider benzene (\(\mathrm{C_6H_6}\)) again. It has zero dipole moment and experiences no dipole-dipole forces, and yet it exists in the liquid stage at room temperature. Something must be holding those molecules together.
In the case of nonpolar molecules and inert gases, only dispersion forces exist. Dispersion forces are also called London forces, after the idea of a momentary dipole which was proposed by the German physicist Fritz London in 1930. These forces are also called van der Waals forces. It is the weakest intermolecular force, and it develops due to the interaction between two nonpolar molecules.
In general, all atoms and molecules experience London dispersion forces, which result from the motion of electrons. At any given instant of time, the electron distribution in an atom may be asymmetrical, giving the atom a short-lived dipole moment. This momentary dipole on one atom can affect the electron distribution in the neighbouring atoms and induce momentary dipoles in them. As a result, a weak attractive force develops.
It is worth being clear about the difference from the previous case. In a dipole-induced dipole interaction, a permanently polar molecule does the inducing. In a London dispersion force, there is no permanent dipole anywhere — the inducing dipole is itself only a flicker in an electron cloud, and yet the flickers of neighbouring particles fall into step and produce a real, if small, attraction.
For example, substances composed of molecules such as \(\mathrm{O_2}\), \(\mathrm{CO_2}\), \(\mathrm{N_2}\), halogens, methane gas, helium and other noble gases show van der Waals force of attraction. These are precisely the substances in which no other intermolecular force is available.
Polarizability
The strength of London forces increases with an increase in molecular size, molecular mass and the number of electrons present in an atom or molecule.
When two nonpolar molecules approach each other, attractive and repulsive forces between their electrons and the nuclei will lead to distortions in the size of the electron cloud, a property referred to as polarizability.
Polarizability is a measure of how easily an electron cloud of an atom is distorted by an applied electric field. It is a property of the atom. The ability to form momentary dipoles — that is, the ability of another molecule to become polar by redistributing its electrons — is known as polarizability of the atom or molecule.
The more the number of electrons in a molecule, the higher is its ability to become polar. A large electron cloud is held less tightly by the nuclei, so it is easier to push out of shape.
Why molecular shape matters
Similarly, the more spread out the shapes of the molecules, the higher the dispersion forces present between the molecules. London dispersion forces are stronger in a long chain of atoms where molecules are not compact, because an extended molecule can lie alongside its neighbour over a much greater length and touch it at many points at once.
This can affect a physical property such as B.P. For example, n-pentane boils at 309.4 K, whereas neo-pentane boils at 282.7 K. Here both the substances have the same molecular formula, \(\mathrm{C_5H_{12}}\), but n-pentane is longer and somewhat spread out, whereas neo-pentane is more spherical and compact (see Fig. 10.3). A compact, ball-shaped molecule can only make contact with a neighbour over a small patch of its surface, so the total dispersion attraction is smaller and less energy is needed to separate the molecules.
Diagram
Fig. 10.3 – (a) n-Pentane as a space-filling model of five carbons in an extended chain beside its structural formula, labelled “longer, less compact molecule, b.p. = 309.4 K”; (b) neo-pentane as a compact roughly spherical space-filling model beside its structural formula with a central carbon carrying four CH₃ groups, labelled “more compact molecule, b.p. = 282.7 K”
v. Hydrogen Bonding
A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom is bonded to a strongly electronegative atom, or to an atom with a lone pair of electrons. Hydrogen bonds are generally stronger than usual dipole-dipole and dispersion forces, and weaker than true covalent or ionic bonds. They therefore sit in a category of their own, between ordinary intermolecular attraction and genuine chemical bonding.
Why must the electronegative partner be so strongly electronegative? Because the hydrogen atom is tiny and has only one electron. When that electron is pulled well away towards a very electronegative atom, the hydrogen nucleus is left almost bare, and a nearly naked proton can approach a lone pair on a neighbouring molecule far more closely than any other atom could.
The strong electronegative atoms that form hydrogen bonds are nitrogen, oxygen and fluorine. A hydrogen bond is denoted by a dotted line (….), to distinguish it from a covalent bond, which is drawn as a solid line.
A hydrogen bond that occurs within the same molecule represents an Intramolecular Hydrogen bond. A hydrogen bond present between two like or unlike molecules represents an Intermolecular Hydrogen bond (see Fig. 10.4 (a) and (b)).
i. Intermolecular H-bonding
H-bonding in H−F may be shown as a chain in which each molecule links to the next:
\[\mathrm{H^{\delta+}}\!-\!\mathrm{F^{\delta-}}\cdots\mathrm{H^{\delta+}}\!-\!\mathrm{F^{\delta-}}\cdots\mathrm{H^{\delta+}}\!-\!\mathrm{F^{\delta-}}\cdots\mathrm{H^{\delta+}}\!-\!\mathrm{F^{\delta-}}\]
Each dotted link in that chain is a hydrogen bond. The same kind of linking occurs in water and in ammonia, as drawn in Fig. 10.4.
Diagram
Fig. 10.4 – H-bonding in H₂O (above) and NH₃ (below): two water molecules with a δ- oxygen lone pair of one accepting a dotted bond from the δ+ hydrogen of the other, and below it the same arrangement between two ammonia molecules through N; solid lines denote covalent bonds and dotted lines denote hydrogen bonds
ii. Intramolecular H-bonding
H-bonding in ethylene glycol takes place inside a single molecule, between the hydrogen of one −OH group and the oxygen of the other −OH group on the neighbouring carbon.
Diagram
Fig. 10.4 (c) – Intramolecular H-bonding in ethylene glycol: the two-carbon skeleton written vertically as CH₂−Oδ-−Hδ+ above CH₂−Oδ-−Hδ+, with a dotted line curving from the hydrogen of the lower OH group to the oxygen of the upper OH group inside the same molecule
H-bonding and boiling point
Due to the presence of H-bonding in the compounds, more energy is required to break the bonds. Therefore boiling point is more in the case of liquid molecules containing H-bonds. Hydrogen bonds can be quite strong, with energies up to 40 kJ/mol — roughly ten times a typical dipole-dipole attraction.
Water in particular is able to form a vast three dimensional network of hydrogen bonds, as shown in Fig. 10.5. This is because each \(\mathrm{H_2O}\) molecule has two hydrogen atoms and two electron pairs on the oxygen atom. Each molecule can therefore donate two hydrogen bonds and accept two more, giving four links per molecule and a structure that extends through the whole liquid.
Diagram
Fig. 10.5 – Liquid water containing a vast three-dimensional network of hydrogen bonds: many water molecules with red oxygen and white hydrogen atoms, each linked to about four neighbours by dotted hydrogen bonds
The boiling point generally increases with an increase in molecular mass, since heavier molecules carry more electrons and larger dispersion forces. But the hydrides of nitrogen (\(\mathrm{NH_3}\)), oxygen (\(\mathrm{H_2O}\)) and fluorine (HF) have abnormally high boiling points due to the presence of hydrogen bonding between the molecules. Their behaviour breaks the mass rule because a second, much stronger force has entered the picture.
The comparison below makes the anomaly hard to miss.
Among the three compounds H₂O, H₂S and H₂Se, the first one, H₂O, has the smallest molecular mass, and on the mass argument alone it should boil lowest of all. Instead it has the highest B.P. of 100 °C, while the B.P. of H₂S is -60 °C and that of H₂Se is -41.25 °C. H₂S and H₂Se follow the expected pattern between themselves — the heavier H₂Se boils higher than H₂S. The extraordinarily high B.P. of H₂O is due to very strong hydrogen bonding, even though it has the lowest molecular mass. Sulphur and selenium are simply not electronegative enough to make their hydrogens available for hydrogen bonding.
Similarly, due to the presence of the H-bond, the viscosity of a liquid increases, because molecules that are tied to their neighbours cannot slide past one another easily. Hydrogen bonds also play a vital role in determining the structure and properties of proteins and nucleic acids present in all living organisms — the double helix of DNA is held together by hydrogen bonds between its base pairs. In the case of gases, intermolecular forces of attraction are very weak.
10.2.2 Intermolecular Forces and Thermal Energy
Intermolecular forces are only half of the story. They are always working to pull molecules together, but something is always working against them.
Thermal energy is the origin of the kinetic energy of the particles of matter that arises due to the movement of particles, an idea you must have met in Physics. It is directly proportional to the temperature; that means thermal energy increases with an increase in temperature and vice versa. Heat a substance and its particles move faster; cool it and they slow down.
The three states of matter are the consequence of a balance between the intermolecular forces of attraction and the thermal energy of the molecules. One tends to gather the particles into an ordered, close-packed arrangement; the other tends to scatter them. Whichever wins decides the state.
If the intermolecular forces are very weak, molecules do not come together to make a liquid or a solid unless the thermal energy is decreased by lowering the temperature. This is exactly why gases such as oxygen and nitrogen have to be cooled, and not merely squeezed, before they will liquefy.
When a substance is to be converted from its gaseous state to the solid state, its thermal energy (or temperature) has to be reduced. At this stage, the intermolecular forces become more important than the thermal energy of the particles. The whole progression is set out in FC 10.2.
Flowchart
FC 10.2 – Balance between attraction and thermal energy across the states: gas → liquid → solid, with one arrow running left to right labelled “intermolecular force increases” and a second arrow running left to right labelled “thermal energy decreases”
A comparison of the various kinds of intermolecular forces discussed in this section is made in Table 10.3. Read the strength column together with the characteristics column, because the two explain each other: ion-dipole heads the list precisely because it is the only entry in which a full ionic charge is involved.
Table 10.3: Comparison of Intermolecular Forces
| Force | Strength | Characteristics |
|---|---|---|
| Ion-dipole | Moderate (10 - 50 kJ/mol) | Occurs between ions and polar solvents |
| Dipole-dipole | Weak (3 - 4 kJ/mol) | Occurs between polar molecules |
| London dispersion | Weak (1 - 10 kJ/mol) | Occurs between all molecules; strength depends on size, polarizability |
| Hydrogen bond | Moderate (10 - 40 kJ/mol) | Occurs between molecules with O-H, N-H, and H-F bonds |
Two entries in this table deserve a second look for exam purposes. First, London dispersion is the only force that occurs between all molecules, polar or not — so it is never absent, merely sometimes overshadowed. Second, a hydrogen bond needs one of the three specific arrangements O−H, N−H or H−F; a molecule containing hydrogen but no such bond, for example \(\mathrm{CH_4}\), cannot hydrogen bond at all. Questions that ask you to name the forces present in a given compound are testing exactly these two points.
10.3 Characteristic Properties of Gases
Of the three states of matter, the gaseous state is the simplest to describe and the easiest to measure, because its molecules are so far apart that they hardly interfere with one another. That is why the study of chemistry so often begins with gases.
Under normal conditions, out of the 118 elements in the periodic table, only a few — eleven elements — exist as gases. Every other element is a solid or a liquid at ordinary temperature and pressure. Yet these few gaseous elements, together with the gaseous compounds, make up the atmosphere on which all life depends.
The Physical Properties of the Gaseous State
The gaseous state is characterized by the following physical properties.
Gases are lighter than solids and liquids, that is, they possess lower density. The same mass of substance is spread over a far greater volume.
Gases do not possess a fixed volume and shape. They occupy the entire space available and take the shape of the container.
Gas molecules are widely separated and are in continuous, random motion. Therefore gases exert pressure equally in all directions, due to the collision of gas molecules on the walls of the container.
In the case of gases, intermolecular forces are weakest. This follows directly from the wide separation of the molecules, since the attractive force falls off as the distance between molecules increases.
Gases possess the property of diffusion, which is a spontaneous homogeneous intermixing of two or more gases. No stirring is needed — the random motion of the molecules does the mixing by itself.
Gases are highly compressible, because there is a great deal of empty space between the molecules that can be squeezed out.
What Is Air?
Air is a mixture of various gases. We cannot see air, but we feel it as a cool breeze. The composition of air by volume is around 78 percent \(\mathrm{N_2}\), 21 percent \(\mathrm{O_2}\) and 1 percent other gases including \(\mathrm{CO_2}\). The proportions are drawn out in Fig. 10.3A.
