9702 · 15.3Kinetic theory of gases

Kinetic theory assumptions

0:00Right now, the air around you is hitting your face. Not wind, just still air. Its molecules are flying around at roughly five hundred metres per second, which is faster than the speed of sound, and billions upon billions of them bounce off every square centimetre of your skin every instant. Each single collision is unbelievably tiny. But add them all up and you get atmospheric pressure, about a hundred thousand pascals.

0:21That's the same as the weight of roughly ten tonnes pressing on every square metre. In the last video, we used p V equals N k T as a rule that gases obey. In this video, we'll find out where pressure actually comes from. Starting from a simple model of molecules bouncing around in a box, and using only Newton's laws, we'll derive an equation for the pressure of a gas.

0:40Then we'll compare it with p V equals N k T, and discover what temperature really means. We need just a few ideas from earlier in the course. Momentum is mass times velocity, and it's a vector, so a molecule moving left has the opposite sign of momentum to one moving right. Newton's second law says force equals the rate of change of momentum. Pressure is force per unit area.

1:02And from the last video: p V equals N k T, where N is the number of molecules and k is the Boltzmann constant. Real gases are messy. To make the physics possible, the kinetic theory uses a simplified model of a gas, built on a set of assumptions. You need to be able to state these, so let's see why each one is there. Assumption one: a gas contains a very large number of molecules.

1:24That lets us take reliable averages, even though each molecule on its own behaves unpredictably. Assumption two: the molecules are in constant, random motion. They move in all directions, with a range of speeds, and no direction is preferred over any other. Assumption three: all collisions, between molecules and with the walls of the container, are perfectly elastic. That means no kinetic energy is lost in collisions. Otherwise, a sealed, insulated gas would slowly cool down by itself.

1:52Assumption four: the volume of the molecules themselves is negligible compared with the volume of the container. In other words, the molecules are treated as tiny points, and the gas is almost entirely empty space. Assumption five: there are no forces between the molecules, except during collisions. Between collisions, each molecule moves in a straight line at constant speed. Assumption six: the time a collision lasts is negligible compared with the time between collisions.

2:18A gas that obeys all of these assumptions is exactly the ideal gas from the last video. Real gases get close to this at low pressures and reasonably high temperatures, when the molecules are far apart and moving fast.

Deriving the kinetic theory equation

2:30Before any equations, here's the physical picture. A molecule flies towards a wall, hits it, and bounces back. Its velocity has reversed, so its momentum has changed. By Newton's second law, something must have exerted a force on the molecule to change its momentum, and that something is the wall. By Newton's third law, the molecule exerts an equal and opposite force on the wall. One molecule gives the wall a tiny kick.

2:52But an enormous number of molecules are hitting the wall all the time, and those tiny kicks add up to a steady force spread over its area. Force per unit area is pressure. Now let's turn that picture into an equation, starting with one molecule moving in one direction, and building up to a whole gas in three dimensions. Step one: set up the model. Picture a cube-shaped box with sides of length L.

3:14Inside, there's a single molecule of mass m. To start with, let's say it's moving straight along the x-direction, parallel to one edge of the box, with speed u. It's heading towards the right-hand wall, which we'll call wall A. Step two: the change of momentum in one collision. Before hitting wall A, the molecule's momentum is m u. The collision is elastic, so it bounces straight back with the same speed, u, but now in the opposite direction.

3:38Its momentum after the collision is minus m u. The change in momentum is final minus initial: minus m u, minus m u, which is minus two m u. So the wall receives two m u of momentum, outwards. It's two m u, not m u, because the molecule doesn't just stop. It stops and then goes back the other way. Step three: the time between collisions with wall A.

4:00After bouncing off wall A, the molecule travels across the box to the opposite wall, bounces off that, and comes back to wall A. That's a distance of L there and L back, so two L. At speed u, the time taken is distance divided by speed: two L over u. Step four: the average force from one molecule. Force is the rate of change of momentum. The molecule gives wall A two m u of momentum, once every two L over u seconds.

4:26So the average force on the wall is two m u divided by two L over u. Dividing by a fraction means multiplying by its flip, so that's two m u times u over two L. The twos cancel, and we get m u squared over L. Step five: the pressure from one molecule. Wall A is a square of side L, so its area is L squared. Pressure is force divided by area: m u squared over L, divided by L squared, which is m u squared over L cubed.

4:52And L cubed is just the volume of the box, V. So the pressure from one molecule is m u squared over V. Step six: add up all N molecules. A real gas has N molecules, not one, and they don't all have the same speed. So we add up the contribution of each one. Molecule one has x-velocity u one, molecule two has u two, and so on. The total pressure is m over V, times u one squared plus u two squared plus all the way up to u N squared.

5:20That long sum is just N times the average value of u squared, which we write as c x squared inside angle brackets. The angle brackets mean "the average of". So the pressure becomes N m times the mean of c x squared, all over V. What about molecules colliding with each other on the way? Those elastic collisions just shuffle velocities between molecules. They don't change the overall average, so the result still holds.

