PHYS 7.5: Kinetic theory – an example of microscopic modelling	PPLATO @
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1.1 Module introduction

Modern physics tries to understand matter at a microscopic level, where bulk properties are explained in terms of atoms or molecules behaviour. One such theory is the kinetic theory which attempts to explain the bulk properties of gases in terms of a microscopic picture of the motion of atoms and molecules, interacting according to Newton’s laws of motion.

It is not feasible to attempt to apply Newton’s laws to each atom, since even a small macroscopic system of a few grams involves of the order of 10²³ atoms. Not only would the mathematical task of solving the equations be impossible, but even a minute uncertainty in the initial position and velocity of each individual atom would, after a very short time, produce a prediction which bore no resemblance to the real situation. To deal with these enormous numbers, we have to use statistical methods that describe the average behaviour of the atoms. These statistical methods are based on the idea that the individual motions of atoms are random. It is the combination of Newton’s laws (to describe the individual atomic behaviour) and statistical reasoning (to describe the average behaviour of many atoms) that makes up the study of kinetic theory.

Section 2 begins the study of kinetic theory by considering an ideal gas. This is an ‘idealization’ of a real gas (one to which real gases approximate quite closely) in which the molecules are assumed to be vanishingly small and to interact with each other only during collisions. Using simple statistical ideas the macroscopic properties of the gas such as temperature, pressure, internal energy and specific heat can all be related to average molecular motion and this allows relationships between the macroscopic properties, such as the ideal gas law, to be derived. This section concludes with a discussion of the characteristic distances between collisions, the mean free path, and the characteristic time between collisions, the mean free time, relating both to molecular size and the average speed of the molecules.

Section 3 considers how individual molecular speeds are distributed about the mean molecular speed. This can be characterized in terms of a probability distribution function, the Maxwell–Boltzmann speed distribution function. This function is quoted and discussed, but its derivation lies beyond the scope of FLAP. The discussion of the speed distribution leads to the identification of three characteristic speeds for the distribution – the most probable speed, the average speed and the root-mean-squared speed. The distribution of molecular kinetic energies can then be derived from the speed distribution. This section concludes with brief discussions of the experimental validation of the speed distribution, and some of the transport processes in gases, e.g. diffusion, viscosity and thermal conduction.

In Section 4 we consider refinements to the simplest model of the ideal gas so as to describe real gases more accurately. This is done by incorporating molecular excluded volumes and intermolecular forces. This leads us to the introduction and discussion of the van der Waals equation of state.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? section.

1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

1.3 Ready to study?

Study comment In order to study this module you should have a clear understanding of the following terms: absolute temperature, atom, atomic_mass_unitatomic mass, atomic mass unit, conservation of energy, conservation of momentum, density, displacement, elastic collision, force, kinetic energy (translational_kinetic_energytranslational, rotational_kinetic_energyrotational and vibrational_kinetic_energyvibrational), mole, molecule, Newton’s laws of motion, potential energy, pressure, relative molecular mass, speed, velocity, volume and work. You should be familiar with vectors, components_of_a_vectorvector components and vector addition and also with calculuscalculus notation, although the only derivative used is of a quadratic function. Ability to manipulate simple algebraic equations, including exponential functions, is assumed. If you are uncertain about any of these terms, review them by reference to the Glossary, which will also indicate where in FLAP they are developed. The following Ready to study question will allow you to establish whether you need to review some of the topics before embarking on this module.

2 Kinetic theory of ideal gases

2.1 The ideal gas model i

The goal of the kinetic theory of gases is to understand the bulk (macroscopic) properties of gases in terms of the constituent atoms or molecules – that is, to develop a microscopic model of these properties. When a physicist develops a model for a phenomenon, the first attempt always involves as much simplification as possible, keeping only the essential physical attributes and minimizing the mathematical complication. Our attempt to understand thermal phenomena at the microscopic level begins with the simplest system. We anticipate that the interactions between gas molecules are much weaker than those between molecules in solids or liquids, because in a gas the molecules are much further apart. We hope that this may allow us to describe the behaviour of gas molecules without having to know or calculate the detailed microscopic forces involved. We will ignore these forces, except when molecules collide, and assume all collisions are elastic collisions, conserving total kinetic energy.

To simplify the model further we will describe real gases without regard to their detailed molecular structure. This approach seems reasonable since we know that the behaviour of many gases approximates that of an ideal gas, characterized by:

where P is the pressure, V the volume, n the number of moles of gas i and R the universal molar gas constant of magnitude R = 8.314 J K⁻¹ mol⁻¹ and T is the absolute temperature.

The first task for our microscopic model is to explain this ideal gas equation, i.e. to develop the kinetic_theory_of_ideal_gaseskinetic theory of an ideal gas. This we will do in Subsection 2.2, but first try the following question.

2.2 The assumptions of the kinetic theory of an ideal gas

Since the ideal gas equation works well for most real gases, irrespective of whether the molecules are monatomic_ideal_gasmonatomic (single atom), diatomic_ideal_gasdiatomic (two atoms), or triatomic_ideal_gaspolyatomic (more than two atoms), it is likely that an adequate model can be based upon the simplest form of microscopic particle – a point particle.

Study comment From now on we will use the word ‘molecule’ as a general term to describe the constituent particles of all gases, whether they be individual atoms, or diatomic molecules or polyatomic molecules. At some stages we will have to distinguish between these possibilities.

Now we will set out the five basic assumptions on which we can base the simplest microscopic model for gas behaviour – these form the foundation of the kinetic theory model of an ideal gas.

Implication  Random motion means that the molecules are equally likely to be travelling in any direction. Although individual molecules undergo collisions and change direction frequently, the fact that there is a large number of them means that at any time there will be, to a very good approximation, equal numbers travelling in all directions, illustrated schematically in Figure 1. Molecules will also be uniformly distributed in space. i

Later we will discuss more implications of this randomness, but it is the crucial assumption of kinetic theory. We will use this randomness to discuss the properties of the gas in terms of the average molecule, which is all we will need to characterize much of the macroscopic behaviour of the gas.

Implication  You might guess that this is so since a gas isolated from external influences maintains its pressure and hence the molecules must maintain their kinetic energy. From this we can assume that kinetic energy must be conserved on a microscopic scale. You may be puzzled why perfectly elastic collisions are assumed between molecules, yet they never occur in the macroscopic world. The reason is that on a microscopic scale the walls also are molecular and these molecules are also in random vibrational motion. Therefore Assumption 2 implies that on average a molecule neither gains nor loses kinetic energy when it collides with another molecule – whether it is in the gas or in the wall.