Diagram
Fig. 10.3A – Pie chart of the composition of air by volume: nitrogen 78%, oxygen 21%, inert gases (mainly argon) 0.97%, carbon dioxide and other gases 0.03%, plus a small wedge labelled water vapour
The chemistry of atmospheric gases is an important subject of study, as it involves air pollution. The \(\mathrm{O_2}\) in air is essential for the survival of aerobic life, that is, of every organism that breathes oxygen.
Measurable Properties of Gases
To describe the state of a gas quantitatively we need numbers, and each number has to be measured in a defined unit. Some of the important measurable properties of gases are given below. You have already met most of them in Chapter 1; here we look at them specifically as they apply to gases.
Mass
The mass, \(m\), of a gas sample is a measure of the quantity of matter it contains. It can be measured experimentally, usually by weighing the container full and then empty. The SI unit of mass of a gas is the kilogram (kg), where \(1\ \text{kg} = 10^{3}\ \text{g}\).
In chemistry we more often need the amount of gas in moles than in grams, because chemical behaviour depends on the number of particles present. The mass of a gas is related to the number of moles (\(n\)) by the expression:
\[n = \frac{\text{mass in grams}}{\text{molar mass in grams}} = \frac{m}{M}\]
The following expressions are also useful in calculations:
\[n = \frac{\text{Number of molecules}}{\text{Avogadro Number}} = \frac{N}{N_A} = \frac{N}{6.022\times10^{23}}\]
\[n = \frac{\text{Volume of a given gas in litres at STP}}{22.414\ \text{litres at STP}}\]
All three expressions give the same quantity, \(n\). Which one you use depends only on what the question hands you — a mass, a number of molecules, or a volume at STP.
Volume
The volume (\(V\)) of a sample of gas is the amount of space it occupies. Since a gas fills its container completely, the volume of a gas is simply the volume of the vessel holding it.
It is expressed in terms of different units like litres (L), millilitres (mL), cubic centimetre (cm³), cubic metre (m³) or decimetre cube (dm³). The SI unit of volume is cubic metre (m³). The most commonly used unit to measure the volume of a gas is decimetre cube or litre, because a cubic metre is an inconveniently large box for laboratory work.
\[1\ \text{L} = 1000\ \text{mL} = 1000\ \text{cm}^3 = 1\ \text{dm}^3\]
\[1\ \text{m}^3 = 10^{3}\ \text{dm}^3 = 10^{3}\ \text{L} = 10^{6}\ \text{cm}^3 = 10^{6}\ \text{mL}\]
The second line is worth memorising in the form “a cubic metre is a thousand litres”, since gas-law problems frequently mix m³ with dm³ and an error here shifts an answer by a factor of a thousand.
Pressure
Pressure (\(P\)) is defined as force per unit area.
\[\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{f}{a}\]
The pressure of a gas is measured with a ‘manometer’, and atmospheric pressure is measured by a ‘barometer’. Both instruments are shown in Fig. 10.3B.
Diagram
Fig. 10.3B – Side by side: (a) a simple mercury barometer, an inverted tube standing in a mercury trough with a vacuum above the mercury column and the column height marked 760 mm at 1 atm; (b) a U-tube manometer connected to a gas bulb, showing the difference in mercury levels in the two arms as the measure of gas pressure
The SI unit of pressure is the pascal (Pa), or newton per metre square (N m⁻²).
\[1\ \text{Pa} = 1\ \text{N m}^{-2} = 1\ \text{kg m}^{-1}\text{s}^{-2}\]
\[1\ \text{bar} = 1.00\times10^{5}\ \text{Pa}\]
The bar is now replacing the standard atmosphere (atm) as the most convenient unit of pressure, largely because it is a round multiple of the SI unit.
\[1\ \text{atm} = 76\ \text{cm of Hg} = 760\ \text{mm of Hg} = 760\ \text{torr},\quad \text{as }1\ \text{torr} = 1\ \text{mm Hg}\]
\[1\ \text{atm} = 101.325\ \text{kPa} = 101325\ \text{Pa} = 1.01325\times10^{5}\ \text{N m}^{-2} = 1.01325\ \text{bar}\]
Temperature
Temperature is the property of an object that determines the direction in which energy will flow when that object is in contact with another object. Heat always flows from the object at the higher temperature to the one at the lower temperature.
In scientific measurements, temperature (\(T\)) is measured either on the celsius scale (°C) or on the Kelvin scale (K). Note that the degree sign is not used with the kelvin unit — we write 300 K, not 300 °K.
The SI unit of temperature of a gas is kelvin (K). The celsius and kelvin scales are related by the expression:
\[T\ \text{(K)} = t\ ^\circ\text{C} + 273.15\]
Density
Density is the mass per unit volume.
\[d = \frac{m}{V}\]
Therefore the SI unit of density is kg m⁻³.
In the case of gases, relative density is measured with respect to hydrogen gas and is called vapour density. Since the molar mass of hydrogen gas, \(\mathrm{H_2}\), is \(2\ \text{g mol}^{-1}\), comparing a gas with hydrogen amounts to dividing its molar mass by two:
\[\text{Vapour density} = \frac{\text{molar mass}}{2}\]
Vapour density is thus a pure number with no unit, unlike density. Remember also that the density of a gas is not a fixed property of the substance the way the density of a solid nearly is: squeeze a gas or heat it and its density changes at once, because its volume changes.
Diffusion
In the case of gases, the rate of diffusion of two or more gases is measured.
\[\text{Rate of diffusion} = \frac{\text{Volume of a gas diffused}}{\text{time required for diffusion}}\]
Therefore the SI unit for rate of diffusion is dm³ s⁻¹ or cm³ s⁻¹.
10.4 Gas Laws
The behaviour of gases can be studied by four variables, namely pressure, volume, temperature and the number of moles. These variables and the measurable properties of gases are related with one another through the different gas laws.
Think of a gas in a cylinder or a sealed container. We can measure the number of moles (\(n\)) of gas inside, the pressure (\(P\)) of the gas, the volume (\(V\)) of the gas — which is equal to the volume of the container — and the temperature (\(T\)) of the gas. The observed relationships between \(P\), \(V\), \(n\) and \(T\) are summarized by five gas laws: Boyle’s law, Charles’ law, Gay Lussac law, Avogadro law and Dalton’s law.
Each of these laws works in the same way. Two of the four variables are held constant, and the relationship between the remaining two is measured. This is what makes the laws so easy to apply and so easy to misapply: before using any of them, always check which quantities the problem is keeping fixed.
The gas laws are based on the experimental observations made by scientists. They were not deduced from theory but read off from measurements, and these gas laws have played a significant role in the development of chemistry.
10.4.1 Boyle’s Law: (Pressure-Volume Relationship)
Boyle’s law can be stated in either of two equivalent ways.
Notice the two conditions built into both statements: the amount of gas must be fixed, and the temperature must be constant. Drop either condition and the law no longer applies. The law is illustrated diagrammatically in Fig. 10.6, where the same sample of gas is squeezed to half and then to a quarter of its volume.
Diagram
Fig. 10.6 – Schematic illustration of Boyle’s law: three cylinders with pistons at constant temperature 298 K; first at P = 1 atm with V = 1 L, second at P = 2 atm with V = 0.50 L, third at P = 4 atm with V = 0.25 L; arrows between them labelled “volume decreases, pressure increases” and “volume increases, pressure decreases”
Mathematical Expression
Mathematically, Boyle’s law is expressed as:
\[P \propto \frac{1}{V} \quad (\text{at constant }T\text{ and }n)\]
\[\therefore\ P = k_1\frac{1}{V} \qquad (10.1)\]
where \(k_1\) is the proportionality constant. On rearranging, Eq. (10.1) becomes:
\[\therefore\ PV = k_1 \qquad (10.2)\]
It means that at constant temperature, the product of the pressure and the volume of a fixed amount of a gas is constant.
Thus, when a fixed amount of a gas at constant temperature (\(T\)) occupying volume \(V_1\) initially at pressure \(P_1\) undergoes expansion or compression, the volume of the gas changes to \(V_2\) and the pressure to \(P_2\). According to Boyle’s law,
\[P_1V_1 = P_2V_2 = \text{constant} \qquad (10.3)\]
Equation (10.3) is the working form of the law, and it is the one you will use in nearly every numerical problem. It follows from Eq. (10.2) simply because both products equal the same constant \(k_1\).
An actual experiment of this kind is drawn in Fig. 10.7. Figure 10.7 illustrates the Boyle’s law experiment when the applied pressure increases from 1 atm to 2 atm and the volume of the gas decreases from 200 cm³ to 100 cm³ — doubling the pressure has halved the volume, exactly as Eq. (10.3) requires.
Diagram
Fig. 10.7 – Schematic of the experiment on Boyle’s law: two gas syringes with pressure gauges, the upper one reading 2 atm with the trapped gas compressed to the 100 cm³ mark and labelled “high pressure, low volume”, the lower one reading 1 atm with the gas at the 200 cm³ mark and labelled “low pressure, high volume”
Graphical Representation
Figure 10.8 shows how the pressure-volume relationship can be expressed graphically, and each of its three parts tells you something different.
Figure 10.8 (a) shows a graph of the equation \(PV = k\) at a given constant temperature, known as an Isotherm (a constant temperature plot). It is a curve, not a straight line, because \(P\) and \(V\) are inversely related. For a given mass of a gas, the value of \(k\) varies only with temperature, which is why a separate isotherm is drawn for each of \(T_1\), \(T_2\) and \(T_3\).
If the product of pressure and volume (\(PV\)) is plotted against pressure (\(P\)), a straight line is obtained parallel to the x-axis (pressure axis), as in Fig. 10.8 (b). This is the most direct test of the law: if \(PV\) really is a constant, its graph against \(P\) can only be a horizontal line.
When the pressure (\(P\)) of the gas is plotted against \(1/V\), a straight line is obtained passing through the origin, as shown in Fig. 10.8 (c), provided \(T\) and \(n\) are constant. Plotting against \(1/V\) converts the inverse relationship into a direct one, and any direct proportionality gives a straight line through the origin.
Diagram
Fig. 10.8 – Boyle’s law, graphical representation, three panels: (a) pressure P on the y-axis against volume V on the x-axis, three hyperbolic isotherms labelled T₁, T₂, T₃ with T₃ highest; (b) product PV on the y-axis against pressure P, three horizontal straight lines labelled T₁, T₂, T₃; (c) pressure P on the y-axis against 1/V on the x-axis, three straight lines through the origin of different slopes labelled T₁, T₂, T₃
However, at high pressure, deviation from Boyle’s law is observed in the behaviour of gases. Why this happens is taken up in Section 10.7.
Boyle’s Law in Terms of Density of Gas
With an increase in pressure, gas molecules get closer and the density (\(d\)) of the gas increases. Hence, at constant temperature, pressure is directly proportional to the density of a fixed mass of gas. This can be shown from Eq. (10.2).
\[PV = k_1 \qquad (10.2)\]
\[\therefore\ V = \frac{k_1}{P} \qquad (10.4)\]
But
\[d = \frac{m}{V}\]
On substituting \(V\) from Eq. (10.4),
\[d = \left(\frac{m}{k_1}\right)\times P\]
\[\therefore\ d = k'P \qquad (10.5)\]
where \(k'\) = New Constant. Since \(m\) and \(k_1\) are both fixed for a given mass of gas at a given temperature, their ratio is just another constant.
\[\therefore\ d \propto P\]
The above equation shows that at constant temperature, the pressure is directly proportional to the density of a fixed mass of the gas. This makes physical sense: the same mass squeezed into a smaller volume must be denser.
10.4.2 Charles’ law: (Temperature -Volume Relationship)
J. Charles (1746-1823) and Gay Lussac (1778-1850) worked independently on the influence of temperature on the volume of a gas. Their experiments showed that for a given mass of a gas at constant pressure, its volume increases with an increase in temperature.
It was found that for an increase of every degree of temperature, the volume of the gas increases by \(\dfrac{1}{273.15}\) of its original volume at 0 °C. This observation is expressed mathematically as follows:
\[V_t = V_0 + \frac{t}{273.15}\,V_0 \qquad (10.6)\]
where \(V_t\) and \(V_0\) are the volumes of the given mass of gas at the temperatures \(t\) °C and 0 °C respectively.
