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1. The Kinetic Theory of Gases

1.1. Avogadro constant

The laws of classical thermodynamics do not show the direct dependence of the observed

macroscopic variables on microscopic aspects of the motion of atoms and molecules. It is

however clear that the pressure exerted by a gas is related to the linear momentum of the atoms

and molecules, and that the temperature of the gas is related to the kinetic energy of the atoms

and molecules. In relating the effects of the motion of atoms and molecules to macroscopic

observables like pressure and temperature, we have to determine the number of molecules in the

gas. The mole is a measure of the number of molecules in a sample, and it is defined as

" the amount of any substance that contains as many atoms/molecules as there are atoms

in a 12-g sample of12C "

Laboratory experiments show that the number of atoms in a 12-g sample of12C is equal to 6.02 x

1023 mol-1. This number is called the Avogadro constant, NA. The number of moles in a

sample, n, can be determined easily:

1.2. The Ideal Gas

Avogadro made the suggestion that all gases - under the same conditions of temperature

and pressure - contain the same number of molecules. Reversely, if we take 1 mole samples of

various gases, confine them in boxes of identical volume and hold them at the same temperature,

we find that their measured pressures are nearly identical. Experiments showed that the gases

obey the following relation (the ideal gas law):

where n is the number of moles of gas, and R is the gas constant. R has the same value for all

gases:


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The temperature of the gas must always be expressed in absolute units (Kelvin).

Using the ideal gas law we can calculate the work done by an ideal gas. Suppose a

sample of n moles of an ideal gas is confined in an initial volume Vi. The gas expands by

moving a piston. Its final volume is Vf. During the expansion the temperature T of the gas iskept constant (this process is called isothermal expansion). The work done by the expanding

gas is given by

The ideal gas law provides us with a relation between the pressure and the volume

Since T is kept constant, the work done can be calculated easily

Note:

Sample Problem 18-1

A cylinder contains oxygen at 20C and a pressure of 15 atm. at a volume of 12 l. The

temperature is raised to 35C , and the volume is reduced to 8.5 l. What is the final pressure of

the gas ?

The ideal gas law tells us that


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The initial state of the gas is specified by Vi, pi and Ti; the final state of the gas is specified by

Vf, pfand Tf. We conclude that

Thus

The temperature T in this formula must be expressed in Kelvin:

Ti = 293 K

Tf = 308 K

The units for the volume and pressure can be left in l and atm. since only their ratio enter the

equation. We conclude that pf = 22 atm.

1.3. Pressure and Temperature: A Molecular View

Let n moles of an ideal gas be confined to a cubical box of volume V. The molecules in

the box move in all directions with varying speeds, colliding with each other and with the walls

of the box. Figure 18.1 shows a molecule moving in the box. The molecule will collide with the

right wall. The result of the collision is a reversal of the direction of the x-component of the

momentum of the molecule:

The y and z components of the momentum of the

molecule are left unchanged. The change in themomentum of the particle is therefore

After the molecule is scattered of the right wall, it will

collide with the left wall, and finally return to the rightFigure 18.1. Molecule moving in box.


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wall. The time required to complete this path is given by

Each time the molecule collides with the right wall, it will change the momentum of the wall by

p. The force exerted on the wall by this molecule can be calculated easily

For n moles of gas, the corresponding force is equal to

The pressure exerted by the gas is equal to the force per unit area, and therefore

The term in parenthesis can be rewritten in terms of the average square velocity:

Thus, we conclude that

where M is the molecular weight of the gas. For every molecule the total velocity can be

calculated easily

Since there are many molecules and since there is no preferred direction, the average square of

the velocities in the x, y and z-direction are equal


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and thus

Using this relation, the expression for the pressure p can be rewritten as

where vrms is called the root-mean-square speed of the molecule. The ideal gas law tells us that

Combining the last two equations we conclude that

and

For H at 300 K vrms = 1920 m/s; for14N vrms = 517 m/s. The speed of sound in these two gases

is 1350 m/s and 350 m/s, respectively. The speed of sound in a gas will always be less than

vrms since the sound propagates through the gas by disturbing the motion of the molecules.

The disturbance is passed on from molecule to molecule by means of collisions; a sound wave

can therefore never travel faster than the average speed of the molecules.


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1.4. Translational Kinetic Energy

The average translational kinetic energy of the molecule discussed in the previous section

is given by

Using the previously derived expression for vrms, we obtain

The constant k is called the Boltzmann constant and is equal to the ratio of the gas constant R

and the Avogadro constant NA

The calculation shows that for a given temperature, all gas molecules - no matter what their

mass - have the same average translational kinetic energy, namely (3/2)kT. When we

measure the temperature of a gas, we are measuring the average translational kinetic energy of its

molecules.

