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 of 12C "

Laboratory
experiments show that the number of atoms in a 12-g sample of 12C 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:

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:

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 is kept 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 20¡C and a pressure of 15 atm. at a volume of 12 l. The temperature is raised to 35¡C , and the volume is reduced to 8.5 l. What is the final pressure of the gas ?

The ideal gas law tells us that

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

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:

Figure 18.1. Molecule moving in box. |

The y and z components of the momentum of the molecule are left unchanged. The change in the momentum 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 right 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 Æp. 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

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; for 14N 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 v****rms**** 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**.

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.

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** l. The
mean free path l 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 between collisions is Æt. During this time, the molecule travels a distance v . Æt, and sweeps a volume equal to

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 Æt defined in this manner is the mean time between collisions, and the mean free path l 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 molecular motion 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

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.

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 of such 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 N****2**** 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**.

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**, ÆQ is the heat added, and ÆT is the resulting increase
in the temperature of the system. The first law of thermodynamics
shows that

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

and

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

and thus

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 ÆV to ÆT, 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, ÆU/ÆT = 3/2 n R = n CV. The previous equation can therefore be rewritten as

or

We see that Cp CV.

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 ÆV:

or

The specific heat capacity CV is related to ÆU

Using the first law of thermodynamics we can write

or

Combining the two expressions obtained for n . ÆT we obtain

or

This expression can be rewritten as

For small changes this can be rewritten as

where we have defined g as (Cp/CV). After integrating this expression we obtain

or

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

Thus

Send comments, questions and/or suggestions via email to **wolfs@pas.rochester.edu** and/or
visit the home page of Frank
Wolfs.