Rearranging Eq. (10.6) gives — first taking \(V_0\) common —
\[V_t = V_0\left(1 + \frac{t}{273.15}\right)\]
\[\therefore\ V_t = V_0\left(\frac{t + 273.15}{273.15}\right) \qquad (10.7)\]
The step from the first line to the second is only the addition of two fractions: writing 1 as \(\dfrac{273.15}{273.15}\) and adding \(\dfrac{t}{273.15}\) to it.
A New Temperature Scale
At this stage, a new scale of temperature was introduced, namely, the absolute temperature scale. This absolute temperature (\(T\) K) was defined as
\[T\ \text{K} = t\ ^\circ\text{C} + 273.15\]
The reason for inventing it is visible in Eq. (10.7). The awkward combination \((t + 273.15)\) keeps appearing, so it is far simpler to treat that whole quantity as the temperature itself.
This is also called the thermodynamic scale of temperature. The unit of this absolute temperature scale is called the kelvin (K), in honour of Lord Kelvin, who determined the accurate value of absolute zero as -273.15 °C in 1854.
Equation (10.7) is now rewritten by replacing the celsius temperatures by absolute temperatures as follows:
\[V_t = V_0\left(\frac{T_t}{T_0}\right) \qquad (10.8)\]
where \(T_t = t + 273.15\) and \(T_0 = 273.15\).
Equation (10.8) on rearrangement takes the following form:
\[\frac{V_t}{T_t} = \frac{V_0}{T_0}\]
The subscripts here are not special. State 0 and state \(t\) are just two states of the same gas, so from this a general equation can be written as follows:
\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \qquad (10.9)\]
\[\therefore\ \frac{V}{T} = \text{constant} = k_2\]
\[\therefore\ V = k_2T \qquad (10.10)\]
Equation (10.10) is the mathematical expression of Charles law, which is stated as follows.
Two conditions again: the pressure must be constant and the mass of gas fixed. And the temperature must be in kelvin — a point worth dwelling on. If you used celsius, doubling the temperature from 10 °C to 20 °C would have to double the volume, which experiment flatly denies. Only the absolute scale makes the proportionality true.
The law can be illustrated diagrammatically as shown in Fig. 10.9, where doubling the absolute temperature from 300 K to 600 K at 1 atm doubles the volume from 2 litres to 4 litres.
Diagram
Fig. 10.9 – At constant pressure, volume changes with temperature: two cylinders each closed by a free piston loaded at 1 atm; the left one holds 2 litres of gas at 300 kelvin, an arrow marked “heat” leads to the right one holding 4 litres at 600 kelvin
Absolute Zero Temperature and Charles’ Law
According to Charles law, Eq. (10.10), the graph of the volume of a gas (at a given constant pressure, say \(P_1\)) plotted versus its temperature in celsius is a straight line with a positive slope. See Fig. 10.10.
On extending the line to zero volume, the line intercepts the temperature axis at -273.15 °C. At any other value of pressure, say \(P_2\), a different straight line for the volume-temperature plot is obtained, but we get the same zero-volume temperature intercept at -273.15 °C. That every pressure gives the same intercept is the striking part: the number -273.15 °C is a property of gases in general, not of one particular sample.
Diagram
Fig. 10.10 – Graph of volume against temperature: volume on the y-axis, temperature in °C on the x-axis with 0 °C and -273.15 °C marked (and 273.15 K and 0 K marked beneath them); two straight lines of different slopes labelled P₁ and P₂, both extrapolated backwards as dashed lines meeting the temperature axis at the same point, -273.15 °C
The straight line of the volume versus temperature graph at constant pressure is called an isobar.
Zero volume for a gas sample is a hypothetical state. In practice all gases get liquified at a temperature higher than -273.15 °C, and a liquid certainly does not have zero volume, so the dashed part of the graph describes something that never actually happens. This temperature is the lowest temperature that can be imagined but practically cannot be attained. It is the absolute zero temperature on the kelvin scale (0 K).
In short, Charles’ law explains that at constant pressure, gases expand on heating and contract on cooling. Thus hot air is less dense than cold air, since the same mass of air now occupies a larger volume.
10.4.3 Gay-Lussac’s Law: (Pressure-Temperature Relationship)
The two laws so far have each left one variable out. Gay-Lussac’s law fills the gap: it relates the pressure and absolute temperature of a fixed mass of a gas at constant volume — that is, in a rigid container that cannot expand.
Mathematical Expression
Mathematically, the law can be expressed as:
\[P \propto T\]
\[\therefore\ P = k_3T \qquad (10.11)\]
\[\therefore\ \frac{P}{T} = \text{constant} \quad (\text{at constant }V\text{ and }m)\]
As with Charles’ law, comparing two states of the same gas gives the working form \(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\), since both ratios equal the same constant \(k_3\).
Why should heating a gas in a rigid container raise its pressure? Because raising the temperature raises the average kinetic energy of the molecules. They strike the walls both more often and harder, and since the wall area cannot change, the force per unit area — the pressure — must rise.
Graphical Representation
When a graph is plotted between pressure (\(P\)) in atm and temperature (\(T\)) in kelvin, a straight line is obtained as shown in Fig. 10.11. It is known as an isochore. With the help of this law, one can understand the relation between pressure and temperature in day to day life.
Diagram
Fig. 10.11 – Pressure versus temperature graph of a gas: pressure on the y-axis, temperature in K on the x-axis; four straight lines of decreasing slope labelled V₁ < V₂ < V₃ < V₄, each extrapolated backwards by a dashed line towards the origin (0,0)
10.4.4 Avogadro Law: (Volume-Amount Relationship)
In 1811 the Italian scientist Amedeo Avogadro combined Dalton’s atomic theory with Gay Lussac’s Law of combining volumes (Chapter 1) to put forth what is known as Avogadro law.
This means that at a fixed temperature and pressure, the volume of a gas depends upon the number of molecules of the gas, that is, the amount of the gas. Remarkably, it does not depend on which gas it is. A litre of chlorine and a litre of methane at the same temperature and pressure hold the same number of molecules, even though a chlorine molecule is much the heavier of the two.
Avogadro law can be expressed mathematically as:
\[V \propto n\]
where ‘\(n\)’ is the number of moles of the gas in volume ‘\(V\)’. On converting the proportionality into an equation we get
\[V = k_4n \qquad (10.12)\]
\[\text{or}\quad \frac{V}{n} = \text{constant} \quad (\text{at constant temperature and pressure})\]
Avogadro law can be well represented from Fig. 10.12.
Diagram
Fig. 10.12 – Avogadro law: four containers side by side, all at 0 °C and 1 atm pressure, holding respectively 6.022 × 10²³ molecules of Cl₂, 6.022 × 10²³ molecules of O₂, 6.022 × 10²³ atoms of Ne and 6.022 × 10²³ molecules of CH₄; a single caption underneath states that the volume of each gas is 22.414 L mol⁻¹
Molar Volume and STP
As the volume of a gas is directly proportional to the number of moles, one mole of any gas at STP occupies 22.414 L mol⁻¹ volume. This volume is known as molar volume.
STP means Standard Temperature and Pressure. Because the chosen standard has been revised, two sets of STP conditions are in use, and you must know which one a problem intends.
IUPAC has set STP as standard pressure = 1 bar = \(10^5\) Pa and standard temperature = 273.15 K = 0 °C. Under these STP conditions, the molar volume of an ideal gas or of a mixture of two or more gases = 22.71 L mol⁻¹.
The old STP conditions are also in use, where standard pressure = 1 atm = 1.013 bar and standard temperature = 273.15 K = 0 °C. Under the old STP conditions, the molar volume of an ideal gas or of a mixture of two or more gases = 22.414 L mol⁻¹.
The temperature is the same in both; only the pressure differs, and the slightly lower standard pressure of 1 bar gives the slightly larger molar volume. This is Boyle’s law showing up in the definition itself.
Relation Between Molar Mass and Density of a Gas
We have learnt how to find the number of moles of a gas (refer to Chapter 1). We know,
\[n = \frac{m}{M}\]
where \(m\) = mass of a gas and \(M\) = molar mass of a gas. Substituting in equation (10.12) we get,
\[V = k_4 \times \frac{m}{M}\]
\[\therefore\ M = k_4\frac{m}{V} \qquad (10.13)\]
But
\[\frac{m}{V} = d\]
Substituting this in Eq. (10.13) we get
\[M = k_4d \qquad (10.14)\]
where \(d\) = density of a gas, \(M\) = molar mass and \(m\) = mass of gas.
\[\therefore\ d \propto M\]
From Eq. (10.14), we can conclude that the density of a gas is directly proportional to its molar mass. Two gases measured at the same temperature and pressure hold the same number of molecules per litre, so the denser gas can only be the one whose molecules are individually heavier.
10.5 Ideal Gas Equation
A gas that follows strictly all the three laws — Boyle’s law, Charles’ law and Avogadro law — is an ideal gas. Practically, an ideal gas does not exist. Real gases show ideal behaviour under certain specific conditions, when intermolecular interactions are practically absent.
Why bother with a substance that does not exist? Because the three gas laws, and the ideal gas equation obtained by treating them mathematically, are found to be very valuable in Chemistry. Most common gases behave almost ideally at ordinary temperatures and low pressures, so the equation gives good answers for real work while remaining simple enough to handle.
10.5.1 Derivation of Ideal Gas Equation
The three gas laws, namely Boyle’s law, Charles law and Avogadro law, are combined mathematically to obtain what is called the ideal gas equation. Each law tells us how the volume responds to one variable while the others are held constant; putting them together gives a single statement covering all of them at once. The steps are collected in FC 10.4.
Flowchart
FC 10.4 – Building the ideal gas equation: Boyle’s law (V ∝ 1/P at constant T, n) + Charles’ law (V ∝ T at constant P, n) + Avogadro law (V ∝ n at constant P, T) → combine the three proportionalities into V ∝ nT/P → introduce the proportionality constant R → V = RnT/P → rearrange → PV = nRT
Let us write the proportionalities of the three gas laws:
- At constant \(T\) and \(n\), \(\ V \propto \dfrac{1}{P}\) (Boyle’s law)
- At constant \(P\) and \(n\), \(\ V \propto T\) (Charles’ law)
- At constant \(P\) and \(T\), \(\ V \propto n\) (Avogadro law)
Combining all the above three mathematical proportionalities we get
\[V \propto \frac{nT}{P}\]
A proportionality can be turned into an equation by putting in a constant. Converting this proportionality into an equation by introducing a constant of proportionality we get
\[\therefore\ V = R\left(\frac{nT}{P}\right)\]
On rearranging the above equation, we obtain
\[\therefore\ PV = nRT \qquad \text{...... (10.15)}\]
This equation is known as the Ideal Gas Equation. In the ideal gas equation, if three variables are known, the fourth can be calculated. It describes the state of an ideal gas, and therefore it is also called the equation of state.
Here R is called the Gas constant. The value of R is the same for all the gases. Therefore it is called the Universal gas constant.