5:43Step seven: extend to three dimensions. So far, we've pretended the molecules only move along x. Really, each molecule moves in some random direction in three dimensions. Its velocity has three components: c x, c y and c z. By Pythagoras in three dimensions, the square of the speed is the sum of the squares of the components: c squared equals c x squared plus c y squared plus c z squared.

6:06Averaging over all the molecules, the mean of c squared equals the mean of c x squared, plus the mean of c y squared, plus the mean of c z squared. Now here's where the random-motion assumption does its job. Because the motion is random, there's no preferred direction. The molecules are just as likely to move along x as along y or z. So the three averages must be equal.

6:27If three equal things add up to the mean of c squared, each one is a third of it. So the mean of c x squared equals one third of the mean of c squared. Step eight: put it together. Replace the mean of c x squared with one third of the mean of c squared. The pressure becomes one third N m times the mean of c squared, over V.

6:47Multiply both sides by V, and we get the result: p V equals one third N m times the mean-square speed. Here, N is the number of molecules, m is the mass of one molecule, and the mean of c squared is the mean-square speed. So N times m is the total mass of gas. It makes sense. More molecules means more collisions, so more pressure. Faster molecules hit harder and more often, so more pressure.

7:10And the one third appears because only one of the three directions pushes on any particular wall. We used a cube, but the result holds for a container of any shape.

Mean-square and r.m.s. speed

7:19That equation contains a strange-looking quantity: the mean-square speed. Why not just use the average speed? Because squares are what came out of the physics. The force from each molecule depended on u squared: one u from the momentum change, and one u from how often it hits the wall. So the quantity that naturally appears is the average of the squares. Let's see exactly what mean-square means with a tiny example.

7:43Imagine just three molecules, with speeds of two, four and six units. The ordinary mean speed is two plus four plus six, which is twelve, divided by three. That's four. The mean-square speed means: square each speed first, then take the mean. Two squared is four, four squared is sixteen, six squared is thirty-six. Add them to get fifty-six, and divide by three. The mean-square speed is fifty-six over three, which is about eighteen point seven.

8:11The mean-square speed isn't a speed. Its units are metres squared per second squared. To get back to a speed, we take the square root. That gives the root-mean-square speed, or r.m.s. speed, written c r.m.s. The name tells you the order of operations, read from right to left. Square each speed, take the mean, then take the root. For our three molecules, the r.m.s. speed is the square root of fifty-six over three, which is about four point three two units.

8:38Notice that the r.m.s. speed is a bit bigger than the ordinary mean of four, because squaring gives extra weight to the faster molecules. And notice that the square of the mean speed is sixteen, while the mean of the squares is about eighteen point seven. They are not the same thing.

Temperature and molecular kinetic energy

8:52Now for the payoff. We have two equations for the same quantity, p V. From the kinetic theory: p V equals one third N m times the mean of c squared. From experiment, in the last video: p V equals N k T. If both describe the same ideal gas, the right-hand sides must be equal. One third N m times the mean of c squared equals N k T.

9:15N appears on both sides, so it cancels. Multiply both sides by three, and m times the mean of c squared equals three k T. Now, we want kinetic energy, which is a half m v squared. So divide both sides by two. A half m times the mean of c squared equals three-halves k T. The left-hand side, a half m times the mean-square speed, is the average translational kinetic energy of one molecule.

9:38Translational means the energy of moving from place to place, as opposed to spinning or vibrating. So the average translational kinetic energy of a molecule is three-halves k T. This is a remarkable result. It says that thermodynamic temperature is a direct measure of the average kinetic energy of the molecules. Double the temperature in kelvin, and you double the average kinetic energy. And look what's missing from three-halves k T.

10:01There's no mass. So at the same temperature, every ideal gas has the same average kinetic energy per molecule, whether it's light hydrogen or heavy xenon. The heavier molecules make up for it by moving more slowly. Rearranging also gives a useful formula for the r.m.s. speed: the mean of c squared equals three k T over m, so c r.m.s. equals the square root of three k T over m.

10:22This result, three-halves k T, is exactly what you'll use in the next video, on internal energy.

Worked examples with gas molecules

10:28Let's use p V equals one third N m times the mean of c squared directly. A container of volume 0.10 cubic metres holds 0.12 kilograms of gas at a pressure of 1.0 times ten to the five pascals. Find the r.m.s. speed of the gas molecules. The key is to spot that N times m is the total mass of the gas, 0.12 kilograms. We don't need to know N or m separately.

10:51Rearranging, the mean of c squared equals three p V over N m. That's three, times 1.0 times ten to the five, times 0.10, divided by 0.12. Three times ten to the five times 0.10 is thirty thousand. Divided by 0.12, that's two hundred and fifty thousand, or 2.5 times ten to the five metres squared per second squared. Take the square root: the r.m.s. speed is five hundred metres per second.

11:18Now let's check the claim from the start of the video. Air is mostly nitrogen. Find the average translational kinetic energy and the r.m.s. speed of nitrogen molecules at 300 kelvin. The molar mass of nitrogen gas is 0.028 kilograms per mole. Use the Boltzmann constant, 1.38 times ten to the minus twenty-three joules per kelvin, and the Avogadro constant, 6.02 times ten to the twenty-three per mole. Step one: the average kinetic energy.

