**21.1 Molecular Model of an Ideal Gas**

We begin this chapter by
developing a microscopic model of an ideal gas. The model

shows that the pressure that a
gas exerts on the walls of its container is a consequence

of the collisions of the gas
molecules with the walls and is consistent with the macroscopic

description of Chapter 19. In
developing this model, we make the following assumptions:

- The number of molecules in the gas is large, and the average separation between them is large compared with their dimensions. This means that the molecules occupy a negligible volume in the container. This is consistent with the ideal gas model, in which we imagine the molecules to be point-like.
- The molecules obey Newton’s laws of motion, but as a whole they move randomly. By “randomly” we mean that any molecule can move in any direction with any speed. At any given moment, a certain percentage of molecules move at high speeds, and a certain percentage move at low speeds.
- The molecules interact only by short-range forces during elastic collisions. This is consistent with the ideal gas model, in which the molecules exert no longrange forces on each other.
- The molecules make elastic collisions with the walls.
- The gas under consideration is a pure substance; that is, all molecules are identical.

Although
we often picture an ideal gas as consisting of single atoms, we can assume that
the behavior of molecular gases approximates that of ideal gases rather well at
low pressures. Molecular rotations or vibrations have no effect, on the
average, on the

motions
that we consider here.

*N*molecules of an ideal gas in a container of volume

*V*in terms of microscopic quantities. The container is a cube with edges of length

*d*

Because the momentum

component

*pxi*of the molecule is*mvxi*before the collision and -*mvxi*after the collision, the change in the*x*component of the momentum of the molecule is
Because the molecules obey Newton’s laws (assumption
2), we can apply the impulsemomentum theorem
(Eq. 9.8) to the molecule to give us

The force that causes the
change in momentum of the molecule in the collision with

the wall occurs only during the
collision. However, we can average the force over the

time interval for the molecule
to move across the cube and back. Sometime during this time interval, the
collision occurs, so that the change in momentum for this time interval is the
same as that for the short duration of the collision. Thus, we can rewrite the impulse-momentum
theorem as

Now, by Newton’s third law, the average

*x*component of the force exerted by the molecule on the wall is equal in magnitude and opposite in direction:
Consider an ideal gas
undergoing several processes such that the change in temperature

is ∆

*T*=*Tf*-*Ti*for all processes. The temperature change can be achieved by taking
a variety of paths from one
isotherm to another, as shown in Figure 21.3. Because

∆

*T*is the same for each path, the change in internal energy ∆*E*int is the same for all
paths. However, we know from
the first law,

*Q*= ∆*E*int -*W*, that the heat*Q*is different
for each path because

*W*(the negative of the area under the curves) is different for
each path. Thus, the heat
associated with a given change in temperature does

*not*have
where

*CV*is the molar specific heat at constant volume and*CP*is the molar specific
heat at constant pressure. When
we add energy to a gas by heat at constant pressure, not only does the internal
energy of the gas increase, but work is done on the gas because of the change
in volume. Therefore, the heat

*Q*constant*P*must account for both the increase in internal energy and the transfer of energy out of the system by work. For this reason,*Q*constant*P*is greater than*Q*constant*V*for given values of*n*and ∆*T*. Thus,*CP*is greater than*CV*. In the previous section, we found that the temperature of a gas is a measure of the average translational kinetic energy of the gas molecules. This kinetic energy is associated with the motion of the center of mass of each molecule. It does not include the energy associated with the internal motion of the molecule—namely, vibrations and rotations about the center of mass. This should not be surprising because the simple kinetic theory model assumes a structureless molecule. In view of this, let us first consider the simplest case of an ideal monatomic gas, that is, a gas containing one atom per molecule, such as helium, neon, or argon. When energy is added to a monatomic gas in a container of fixed volume, all of the added energy goes into increasing the translational kinetic energy of the atoms. There is no other way to store the energy in a monatomic gas. Therefore, from Equation 21.6, we see that the internal energy*E*int of*N*molecules (or*n*mol) of an ideal monatomic gas is**21.3 Adiabatic Processes for an Ideal Gas**

As we noted in Section 20.6, an
adiabatic process is one in which no energy is transferred by heat between a
system and its surroundings. For example, if a gas is compressed (or expanded)
very rapidly, very little energy is transferred out of (or into) the system by
heat, and so the process is nearly adiabatic. Such processes occur in the cycle
of a gasoline engine, which we discuss in detail in the next chapter. Another
example of an adiabatic process is the very slow expansion of a gas that is
thermally insulated from its surroundings. Suppose that an ideal gas undergoes
an adiabatic expansion. At any time during the process, we assume that the gas
is in an equilibrium state, so that the equation of state

*PV*=*nRT*is valid. As we show below, the pressure and volume of an ideal gas at any time during an adiabatic process are related by the expression
where γ=

*CP*/*CV*is assumed to be constant during the process. Thus, we see that all
three variables in the ideal
gas law—

*P*,*V*, and*T*—change during an adiabatic process.**Proof That PV**γ =

**Constant for an Adiabatic Process**

When a gas is compressed
adiabatically in a thermally insulated cylinder, no energy is

transferred by heat between the
gas and its surroundings; thus,

*Q*= 0. Let us imagine
an infinitesimal change in
volume

*dV*and an accompanying infinitesimal change in
temperature

*dT*. The work done on the gas is -*P dV*. Because the internal energy of
an ideal gas depends only on
temperature, the change in the internal energy in an

adiabatic process is the same
as that for an isovolumetric process between the same

temperatures,

*dE*int =*nCV dT*(Eq. 21.12). Hence, the first law of thermodynamics,
∆

*E*int =*Q*+*W*, with*Q*= 0 becomes
Taking the total differential
of the equation of state of an ideal gas,

*PV*=*nRT*, we see that
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