Implication  We know today that quantum mechanics replaces the classical mechanics of Newton’s laws at the atomic level, but a classical model still gives many correct insights into the properties of gases. It is fortunate that the assumption that classical physics is applicable turns out to be justified in most of kinetic theory – there are only a few places where quantum mechanics is required, and we will point these out as we come to them.

Implication  We ignore the presence of long–range forces between the molecules (such as electrical forces). We can be fairly certain that some long–range forces exist, but ignoring them is reasonable if their magnitude is small. This assumption allows us to take the motions of molecules as being along straight lines between collisions.

Implication  This is another way of saying that the molecular diameter is small compared to the average distance between molecules and that the molecules may be treated as particles.

2.3 The molecular basis of temperature, pressure and internal energy

Before making detailed calculations let us consider what our model and our common experience qualitatively suggests about the temperature, pressure and internal energy of a gas. In general terms we may expect the internal energy of a system to increase as its temperature rises. For an ideal gas the only form of internal energy possible is the kinetic energy of the translational motion of its molecules. On this basis, we would expect the speed of our molecules to increase as the temperature of the gas increases. While we do not know the precise relationship, we would expect to be able to associate the molecular speed in some sense with the temperature, if our kinetic theory model is to make sense.

At the same level of understanding, the pressure exerted by the ideal gas must be the result of the collisions of the molecules with the walls of the container, so we would expect the pressure to increase as we add more molecules to the system or if we increase their speed by raising the temperature. This is in accordance with Equation 1,

PV = nRT(Eqn 1)

the pressure increases as the temperature increases for a fixed volume and a fixed number of moles of a gas; for a fixed temperature and volume, the pressure increases with the number of moles of gas.

On the basis of our simplified ideal gas point particle model, the molecules in the gas can have only translational kinetic energy. We might therefore associate the internal energy of a gas with the average translational kinetic energy per molecule. On the other hand, if the molecules have some internal structure, for instance if they are diatomic or polyatomic, then there could be additional energy associated with the rotational and vibrational motions of molecules For more complicated molecules, we might expect to associate the internal energy with the average total energy per molecule i, even though the temperature and pressure are associated with only translational kinetic energy. These are reasonable extensions to the simplified ideal gas point particle model, which allows no internal structure of the molecules.

These ideas about the molecular basis of temperature, pressure and internal energy are merely qualitative, but it is encouraging that the kinetic theory model seems to be consistent with our understanding of bulk properties at this level. However, for the model to be truly useful, we must be able to use it to make calculations, which in turn requires equations. These will be derived in the next subsection.

2.4 Kinetic theory and the ideal gas equation of state

We will derive the ideal gas equation of state (Equation 1),

PV = nRT(Eqn 1)

from the assumptions listed in Subsection 2.2, treating the molecules as point particles. We will start by using our kinetic theory model to calculate the pressure. The pressure is the net force exerted on unit area of the wall, so we need to calculate the force exerted by a molecule when it collides with the wall. The force exerted by the molecule on the wall is equal and opposite to the force exerted by the wall on the molecule, so we can work out either. The force on the wall at an impact can be calculated from Newton’s second law as the rate of change of momentum of the molecule. We take as our model a cubical box of side L, filled with N molecules each of mass m (see Figure 1). We will calculate the force exerted on the wall at x = L, shown by hatching on Figure 1.

Consider a typical molecule with velocity vector υ = (υ_x, υ_y, υ_z). Collisions of this molecule with the chosen wall arise because of the component υ_x and after the elastic wall collision υ_x is reversed, with υ_y and υ_z unchanged. Thus, if p_x represents the x–component of the molecule’s momentum and ∆p_x the change caused by a collision we can write:

initial value of p_x + ∆p_x = final value of p_x

i.e.mυ_x + ∆p_x = −mυ_x

so∆p_x = −mυ_x −(mυ_x) = −2mυ_x

Assuming no further intermolecular collisions, the molecule will return to impact on this wall again after rebounding from the wall at x = 0 – a round trip along x of length 2L that will require a time ∆t = 2L/υ_x. i The chosen molecule will therefore impact on the chosen wall once in each time interval ∆t, and from Newton’s second law, the average component of force in the x–direction $\langle f_x\rangle$ exerted by the wall, is given by the rate of change of momentum of the molecule in the x–direction, ∆p_x/∆t,

i.e.$\langle f_x\rangle = \dfrac{\Delta p_x}{\Delta t} = \dfrac{-2m\upsilon_x}{2L/\upsilon_x} = \dfrac{-m\upsilon_x^2}{L}$

This is the average force exerted by the wall on the molecule; the force exerted by the molecule on the wall will be just the negative of this. The effects of N molecules impacting on this same wall, each molecule with its own υ_x component, causes an average force $\langle F_x\rangle$ exerted on the wall given by:

$\langle F_x\rangle = -N\langle f_x\rangle = \dfrac{Nm}{L}\langle\upsilon_x^2\rangle$

where $\langle\upsilon_x^2\rangle$ is the average or mean squared υ_x component of all the N molecules,

i.e.$\displaystyle \langle\upsilon_x^2\rangle = \dfrac1N\sum_{i=1}^N\upsilon_{x_i}^2$ i

However, we want to know the pressure rather than the force, so we must divide the force by the wall area L². The total pressure on the wall from the molecules is

$P_x = \dfrac{\langle F_x\rangle}{L^2}$

Substituting for $\langle F_x\rangle$ gives us

$P_x = \dfrac{Nm}{L^3}\langle\upsilon_x^2\rangle = \dfrac{Nm}{V}\langle\upsilon_x^2\rangle$(2)

where V is the volume of the gas. If we now invoke Assumption 1 we see that there is nothing special about our choice of x–axis – we could have reached similar conclusions about P_y and P_z, so:

$P_y = \dfrac{Nm}{V}\langle\upsilon_y^2\rangle\quad\text{and}\quad P_z = \dfrac{Nm}{V}\langle\upsilon_z^2\rangle$

Assumption 1 implies

$\langle\upsilon_x^2\rangle = \langle\upsilon_y^2\rangle = \langle\upsilon_z^2\rangle$(3)

Thus P_x = P_y = P_z = P, where P is the pressure exerted by the gas, which is equal in all directions. The velocity components of an individual molecule are related to the speed υ by the expression:

υ² = υ_x² + υ_y² + υ_z²

This relationship also must hold for the averages of these components and so the mean–squared speed is:

$\langle\upsilon^2\rangle = \langle\upsilon_x^2\rangle + \langle\upsilon_y^2\rangle + \langle\upsilon_z^2\rangle$

Equation 3 then gives

$\langle\upsilon_x^2\rangle + \langle\upsilon_y^2\rangle + \langle\upsilon_z^2\rangle = \frac13\langle\upsilon^2\rangle$(4)

You may have noticed that we have ignored collisions between molecules as they bounce back and forth between the walls. Surprisingly enough, this omission does not matter. The large number of molecules means that at any one time there will be as many molecules travelling towards the wall as away from it and because the collisions are elastic, the distribution of speeds will not vary with time (this is considered in more detail in Section 3). Thus the average effect on the wall will be the same as if the molecules never collided with each other.