1.5. Mean Free Path

The motion of a molecule in a gas is complicated. Besides colliding with the walls of the

confinement vessel, the molecules collide with each other. A useful parameter to describe this

motion is the mean free path. The mean free path is the average distance traversed by a

molecule between collisions. The mean free path of a molecule is related to its size; the larger its

size the shorter its mean free path.

Suppose the gas molecules are spherical and have a diameter d. Two gas molecules will

collide if their centers are separated by less than 2d. Suppose the average time betweencollisions is t. During this time, the molecule travels a distance v .t, and sweeps a volume

equal to


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If on average it experiences one collision, the number of molecules in the volume V must be 1.

If N is the number of molecules per unit volume, this means that

or

The time interval t defined in this manner is the mean time between collisions, and the mean

free path is given by

Here we have assumed that only one molecule is moving while all others are stationary. If we

carry out the calculation correctly (all molecules moving), the following relation is obtained for

the mean free path:

The relation derived between the macroscopic pressure and the microscopic aspects of molecularmotion only depend on the average root-mean-square velocity of the molecules in the gas. Quit

often we want more information than just the average root-mean-square velocity. For example,

questions like what fraction of the molecules have a velocity larger than v0 can be important

(nuclear reaction cross sections increase dramatically with increasing velocity). It can be shown

that the distribution of velocities of molecules in as gas is described by the so-called Maxwell

velocity distribution

The product P(v)dv is the fraction of molecules whose speed lies in the range v to v + dv. The

distribution is normalized, which means that


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The most probable speed, vp, is that velocity at which the speed distribution peaks. The most

probable speed is obtained by requiring that dP/dv = 0

We conclude that dP/dv = 0 when

and thus

The average speed of the gas molecules can be calculated as follows

The mean square speed of the molecules can be obtained in a similar manner.

The root-mean-square speed, vrms, can now be obtained

We observe that vp < vav < vrms.


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1.6. Heat Capacity of Ideal Gas

The internal energy of a gas is related to the kinetic energy of its molecules. Assume for

the moment that we are dealing with a monatomic gas. In this case, the average translational

kinetic energy of each gas molecule is simply equal to 3kT/2. If the sample contains n moles ofsuch a gas, it contains nNA molecules. The total internal energy of the gas is equal to

We observe that the total internal energy of a gas is a function of only the gas temperature, and is

independent of other variables such as the pressure and the density. For more complex

molecules (diatomic N2 etc.) the situation is complicated by the fact that the kinetic energy

of the molecules will consist not only out of translational motion, but also out of rotational

motion.

1.6.1. Molar Heat Capacity at Constant Volume

Suppose we heat up n moles of gas while keeping its volume constant. The result of

adding heat to the system is an increase of its temperature

Here, CV is the molar heat capacity at constant volume, Q is the heat added, and T is theresulting increase in the temperature of the system. The first law of thermodynamics shows that

Since the volume is kept constant (V = 0) we conclude that

and

Using the previously derived equation for U in terms of T we can show that


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and thus

1.6.2. Molar heat Capacity at Constant Pressure

Suppose that, while heat is added to the system, the volume is changed such that the gas

pressure does not change. Again, the change in the internal energy of the system is given by

where Cp is the molar heat capacity at constant pressure. This expression can be rewritten as

For an ideal gas (pV = nRT) we can relate V to T, if we assume a constant pressure

Using this relation, the first law of thermodynamics can be rewritten as

or

However, the internal energy U depends only on the temperature and not on how the volume

and/or pressure is changing. Thus, U/T = 3/2 n R = n CV. The previous equation can

therefore be rewritten as

or


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We see that Cp CV.

1.7. The Adiabatic Expansion of an Ideal Gas

During an adiabatic expansion of an ideal gas no heat is added or extracted from the

system. This can be achieved by either expanding the gas very quickly (such that there is not

time for the heat to flow) or by very well insulating the system. The first law of thermodynamics

tells us that

The ideal gas law can be used to rewrite p V:

or

The specific heat capacity CV is related to U

Using the first law of thermodynamics we can write

or

Combining the two expressions obtained for n .T we obtain


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or

This expression can be rewritten as

For small changes this can be rewritten as

where we have defined as (Cp

/CV

). After integrating this expression we obtain

or

Using the ideal gas law to eliminate p from the expression we obtain

Thus

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