Solids, Liquids, and Gases By Robert A. Edwards

We usually think of oxygen as a gas, but oxygen will form a liquid and even a

solid if it is cooled down enough (or compressed enough). This is also true of

other gases as well, including the nitrogen and argon found in the air that we

breathe. For example, argon (0.94% of the air) forms a liquid when cooled to -

185.7C and then becomes a solid below -189.2C. The differences between

solids, liquids, and gases are illustrated by the simple model shown in Figure

17.1. The molecules (or atoms) are touching one another in solids. Liquid,

on the other hand, have a small amount of empty space between the molecules.

Gases have large amounts of empty space between molecules which are

travelling through space. The gas molecules exert pressure on the walls of the

container by colliding with the walls.

Figure 17.1The left hand panel shows a model of a solid with spherical atoms

(or molecules) touching one another. The central panel shows a model of a

liquid with some atoms touching one another but small spaces and holes

between molecules are present. The right hand panel shows a model of a gas

with atoms bouncing around in mostly empty space.

Example

Using the model in Figure 17.1, describe how the liquid water exerts pressure

on the sides of an open graduated cylinder which is filled with water?

ANSWER

As a gaseous air molecule strikes the top water in the cylinder it compresses the

molecule which it struck (call that “water molecule-X”) very slightly which exerts force on

any molecule touching water molecule-X. This force is then transferred through a

sequence of molecules which were touching one another to the wall of the cylinder. If

water molecule-X happened to have a space next to it when it was struck by the air

molecule, the water molecule-X might move into the space, but that also transfers the

pressure, because it jostles other molecules as it moves into the space.

Example

The density of solid argon near 0k is about 1.77 g/cm3 (Dobbs ER, BF Friggins,

and GO Jones, Properties of solid argon, del nuovo cimento, 1958, supplemento

IX (X), 32-35). What is the volume and radius of an argon atom? Assume

that the argon atoms in a solid undergo closest packing (as is illustrated in

Figure 17.1). (Hint it can be shown geometrically that for closet packing the

spheres occupy 74.0% of the space.)

ANSWER

Volume (atom + empty space) in the bulk per mole of atoms

39.948 g/mol / 1.77 g/cm3 = 22.57 cm

3/mol (1L/1000cm

3) = 0.02257 L/mol

Volume & radius actually occupied by the hard sphere atoms is only 74.0%:

0.02257 L/mol 0.740 = 0.0167 L/mol

0.0167 10-3

m3/mol / 6.02 10

23 molecules/mol = 2.77 10

-26 m

3/molecule

2.77 10-26

m3/molecule = (4/3)r

3 r = 1.88 10

-9 m

By increasing the pressure on a gas in a closed container compresses the gas

into a smaller volume. Boyle‟s Law (VmP = k) mathematically describes the

change in molar volume as pressure increases. The same mathematically

relationship between volume and pressure is found in the ideal gas law (PVm =

RT). These „laws‟ are a fairly good description of how the volume of most

gases at relatively low pressures (<1 atm) respond when compressed.

However, these laws are a poor description of the pressures above 10

atmospheres.

Liquids and solids also decrease their volume when compressed at room

temperature, but not nearly as much as for a gas. Qualitatively the models in

Figure 17.1 helps us to understand this behavior, because there is very little

space to squeeze out when a liquid or solid are compressed; whereas, for a gas

there is a lot of space between the atoms. For solids and liquids, the equation

used to express the volume as a function of pressure (Vm = Vmo {1-P}), has a

constant () which is called the “isothermal compressibility”. This constant is

positive so the negative sign in the equation implies that the volume decreases

with increased pressure. This equation does not include an explicit description

of how the volume will change as temperature changes. However, the

isothermal compressibility is a constant only at a fixed temperature and in many

cases tables can be found of the isothermal compressibility at many

temperatures.

There are several kinds of attractive forces between molecules which pull the

molecules of a solid together to form a crystal or a tightly packed amorphous

solid. These attractive forces are also operating in the liquid phase but the

temperature is high enough that the molecules roll around each other so that

there is some empty space between the molecules. In the gaseous phase most

of the molecules are not in contact with one another; instead, they are bouncing

around in the large volume of the container but the do exert small attractive

forces on one another.