Combining Equations 2 and 4 produces our required equation, which relates the macroscopic properties of a gas (on the left–hand side of Equation 5) and the average microscopic quantities (on the right–hand side of Equation 5):

Here $\langle\upsilon^2\rangle$ is the mean–squared speed of the molecules. The positive square root of this is known as the root_mean_square_r.m.s_speedroot–mean–squared (rms) speed υ_rms:

$\upsilon_{\rm rms} = \sqrt{\langle\upsilon^2\rangle}$(6)

So Equation 5 can be written as:

$PV = \frac13Nm\upsilon_{\rm rms}^2$

We can also write Equation 5 in terms of the aυerage molecular translational kinetic energy, $\langle\varepsilon_{\rm tran}\rangle$, i by noting that:

$\frac13m\langle\upsilon^2\rangle = \frac13\langle m\upsilon^2\rangle = \frac23\left\langle\frac12m\upsilon^2\right\rangle = \frac23\langle\varepsilon_{\rm tran}\rangle$

Equation 5 then becomes

which relates the product PV to the average translational kinetic energy of the molecules in the gas. This is the result of our kinetic theory model calculation, and that is as far as we can take the equation simply on the basis of kinetic theory.

If you compare Equations 1,

PV = nRT(Eqn 1)

and Equation 7 you will see that the macroscopic ideal gas law and the kinetic theory result imply a particular association between temperature and the mean molecular kinetic energy. This relationship was anticipated qualitatively in Subsection 2.3. The relationship is:

$PV = nRT =\frac 2 N\langle\varepsilon_{\rm tran}\rangle$

or, taking the last two parts of the chain and rewriting them slightly we obtain:

$\langle\varepsilon_{\rm tran}\rangle = \dfrac32\dfrac{nR}{N}T = \dfrac32{nR}{nN_A}T = \dfrac32\dfrac{R}{N_A}T$(8)

where we have written the total number of molecules N in n moles as nN_A, where N_A is Avogadro’s constant, the number of molecules in one mole.

Equation 8 relates the macroscopic idea of absolute temperature to the average molecular (microscopic) translational kinetic energy of the molecules

Equation 9 gives us a microscopic definition of temperature in terms of the average molecular translational kinetic energy $\langle\varepsilon_{\rm tran}\rangle$.

Notice that the average molecular translational kinetic energy is determined only by the temperature and is independent of the molecular mass.

2.5 Boltzmann’s constant

Equation 8,

$\langle\varepsilon_{\rm tran}\rangle = \dfrac32\dfrac{nR}{N}T = \dfrac32{nR}{nN_A}T = \dfrac32\dfrac{R}{N_A}T$(Eqn 8)

is usually written as

where the new constant k, defined by the relationship

$k = \dfrac{R}{N_A}$(11)

is known as Boltzmann’s constant. i This constant appears frequently in equations relating macroscopic parameters to microscopic phenomena. From Equation 11 it is apparent that one can view k as the gas constant for one molecule in contrast with R, as the gas constant for one mole. The currently accepted value for k is 1.380 662 × 10⁻²³ J K⁻¹, normally rounded to 1.381 × 10⁻²³ J K⁻¹.

We can express the root-mean-squared speed,

$\upsilon_{\rm rms} = \sqrt{\langle\upsilon^2\rangle}$(Eqn 6)

in terms of Boltzmann’s constant using Equation 10. Since m is a constant:

$\langle\varepsilon_{\rm tran}\rangle = \langle\frac12m\upsilon^2\rangle = \frac12m\langle\upsilon^2\rangle = \frac12m\upsilon_{\rm rms}^2 = \frac32kT$

Notice here that the root-mean-squared speed is determined by both the temperature and the molecular mass; this is in contrast to the average molecular kinetic energy which is determined only by the temperature.

Here k has arisen in relating the ideal gas law to kinetic theory. Another way of obtaining Boltzmann’s constant is by examining the behaviour of the quantity PV/NT as a function of pressure, as shown in Figure 2. Here we can see that as P approaches zero, the value of PV/NT approaches a single value for a range of simple gases, and that this number is Boltzmann’s constant. This is just what we would expect from the ideal gas law since PV = nRT = nN_AkT. Thus the ideal gas law can be written as

with k = PV/NT.

Figure 2 also shows how real gases approach ideal gas behaviour at low pressures. As you can see from Figure 2, helium approximates better than others to the ideal gas law, but many more common gases deviate from this by no more than 0.1% even at atmospheric pressure.

2.6 Kinetic theory and the internal energies of molecules

We have already shown that the pressure and temperature of an ideal gas are determined by the average translational kinetic energy of the molecules in the gas. If we allow our molecules only to be monatomic hard spheres (as we have done so far), then that is the only significant form the energy can take. However, if we allow for more complex diatomic or polyatomic molecules then energy may also be stored as kinetic and potential energy within individual molecules as a molecule can vibrate or rotate about its centre of mass. When energy is added to a sample of such a gas the translational motion of its molecules will increase, but also they may store some of this energy in additional rotational or vibrational motion. The total energy of such a molecule would exceed that due to translational motion alone. i

For the monatomic gas

$\langle\varepsilon_{\rm tran}\rangle = \frac23kT$(Eqn 10)

and we assumed that the molecules were free to move in three dimensions, but that they had no internal structure and hence no rotational or vibrational energy. This freedom to move in three dimensions can be expressed by saying that the molecules have three degrees of freedom. If we were to allow for rotational or vibrational motion then this would give additional degrees of freedom, in which additional energy may be present.

Where there are several degrees of freedom available statistical mechanics i leads to the result that, on average, the available energy will be shared equally between all degrees of freedom.

The proof of this lies beyond the scope of FLAP but it is another manifestation of the assumption of total randomness, and it is known as the equipartition of energy theorem.

So, with three degrees of freedom each molecule has an average energy of $3\times\frac12kT = \frac23kT$ and this average total energy per molecule, $\langle\varepsilon_{\rm tot}\rangle$, is purely translational.

In general, if a molecule has q degrees of freedom then:

Equation 14b provides an alternative to Equation 9,

$T = \dfrac23\dfrac{N_A}{R}\langle\varepsilon_{\rm tran}\rangle$(Eqn 9)

as a microscopic definition of temperature.

