Page 26 - Physical chemistry eng

1.2 THE MACROSCOPIC VARIABLES VOLUME, PRESSURE, AND TEMPERATURE 3

2
The particle kinetic energy is 1>2 mv such that
1 2 1 2 1 2 1 2
e Tr = mv = mv + mv + mv = e + e + e (1.2)
z
x
y
2 2 2 2 Tr x Tr y Tr z
where the subscript Tr indicates that the energy corresponds to translational motion of
the particle. Furthermore, this equation states that the total translational energy is the
sum of translational energy along each Cartesian dimension.
Pressure arises from the collisions of gas particles with the walls of the container;
therefore, to describe pressure we must consider what occurs when a gas particle col-
lides with the wall. First, we assume that the collisions with the wall are elastic
collisions, meaning that translational energy of the particle is conserved. Although the
collision is elastic, this does not mean that nothing happens. As a result of the collision,
linear momentum is imparted to the wall, which results in pressure. The definition of
pressure is force per unit area and, by Newton’s second law, force is equal to the product
of mass and acceleration. Using these two definitions, the pressure arising from the col-
lision of a single molecule with the wall is expressed as
F ma i m dv i 1 dmv i 1 dp i
P = = = ¢ ≤ = ¢ ≤ = ¢ ≤ (1.3)
A A A dt A dt A dt
In Equation (1.3), F is the force of the collision, A is the area of the wall with which the
particle has collided, m is the mass of the particle, v is the velocity component along
i
the i direction (i = x , y, or z), and p is the particle linear momentum in the i direction.
i
Equation (1.3) illustrates that pressure is related to the change in linear momentum with mv x
respect to time that occurs during a collision. Due to conservation of momentum, any
change in particle linear momentum must result in an equal and opposite change in mv x
momentum of the container wall. A single collision is depicted in Figure 1.2. This fig-
ure illustrates that the particle linear momentum change in the x direction is -2mv x
(note there is no change in momentum in the y or z direction). Given this, a correspon-
ding momentum change of 2mv x must occur for the wall.
The pressure measured at the container wall corresponds to the sum of collisions
involving a large number of particles that occur per unit time. Therefore, the total x
momentum change that gives rise to the pressure is equal to the product of the momen-
FIGURE 1.2
tum change from a single particle collision and the total number of particles that collide
Collision between a gas particle and a wall.
with the wall: Before the collision, the particle has a
¢p momentum of mv x in the x direction, and
¢p = * (number of molecules) (1.4) after the collision the momentum is -mv x .
Total
molecule Therefore, the change in particle momen-
How many molecules strike the side of the container in a given period of time? To tum resulting from the collision is -2mv x .
By conservation of momentum, the change
answer this question, the time over which collisions are counted must be considered.
in momentum of the wall must be 2mv x .
Consider a volume element defined by the area of the wall A times length ¢x as illus-
The incoming and outgoing trajectories are
trated in Figure 1.3. The collisional volume element depicted in Figure 1.3 is given by offset to show the individual momentum
components.
V = A¢x (1.5)
The length of the box ¢x is related to the time period over which collisions will be
counted ¢t and the component of particle velocity parallel to the side of the box (taken
to be the x direction):
¢x = v ¢t (1.6)
x
In this expression, v is for a single particle; however, an average of this quantity
x
will be used when describing the collisions from a collection of particles. Finally, the
number of particles that will collide with the container wall N coll in the time interval ¢t
'
N
is equal to the number density . This quantity is equal to the number of particles in
Area A
the container N divided by the container volume V and multiplied by the collisional vol-
Δx v Δt
x
ume element depicted in Figure 1.3:
FIGURE 1.3
' 1 nN A 1
N coll = N * 1Av x ¢t2¢ ≤ = 1Av x ¢t2¢ ≤ (1.7) Volume element used to determine the num-
2 V 2 ber of collisions with the wall per unit time.