10.5.2 Values of ‘R’ in Different Units
The value of R depends upon the units used to express \(P\), \(V\) and \(T\). Hence we recall STP conditions for determining values of ‘R’, since at STP we know the pressure, temperature and molar volume of one mole of gas exactly.
i. R in SI Unit (in Joules)
The value of R can be calculated by using the SI units of \(P\), \(V\) and \(T\). Pressure \(P\) is measured in N m⁻² or Pa, volume \(V\) in metre cube (m³) and temperature \(T\) in kelvin (K). Taking one mole of an ideal gas at the current IUPAC STP, where the molar volume is 22.71 L mol⁻¹, that is \(22.71 \times 10^{-3}\) m³ mol⁻¹:
\[\mathrm{R} = \frac{PV}{nT}\]
\[\therefore\ \mathrm{R} = \frac{10^{5}\ \text{Pa} \times 22.71 \times 10^{-3}\ \text{m}^3}{1\ \text{mol} \times 273.15\ \text{K}}\]
\[\therefore\ \mathrm{R} = 8.314\ \text{Pa m}^3\text{K}^{-1}\text{mol}^{-1}\]
then
\[\mathrm{R} = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\]
The last step is only a change of name, not of value: a pascal is a newton per square metre, so Pa m³ is N m, which is the joule.
ii. R in Litre Atmosphere
If pressure (\(P\)) is expressed in atmosphere (atm) and volume in litre (L) or decimetre cube (dm³) and temperature in kelvin (K) — that is, the old STP conditions, where the molar volume is 22.414 L mol⁻¹ — then the value of R is:
\[\mathrm{R} = \frac{1\ \text{atm} \times 22.414\ \text{L}}{1\ \text{mol} \times 273.15\ \text{K}}\]
\[\therefore\ \mathrm{R} = 0.08206\ \text{L atm K}^{-1}\text{mol}^{-1}\]
OR
\[\mathrm{R} = 0.08206\ \text{dm}^3\text{atm K}^{-1}\text{mol}^{-1}\]
Since 1 L and 1 dm³ are the same volume, the two forms are identical.
iii. R in Calories
We know that 1 calorie = 4.184 Joules. Therefore
\[\therefore\ \mathrm{R} = \frac{8.314}{4.184} = 1.987 \cong 2\ \text{cal K}^{-1}\text{mol}^{-1}\]
This is the value worth remembering as “about 2 calories per kelvin per mole”, and it appears often in thermodynamic calculations.
Note: Often pressure (\(P\)) is measured in atmosphere (atm). The conversion between the units is done by noting that 1 atm = 1.013 × 10⁵ Nm⁻², or 1 atm = 101.3 kPa.
While using the ideal gas equation one should use consistent units. If the pressure is in atm, the volume must be in dm³ or L and R must be the 0.0821 value; mixing a pressure in kPa with the litre-atmosphere value of R is the single commonest source of wrong answers in this topic. The most common units are shown in Table 10.4.
Table 10.4 Unit of R
| Pressure (\(P\)) | Volume (\(V\)) | Number of moles (\(n\)) | Temperature (\(T\)) | Gas constant (R) |
|---|---|---|---|---|
| Pa (pascals) | m³ | mol | K | 8.314 J K⁻¹ mol⁻¹ |
| atm | dm³ | mol | K | 0.0821 atm dm³ K⁻¹ mol⁻¹ |
| atm | L | mol | K | 0.0821 L atm K⁻¹ mol⁻¹ |
10.5.3 Expression for Molar Mass
The ideal gas equation can be used to determine the molar mass (\(M\)) of a compound, which is one of its most useful applications: weighing a gas and measuring its pressure, volume and temperature is enough to identify it.
Rearranging the equation (10.15),
\[n = \frac{PV}{\mathrm{R}T} \qquad \text{...... (10.16)}\]
But for a known mass of a gas ‘\(m\)’, the number of moles is
\[n = \frac{m}{M}\]
On substituting this in equation (10.16) we get
\[\frac{m}{M} = \frac{PV}{\mathrm{R}T}\]
\[\therefore\ M = \frac{m\mathrm{R}T}{PV} \qquad \text{...... (10.17)}\]
10.5.4 Combined Gas Law
The ideal gas equation is written as
\[PV = n\mathrm{R}T \qquad \text{...... (10.15)}\]
On rearranging Eq. (10.15) we get
\[\therefore\ \frac{PV}{T} = n\mathrm{R} = \text{constant}\]
For a fixed amount of gas, the whole right-hand side is a constant, so the quantity \(PV/T\) must have the same value in any two states of that gas. Hence
\[\therefore\ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \qquad \text{...... (10.18)}\]
The ideal gas equation used in this form is called the combined gas law. We have to assume a gas under two different sets of conditions, where pressure, volume and temperature are written for one state as \(P_1\), \(V_1\), \(T_1\) and for the other state as \(P_2\), \(V_2\), \(T_2\), respectively.
This one relation contains Boyle’s, Charles’ and Gay-Lussac’s laws inside it. Hold \(T\) constant and it becomes \(P_1V_1 = P_2V_2\); hold \(P\) constant and it becomes \(V_1/T_1 = V_2/T_2\); hold \(V\) constant and it becomes \(P_1/T_1 = P_2/T_2\). That is why it is so useful when two variables change at the same time, which no single gas law can handle.
10.5.5 Relation Between Density, Molar Mass and Pressure
The ideal gas equation can be expressed in terms of density as follows:
\[PV = n\mathrm{R}T \qquad \text{...... (10.15)}\]
On rearranging it gives
\[\frac{n}{V} = \frac{P}{\mathrm{R}T}\]
But we know
\[n = \frac{m}{M}\]
On substituting the value of ‘\(n\)’, the rearranged equation becomes
\[\therefore\ \frac{m}{MV} = \frac{P}{\mathrm{R}T} \qquad \text{(at constant temperature)}\]
\[\therefore\ \frac{d}{M} = \frac{P}{\mathrm{R}T} \qquad \text{...... (10.19)}\]
where \(\dfrac{m}{V} = d\) = density of a gas.
On rearranging equation (10.19) we get the expression to calculate the molar mass of a gas in terms of its density:
\[M = \frac{d\mathrm{R}T}{P} \qquad \text{...... (10.20)}\]
Equation (10.20) is worth noticing for a practical reason: it lets you find the molar mass of a gas without weighing out any particular mass at all, since a density measurement already carries the mass and volume together.
From equation (10.19), Boyle’s law can be stated in terms of density as: at constant temperature, the pressure of a given mass of gas is directly proportional to its density. This is the same conclusion that was reached in Section 10.4.1 by a different route, and it is reassuring that the ideal gas equation delivers it again.
10.5.6 Dalton’s Law of Partial Pressure
So far every calculation has dealt with a single pure gas. But most gases you actually meet — air, exhaled breath, the gas collected over water in a laboratory experiment — are mixtures. Dalton’s law tells us how to handle them.
This law is applicable for those gases which do not react chemically on mixing. If the gases reacted, the number of molecules would change and no simple addition would work.
The pressure exerted by an individual gas in a mixture of two or more gases is called partial pressure. It is also the pressure that each gas would exert if it were present alone in the same container at the same temperature.
Dalton’s law of partial pressure is stated as:
Mathematically, Dalton’s law may be expressed as
\[P_{\text{Total}} = P_1 + P_2 + P_3 + \dots \qquad \text{(at constant } V \text{ and } T)\]
where \(P_{\text{Total}}\) is the total pressure of the mixture and \(P_1\), \(P_2\), \(P_3\), ……… are the partial pressures of the individual gases 1, 2, 3, ….. in the mixture. The situation is shown in Fig. 10.13.
Diagram
Fig. 10.13 – Schematic illustration of Dalton’s law of partial pressures: at constant volume and temperature, one cylinder of gas with its gauge reading P₁, plus a second cylinder of a different gas with its gauge reading P₂, with an arrow labelled “combining the gases” leading to a third identical cylinder containing both gases whose gauge reads P_T = P₁ + P₂
Partial Pressure and Mole Fraction
We know that the pressure of a pure gas is given by the ideal gas equation
\[P = \frac{n\mathrm{R}T}{V} \qquad \text{...... (10.21)}\]
At constant \(V\) and \(T\), everything on the right except \(n\) is fixed, so
\[\therefore\ P \propto n\]
Consequently, the pressure of an individual gas in a mixture of gases is proportional to its amount in that mixture. This implies that the total pressure (\(P_T\)) of a mixture of gases at constant volume (\(V\)) and temperature (\(T\)) is equal to the sum of the pressures that each individual gas exerts in the mixture. This is Dalton’s law of partial pressures.
The partial pressures of individual gases can be written in terms of the ideal gas equation as follows:
\[P_1 = n_1\left(\frac{\mathrm{R}T}{V}\right),\qquad P_2 = n_2\left(\frac{\mathrm{R}T}{V}\right),\qquad P_3 = n_3\left(\frac{\mathrm{R}T}{V}\right),\ \dots \text{and so on} \qquad \text{...... (10.21)}\]
\[\therefore\ P_{\text{Total}} = n_1\left(\frac{\mathrm{R}T}{V}\right) + n_2\left(\frac{\mathrm{R}T}{V}\right) + n_3\left(\frac{\mathrm{R}T}{V}\right) + \dots\]
Taking the common factor out,
\[= \left(\frac{\mathrm{R}T}{V}\right)(n_1 + n_2 + n_3 \dots) = \frac{\mathrm{R}T}{V}\,n_{\text{Total}} \qquad \text{...... (10.22)}\]
The mole fraction of any individual gas in the mixture is given by the equation
\[X_1 = \frac{n_1}{n_1 + n_2 + n_3 \dots} = \frac{n_1}{n_{\text{Total}}} \qquad \text{...... (10.23)}\]
From Eq. (10.21) and Eq. (10.23) we get
\[n_1 = P_1 \Big/ \left(\frac{\mathrm{R}T}{V}\right) \qquad \text{...... (10.24)}\]
and
\[n_{\text{Total}} = P_{\text{Total}} \Big/ \left(\frac{\mathrm{R}T}{V}\right) \qquad \text{...... (10.25)}\]
By combining equation Eq. (10.24) and Eq. (10.25) we get
\[\frac{n_1}{n_{\text{Total}}} = X_1 = \frac{P_1 \Big/ \left(\dfrac{\mathrm{R}T}{V}\right)}{P_{\text{Total}} \Big/ \left(\dfrac{\mathrm{R}T}{V}\right)} = \frac{P_1}{P_{\text{Total}}} \qquad \text{...... (10.26)}\]
The factor \(\mathrm{R}T/V\) has cancelled from top and bottom, which is the key step: it is the same for every gas in the mixture, because they all share the same container and the same temperature.
Therefore it follows that
\[P_1 = X_1 \cdot P_{\text{Total}} \qquad \text{...... (10.27)}\]
Similarly,
\[P_2 = X_2 \cdot P_{\text{Total}},\qquad P_3 = X_3 \cdot P_{\text{Total}}\quad \text{and so on.}\]
Thus the partial pressure of a gas is obtained by multiplying the total pressure of the mixture by the mole fraction of that gas. In exam problems this is almost always the quickest route, because mole fractions can be found from masses without ever needing \(V\), \(T\) or R.
Aqueous Tension
A vapour is a gas in contact with a liquid of the same substance. For example, the ‘gas’ above the surface of liquid water is described as water vapour. The pressure of the water vapour is known as its vapour pressure.
Suppose liquid water is placed into a container, the air above it is pumped away and the container is sealed. Then the liquid water evaporates and only water vapour remains in the space above. After sealing, the vapour pressure increases initially, then slows down as some water molecules condense back to form liquid water. After a few minutes the vapour pressure reaches a maximum, called the saturated vapour pressure; at this stage evaporation and condensation are going on at equal rates, and this idea is taken further in Section 10.9.1. The pressure exerted by saturated water vapour is called Aqueous Tension (\(P_{aq}\)).
When a gas is collected over water in a closed container, it gets mixed with the saturated water vapour in that space. The measured pressure, therefore, corresponds to the pressure of the mixture of that gas and the saturated water vapour in that space — a direct application of Dalton’s law, since the gas and the water vapour do not react.
The pressure of pure and dry gas can be calculated by using the aqueous tension. It is obtained by subtracting the aqueous tension from the total pressure of the moist gas.
\[\therefore\ P_{\text{Dry gas}} = P_{\text{Total}} - P_{aq}\]
that is,
\[P_{\text{Dry gas}} = P_{\text{Total}} - \text{Aqueous Tension}\]
Table 10.5: Aqueous Tension of Water (Vapour Pressure) as a function of Temperature
| Temp. (K) | Pressure (bar) | Temp. (K) | Pressure (bar) |
|---|---|---|---|
| 273.15 | 0.0060 | 295.15 | 0.0260 |
| 283.15 | 0.0121 | 297.15 | 0.0295 |
| 288.15 | 0.0168 | 299.15 | 0.0331 |
| 291.15 | 0.0204 | 301.15 | 0.0372 |
| 293.15 | 0.0230 | 303.15 | 0.0418 |
The above table reflects that aqueous tension increases with increase in temperature. The reason is that warmer water molecules have more kinetic energy, so a greater fraction of them can escape from the liquid surface before equilibrium is reached.