We can now relate the total internal energy, E_int, for the whole molecular system to a key macroscopic parameter, the absolute temperature. For a gas of N molecules with q degrees of freedom, we can write

$E_{\rm int} = N\langle\varepsilon_{\rm tot}\rangle = \dfrac q2NkT$(15)

2.7 Kinetic theory and the specific heat of gases

The specific heat is defined as the energy required to raise the temperature of a specified amount of material by one degree absolute. Normally the specified amount is 1 kg, with the specific heat measured in units of J K⁻¹ kg⁻¹. We can also define a molar specific heat, which has units of J K⁻¹ mol⁻¹, where the specified amount is one mole. From the general definition of specific heat we can define the molar specific heat by the equation:

where C is the molar specific heat, ∆Q_m is the energy supplied per mole of gas, in the form of heat, i causing a change in temperature ∆T. i For a gas, the conditions under which the energy is added must be carefully specified, (e.g. at constant pressure, or at constant volume), since this will affect the value of the resulting specific heat. We will consider now what our kinetic theory model for the ideal gas tells us about these specific heats.

First we will consider C_V, the molar specific heat at constant volume. When heat is added to the system, the energy transferred to the gas must increase the energy of the gas molecules. This is all that can happen if the process is at constant volume. For a monatomic ideal gas, we therefore see from Equation 15,

$E_{\rm int} = N\langle\varepsilon_{\rm tot}\rangle = \dfrac q2NkT$(Eqn 15)

that the increase in internal energy corresponding to a temperature increase ∆T is

$\Delta E_{\rm int} = \dfrac32Nk\Delta T = \dfrac32nR\Delta T$

where n is the number of moles present. Thus, to cause the temperature rise, the heat that must be supplied per mole is

$\Delta Q_{\rm m} = \dfrac{\Delta E_{\rm int}}{n} = \dfrac32\dfrac{Nk\Delta T}{n} = \dfrac32R\Delta T$

so$C_V = \left(\dfrac{\Delta Q_{\rm m}}{\Delta T}\right)_V = \dfrac32N_Ak = \dfrac32R$(17) i

where R is the gas constant (Equation 11),

$k = \dfrac{R}{N_A}$(Eqn 11)

If we have polyatomic molecules with q degrees of freedom, this is simply generalized to:

We also need to derive an expression for the molar specific heat at constant pressure, C_P. The difference between the two specific heats, in terms of bulk properties, is that for the constant pressure process, the gas will expand as heat is added. This means that part of the energy transferred to the system as heat will be used to do the work of expansion and there will be a correspondingly smaller increase in internal energy than for the constant volume system. Since the internal energy of the ideal gas is proportional to the absolute temperature, there will be a smaller rise in temperature for the same amount of heat transferred, which implies that C_P > C_V. We will now try to understand this in terms of our kinetic theory model.

As heat is added to the system, the volume increases to maintain constant pressure. This could be done by having the gas contained in a cylinder with a frictionless piston (see Figure 4), with a fixed reference pressure (say atmospheric pressure) on top of the piston. As the gas in the cylinder is heated, the piston will rise, maintaining constant pressure, so allowing the volume to increase. How can we interpret this microscopically?

In Subsection 2.4 we assumed that the wall of the container was held rigidly in place by external forces, so that the elastic collisions of the molecules caused no recoil of the wall itself. Now we are going to allow the wall of the piston to move in response to the collisions, by making the piston frictionless, so we need to re–analyse the situation. In an elastic collision where both masses are free to move, the energy will be shared out between them. This means that the molecules will, on average, transfer some of their kinetic energy to the recoiling piston as the gas expands and more heat will therefore be required to produce the same rise in temperature, compared to the constant volume case.

In Figure 4 the gas initially has volume V, pressure P, temperature T and is contained in a cylinder with a piston of cross–sectional area A. An amount of heat ∆Q is added to the gas, its temperature rises to (T + ∆T) and it also expands under constant pressure P to volume (V + ∆V). To raise the piston, the gas must apply an upwards force of magnitude F given by F = PA. The gas does work by raising the piston a distance ∆l.

This work done by the gas is given by

∆W = F∆l = PA∆l = P∆V

Using the principle of energy conservation

heat added = change in internal energy of gas + work done by the gas

we obtain the first law of thermodynamics:

We can calculate P∆V from Equation 1,

PV = nRT(Eqn 1)

by considering the system before and after the expansion. For an ideal gas:

initiallyPV = nRT

and finallyP(V + ∆V) = nR(T + ∆T)

If we subtract these two, we find

P∆V = nR∆T(20)

∆E_int for the system can be obtained from Equation 15,

$E_{\rm int} = N\langle\varepsilon_{\rm tot}\rangle = \dfrac q2NkT$(Eqn 15)

and using Equation 18 we may write this as

∆E_int = nC_V∆T(21)

If we now substitute Equations 20 and 21 into Equation 19,

∆Q = ∆E_int + P∆V(Eqn 19)

and divide both sides by n we find that at constant pressure the heat per mole required to raise the temperature by ∆T is

∆Q_m = C_V∆T + R∆T = (C_V + R)∆T

so$C_P = \left(\dfrac{Q_{\rm m}}{\Delta T}\right)_P = C_V + R$ i

Equation 22 gives the difference in the molar specific heats for an ideal gas. As expected from our earlier discussion, C_P exceeds C_V.

The ratio of the specific heats, C_P/C_V is often written as γ, i and using Equations 22 and 18,

$C_V = \dfrac q2N_Ak = \dfrac q2R$(Eqn 18)

we see that

Notice that the difference between the two specific heats is independent of the number of degrees of freedom q of the molecule, but the two specific heats themselves and their ratio γ depend on q.

We may collect the result from Question T6 into our set of equations by writing, for a gas with q degrees of freedom:

The conclusion from this subsection is that our microscopic kinetic theory model allows the number of degrees of freedom of the molecules in a gas to be inferred from macroscopic measurements of the gas specific heats.

2.8 The mean free path of an ideal gas molecule

In Subsection 2.4, it was claimed that intermolecular collisions did not invalidate the derivation of Equation 5,

$PV = \frac13Nm\langle\upsilon^2\rangle$(Eqn 5)

However, it is still of interest to know something about the frequency of these collisions and how far molecules travel between them.

The mean free path λ i of a molecule is defined as the average distance it travels between collisions. We might expect the mean free path to vary strongly with temperature and pressure as these properties determine the speed and the closeness of the molecules, respectively. We can get a feel for the kind of distances involved by thinking about a specific example. Let us consider one mole of a gas at a temperature of 300 K and a pressure of 1.0 × 10⁵ Pa – these were the conditions for the gas in Question T2.

In Question T2 we calculated the average separation between molecules under these conditions as 3.46 nm. You might expect that the mean free path would be very similar to the intermolecular separation – in fact it is considerably different, as we will see in a moment – mainly because it also depends on the size of the molecules as well as on their mean separation.

