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.
