Good morning. In this video, we're going to start
our discussion of the transport properties
of ideal gases. But first, we have to
lay some groundwork, and we're going to do this
in a very simple fashion. Much of this follows concerning in Krueger's
Physical Gas Dynamics course that I took from Charles Krueger, when I was in graduate school. So we're going to start out with this molecular picture
of dynamic behavior. So consider a gas in a
state of equilibrium, inside a cubical box. We know that the molecules
are zooming around, banging, perhaps into each other, and with the walls. But in this first
little instance, we're just going to assume
that the molecules do not collide with each other and the collisions with the
wall take place specularly, which means that energy is completely conserved and
with speed unchanged. These assumptions are untrue
but for an ideal gas, it doesn't really matter. So given these assumptions, the situation as
is in this figure. So we're just following a
molecule that has a speed C regardless of its direction
and it zooms across the box, bounces into the wall, reflects off in an equal but
opposite angle and someone. When it collides with the wall, its velocity perpendicular
to the wall is reversed, but it's speed is unchanged. The magnitudes of
the components have fixed values C_1, C_2, and C_3. By the way, it is typical
to use the term C for speed or velocity
of a molecule. At the macroscopic level, we usually use the term U or V, for speed or velocity and we talk about the
velocity components U_ x, U_ y, U_ z etc. Anyway, so the magnitude of the components
have fixed values, and there are three components. That would be C_1, C_2, C_3, for the three
coordinate directions. The sum of the squares
of the three components, is equal to the square
of the absolute speed. So we can do a little exercise
by noting that the force on a given wall depends on the number of
collisions per unit time, and the change in
momentum per collision. If l is the length of the box, the time to travel
between walls is, say in the one direction, l over the absolute value of C_1. Since there are two traverses per collision with one wall, the number of collisions
per unit time is absolute value of
C_1 divided by 2l. The change in the
x_1 momentum per collision is 2m times the
absolute value of C_1. The magnitude of
the total change in x_1 momentum per unit time, which is the force exerted
on the wall by the molecule, is the change in momentum
times the collision frequency, and that turns out to
be mC squared over l. Since the area of
the wall is l squared, then the pressure on the
wall is mmC_1 squared, divided by l cubed or mC_1
squared divided by the volume. Now, the gas mixture is
composed of a large number of molecules of different mass, m_a, m_b and so on and
speed C_a, C_b and so on. So that the total pressure
exerted on the wall is equal to 1 over the volume times a
sum for a given direction, for a given wall of m_zC_z
squared in that one direction. Since the gas is in equilibrium, this will also be pressure
on all the other walls. Taking the sum and
dividing by three, we get that the pressure is one-third or 1 over
3 times the volume, times the sum of
the mC_z squares. Since the total energy of translation of the
molecules is one-half, mC squared sum of all
of the molecules, we can write that pV is equal to two-thirds E translation. If we consider the
perfect gas law, where n is the number of moles. We know that pV equals NRT where R is the
universal gas constant. If we compare the two, we get E translation is
equal to three-halves NR universal T. We'll get to
that in this next slide. R_u is the universal
gas constant per mole, 8.3144621 joule per
mole degree Kelvin. The gas constant per molecule
is Boltzmann's constant, which is 1.38 times 10 minus
23 joules per degree Kelvin. So that the average kinetic
energy per mole or molecule, is either three-halves times
the universal gas constant, times T or three-halves kT. Now, we've ignored any
internal molecular structure, so translation provides the entire internal
energy of the gas. Therefore, the specific heat is equal to the partial
of v with T and that turns out to be
three-halves times the universal gas
constant in molar units. For an ideal gas,
C_p minus C_v is R, and C_ p is five-halves
times the gas constant. So the ratio of specific heats
is Gamma is five-thirds. So we can estimate
molecular speeds, the mean square speed is C squared bar is equal to the sum of mC squared over the
sum of the masses. Noting that the density is
mass divided by volume, then P over Rho is equal to
one-third C squared bar, P over Rho is also equal to RT. So we can get C
squared from this and it turns out that the root mean square of the molecular
speed for air is about 506.5 meters per second. So now let's talk
about collisions, collisions between molecules
which we ignored so far. The average spacing
is going to be equal to the cube root divided by one over the number density. However, the average distance
between collisions can be quite large or it can
be larger in an event. The average distance is
called the mean free path. So first of all, consider the
following definition, the sphere of influence. So if we have two
molecules that touch each other and they're the same size, then the center to center spacing is equal to the
diameter of the molecules. But another way to look at it is, if we take twice the diameter
that forms a sphere, and if another molecule's
center enters that sphere, then a collision can be
said to have taken place. In other words, it's
entered the sphere of influence of the molecule. Now, the actual path followed by molecules is quite random. So we can do a simple back of the envelope calculation by
straightening out the path, like just some of the
molecules are going straight. Let's assume that the speed of the molecules is equal to
the average molecular speed. Then the sphere of
influence sweeps out a cylindrical volume
per unit time of Pi d squared times the
velocity, times the speed. So the number of collisions per unit time turns out to be
equal to Pi d squared, C-bar times the number density. So that's a number of
collisions per unit time, and we call that Theta. So the average
distance traveled per collision is going to
be the average speed, divided by Theta of the
collision frequency, and that is we'll call l and it's equal to 1 over Pi d squared
times the number density. Now it turns out
that better theory, collision theory, throws
in a square root of two, so that the mean
free path is 1 over the square root of 2
Pi d squared times n, and the collision frequency
is the square root of 2Pi d squared C-bar
n. The collision, the quantity Sigma equals Pi d squared can be thought of as a cross section
for the collision. So that we can write the mean free path is 1
over the square root of 2 Sigma n or Theta is
equal to the square root 2 Sigma C-bar n. Now we can calculate the ratio of the mean free path to
the average spacing, that is going to be equal
to the cube root of the number density divided by 2Pi square root of 2Pi d squared, and you run the numbers. It's 1 over the square root of 2Pi times the average spacing
divided by the diameter. So here are some numbers. These numbers happened
to be for hydrogen. The number density at standard temperature
and pressure from the ideal gas law is 2.7 times 10 to the 24 per meter cube. The collision diameter
is 2.7 angstroms. So the average spacing between molecules is 34.45 angstroms. Angstrom is 10 to
the minus 10 meters. But the mean free path on the other hand is 1262 angstrom. In other words, the ratio
of the mean free path to the average spacing is 36.6. A way to think about this is you're walking on a sidewalk. There's a fair number of people, but you don't happen
to run into any of them because you can fit through
the spaces between them, and that's exactly
what happens on average in a colliding
gas mixture. So here are some other numbers. The average speed for hydrogen
is 1782 meters per second. This is a very light molecule
hence a higher average speed and the collision frequency is 1.4 times 10 to
the 10 per second. Collision frequencies
are typically quite large on the order of 10
to the 10 per second. So that's the end of this video. Thanks for listening
and have a great day.