To calculate λ, we need to think about the average volume per molecule. This must also be the volume a molecule will have to sweep through before it makes a collision. A molecule will contact another identical molecule if (assuming they are spherical) the distance between their centres is equal to twice the molecular radius or 2r (see Figure 5).

We are only interested in centre-to-centre intermolecular distances, so the situation is equivalent to one where all the other molecules are considered to be points, and our reference molecule has a radius of 2r.

In a time ∆t our reference molecule, assumed to be travelling with the average speed $\langle\upsilon\rangle$, sweeps out a volume equal to that of a cylinder of length $\langle\upsilon\rangle\Delta t$ and radius 2r. This volume is $\Delta V = \pi(2r)^2\langle\upsilon\rangle\Delta t = 4\pi r^2\langle\upsilon\rangle\Delta t$. All other molecules whose centres lie within this volume will be struck by the reference molecule within the time ∆t. This number is n_ρ∆V where n_ρ is the number density of the molecules. So, in a time ∆t the number of collisions with the reference molecule is $4\pi r^2n_\rho\langle\upsilon\rangle\Delta t$. We can define the mean collision frequency f_coll (mean number of collisions per second) as

$f_{\rm coll} = \dfrac{\text{mean number of collisions in }\Delta t}{\Delta t} = \dfrac{4\pi r^2n_\rho\langle\upsilon\rangle\Delta t}{\Delta t} = 4\pi r^2n_\rho\langle\upsilon\rangle$(25)

The mean time between collisions, or the mean free time τ_coll i is

$\tau_{\rm coll} = \dfrac{1}{f_{\rm coll}} = \dfrac{1}{4\pi r^2n_\rho\langle\upsilon\rangle}$(26)

and the mean distance between collisions, or the mean free path λ, is

$\lambda = \langle\upsilon\rangle\tau_{\rm coll} = \dfrac{1}{4\pi r^2n_\rho}$(27)

Notice that the mean free path is independent of the average molecular speed. You may be concerned that this simple calculation has ignored the effects of collisions on our calculated volumes. Again we are justified in doing this because of Assumption 1. Since we are concerned with random motion, we only have to use averages. In fact, a more careful calculation, which takes into account the motion of all the molecules, changes these results only by a factor of 2, so that for example, the expression for mean free path becomes

2.9 Summary of Section 2

The kinetic theory model of an ideal gas is based on Newton’s laws and certain simplifying assumptions. The simplifying assumptions are as follows:

Assumption 1

The ideal gas consists of a large number of identical molecules in high speed random motion.

Assumption 2

All collisions between molecule and molecule, or between molecule and wall, are elastic.

Assumption 3

The individual molecules obey Newton’s laws of motion.

Assumption 4

The molecules only experience contact forces. They interact like hard spheres and this only when they touch.

Assumption 5

The volume occupied by the molecules themselves is very small compared to the volume occupied by the gas as a whole.

This model provides a derivation of the ideal gas equation of state (PV = nRT) and shows that the absolute temperature can be expressed in terms of the average kinetic energy per molecule:

$T = \dfrac23\dfrac{N_A}{R}\langle\varepsilon_{\rm tran}\rangle = \dfrac{2}{3k}\langle\varepsilon_{\rm tran}\rangle$(Eqns 9 and 10)

Boltzmann’s constant is a useful parameter, it is defined by the equation k = R/N_A and can be interpreted as the gas constant per molecule.

The kinetic theory model can be used to derive theoretical expressions for the specific heat of an ideal gas at constant volume and at constant pressure and for the mean free path of a molecule. In a monatomic ideal gas:

$C_V = \dfrac{3R}{2}\quad C_P = \dfrac{5R}{2}\quad\text{and}\quad\lambda = \dfrac{1}{4\sqrt{2\os}\pi r^2n_\rho}$

3 The Maxwell–Boltzmann speed distribution

Study comment This is a rather mathematical and even abstract section. Do not worry too much about the details of the mathematics if you are uncomfortable with them. Concentrate on the basic physical concepts of the Maxwell–Boltzmann distribution and on the physical meaning of the equation that characterizes it. The material in Section 4 does not depend on the details of what is described in Section 3.

3.1 The distribution of molecular speeds

So far we have been able to base our discussions on the average properties of the molecules, but to understand the detailed behaviour properly, we need to have some idea about how the properties of the molecules vary about the average. In particular, we would like to know the way the speeds of the molecules are spread or distributed around the average speed. Are most of the speeds within a per cent or so of this average or are they spread much wider than this? The distribution of molecular speeds in an ideal gas was first obtained by James Clerk Maxwell (1831–1879), using arguments based on statistical mechanics. This speed distribution is known as the Maxwell–Boltzmann speed distribution. We will make no attempt to derive it here but we will comment on its origins, specify it explicitly and finally explore some of its implications.

The speeds of individual molecules in an ideal gas will be constantly changing as the molecules collide. However, for a given sample of gas at a fixed temperature, containing a given number of molecules, the effect of the collisions is to establish and maintain a particular distribution of speeds – the Maxwell–Boltzmann distribution. This distribution will determine the average number (or proportion) of the molecules that have speeds in any specified range of values. If the actual speeds of the molecules depart from this distribution (as they will) the overall effect of the energy and momentum exchanges that take place in intermolecular collisions will tend to restore the actual speed distribution to that described by the Maxwell–Boltzmann distribution. Thus, the Maxwell–Boltzmann speed distribution represents an average distribution of molecular speeds, about which the true distribution fluctuates.

There are two commonly used methods of specifying the distribution of molecular speeds within a sample of ideal gas. The first is to specify the number of molecules in the gas that have speeds in any narrow range between υ and υ + ∆υ. The second is to specify the relative likelihood (i.e. probability) i that an individual molecule, chosen at random, will have a speed in the range υ to υ + ∆υ. Though different, these two methods are deeply related since the greater the number of molecules with speed in a given range the greater the likelihood that a randomly chosen molecule will have its speed in that range. We will present these two specifications of the Maxwell–Boltzmann distribution in turn.

Suppose you have a sample of ideal gas at temperature T that consists of N identical molecules of mass m. If the number of molecules that have speeds in the narrow range υ to υ + ∆υ is represented by the quantity n(υ)∆υ then:

where k is Boltzmann’s constant.

Note that since n(υ)∆υ represents a number of molecules, the quantity n(υ) must have the dimensions of (speed)⁻¹, so it might be measured in units of s m⁻¹, and it should be interpreted as the number of molecules per unit speed interval with speeds in the range υ to υ + ∆υ.