Diagram
Fig. 10.5A – Graph of aqueous tension of water against temperature, using the values of Table 10.5: temperature in K from 273.15 to 303.15 on the x-axis, pressure in bar from 0 to 0.045 on the y-axis, plotted points joined by a smooth upward-curving line
10.6 Kinetic Molecular Theory of Gases
The gas laws tell us what gases do. They do not tell us why. Boyle’s law was read off from measurements, not deduced from any picture of what a gas is made of. To explain the observed behaviour we need a model of the gas at the molecular level.
The kinetic molecular theory of gases is a theoretical model put forth to explain the behaviour of gases. It starts from a small set of assumptions about the molecules, and from these every gas law we have met can be recovered.
The word “theoretical” deserves emphasis. These assumptions are deliberate simplifications, not descriptions of a real sample of oxygen. That is exactly why they define an ideal gas rather than a real one, a point taken up in Section 10.7.
Assumptions of Kinetic Molecular Theory of Gases
Gases consist of tiny particles (molecules or atoms). For a monoatomic gas such as a noble gas these particles are single atoms; for most other gases they are molecules.
On an average, gas molecules remain far apart from each other. Therefore the actual volume of the gas particles is negligible as compared to the volume of the container. That is why gases are highly compressible — squeezing a gas mostly squeezes out empty space, not the particles themselves.
The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, a gas expands to occupy the entire volume of the container. Nothing holds the molecules back, so they spread until the walls stop them.
Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container. “Random” here means that no direction is preferred over any other, which is why a gas does not drift to one side or settle at the bottom of a vessel.
Pressure of the gas is due to the collision of gas particles with the walls of the container. Each collision delivers a small push to the wall; the enormous number of collisions every second adds up to a steady force, and that force divided by the wall area is what we measure as pressure.
The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after collision. A molecule may leave a collision faster or slower than it arrived, but the energy is only passed around, never lost.
The different molecules of a gas move with different velocities at any instant and hence have different kinetic energies. But the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.
\[\text{Average K.E.} \propto T\]
Diagram
Fig. 10.6A – Kinetic molecular model of a gas: a closed box containing about fifteen small spheres widely separated, each carrying an arrow of a different length and direction to show random motion at different speeds; show one sphere striking a wall and rebounding at an angle, with the incoming and outgoing arrows of equal length to indicate a perfectly elastic collision, and label the wall “pressure = force of collisions per unit area”
How the Assumptions Explain Pressure and Temperature
Assumptions 4, 5 and 7 taken together form the heart of the theory, because between them they explain the two quantities we measure most often. The chain of reasoning is set out in FC 10.3.
Flowchart
FC 10.3 – From molecular motion to measured quantities: random straight-line motion → collisions with the container walls → force delivered per unit area → pressure of the gas; branch from the same starting point: rise in temperature → rise in average kinetic energy → collisions become more frequent and harder → (if volume is fixed) pressure rises, or (if pressure is fixed) volume rises
This chain is worth reading in both directions. Read forwards, it explains Gay-Lussac’s law: heat a gas in a rigid vessel, the molecules strike the walls more often and more forcefully, and the pressure climbs. Read the same chain with the volume free to change and it explains Charles’ law: the gas must expand to keep the pressure matched to the surroundings.
10.7 Deviation from Ideal Behaviour
An ideal gas is the one that exactly follows the ideal gas equation. On rearranging the ideal gas equation, we get \(n = PV/\mathrm{R}T\). If we use 1 mole of any gas, then the ratio \(PV/\mathrm{R}T\) is predicted to have the numerical value of 1 at all pressures.
That prediction can be tested, and this is exactly what makes it valuable. If a gas does not obey the ideal gas law, the ratio will be either greater than 1 or less than 1. Such a gas is said to be a Real Gas.
The Ideal Gas, in Summary
Before examining the deviations, it is worth setting down clearly what an ideal gas was assumed to be.
- A gas that obeys all the gas laws over the entire range of temperature and pressure.
- There are no interactive forces between the molecules.
- The molecular volume is negligibly small compared to the volume occupied by the gas. These are point particles.
The second and third statements are the ones that will fail for a real gas, and both are traceable to the kinetic molecular theory of Section 10.6, where they appeared as assumptions.
The Real Gas
A real gas does not obey Boyle’s law and Charles’ law under all the conditions of temperature and pressure. It obeys them well under some conditions and poorly under others, so the useful question is not “does it deviate?” but “when does it deviate?”
A deviation from the ideal behaviour is observed at high pressure and low temperature. It is due to two reasons.
The intermolecular attractive forces are not negligible in real gases. These do not allow the molecules to collide with the container wall with full impact, because a molecule heading for the wall is being tugged backwards by its neighbours behind it. This results in a decrease in the pressure, so the observed pressure is lower than the ideal gas equation predicts.
At high pressure, the molecules are very close to each other. The short range repulsive forces start operating and the molecules behave as small but hard spherical particles. The volume of the molecule is not negligible. Therefore less volume is available for molecular motion than the measured volume of the container suggests.
Notice that these two reasons pull in opposite directions. Attraction makes a gas easier to compress than an ideal gas, while the finite size of the molecules makes it harder. Which effect wins depends on the conditions, and that is why deviations can go either way.
At very low temperature, the molecular motion becomes slow and the molecules are attracted to each other due to the attractive force. Hence, here again, the behaviour of the real gas deviates from the ideal gas behaviour. A slow-moving molecule spends longer in the neighbourhood of another molecule, so a weak attraction that a fast molecule would simply fly past now has time to take effect.
Flowchart
FC 10.6 – When and why a real gas deviates: start with the conditions of the gas → if pressure is low and temperature is high → molecules far apart and fast moving → both attraction and molecular volume negligible → gas behaves ideally, Z ≈ 1; if temperature is low → molecular motion slow → attractive forces dominate → observed pressure less than ideal → Z < 1; if pressure is very high → molecules very close → repulsive forces operate and molecular volume is significant → gas hard to compress → Z > 1
Deviation Seen on a Pressure–Volume Plot
Deviation with respect to pressure can be studied by plotting a pressure (\(P\)) versus volume (\(V\)) curve at a given temperature, and comparing it with the curve an ideal gas would give under the same conditions. See Fig. 10.14 and Table 10.6.
Diagram
Fig. 10.14 – Plot of pressure versus volume for a real gas and an ideal gas at the same temperature: pressure on the y-axis, volume on the x-axis, showing two hyperbola-like curves labelled “real gas” and “ideal gas” which coincide at large volume and low pressure but separate at low volume and high pressure, with the real-gas curve lying at the larger volume
At very high pressure, the measured volume is more than the theoretically calculated volume assuming ideal behaviour. This is the finite size of the molecules showing itself: the molecules themselves take up room that cannot be squeezed away. But at low pressure, measured and calculated volumes approach each other, and the two curves in Fig. 10.14 become indistinguishable.
Table 10.6: Difference between Ideal Gas and Real Gas
| Ideal gas | Real Gas |
|---|---|
| 1. Strictly obeys Boyle’s and Charles’ law. \(\dfrac{PV}{n\mathrm{R}T} = 1\) | 1. Shows deviation from Boyle’s and Charles’ law at high pressure and temperature, i.e. obeys Boyle’s law and Charles’ law at low pressure and high temperature. \(\dfrac{PV}{n\mathrm{R}T} \neq 1\) |
| 2. Molecules are perfectly elastic. | 2. Molecules are not perfectly elastic. |
| 3. No attraction or repulsion between the gas molecules, i.e. collision without loss of kinetic energy (K.E.) | 3. Intermolecular attraction is present, hence collision takes place with loss of kinetic energy. |
| 4. Actual volume of the gas molecules is negligible as compared to total volume of the gas. | 4. Actual volume of individual gas molecule is significant at high pressure and low temperature. |
| 5. Cannot liquify even at low temperature but continues to obey Charles’ law and finally occupies zero volume at -273 °C. | 5. Undergo liquefaction at low temperature when cooled and compressed. |
| 6. Such a gas does not exist. | 6. Gases that exist in nature like \(\mathrm{H_2}\), \(\mathrm{O_2}\), \(\mathrm{CO_2}\), \(\mathrm{N_2}\), He, etc. |
Row 5 of this table is the most important one for what follows. An ideal gas could never be turned into a liquid, because liquefaction needs molecules to attract one another and an ideal gas has no attractions at all. Every gas you can actually liquefy is, by that very fact, a real gas.
Compressibility Factor (Z)
It is inconvenient to describe deviations by drawing a fresh \(P\)–\(V\) curve for every gas, so a single number is used instead.
Real gases show ideal behaviour when pressure approaches zero. Deviation from ideal behaviour can be measured in terms of the compressibility factor Z, which is the ratio of the product \(PV\) and \(n\mathrm{R}T\):
\[Z = \frac{PV}{n\mathrm{R}T}\]
Figure 10.15 shows the graph of \(Z\) vs \(P\). For an ideal gas it is a straight line parallel to the x-axis (pressure axis) where \(Z = 1\). For an ideal gas \(Z = 1\) at all values of temperature and pressure, which is simply the ideal gas equation restated.
Diagram
Fig. 10.15 – Graph of compressibility factor Z = PV/nRT on the y-axis, from 0.4 to 1.8, against pressure P/bar on the x-axis from 0 to 1000, showing a horizontal dashed line at Z = 1 labelled “ideal gas”, curves for N₂, H₂, O₂ and CH₄, and a curve for CO₂ that dips well below Z = 1 at low pressure before rising steeply; all curves converge towards Z = 1 as the pressure approaches zero
The significance of \(Z\) is better understood from the following derivation. We can write two equations, (10.28) and (10.29), for a real gas and an ideal gas respectively:
\[Z = \frac{PV_{real}}{n\mathrm{R}T} \qquad \text{...... (10.28)}\]
and
\[1 = \frac{PV_{ideal}}{n\mathrm{R}T} \qquad \text{...... (10.29)}\]
From Eq. (10.29), rearranging to make the common factor \(P/n\mathrm{R}T\) the subject,
\[\therefore\ \frac{P}{n\mathrm{R}T} = \frac{1}{V_{ideal}}\]
Substituting the value of \(\dfrac{P}{n\mathrm{R}T}\) in Eq. (10.28), we get
\[Z = \frac{V_{real}}{V_{ideal}} \qquad \text{...... (10.30)}\]
Equation (10.30) implies that the compressibility factor is the ratio of the actual molar volume of a gas to its molar volume if it behaved ideally, at that temperature and pressure. This second form is often the more revealing of the two, because it turns an abstract ratio into a direct comparison of volumes.
Reading the Value of Z
A positive deviation in \(Z\), that is \(Z > 1\), means that the volume of a molecule cannot be neglected and the gas is difficult to compress. The actual volume is larger than the ideal one because the molecules themselves occupy space.
At lower pressure, the gases have \(Z = 1\) or \(Z < 1\). Under this condition the molecular volume is negligible and the gases are compressible. Here, the gases show ideal behaviour.
10.8 Liquefaction of Gases and Critical Constant
Most gases behave like ideal gases at high temperature. For example, the \(PV\) curve of \(\mathrm{CO_2}\) gas at 50 °C is like the ideal Boyle’s law curve. As the temperature is lowered, the \(PV\) curve shows a deviation from the ideal Boyle’s law curve.
At a particular value of low temperature the gas gets liquified at a certain increased value of pressure. For example, \(\mathrm{CO_2}\) gas liquifies at 30.98 °C and 73 atmosphere pressure (see Fig. 10.16).
This is the highest temperature at which liquid \(\mathrm{CO_2}\) can exist. Above this temperature, howsoever large the pressure may be, liquid \(\mathrm{CO_2}\) cannot form. Other gases also show similar behaviour, each with its own such temperature.
Reading the Isotherms of Carbon Dioxide
The whole story is contained in one set of curves. Figure 10.16 shows isotherms of \(\mathrm{CO_2}\) measured at several temperatures, one above \(T_c\), one at \(T_c\) and two below it.