11:46This only needs the temperature. Three-halves k T is 1.5, times 1.38 times ten to the minus twenty-three, times 300. That's 6.21 times ten to the minus twenty-one joules. That's the average kinetic energy of one molecule, which is why it's so tiny. Step two: the mass of one molecule. This is the step people get wrong. The m in our equations is the mass of a single molecule, in kilograms.

12:11The molar mass is the mass of one mole, which is the Avogadro constant's worth of molecules. So divide the molar mass by the Avogadro constant. 0.028 divided by 6.02 times ten to the twenty-three is 4.65 times ten to the minus twenty-six kilograms. Step three: the mean-square speed. From a half m times the mean of c squared equals 6.21 times ten to the minus twenty-one, the mean of c squared equals two times 6.21 times ten to the minus twenty-one, divided by 4.65 times ten to the minus twenty-six.

12:43That's 2.67 times ten to the five metres squared per second squared. Step four: take the square root. The r.m.s. speed is about 517 metres per second. So nitrogen molecules at room temperature really do move at around five hundred metres per second. And worked example one had a density of 1.2 kilograms per cubic metre, about the density of air, and gave five hundred metres per second too. One route from pressure, one from temperature, and they agree.

Exam mistakes and question strategies

13:10Let's go through the mistakes that cost the most marks. Mistake one: squaring the mean instead of taking the mean of the squares. With speeds of two, four and six, those gave sixteen and eighteen point seven. Always square each speed first, then average. Mistake two: using the molar mass as m. In p V equals one third N m times the mean of c squared, and in three-halves k T, m is the mass of one molecule in kilograms.

13:36If you're given molar mass in grams per mole, first convert it to kilograms per mole, then divide by the Avogadro constant. Mistake three: in the derivation, writing the momentum change as m u instead of two m u, or the time between collisions as L over u instead of two L over u. The velocity reverses, and the molecule has to go there and back. Mistake four: forgetting to use kelvin.

13:59Three-halves k T only works with thermodynamic temperature. At twenty-seven degrees Celsius, T is three hundred kelvin, not twenty-seven. Mistake five: thinking that doubling the temperature doubles the speed. Doubling the kelvin temperature doubles the mean kinetic energy, and so doubles the mean-square speed. But the r.m.s. speed is a square root, so it only increases by a factor of root two, about 1.41. Mistake six: thinking every molecule has kinetic energy three-halves k T.

14:29It's an average. Some molecules are much faster and some much slower. And mistake seven: saying gas pressure is caused by molecules colliding with each other. Pressure comes from molecules colliding with the walls. This topic is tested in three typical ways. First, you may be asked to state assumptions of the kinetic theory. Use precise wording: say "collisions are perfectly elastic", not "molecules bounce", and "the volume of the molecules is negligible compared with the volume of the container", not "molecules are small".

15:00Second, you may be asked to explain how molecular movement causes pressure. Here's the chain of reasoning a good answer follows. Molecules collide with the walls. Each collision causes a change in momentum of the molecule. A change in momentum requires a force, so the wall exerts a force on the molecule, and by Newton's third law, the molecule exerts an equal and opposite force on the wall. Many molecules collide per second, producing a steady average force.

15:25Pressure is this force per unit area of the wall. Third, a calculation, often with a twist. Here's a typical one. "Helium and nitrogen are at the same temperature. The molar mass of helium is 0.004 kilograms per mole, and the molar mass of nitrogen is 0.028 kilograms per mole. Calculate the ratio of the r.m.s. speed of helium molecules to the r.m.s. speed of nitrogen molecules." Don't reach for your calculator yet.

15:51Same temperature means same average kinetic energy, three-halves k T. So a half m times the mean of c squared is the same for both gases. That means the mean of c squared is inversely proportional to m, and the r.m.s. speed is inversely proportional to the square root of m. The mass of each molecule is proportional to the molar mass. So the ratio of the r.m.s. speeds, helium to nitrogen, is the square root of 0.028 over 0.004, which is the square root of seven.

16:18That's about 2.65. So helium molecules move about 2.65 times faster than nitrogen molecules at the same temperature. At 300 kelvin, that's about 1370 metres per second for helium. Temperature fixes the average kinetic energy, not the speed. Once you see that, the question becomes a single line. Let's bring it all together. The kinetic theory models a gas as a very large number of randomly moving molecules, with elastic collisions, negligible molecular volume, and no forces between molecules except during collisions.

16:50Pressure is caused by molecules colliding with the walls and changing momentum. Following one molecule in a cube, then adding up all N molecules and sharing the motion equally between three directions, we derived p V equals one third N m times the mean-square speed. The mean-square speed is the average of the squared speeds. Its square root is the r.m.s. speed, c r.m.s. Comparing with p V equals N k T shows that the average translational kinetic energy of a molecule is three-halves k T.

17:17Temperature, measured in kelvin, is a direct measure of the average kinetic energy of the molecules.