The ideal gas law is not a good description of the behavior of gases at moderate

to high pressures (>10 atm). The attractive forces between gas molecules as

well as the volume occupied by the hard spheres cause them to deviate from

ideal behavior. Van der Waals modified the ideal gas law to take account for

these two effects and came up with the first equation below. Two “Van der

Waals parameters” designated “a” and “b” are included in the equations.

These parameters are different for different gases. The second equation, which

is derived for the first one with a small approximation, allows the calculation of

volume when pressure and temperature are known.

Where:

“a” is a parameter for the attractive forces between gas molecules.

“b” is a parameter which accounts for the volume occupied by the hard

spheres called the “Van der Waals volume”.

Example

How many moles of oxygen would be present in an M60 oxygen cylinder which

has a volume of 164 L and is filled to a pressure of 2200 psi at room

temperature? Assume (a) the oxygen is an ideal gas and (b) that the oxgyen is

a Van der Waals‟ gas with a=1.378 L2atm/mol

2 and b = 0.13038 L/atm.

ANSWER

2200 psi / (14.69595 psi/atm) = 149.7 atm

Ideal Gas: n = PV/RT = 149.7 atm 164 L/ 0.08206 Latm/mol K 298.15K = 1003 moles

Van der Waals Gas: Vm ≈ 0.08206 Latm/mol K 298.15K/149.7 atm

+ 0.13038 L/atm – 1.378 L2atm/mol2 /0.08206 Latm/mol K 298.15K

Vm ≈ 0.2375 L

n = V / Vm = 690.5 moles

(Vm)2

P = RT - a

(Vm-b)

RT

Vm ≈ RT + b - a

P

The Van der Waals constants for the inert gases are tabulated on Table 17.1.

They show that the Van der Waals volume increases as the atoms get larger from

helium to xenon. In general, the attractive forces also get stronger as the atoms

get larger. The attractive forces operating between these atoms are induced

dipole – induced dipole (London dispersion) force. As two atoms approach

one another there may be a temporary fluctuation of the electron cloud in one of

the atoms which produces a temporary dipole moment (slight positive charge on

one side and slight negative charge on the other). The electron cloud in the

other atoms instantaneously fluctuates in the opposite direction so that there is

an electrical attraction between the two atoms.

Table 17.1 Van der Waals parameters for the inert gases from Weast. R. C. (Ed.),

Handbook of Chemistry and Physics (53rd Edn.), Cleveland:Chemical Rubber

Co., 1972.

Helium Neon Argon Xenon

a (L2atm/mol

2) 0.03457 0.2135 1.363 4.250

b (L/mol) 0.0237 0.01709 0.03219 0.05105

The amount of this interaction between induced dipoles depends upon how large

and flexible the electron clouds are so that they can shift. The “polarizability”

of any atom or molecule is a measure of how easily it electron cloud can be

shifted. Xenon is a large atom with its valence electrons far from the nucleus

and more able to be polarized than the other atoms in this series. Hence,

Xenon has a greater polarizability than the other inert gas atoms.

For molecules with permanent dipole moments there are also dipole-dipole and

dipole-induced dipole attractive forces between them, even in the gas phase.

Thus the Van de Waals‟ parameters in Table 17.2 show greater attraction

between water molecules than between ethane molecules, even though they

have about the same size. All three types of Van der Waals forces (dipole-

dipole, dipole-induced dipole, and induced dipole – induced dipole) exist

between water molecules; whereas, between the only type of Van der Waals

forces between ethane molecules are induced dipole – induced dipole forces.

Table 17.2 Van der Waals parameters for several small molecules from Weast. R.

C. (Ed.), Handbook of Chemistry and Physics (53rd Edn.), Cleveland:Chemical

Rubber Co., 1972.

Water Ammonia Methane Ethane

a (L2atm/mol

2) 5.536 4.225 2.283 5.562

b (L/mol) 0.03049 0.03707 0.04278 0.0638

Example

Describe the dipole – induced dipole forces that occur between water molecules

in the gas phase.