Diagram
Fig. 10.16 – Liquefaction of CO₂, isotherms at various temperatures: pressure on the y-axis, volume on the x-axis; four curves labelled 50 °C, 30.98 °C (Tc), 21.50 °C and 13.10 °C; the two lower isotherms have flat horizontal portions inside a dome-shaped region shaded and labelled “liquid + gas”; the region left of the dome labelled “liquid”, the region right of and above it labelled “gas”; mark Pc on the pressure axis and Vc on the volume axis at the top of the dome, where the Tc isotherm just touches it
Above the critical temperature a substance exists only as gas. The molar volume at the critical temperature is called the critical volume (\(V_c\)), and the pressure at the critical temperature is called the critical pressure (\(P_c\)) of that substance.
Together, \(T_c\), \(P_c\) and \(V_c\) are called the critical constants of a substance. They are as characteristic of a substance as its melting point, because they are decided by the strength of its intermolecular forces. Table 10.7 gives the values of these three critical constants for some common gases.
Table 10.7: Critical constants for common Gases
| Substance | Tc / K | Pc / bar | Vc / dm³ mol⁻¹ |
|---|---|---|---|
| H₂ | 33.2 | 12.97 | 0.0650 |
| He | 5.3 | 2.29 | 0.0577 |
| N₂ | 126.0 | 33.9 | 0.0900 |
| O₂ | 154.3 | 50.4 | 0.0744 |
| CO₂ | 304.10 | 73.9 | 0.0956 |
| H₂O | 647.1 | 220.6 | 0.0450 |
| NH₃ | 405.5 | 113.0 | 0.0723 |
Read the \(T_c\) column alongside what you know of intermolecular forces from Section 10.2, and the numbers stop looking arbitrary. Helium, which has only dispersion forces between single small atoms, has the lowest \(T_c\) of all at 5.3 K. Water, held together by an extensive hydrogen-bonded network, has by far the highest at 647.1 K. The stronger the attraction between the molecules, the warmer a substance can be and still be pulled into a liquid.
Which Gases Are Easy to Liquefy?
From Table 10.7 it is seen that the gases that we come across in everyday life, namely \(\mathrm{N_2}\) and \(\mathrm{O_2}\), have \(T_c\) values much below 0 °C and their \(P_c\) values are high. Consequently, liquefaction of \(\mathrm{O_2}\) and \(\mathrm{N_2}\) — and therefore of air — requires both compression and cooling. Compression alone can never work at room temperature, because room temperature is far above their critical temperatures.
The \(T_c\) value of \(\mathrm{CO_2}\) nearly equals room temperature; however, its \(P_c\) value is very high. Therefore \(\mathrm{CO_2}\) exists as a gas under ordinary conditions, and a cylinder of liquid carbon dioxide has to be kept under considerable pressure.
Flowchart
FC 10.8 – Can this gas be liquefied? Start: note the temperature T of the gas and look up its Tc → if T > Tc → no pressure however large will liquefy it → the gas must first be cooled below Tc → then compress it; if T < Tc → compress the gas → at a definite pressure a liquid surface appears and liquid and gas coexist → continue compressing → the whole sample becomes liquid
Continuity Between the Gaseous and Liquid States
Study of \(PV\) curves of various gases at different temperatures reveals that under certain conditions the substance changes completely from the gaseous to the liquid state, or from liquid to gaseous, without having the coexistence of two states. There is no moment at which you could point to a boundary between liquid and gas.
In other words, there is a continuity between the gaseous and liquid state. This continuity is recognized by using the term Fluid for either gas or liquid.
Under these conditions liquid is just a very dense gas. Liquid and gas can be distinguished from each other only when the fluid is at a temperature below \(T_c\) — only there does a meniscus, a visible surface separating the two, exist at all.
Boiling Point and the Pressure Above the Liquid
When a liquid which is exposed to the atmosphere is heated, its vapour pressure increases. Eventually it becomes equal to the atmospheric pressure. At this temperature the vaporisation takes place throughout the bulk of the liquid, and the vapour formed escapes freely into the surroundings. We call this the boiling point of the liquid.
The boiling temperature (boiling point) therefore depends upon the pressure to which the surface of the liquid is exposed. This is not a small correction but the whole reason boiling points must be quoted with a pressure attached. If the pressure is 1 atm, the boiling temperature is called the normal boiling point. If the pressure is 1 bar, the boiling temperature is called the standard boiling point.
10.9 Liquid State
The liquid state is the intermediate state between the solid state and the gaseous state, and it inherits something from each. Molecules of a liquid are held together by moderately strong intermolecular forces — strong enough to keep the molecules in contact, but not strong enough to lock them into fixed positions — and so they can move about within the boundary of the liquid.
As a result, a liquid possesses properties such as fluidity, a definite volume, and the ability to take the shape of the bottom of the container in which it is placed. It flows like a gas but keeps its own volume like a solid.
Different liquids are characterized by their quantitative properties such as density, boiling point, freezing point, vapour pressure, surface tension and viscosity. We will look at some of these properties in the following sections. Each one of them, as you will see, is a different way of measuring the same underlying thing: how firmly the molecules of that particular liquid hold on to one another.
10.9.1 Vapour Pressure
Molecules of a liquid have a tendency to escape from its surface to form vapour above it. This is called evaporation. Not every molecule can escape — only those at the surface that happen to be moving fast enough to break away from their neighbours.
When a liquid is placed in a closed container, it undergoes evaporation and the vapours formed undergo condensation. At first, escape is the only process going on; but as vapour collects above the liquid, more and more of its molecules strike the surface and return to the liquid. At equilibrium, the rate of evaporation and the rate of condensation are equal. Nothing appears to be happening after that, yet both processes continue at the same speed — this is a dynamic equilibrium, not a standstill.
The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure, or simply vapour pressure.
Flowchart
FC 10.9 – Establishing vapour pressure in a closed vessel: liquid in a sealed container → fast surface molecules escape (evaporation) → vapour collects above the liquid → some vapour molecules strike the surface and return (condensation) → as vapour builds up, the rate of condensation rises → rate of condensation becomes equal to the rate of evaporation → dynamic equilibrium reached → the constant pressure now exerted by the vapour is the saturated vapour pressure; the cycle of escape and return continues at equal rates
The vapour pressure of water is also called Aqueous Tension (refer to Table 10.5 in Section 10.5.6). Schematically, vapour pressure is explained in Fig. 10.17, where a manometer attached to the closed vessel reads the pressure of the vapour above the liquid.
Diagram
Fig. 10.17 – Vapour pressure of a liquid: a closed vessel containing a beaker of liquid, with upward arrows above the liquid surface marked “vapours”, connected by a side tube to a U-shaped mercury manometer whose difference in levels is labelled “vapour pressure”; label the closed vessel, the beaker containing liquid and the mercury manometer
Factors Affecting Vapour Pressure
a. Nature of liquid: Liquids having relatively weak intermolecular forces possess high vapour pressure, because their surface molecules need little energy to escape. These are called volatile liquids. For example, petrol evaporates more quickly than motor oil.
b. Temperature: When the liquid is gradually heated, its temperature rises and its vapour pressure increases. Heating raises the kinetic energy of the molecules, so a larger fraction of them can overcome the attractions at the surface and escape.
This second point connects directly to Section 10.8: a liquid boils when its vapour pressure has been raised until it equals the pressure above its surface.
Unit of Vapour Pressure
Vapour pressure is measured by means of a manometer. The most common unit for vapour pressure is torr, where 1 torr = 1 mm Hg. For example, water has a vapour pressure of approximately 20 torr at room temperature. (Refer to Section 10.5.6 and Table 10.5.)
Twenty torr is a useful figure to hold on to, because atmospheric pressure is 760 torr. At room temperature the vapour pressure of water is only about one-fortieth of atmospheric pressure, which is why an open glass of water evaporates slowly instead of boiling away.
10.9.2 Surface Tension
Many phenomena, such as the capillary rise of liquids and the spherical shapes of liquid drops, are due to the property of liquids called surface tension. The surface of a liquid acts as a stretched membrane, as though a thin invisible skin were pulled tight across it.
The explanation lies in an imbalance of forces. The particles in the bulk of the liquid are uniformly attracted in all directions, and the net force acting on the molecules present inside the bulk is zero — each pull is matched by an equal pull from the opposite side. But the molecules at the surface have neighbours below and beside them and none above, so they experience a net attractive force towards the interior of the liquid. In other words, the forces acting on the molecules on the surface are imbalanced (see Fig. 10.18).
Diagram
Fig. 10.18 – Imbalance of forces at the surface of a liquid: a beaker of liquid drawn with molecules shown as circles; one molecule in the bulk with arrows pulling equally in all directions labelled “molecule in the bulk (balanced force)”, and one molecule at the surface with arrows pulling only sideways and downwards labelled “molecule at the surface (imbalanced force)”
Therefore liquids have a tendency to minimize their surface area, since every molecule dragged into the interior reduces the number left in the unfavourable surface position. A sphere is the shape with the least surface area for a given volume, which is why a free drop of liquid is spherical.
Surface tension is a temperature dependent property. When attractive forces are large, surface tension is large. Surface tension decreases as the temperature increases. The reason is a chain of causes: with an increase in temperature, the kinetic energy of the molecules increases, so the intermolecular forces of attraction become less effective at holding molecules together, and thereby surface tension decreases.
Applications of Surface Tension
i. Cleaning action of soap and detergent is due to the lowering of interfacial tension between water and oily substances. Due to the lower surface tension, the soap solution penetrates into the fibre, surrounds the oily substance and washes it away. Plain water, with its high surface tension, tends to stay as droplets on the fabric instead of soaking in.
ii. Efficacy of toothpastes, mouth washes and nasal drops is partly due to the presence of substances having lower surface tension. This increases the efficiency of their penetrating action, letting the liquid reach into narrow gaps between teeth or into the lining of the nose.
10.9.3 Viscosity
As noted earlier, liquids, that is fluids, have a tendency to flow. But they do not all flow equally readily: honey and water differ in a way that has nothing to do with their densities.
Viscosity measures the magnitude of internal friction in a liquid or fluid to flow, as measured by the force per unit area resisting uniform flow. When a liquid flows, different layers of the liquid flow with different velocity. This is called laminar flow. Here, the layers of molecules in immediate contact with the fixed surface remain stationary, and the subsequent layers slip over one another. Strong intermolecular forces obstruct the layers from slipping over one another, resulting in friction between the layers.
Viscosity is the force of friction between the successive layers of a flowing liquid. It is also the resistance to the flow of a liquid.
When a liquid flows through a tube, the central layer has the highest velocity, whereas the layer along the inner wall of the tube remains stationary. This is a result of the viscosity of a liquid (see Fig. 10.19). Hence a velocity gradient exists across the cross-section of the pipe or tube — the velocity changes steadily as you move from the wall towards the centre.
Diagram
Fig. 10.19 – Laminar flow of a liquid (fluid) through a tube/pipe: a length of tube drawn in section with a series of nested cylindrical layers, arrows of increasing length from the wall towards the axis showing increasing velocity, the axis labelled “centre of tube” and successive layers labelled V - dV, V and V - dV
Viscosity is a temperature dependent property:
\[\text{Viscosity} \propto \frac{1}{\text{Temperature}}\]
Heating a liquid weakens the grip its molecules have on one another, so the layers slide past each other more easily. This is why oil is poured more easily when warm, and why a cold engine is harder to turn over than a warm one.
Viscosity also depends on molecular size and shape. Larger molecules have more viscosity, since they become entangled with one another, and spherical molecules offer the least resistance to flow and are therefore less viscous. Greater the viscosity, slower is the liquid flow.
Unit of Viscosity
Viscosity is expressed in terms of the coefficient of viscosity, ‘\(\eta\)’ (Eta). It is defined as the degree to which a fluid resists flow under an applied force, measured by the tangential frictional force per unit area per unit velocity gradient when the flow is laminar.