It follows from Equation 29 that, on average, the fraction of molecules in the sample with speeds in the range υ to υ + ∆υ will be n(υ)∆υ/N. This fraction, which we will denote by f(υ)∆υ represents the relative likelihood (i.e. probability) that a single molecule chosen at random will have its speed in the range υ to υ + ∆υ. Thus:

In this case the function f(υ), which has the dimensions of 1/υ. and could also be measured in units of s m⁻¹, is called the maxwell_boltzmann_speed_distributionMaxwell–Boltzmann speed distribution function:

$f(\upsilon) = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{3/2}\upsilon^2\exp(-m\upsilon^2/kT)$

Note that this distribution function is independent of angle; the direction of motion is irrelevant as it concerns only the speed. It is also independent of position, because we have assumed a uniform spatial distribution – which means that the probability of finding a molecule in any particular element of volume is constant. i Before we look at the detailed shape of the Maxwell–Boltzmann speed distribution function let us examine Equation 30 qualitatively.

Although the equation looks complicated, let us concentrate first on the exponential (−mυ²/2kT) factor. Like any exponent, this must be dimensionless; we can see that this is so since it is the ratio of two energies, mυ²/2 and kT, where mυ²/2 is the kinetic energy of the molecule with mass m and speed υ, while kT is related to the average translational energy per molecule, 3kT/2. For a molecule which is much slower than average, the kinetic energy is very small compared to kT, and mυ²/2kT is much less than unity.

We now know the shape of the Maxwell–Boltzmann speed distribution at low speeds. What happens at high speeds?

The result of combining the low and high speed behaviours of Equation 30 is that there must be a maximum in the function at some speed, between the quadratic growth regime and the exponential decay regime.

This is illustrated in Figure 6, which shows the shape of the Maxwell–Boltzmann speed distribution function f(υ) for a sample of gas at three different temperatures. The important points to notice here are that as the temperature increases the peak in the distribution becomes lower and broader, and occurs at higher speeds.

The interpretation of f(υ)∆υ as a probability has an important implication for any graph of f(υ) against υ, including the graphs in Figure 6:

The area enclosed between any graph of f(υ) and the υ-axis must be exactly 1 i

(in the scale units appropriate to the graph).

Actually measuring the area under the curves in Figure 6 to verify this would be very time consuming, but you can easily see that it might well be true just by noticing that the 1000 K curve is fairly close to being a triangle with: height 4 × 10⁻⁴ s m⁻¹, base length 5 × 10³ m s⁻¹ and hence area (4 × 10⁻⁴ s m⁻¹ × 5 × 10³ m s⁻¹)/2 = 1.

Keeping this constant area requirement in mind will help you to answer Question T11, which concerns the way the distribution function f(υ) changes shape as the temperature parameter T is altered.

Knowing the Maxwell–Boltzmann distribution function we can answer a variety of questions about an ideal gas. For example, there are at least three characteristic speeds that are likely to be of interest:

1

First, and most obviously, we would like to know the speed corresponding to the peak in the distribution – the most common speed. For obvious reasons we will call this the most probable speed υ_prob.

2

Second, we would like to know the average speed $\langle\upsilon\rangle$.

3

Finally, because of its significance in the kinetic theory link with the ideal gas law, as shown in Subsection 2.4, we would like to know the root-mean-squared speed υ_rms. This would also allow us to calculate the average kinetic energy per molecule mυ_rms²/2.

All of these characteristic speeds i can be calculated from the form of the distribution function implied by (Equation 30),

$f(\upsilon)\Delta\upsilon = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{3/2}\upsilon^2\exp(-m\upsilon^2/2kT)\Delta\upsilon$(Eqn 30)

using calculus.

We will not present the mathematical details of these calculations for the three characteristic speeds, but the results are as follows:

We have already derived the expression for υ_rms in Subsection 2.5, on the basis of a comparison with the ideal gas law; it is reassuring that this result is also derivable from the Maxwell–Boltzmann speed distribution.

Also, Equation 12 confirms that the average kinetic energy per molecule (½ mυ²) is 3kT/2, in agreement with Equation 10,

$\langle\varepsilon_{\rm tran}\rangle = \frac32kT$(Eqn 10)

All three characteristic speeds are proportional to (T/m)^1/2, but it is apparent from the numerical factors involved that υ_rms > $\langle\upsilon\rangle$ > υ_prob.

Figure 7 shows the relationship between υ_prob, $\langle\upsilon\rangle$ and υ_rms for the Maxwell–Boltzmann speed distribution.

For any distribution (where there is a spread of speeds) it is generally true that υ_rms > $\langle\upsilon\rangle$, but the special results shown in Figure 7 and given by Equations 32, 33 and 12 are specific to the Maxwell–Boltzmann speed distribution. These inequalities lead to the interesting point that, for any distribution, the molecule with the average speed is not the molecule with the average kinetic energy!

3.2 The distribution of molecular kinetic energies

We can use the Maxwell–Boltzmann speed distribution to determine how the kinetic energies of the molecules are distributed around the mean kinetic energy, 3kT/2. A molecule with speed υ has translational kinetic energy ε_tran = mυ²/2. The number of molecules, n(υ)∆υ, having speeds within a narrow range between υ and (υ + ∆υ) will also have translational kinetic energies in the range ε_tran to (ε_tran + ∆ε_tran); we will call this number n_ε(ε_tran)∆ε_tran. The numbers in this speed interval and this energy interval are clearly equal, since they are the same molecules.

Therefore we can write

n(υ)∆υ = n_ε(ε_tran)∆ε_tran

Using Equation 29,

$n(\upsilon)\Delta\upsilon = 4\pi N\left(\dfrac{m}{2\pi kT}\right)^{3/2}\upsilon^2\exp(-m\upsilon^2/2kT)\Delta\upsilon$(Eqn 29)

and substituting ε_tran = mυ²/2 we have

$n(\upsilon)\Delta\upsilon = 4\pi N\left(\dfrac{m}{2\pi kT}\right)^{3/2}\left(\dfrac{2\varepsilon_{\rm tran}}{m}\right)\exp(-\varepsilon_{\rm tran}/kT)\Delta\upsilon = n_\varepsilon(\varepsilon_{\rm tran})\Delta\varepsilon_{\rm tran}$

or$n_\varepsilon(\upsilon)\Delta\upsilon\left(\dfrac{\Delta\varepsilon_{\rm tran}}{\Delta\upsilon}\right) = 4\pi N\left(\dfrac{m}{2\pi kT}\right)^{3/2}\left(\dfrac{2\varepsilon_{\rm tran}}{m}\right)\exp(-\varepsilon_{\rm tran}/kT)\Delta\upsilon = n_\varepsilon(\varepsilon_{\rm tran})\Delta\varepsilon_{\rm tran}$

Finding an explicit formula for n_ε(ε_tran) in terms of ε_tran (similar to the expression for n(υ) in terms of υ implied by Equation 30),

$f(\upsilon)\Delta\upsilon = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{3/2}\upsilon^2\exp(-m\upsilon^2/2kT)\Delta\upsilon$(Eqn 30)

requires that we express (∆εtran/∆υ) in terms of ε_tran. This is done by noting that

$\Delta\varepsilon_{\rm tran} = \frac12m(\upsilon + \Delta\upsilon)2 - \frac12m\upsilon^2 = \frac12m\upsilon^2 + m\upsilon\Delta\upsilon + \frac12 m(\Delta\upsilon)^2 - \frac12m\upsilon^2 = m\upsilon(\Delta\upsilon) + \frac12 m(\Delta\upsilon)^2$

But, since the speed range is narrow, ∆υ is small and the term involving (∆υ)² can be ignored.

