18.1. 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
12g sample of 12C "
Laboratory
experiments show that the number of atoms in a 12g sample of 12C is equal to 6.02 x 1023 mol1. This
number is called the Avogadro constant, NA. The number of moles in
a sample, n, can be determined easily:
18.2. 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:
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 181
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.
18.3. 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 xcomponent 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 ∆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 zdirection are equal
and
thus
Using
this relation, the expression for the pressure p can be rewritten as
where
vrms is called the rootmeansquare 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 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.
18.4. 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.
18.5. 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 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 ∆t. During this time, the
molecule travels a distance v . ∆t,
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 ∆t 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 rootmeansquare velocity
of the molecules in the gas. Quit often we want more information
than just the average rootmeansquare 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 socalled 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 rootmeansquare
speed, vrms, can now be obtained
We
observe that vp < vav < vrms.
18.6. 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
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 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.
18.6.1. 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 the resulting 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
and
thus
18.6.2. 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
We
see that Cp ≠ CV.
18.7. 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
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
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