The SI unit of the viscosity coefficient is N s m⁻² (newton second per square metre). In the CGS system, \(\eta\) is measured in poise:
\[1\ \text{poise} = 1\ \text{gm cm}^{-1}\text{s}^{-1} = 10^{-1}\ \text{kg m}^{-1}\text{s}^{-1}\]
Illustrations of Viscosity
i. Lubricating oils are viscous liquids. The gradation of lubricating oils is done on the basis of viscosity. A good quality lubricating oil does not change its viscosity with an increase or decrease in temperature — an important requirement, since an engine oil must not become too thin when hot nor too thick to circulate when cold.
ii. An increase in blood viscosity above the normal value is taken as an indication of cardiovascular disease, because thicker blood flows less easily through narrow vessels.
iii. Glass panes of old buildings are found to become thicker with time near the bottom. This is one evidence which indicates that glass is not a solid but a supercooled viscous liquid.
Read the third row of this table carefully, because it is the key to most questions on this section. A liquid with strong intermolecular forces holds its molecules in, so it evaporates reluctantly (low vapour pressure), pulls its surface tight (high surface tension) and resists flow (high viscosity). All three properties point back to the same cause.
10.10 Summary
Matter exists in three states — solid, liquid and gas — and the same substance can pass from one to another by the exchange of heat, without any change in its chemical identity. Ice, water and water vapour are all \(\mathrm{H_2O}\).
The three states differ in the mean separation of their particles, and every other difference follows from it: arrangement, freedom of movement, definite or indefinite shape and volume, intermolecular space, response of volume to a temperature change, and compressibility.
Intermolecular forces are the attractive and repulsive forces acting between neighbouring molecules. The attraction weakens as the distance between molecules increases, so these forces are strong in solids, less strong in liquids and very weak in gases. Melting overcomes them partially; vaporization overcomes them completely.
There are five types of intermolecular force. Dipole-dipole interactions act between polar molecules and are weak (3 - 4 kJ mol⁻¹). Ion-dipole interactions act between an ion and a polar solvent and strengthen with higher ionic charge and smaller ionic radius, giving the order \(\mathrm{Na^+} < \mathrm{Mg^{2+}} < \mathrm{Al^{3+}}\). Dipole-induced dipole interaction arises when a polar molecule induces a temporary dipole in a non-polar one. London dispersion forces, proposed by Fritz London, arise from momentary dipoles, act between all molecules, and strengthen with molecular size, mass, number of electrons and a more spread-out shape. Hydrogen bonding occurs when hydrogen is attached to N, O or F, and can be as strong as 40 kJ/mol.
Dipole moment, \(\mu = Q \times r\), is a vector quantity measured in debye (D). The larger the dipole moment, the stronger the dipole-dipole attraction and generally the higher the boiling point.
Hydrogen bonds may be intermolecular, between two molecules, or intramolecular, within one molecule. Intermolecular hydrogen bonding gives \(\mathrm{NH_3}\), \(\mathrm{H_2O}\) and HF abnormally high boiling points, raises viscosity, builds the three-dimensional network in liquid water, and holds the structures of proteins and nucleic acids together.
The state of a substance is decided by the balance between intermolecular forces of attraction and the thermal energy of its molecules. Thermal energy is directly proportional to temperature; going from gas to liquid to solid, intermolecular forces become more important as thermal energy is reduced.
Gases are characterized by low density, no fixed shape or volume, wide separation and continuous random motion of molecules, pressure exerted equally in all directions, the weakest intermolecular forces, diffusion, and high compressibility.
The measurable properties of a gas are mass, volume, pressure, temperature, density and rate of diffusion. Number of moles may be found from mass, from the number of molecules or from the volume at STP:
\[n = \frac{m}{M} = \frac{N}{6.022\times10^{23}}\]
Pressure is force per unit area, with the SI unit pascal; \(1\ \text{bar} = 1.00\times10^{5}\ \text{Pa}\) and \(1\ \text{atm} = 760\ \text{mm Hg} = 101325\ \text{Pa} = 1.01325\ \text{bar}\). Temperature must be used in kelvin, with \(T\ \text{(K)} = t\,^\circ\text{C} + 273.15\). Vapour density is molar mass divided by 2.
The gas laws relate \(P\), \(V\), \(T\) and \(n\), each holding the remaining variables constant.
\[P_1V_1 = P_2V_2 \quad \text{(Boyle's law, constant } T,\ n)\]
\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(Charles' law, constant } P,\ n)\]
\[\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad \text{(Gay-Lussac's law, constant } V,\ n)\]
\[\frac{V_1}{n_1} = \frac{V_2}{n_2} \quad \text{(Avogadro law, constant } P,\ T)\]
A plot of \(P\) against \(V\) at constant temperature is an isotherm, of \(V\) against \(T\) at constant pressure an isobar, and of \(P\) against \(T\) at constant volume an isochore. Extrapolating isobars to zero volume gives the same intercept, -273.15 °C, which is absolute zero (0 K) — a temperature that cannot be attained.
Boyle’s law may also be written in terms of density, since \(d \propto P\) at constant temperature for a fixed mass of gas. By Avogadro’s law, \(d \propto M\) at the same temperature and pressure.
One mole of any gas occupies its molar volume at STP: 22.71 L mol⁻¹ at the IUPAC STP of 1 bar and 273.15 K, and 22.414 L mol⁻¹ at the old STP of 1 atm and 273.15 K.
Combining Boyle’s, Charles’ and Avogadro’s laws gives the ideal gas equation, also called the equation of state:
\[PV = n\mathrm{R}T\]
R is the universal gas constant, the same for all gases, with the value \(8.314\ \text{J K}^{-1}\text{mol}^{-1}\), or \(0.0821\ \text{dm}^3\text{atm K}^{-1}\text{mol}^{-1}\), or about \(2\ \text{cal K}^{-1}\text{mol}^{-1}\). The units of \(P\) and \(V\) in a problem decide which value must be used.
From the ideal gas equation follow the expressions for molar mass, the combined gas law, and the density relation:
\[M = \frac{m\mathrm{R}T}{PV},\qquad \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2},\qquad M = \frac{d\mathrm{R}T}{P}\]
- Dalton’s law of partial pressure states that the total pressure of a mixture of non-reactive gases is the sum of the partial pressures of the individual gases at constant volume and temperature. The partial pressure of a gas equals its mole fraction times the total pressure:
\[P_{\text{Total}} = P_1 + P_2 + P_3 + \dots,\qquad P_1 = X_1 \cdot P_{\text{Total}}\]
The pressure of saturated water vapour is called aqueous tension, and it increases with temperature. For a gas collected over water, \(P_{\text{Dry gas}} = P_{\text{Total}} - \text{Aqueous Tension}\).
The kinetic molecular theory explains gas behaviour by assuming tiny particles far apart with negligible volume, negligible attractive forces at ordinary temperature and pressure, constant random straight-line motion, pressure caused by collisions with the walls, perfectly elastic collisions, and an average kinetic energy directly proportional to the absolute temperature (Average K.E. \(\propto T\)).
A real gas obeys the ideal gas equation only at low pressure and high temperature. Deviations appear at high pressure and low temperature, because intermolecular attractions are no longer negligible and the volume of the molecules themselves becomes significant.
Deviation is measured by the compressibility factor:
\[Z = \frac{PV}{n\mathrm{R}T} = \frac{V_{real}}{V_{ideal}}\]
For an ideal gas \(Z = 1\) at all temperatures and pressures. \(Z > 1\) means the molecular volume cannot be neglected and the gas is difficult to compress; \(Z \leq 1\) at lower pressure, where gases are compressible and behave nearly ideally.
Every substance has a critical temperature \(T_c\), above which it cannot be liquefied by pressure alone; the molar volume and pressure at that temperature are the critical volume \(V_c\) and critical pressure \(P_c\). \(\mathrm{CO_2}\), for example, liquefies at 30.98 °C and 73 atmosphere. Since \(\mathrm{N_2}\) and \(\mathrm{O_2}\) have \(T_c\) far below 0 °C, liquefying air needs both compression and cooling.
Above \(T_c\) liquid and gas cannot be distinguished, and either is referred to as a fluid; a liquid is then simply a very dense gas.
A liquid boils when its vapour pressure equals the external pressure. At 1 atm this temperature is the normal boiling point (100 °C for water) and at 1 bar the standard boiling point (99.6 °C for water). Raising the external pressure raises the boiling point; lowering it lowers the boiling point.
The liquid state is intermediate between solid and gas, with moderately strong intermolecular forces giving fluidity, a definite volume and the shape of its container.
Vapour pressure is the pressure of the vapour in equilibrium with its liquid, where the rates of evaporation and condensation are equal. It is higher for liquids with weak intermolecular forces, which are called volatile liquids, and it increases with temperature. It is commonly measured in torr, where 1 torr = 1 mm Hg.
Surface tension is the force per unit length acting perpendicular to a line drawn on the liquid surface, with the SI unit N m⁻¹ and the symbol \(\gamma\). It arises because surface molecules experience an unbalanced inward pull, so liquids minimize their surface area. It decreases as temperature rises, and its lowering explains the cleaning action of soaps and detergents and the penetrating action of toothpastes, mouth washes and nasal drops.
Viscosity is the internal friction between the successive layers of a liquid in laminar flow, expressed through the coefficient of viscosity \(\eta\), with the SI unit N s m⁻² and the CGS unit poise, where \(1\ \text{poise} = 1\ \text{gm cm}^{-1}\text{s}^{-1} = 10^{-1}\ \text{kg m}^{-1}\text{s}^{-1}\). It falls as temperature rises, increases with molecular size, and is least for spherical molecules. It explains the grading of lubricating oils, the use of blood viscosity as a clinical indicator, and the thickening of old glass panes near the bottom.
Flowchart
FC 10.9 – The logic of the whole chapter: strength of intermolecular forces versus thermal energy → decides whether a substance is solid, liquid or gas → for gases, the measurable properties P, V, T, n → related pairwise by Boyle’s, Charles’, Gay-Lussac’s and Avogadro’s laws → combined into the ideal gas equation PV = nRT, and extended to mixtures by Dalton’s law of partial pressure → explained molecularly by the kinetic molecular theory → real gases deviate at high pressure and low temperature, measured by Z → cooling below Tc and compressing gives liquefaction → the liquid state, described by vapour pressure, surface tension and viscosity
10.11 Check Your Concepts
Assertion: HF has a much higher boiling point than HCl. Reason: HF molecules are linked to one another by hydrogen bonds. Decide whether the assertion and the reason are each correct, and whether the reason correctly explains the assertion.
Assertion: Hydrogen shows a compressibility factor greater than 1 over the whole pressure range shown in Fig. 10.15. Reason: The intermolecular attractive forces in hydrogen are extremely weak. Decide whether the assertion and the reason are each correct, and whether the reason correctly explains the assertion.
Why is the dipole-dipole interaction said to be significant only in the solid and liquid states, even though polar molecules keep their dipole moments in the gaseous state as well?
State, with a reason in each case, whether hydrogen bonding is possible between the molecules of:
- CH₄
- NH₃
- PH₃
- CH₃OH
Ethylene glycol shows intramolecular hydrogen bonding while hydrogen fluoride shows intermolecular hydrogen bonding. Explain the difference between the two, and state which type raises the boiling point of a liquid more.
Both H₂S and H₂Se have a larger molecular mass than H₂O, yet water has by far the highest boiling point of the three. Account for this.
Arrange Na⁺, Mg²⁺ and Al³⁺ in increasing order of the strength of their ion-dipole interaction with water, and state the two properties of the ion that decide this order.
Benzene has a zero dipole moment and therefore experiences no dipole-dipole forces, and yet it is a liquid at room temperature. Which intermolecular force holds it together, and how does that force arise in a molecule with no permanent dipole?
Using Table 10.3, explain why an ion-dipole interaction is stronger than a dipole-dipole interaction between molecules of comparable size.
What happens if a gas at a temperature above its critical temperature is subjected to a very high pressure? Explain your answer in terms of thermal energy and intermolecular attraction.
Using the critical temperatures in Table 10.7, arrange He, H₂, N₂ and CO₂ in increasing order of ease of liquefaction, and state which one can be liquefied at 25 °C by pressure alone.