Thus ∆ε_tran ≈ mυ∆υ and so $\dfrac{\Delta\varepsilon_{\rm tran}}{\Delta\upsilon} \approx m\upsilon = \sqrt{2m\varepsilon_{\rm tran}\os}$ i

We therefore have

$n_\varepsilon(\varepsilon_{\rm tran}) = 4\pi N\left(\dfrac{m}{2\pi kT}\right)^{3/2}\left(\dfrac{2\varepsilon_{\rm tran}}{m}\right)\left(\dfrac{1}{2m\varepsilon_{\rm tran}}\right)^{1/2}\exp(-\varepsilon_{\rm tran}/kT)$

for the distribution of molecular translational kinetic energy. As before, we can use our knowledge of n_ε(ε_tran) to write down a maxwell_boltzmann_energy_distributionMaxwell–Boltzmann energy distribution function, f_ε(ε_tran) = n_ε(ε_tran)/N, such that f_ε(ε_tran)∆ε_tran is the probability that a molecule chosen at random will have energy in the narrow range ε_tran to ε_tran + ∆ε_tran.

Figure 8 shows the energy distribution function for the same three temperatures used for the speed distributions shown in Figure 6.

It is interesting to note (from Equation 34) that the energy distribution is determined only by the temperature and is independent of the molecular mass; this extends the observation we made in Subsection 2.4, that the average molecular kinetic energy is determined only by the temperature and is independent of the molecular mass. This is in contrast to the average speeds and the distributions over speed which are dependent on the molecular masses, as expected from Subsection 2.5.

3.3 Applications and verification of the speed distribution

In Section 2 we were able to relate the microscopic molecular behaviour of a gas to its macroscopic properties of pressure and temperature. This relationship did not involve any particular speed distribution but only required there to be an average molecular translational kinetic energy. In contrast, there are other properties of gases which do depend on the details of the speed distribution. Examples of such properties are:

1

The inter–mixing of gases (same or different) when there are local differences of concentrations present; this process is called diffusion.

2

Viscosity or gas friction, in which layers of gas in relative motion tend to inter–mix and thereby reduce the relative motion.

3

Thermal conduction, when there are temperature gradients present in a gas, in which inter–mixing of gas from the hot and cold regions tends to equalize the average speeds and hence the temperatures in each region.

You will realize that these three properties are related in that they are each controlled by the rate of molecular inter-mixing, or how quickly molecules can find their way from one region to another region of the gas. This rate will be limited due to intermolecular collisions, which slow down the rate of transport. It is perhaps not surprising then that these three properties are collectively known as transport processes. These transport processes can each be predicted assuming the speed distribution and the agreement between the predicted and measured properties then gives strong indirect support for the validity of the Maxwell–Boltzmann speed distribution itself. i

The validity of the Maxwell–Boltzmann speed distribution can also be demonstrated from direct measurement of the distribution of molecular speeds in a gas. How can we accomplish this? We cannot measure the speed of individual molecules, so we need to make some measurement that directly depends on their speeds. One method of performing such measurements is shown in Figure 9. We generally begin by taking a container of gas and place a fine slit in the side (termed a nozzle). When molecules from inside strike the slit space they emerge from the slit. If the region beyond the slit is a good vacuum these emerging molecules continue on in the same direction.

If we then interpose a second fine slit some distance downstream we will select only those molecules travelling in a particular direction. Such a beam, with a well–defined direction, is called a molecular beam.

Molecules in such a molecular beam, when launched horizontally, will tend to fall due to gravity and if the vertical drop to a distant detector is measured this can be related to the time of flight and hence to the molecular speed. Careful interpretation i of the numbers of molecules arriving as a function of the distance dropped confirms the form of the speed distribution, but the details of the method need not concern us here.

3.4 Summary of Section 3

The Maxwell–Boltzmann speed distribution for a gas has been introduced but not derived. This distribution gives the numbers of molecules within a narrow speed range as a function of the speed. From this expression it is possible to identify three characteristic speeds for a gas: these are the most probable speed, the average speed and the root-mean-squared speed – arranged in order of increasing magnitude. The distribution of molecular kinetic energies associated with the Maxwell–Boltzmann speed distribution has been derived from the speed distribution and this is seen to be independent of the molecular mass, being determined only by the temperature. Finally, both indirect and direct evidence in support of the Maxwell–Boltzmann speed distribution has been given in terms of transport processes and the speeds in molecular beams, respectively.

4 A more realistic model of a gas

4.1 Van der Waals equation of state

It was discovered through careful measurements in the 19th century that the ideal gas equation of state was only an approximation to the true behaviour of real gases. The Dutch physicist Johannes van der Waals (1837–1923) devised the following generalization to the ideal gas equation of state in order to describe more accurately the behaviour of real gases:

Table 1 Values of van der Waals constants a and b for various gases.

	a/10−1 m6 Pa mol−2	b/10−5 m3 mol−1
nitrogen (N2)	0.035	2.4
oxygen (O2)	1.41	3.91
xenon (Xe)	1.38	3.18
helium (He)	4.25	5.10

In this equation, V_m represents the volume per mole; a and b can be regarded as empirical constants, although they have a theoretical interpretation in terms of a kinetic theory model.

The values of a and b vary from one gas to another, as indicated in Table 1. We want to consider briefly what the physical origin of such terms might be in terms of our kinetic theory model of a gas.

4.2 Excluded volumes

In Subsection 2.4, in deriving the ideal gas equation of state, we ignored the volume taken up by the gas molecules themselves. Although reasonable, this approximation has repercussions. For example, the ideal gas equation of state implies that the volume of the gas approaches zero when the temperature approaches zero, whereas in reality the minimum volume that the gas can occupy must exceed the total molecular volume.

