 We can quantify how fast gas molecules move by their root mean square or RMS velocity. The internal energy for one mole of gas molecules, roughly 22 liters of gas at room temperature, is u equals Avogadro's number times the average kinetic energy of a molecule, one-half m times the mean square velocity. This equals the molar heat capacity, three-halves Avogadro's number times Boltzmann's constant, times temperature T. Avogadro's number times molecular mass is the molar mass, M, and Avogadro's number times Boltzmann's constant is the gas constant R. Substituting these and multiplying the previous equation by two, we get Big M times mean square velocity equals 3RT. Solving for the square root of the mean square velocity and calling this the root mean square velocity, we obtain VRMS equals square root 3RT over Big M. The molar mass of argon is about 40 grams or .04 kilograms. Let's take a temperature of 300 Kelvin, about 27 degrees Celsius or about 80 degrees Fahrenheit, and a nice warm day. Then the RMS velocity of atmospheric argon is 433 meters per second. For comparison, the speed of sound and air at this temperature is 347 meters per second. Air molecules are moving very fast. This presents us with a fascinating picture. When exposed to warm air, our skin is being bombarded with countless molecules traveling at supersonic speeds, yet we don't perceive this bombardment. If the molecules are equally likely to be moving in any direction, our experience will be of nice, warm, still air, and our local wind turbine will sit motionless. Yet if the molecules have a common macroscopic velocity that is only, say, about 1% of their microscopic velocity, we will feel a breeze and our local wind turbine will turn and generate electricity. Kinetic energy varies as the square of velocity, so the microscopic kinetic energy due to the thermal motion of the molecules is on the order of 10,000 times the macroscopic kinetic energy of the wind. Yet our wind turbine converts only the tiny macroscopic component into electricity, and the macroscopic component is only there on breezy days, while the microscopic component is always there. So why don't we build a machine that extracts the vastly greater and always present microscopic kinetic energy of the molecules? The answer is because it's not possible, and the reason why has to do with the second law of thermodynamics, which we'll consider in a future video. Another important characteristic of the microscopic motion is how far, on average, molecules travel between collisions. We call this the mean-free path. Here's how we can get a rough estimate of this. Assume molecules have diameter D. In fact, molecules are not hard spheres with a well-defined surface, but we can take D to be an effective diameter. As a molecule travels through space, it sweeps out a cylinder of diameter 2D and length L. The diameter is 2D because if the center of another molecule lies on this cylinder, the two will just collide. From the gas law we can solve for V over N, the volume per molecule, as kT over P. We set this equal to the volume of the cylinder, L pi D squared, so that there is, crudely, a 50-50 chance of the molecule encountering a second molecule as it sweeps out this volume. Solving for L, we get kT over pi D squared P. Plugging in values, T equals 273.15 Kelvin, which is zero degrees Celsius, standard atmospheric pressure, and an effective diameter of 3 times 10 to the minus 10 meters, or 3 angstroms. We find a mean-free path of a bit more than 10 to the minus 7 meters, or 1 tenth of a micron. Having characterized some of the average microscopic properties of the gas molecules, let's consider the state of the gas system as a whole. We can view this at the molecular level, in which case we talk about the microstate of the system, or at a larger-scale level, in which case we talk about the macrostate of the system. The dynamical state of a single particle is characterized by its position and velocity at a given time. For a system of N particles, we need to specify the position and velocity of each particle. This provides a complete description of the microstate of the system. Then, assuming we have a model for how the particles interact with each other and the walls of the gas container, we can calculate the force on the Ith particle as a function of the positions and velocities of all the particles. By Newton's second law of motion, this gives us the Ith particle's mass times acceleration, denoted by ui dot. Acceleration is the rate of change of velocity, so multiplying it by a small time increment dt gives us the change of velocity, dui. And velocity is the rate of change of position, so multiplying it by dt gives us the change of position, dxi. We can use these increments to update all particle positions and velocities and then repeat the process indefinitely to predict the system's microstate at any time in the future or past. This is the basic recipe of Newtonian mechanics, and it works, in theory. In Newtonian practice, the obvious practical problem is that a macroscopic volume of gas can contain millions of billions of molecules. The bookkeeping and computations required to describe the evolution of so many particles is simply intractable. And, as we'll see shortly, there's a more fundamental problem when trying to track the microstate of a gas through time. Yet, the macrostate of a gas, characterized by quantities such as pressure, volume, and temperature, is describable, even though it arises from the indescribable microstate of its vast number of molecules. Now we come to the more fundamental problem with describing the microstate of a gas, chaotic motion. The motion of colliding balls is chaotic. Specifically, a small change in the initial state grows exponentially in time. Let's explore this with a simple model. Two blue balls are fixed in place. A red ball moves between them, bouncing back and forth. Can we predict the motion of this ball arbitrarily far into the future? To represent gas molecules, we'll use some of the order of magnitude values we've developed so far. The balls are about three angstroms in diameter, about 1,000 angstroms apart, and the red ball travels at about 400 meters per second. The time between collisions is the distance between the fixed balls divided by the speed of the red ball, about 2.5 times 10 to the minus 10 seconds. The inverse of this, about 4 billion, is the number of collisions per second. An atmospheric gas molecule, under typical conditions, undergoes billions of collisions per second. Here is a numerical solution of the dynamics of this system. If the centers of the three balls fall exactly on a common line, and if the velocity of the moving ball is exactly parallel to this line, it will bounce back and forth forever. Now, let's consider what happens if the red ball's velocity is tilted by an angle theta zero. As it travels a distance L, it will drift vertically, so its next collision with a blue ball will not be head-on, but will have a vertical offset. For theta zero very small, the ball's centers will be offset horizontally by the diameter D, and vertically by H equals L times theta zero. The red ball's direction of motion will make an oblique angle, phi, with the line through the ball's centers. When it bounces off the blue ball, it will travel at that same angle, but in the opposite direction relative to the center line. We can determine the angle phi from the top figure as approximately H over D, which is approximately L over D times theta zero. After one collision, the red ball will be traveling at an angle theta one, which is approximately two phi, or two L over D times theta zero. The collision has amplified the initial angle theta zero by a factor two L over D. The distance L is much larger than the diameter D, so two L over D is a larger number. The red ball will now travel at an angle theta one. The next collision will amplify that by a factor of two L over D. After n collisions, it will travel at an angle theta n equals two L over D to the nth power times the initial angle theta zero. The angle grows exponentially with the number of collisions. Eventually, the red ball will fly off into space and no longer collide with the blue balls. This will happen if the vertical offset H is equal to or greater than the diameter D. Let's write D equals the distance L times the angle theta n, which is two L over D to the nth power, times theta zero. We can solve for theta zero as D over L times D over two L to the nth power. This is the initial angle that will cause the red ball to undergo n collisions before flying off into space.