Why must the temperature always be expressed in kelvin, and never in degree celsius, when using Charles’ law or the ideal gas equation? Illustrate your answer with what would go wrong at 0 °C.
Match the following.
| ‘A’ | ‘B’ |
|---|---|
| a. Isotherm | i. plot of P against T at constant volume |
| b. Isobar | ii. plot of P against V at constant temperature |
| c. Isochore | iii. plot of V against T at constant pressure |
- Match the following.
| ‘A’ | ‘B’ |
|---|---|
| a. Debye | i. unit of pressure equal to 1 mm Hg |
| b. Poise | ii. unit of dipole moment |
| c. Torr | iii. SI unit of pressure |
| d. Pascal | iv. CGS unit of coefficient of viscosity |
An ideal gas can never be liquefied, however much it is cooled or compressed. Which two assumptions of the kinetic molecular theory make this so?
Explain, using the kinetic molecular theory, why the pressure of a gas rises when it is heated in a rigid closed vessel, and name the gas law that describes this behaviour.
Why do all real gases approach ideal behaviour as the pressure approaches zero? Answer in terms of both reasons for deviation.
Distinguish between the situations Z < 1 and Z > 1 for a real gas, stating in each case which molecular effect is dominant and whether the gas is easier or harder to compress than an ideal gas.
When a gas is collected over water, why must the aqueous tension be subtracted from the measured pressure? What would be the effect on a molar mass calculated from that gas if the correction were forgotten?
The molar volume of an ideal gas is 22.71 L mol⁻¹ at the IUPAC STP but 22.414 L mol⁻¹ at the old STP, although the temperature is the same in both. Explain the difference.
Half of the liquid in a closed vessel at constant temperature is removed, and the vessel is resealed. What happens to the vapour pressure of the liquid, and why?
Explain why raising the temperature of a liquid increases its vapour pressure but decreases both its surface tension and its viscosity.
Account for each of the following in one or two lines:
- a free drop of a liquid is spherical;
- petrol is described as a volatile liquid but motor oil is not;
- soap solution cleans an oily fabric better than water alone;
- spherical molecules give less viscous liquids than long-chain molecules of similar mass.
A gas is enclosed in a cylinder fitted with a movable piston. Predict what happens to the volume, and name the law you use, if:
- the pressure is doubled at constant temperature;
- the absolute temperature is halved at constant pressure;
- the amount of gas is doubled at constant temperature and pressure.
Name the term used for mixing of different gases by random molecular motion and frequent collision.
The pressure that each individual gas would exert if it were alone in the container — what do we call it?
When a gas is heated the particles move more quickly. What is the change in volume of a heated gas if the pressure is kept constant?
A bubble of methane gas rises from the bottom of the North sea. What will happen to the size of the bubble as it rises to the surface?
Observe the following conversions shown in Fig. 10.20: (a) H₂ + Cl₂ → 2HCl and (b) 2H₂ + Cl₂ → 2HCl + H₂. Which of the above reactions is in accordance with the principle of stoichiometry?
Diagram
Fig. 10.20 – Two rows of molecular models for question A-Q14: row (a) one two-sphere H₂ molecule plus one two-sphere Cl₂ molecule giving two HCl molecules; row (b) two H₂ molecules plus one Cl₂ molecule giving two HCl molecules plus one leftover H₂ molecule; label the rows a and b with the equations beneath
Hot air balloons float in air because of the low density of the air inside the balloon. Explain this with the help of an appropriate gas law.
List the characteristic physical properties of the gases.
Define the terms:
- Polarizability
- Hydrogen bond
- Aqueous tension
- Dipole moment
Would it be easier to drink water with a straw on the top of Mount Everest or at the base? Explain.
Identify the type of intermolecular forces in the following compounds:
- CH₃ - OH
- CH₂=CH₂
- CHCl₃
- CH₂Cl₂
Name the types of intermolecular forces present in Ar, Cl₂, CCl₄ and HNO₃.
Write the statement for:
- Boyle’s law
- Charles’ law
Differentiate between real gas and ideal gas.
State and write the mathematical expression for Dalton’s law of partial pressure and explain it with a suitable example.
Derive the ideal gas equation. Mention the terms involved in it. Also write how it is utilised to obtain the combined gas law.
Write the postulates of the kinetic theory of gases.
Write a short note on —
- Vapour pressure
- Surface tension
- Viscosity
10.12 Practice With Numericals
A fixed mass of a gas occupies 250 cm³ at a pressure of 1.0 bar. Calculate its volume at 2.5 bar, the temperature remaining constant.
A gas occupies 3.0 dm³ at 27 °C. Calculate its volume at 127 °C if the pressure is kept constant.
A gas in a rigid closed vessel exerts a pressure of 2.0 bar at 300 K. Calculate the pressure at 375 K, and name the law used.
A cylinder of a gas at 15 bar and 300 K must be cooled until its pressure falls to 10 bar. Calculate the temperature, in kelvin and in °C, to which it must be cooled at constant volume.
A sample of gas occupies 500 cm³ at 1.0 bar and 300 K. Calculate its volume at 2.0 bar and 450 K.
Calculate the number of moles of an ideal gas that occupies 5.0 dm³ at 2.0 atm and 300 K. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹)
Calculate the volume occupied by 0.25 mol of an ideal gas at 1.5 atm and 320 K. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹)
A glass bulb of volume 250 cm³ contains 0.44 g of a gas at 1.00 atm and 300 K. Calculate the molar mass of the gas. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹)
The density of a gas at 1.00 atm and 273.15 K is 1.96 g dm⁻³. Calculate its molar mass and its vapour density. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹)
A cylinder contains a non-reacting mixture of 8.0 g of dioxygen and 14 g of dinitrogen at a total pressure of 3.0 bar. Calculate the mole fraction and the partial pressure of each gas.
A 4.0 dm³ vessel at 300 K contains 0.20 mol of helium and 0.30 mol of argon, which do not react. Calculate the partial pressure of each gas and the total pressure in the vessel. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹)
A gas is collected over water at 297.15 K and the total pressure of the moist gas is 1.0000 bar. Using Table 10.5, calculate the pressure of the dry gas.
One mole of a gas at 100 atm and 300 K is found to occupy 0.30 dm³. Calculate its compressibility factor and state which molecular effect is dominant under these conditions. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹)
Calculate the number of molecules present in 2.0 dm³ of a gas at 1.00 atm and 273.15 K. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹, N_A = 6.022 × 10²³ mol⁻¹)
A gas occupies 3.0 dm³ at 300 K. At what temperature, in kelvin and in °C, will it occupy 4.5 dm³ at the same pressure?
Convert the following, showing the relation used in each case: a. 2.0 atm into kPa; b. 0.50 bar into Pa; c. 380 torr into atm; d. 1.2 poise into N s m⁻²; e. -33 °C into kelvin.
A gas sample of volume 400 cm³ diffuses out of a porous container in 80 s. Calculate its rate of diffusion and state the unit.
Calculate the mass of methane, CH₄, present in a 10.0 dm³ cylinder at 2.00 atm and 300 K. (R = 0.0821 dm³ atm K⁻¹ mol⁻¹, molar mass of CH₄ = 16 g mol⁻¹)
Convert the following temperatures from degree celsius to kelvin:
- -15 °C
- 25 °C
- -197 °C
- 273 °C
Convert the following pressure values into pascals:
- 10 atmosphere
- 1 kPa
- 107000 Nm⁻²
- 1 atmosphere
Convert:
- Exactly 1.5 atm to pascals
- 89 kPa to newton per square metre (Nm⁻²)
- 101.325 kPa to bar
- -100 °C to kelvin
- 0.124 torr to standard atmosphere
A balloon is inflated with helium gas at a room temperature of 25 °C and at 1 bar pressure, when its initial volume is 2.27 L, and is allowed to rise in air. As it rises in the air the external pressure decreases and the volume of the gas increases till finally it bursts when the external pressure is 0.3 bar. What is the limit at which the volume of the balloon can stay inflated?
A syringe has a volume of 10.0 cm³ at a pressure of 1 atm. If you plug the end so that no gas can escape and push the plunger down, what must be the final volume to change the pressure to 3.5 atm?
The volume of a given mass of a gas at 0 °C is 2 dm³. Calculate the new volume of the gas at constant pressure when —
- the temperature is increased by 10 °C;
- the temperature is decreased by 10 °C.
A hot air balloon has a volume of 2800 m³ at 99 °C. What is the volume if the air cools to 80 °C?
At 0 °C, a gas occupies 22.4 litres. How much hot must the gas be, in celsius and in kelvin, to reach a volume of 25.0 litres?
A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar. What will be the new pressure inside the container if the volume is reduced to 12 L, the temperature is increased to 177 °C and 1.25 mol of additional He gas is added to it?
Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.
At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where the temperature is 10 °C and the volume of the gas is 640 mL?
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If the pressure of the mixture of gases in the cylinder is 25 bar, what is the partial pressure of dioxygen and neon in the mixture?
Calculate the pressure, in atm, of 1.0 mole of helium in a 2.0 dm³ container at 20.0 °C.
Calculate the volume of 1 mole of a gas at exactly 20 °C at a pressure of 101.35 kPa.
Calculate the number of molecules of methane in 0.50 m³ of the gas at a pressure of 2.0 × 10² kPa and a temperature of exactly 300 K.
10.13 Case-Based Questions
- Identify the gas laws from the diagrams in Fig. 10.21:
- a gas at 1 atm compressed to 0.5 atm at constant T;
- a gas at 1 atm heated from T = 200 K to T = 600 K;
- a gas at 1 bar changing from 24.8 L with n = 1 mol to 37.2 L with n = 1.5 mol.
Diagram
Fig. 10.21 – Three pairs of piston-and-cylinder diagrams for question A-Q16: pair (a) marked 1 atm and 0.5 atm with “T constant” below; pair (b) both marked 1 atm with T = 200 K and T = 600 K below; pair (c) both marked 1 bar with 24.8 L, n = 1 mol and 37.2 L, n = 1.5 mol below; leave a blank line beside each pair for the student to name the law
- Consider a sample of a gas in a cylinder with a movable piston, as in Fig. 10.22. Show diagrammatically the changes in the position of the piston if —
- Pressure is increased from 1.0 bar to 2.0 bar at constant temperature;
- Temperature is decreased from 300 K to 150 K at constant pressure;
- Temperature is decreased from 400 K to 300 K and pressure is decreased from 4 bar to 3 bar.
Diagram
Fig. 10.22 – A vertical cylinder closed by a movable piston with the enclosed space labelled “Gas” and the piston labelled “movable piston”, drawn with space beside it for the student to sketch the three new piston positions
- With the help of the graph in Fig. 10.23, answer the following. At constant temperature:
- The graph shows the relation between pressure and volume — represent the relation mathematically.
- Identify the law.
- Write the statement of the law.
Diagram
Fig. 10.23 – Graph for question A-Q28: pressure P on the y-axis against 1/V on the x-axis, showing a straight line of plotted points passing through the origin, marked “T constant”
10.14 Objective Questions
- The unit of viscosity is —
- dynes
- newton
- gram
- poise
- Which of the following is true for 2 moles of an ideal gas? —
- PV = nRT
- PV = RT
- PV = 2RT
- PV = T
- Intermolecular forces in liquid are —
- greater than gases
- less than solids
- both a and b
- greater than solids
- Interactive forces are ………. in ideal gas. —
- nil
- small
- large
- same as that of real gases
- At constant temperature the pressure of 22.4 dm³ volume of an ideal gas was increased from 105 kPa to 210 kPa. New volume could be —
- 44.8 dm³
- 11.2 dm³
- 22.4 dm³
- 5.6 dm³
- If density of a gas is measured at constant temperature and pressure, then which of the following statements is correct? —
- Density is directly proportional to molar mass of the gas.
- Greater the density, greater is the molar mass of the gas.
- If density, temperature and pressure is given, the ideal gas equation can be used to find molar mass.
- All the above statements are correct.
- Match the pairs of the following.
| ‘A’ | ‘B’ |
|---|---|
| a. Boyle’s law | i. at constant pressure and volume |
| b. Charles’ law | ii. at constant temperature |
| iii. at constant pressure |