If we consider the molecules to be perfect spheres of radius r, then on touching, the molecules will be separated by 2r. This means that the centre of a second molecule will be excluded from a sphere of radius 2r, whose volume equals $\frac43\pi(2r)^3$, which is eight times the individual molecular volume. Thus the excluded volume per molecule for spheres is four times the molecular volume.

For one mole of gas, the total volume excluded by the finite size of the molecules will be N_A × 4 × υ_m, where υ_m is the individual molecular volume. We might then predict that the appropriate generalization for the ideal gas would be to replace the volume per mole, V/n, by (V_m − 4υ_mN_A). This identifies the constant b in the van der Waals equation of state with 4υ_mN_A:

$\left(P + \dfrac{a}{V_{\rm m}^2}\right)(V_{\rm m}-b) = RT$(Eqn 35)

4.3 Intermolecular forces

We now test our assumption that the long–range forces between molecules are negligible. Figure 10 shows a typical variation in (a) intermolecular force, and (b) the corresponding potential energy curves for the interaction between two molecules.

Figure 10a shows that there is a large repulsive force between molecules when they are very close to each other, and a relatively weak attractive force at larger separations. i

This is to be expected, since we know that most gases will condense to form a liquid if we cool them enough, and this would only happen if there were an attractive force that pulled the molecules together (this was actually van der Waals motivation in deriving his equation). The short–range strong repulsion accounts for the collisional forces and the success of the hard sphere model. What effect will the weak long–range attractive force have on the ideal gas equation of state? Let us consider a single reference molecule as it moves towards a wall.

In the centre of the volume of gas, there will be (on average) an equal number of gas molecules on all sides of our reference molecule, so that the net force will be approximately zero. When our reference molecule approaches a wall of the container, there will be more gas molecules on the side away from the wall than on the side toward the wall, and ultimately (just before the collision with the wall) all other gas molecules will be on the side away from the wall. The reference molecule will then feel a weak attraction from the remaining molecules in the gas, which will tend to reduce its velocity before impact with the wall and hence reduce the change in momentum on colliding with the wall. This will reduce the effective pressure compared with the gas at the same temperature but without such attractive forces.

What form will this reduction take? The reduction in pressure must depend on the size of the attractive force on each molecular impact and on the number of these impacts per second. The force depends on the number of molecules in the gas which are within range, and for a fixed number of molecules (e.g. per mole) this will depend inversely on the volume. Also, for a fixed number of molecules, the rate of molecular impacts depends on the pressure – which also will depend inversely on the volume. These two multiplying factors, each proportional to 1/V_m, suggest that the effective pressure reduction should be proportional to 1/V_m². Since we have not evaluated any actual forces, we can only say that the pressure should be less by some term proportional to 1/V_m². This effect is reproduced by replacing the term P in the ideal gas equation of state by a term (P + a/V_m²), as in the van der Waals equation of state, to restore the effective pressure reduction.

$\left(P + \dfrac{a}{V_{\rm m}^2}\right)(V_{\rm m}-b) = RT$(Eqn 35)

5 Closing items

5.1 Module summary

1

The macroscopic or bulk properties of an ideal gas, as given by the ideal gas law

PV = nRT(Eqn 1)

can be derived from a simple microscopic model of the behaviour of molecules within the gas, based on certain fundamental assumptions. This model is known as the kinetic theory.

2

The kinetic theory calculation of the pressure due to an ideal gas leads to a form of the equation of state which allows us to associate the average translational kinetic energy per molecule with the absolute temperature. This enables us to give a microscopic interpretation to the macroscopic concept of temperature.

$PV = \frac23N\langle\varepsilon_{\rm tran}\rangle$(Eqn 7)

$\langle\varepsilon_{\rm tran}\rangle = \frac32kT$(Eqn 10)

where$k = \dfrac{nR}{N} = \dfrac{R}{N_A}$ is Boltzmann’s constant.

3

The kinetic theory can also be used to give microscopic interpretations of the internal energy and specific heats of gases, in terms of the number of molecular degrees of freedom q and the equipartition of energy theorem:

$E_{\rm int} = \dfrac q2NkT$(Eqn 15)

$C_V = \dfrac q2R\quad\text{and}\quad C_P = \left(1 + \dfrac q2\right)R$(Eqns 18 and 24)

$C_P - C_V= R\quad\text{and}\quad\gamma = \dfrac{C_P}{C_V} = 1 + \dfrac2q$(Eqns 22 and 23)

4

The mean free path is the average distance travelled by molecules between collisions. The mean free time is the average time between collisions. A value for these, given specified conditions, can also be calculated from kinetic theory.

5

The random nature of molecular motion leads to an expression for the distribution of speeds among the gas molecules; this is the Maxwell–Boltzmann speed distribution:

$f(\upsilon) = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{3/2}\upsilon^2\exp(-m\upsilon^2/kT)$

The distribution can be characterized in terms of a most probable speed, an average speed and a root-mean-squared speed. The Maxwell–Boltzmann speed distribution has been verified experimentally, indirectly through its effect on transport processes and directly by measuring molecular speeds in a molecular beam.

6

The van der Waals equation of state,

$\left(P + \dfrac{a}{V_{\rm m}^2}\right)(V_{\rm m}-b) = RT$(Eqn 35)

provides a more accurate description of real gases than the ideal gas equation. Its empirically determined parameters can be explained in terms of modifications to the simple kinetic theory, through excluded volumes and intermolecular forces.

5.2 Achievements

Having completed this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Describe and explain the basic assumptions underpinning the kinetic theory of gases.

A3

Explain qualitatively how the kinetic theory model of a gas is in accord with the ideal gas equation of state PV = nRT.

A4

Make approximate calculations of mean free paths and mean free times, given appropriate molecular data.

A5

Carry out simple calculations based on the kinetic theory equation,

$PV = \frac23N\langle\varepsilon_{\rm tran}\rangle$(Eqn 7)

A6

Recall and explain the significance of the equation

$\langle\varepsilon_{\rm tran}\rangle = \frac32kT$(Eqn 10)

and use it in calculations.

A7

Explain the interpretation of the internal energy of a gas in terms of the kinetic theory model and calculate internal energies and specific heats in terms of degrees of freedom for different types of molecules.

A8

Recall the equation relating υ_rms to absolute temperature and molecular mass and be able to use it in calculations.

A9

Explain what is meant by the Maxwell–Boltzmann speed distribution, sketch its form and briefly explain how it may be tested.

A10

Distinguish between the quantities υ_prob, $\langle\upsilon\rangle$ and υ_rms for the Maxwell–Boltzmann speed distribution and explain why they differ and how their magnitudes compare (without detailed formulae).

A11

Describe the physical reasoning behind the modifications to the ideal gas equation of state which lead to the van der Waals equation of state.

A12

Use the van der Waals equation of state to calculate physical parameters.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

5.3 Exit test

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