ANSWER

Water molecules have a permanent dipole in which the oxygen is slightly negative and

the hydrogen atoms are slightly positive. When this dipole on one water molecule

causes a shift in the electron cloud of a neighboring water molecule then an induced

dipole in the neighbor is formed. Because this induced dipole has an orientation such

that the slight negative charge is close to the positive of the permanent dipole, there is

an attractive force between the molecules.

Van der Waals constants are determined by fitting the experimental pressure

versus volume graph at constant temperature (referred to as a PV isotherm) of

each gas to the Van der Waals equation. This yields a Van der Waals volume

(b) that is larger than the volume of the hard sphere atoms calculated from the

solid phase. For example, The VdWs volume of argon is 0.03219 L/mol;

whereas, the volume occupied by the atoms was 0.0167 L/mol as calculated in

the example above. So the VdWs volume is about twice the actual volume of

the sheres in a solid. Part of the reason for this is that the atoms are not

completely hard spheres. Instead, they are partially soft spheres that can

overlap slightly in the solid phase.

It is common to calculate the radius of the spheres in the solid phase as was

done in the earlier example and call the “Van der Waals radius” (i.e. the Van der

Waals radius of argon is 188 nm as calculated in the example above). This can

be confusing because the “Van der Waals radius” indicates how closely atoms

can come together in condensed phases like the solid. It is not calculated from

the Van der Waals volume but from the density of the solid.

PRACTICE PROBLEMS

1. The density of argon liquid at its melting point is 1.41 g/cm3 (calculated from

equation in Goldman, K and NG Scrase, Densities of saturated liquid argon,

Physica, 1969, 45, 1-11). Calculate the amount of empty space in argon

liquid at its melting point.

2. Why do attractive forces between atoms reduce the pressure exerted on the

walls of a container by a gas?

3. The density of H2O(l) at 0°C is 1.0454 g /cm3 under 1000. atm of pressure.

(a) How well does the equation Vm = a + bT + cP predict the density value

above? (Hint: calculate the percentage error between the actual and

predicted value.)

(b) Assuming that H2O(l) under these conditions has 30% vacant space,

calculate the Van der Waals' radius of water (assuming that H2O molecules

are spheres)

4. Use calculus to determine the pressure change inside a cylinder with 3.200 kg

of oxygen gas at 298.00°K in a volume of 45.00 L in each case below.

(Assume that O2(g) is an ideal gas so that P = nRT/V.)

a) A decrease of 0.010 moles in the number of moles of gas.

b) A decrease of 0.010K in the temperature.

c) A decrease of 0.010 L in the volume.

5. Consider a cylinder containing 4.48 kg of ethylene gas (C2H4) in a volume of

45.0 L, which gives a pressure of 50.0 atm. at 20C. For ethylene the

VdWs‟ parameters are a=4.471 L2atm/mol

2 and b=0.05714 L/mol.

(a) What would the volume be if this were (i) an ideal gas or (ii) a VdWs gas.

(b) Is this gas closer to an ideal gas or a VdWs‟ gas?

(c) Explain, at a molecular level, why the gas pressure increases then

decreases if a VdWs‟ gas is compressed.

6. In liquid water with oxygen in it (i.e., O2(aq)), which of the Van der Waals'

forces are present between the molecules listed below? Why are liquids more

compressible than solids?

(a) Two water molecules.

(b) A water molecule and an oxygen molecule.

ANSWERS to PRACTICE PROBLEMS

1. The density of argon liquid at its melting point is 1.41 g/cm3 (calculated from

equation in Goldman, K and NG Scrase, Densities of saturated liquid argon,

Physica, 1969, 45, 1-11). Calculate the amount of empty space in argon

liquid at its melting point.

Volume in the bulk per mole of atoms

39.948 g/mol / 1.41 g/cm3 = 28.33 cm3/mol (1L/1000cm3) = 0.02833 L/mol

The volume actually occupied by the atoms

0.01825 L/mol (from the calculation on solid argon in the example)

Therefore:

100 (0.1825/.02833) = 64.4% of the volume is occupied by atoms

100-64.4 = 35.6% is empty space

2. Why do attractive forces between atoms reduce the pressure exerted on the

walls of a container by a gas?

As a molecule approaches the wall it is attracted back into the bulk by the gas molecules

behind it so it will slow down slightly and thus hit the wall with less force. Pressure is

force per unit area, so there will be less pressure on the wall.