This is a statistical treatment of the large ensemble of molecules that make up a gas. We had expressed the ideal gas law as: pv = nrt (1)

Transcription

1 1. Kinetic Theory of Gases This is a statistical treatment of the large ensemble of molecules that make up a gas. We had expressed the ideal gas law as: pv = nrt (1) where n is the number of moles. We can also express it as: pv = NkT () where N is the number of molecules and k is Boltzmann s constant k = R/N A J/K. A volume of air the size of a birthday balloon contains some 10 3 molecules. The very large number of molecules allows us to treat gases using statistics, in which averages of quantities such as speed appear rather than their values for individual molecules.. Pressure and Molecular Motion First, we know that a gas exerts a pressure, and we must clearly understand what this is due to. If our ears were a few times more sensitive, we would hear a perpetual rushing noise. Evolution has not developed the ear to that point, because it would be useless if it were so much more sensitivewe would hear a perpetual racket. The reason is that the eardrum is in contact with the air, and air is a lot of molecules in perpetual motion and these bang against the eardrums. In banging against the eardrums they make an irregular tattooboom, boom, boomwhich we do not hear because the atoms are so small, and the sensitivity of the ear is not quite enough to notice it. The result of this perpetual bombardment is to push the drum away, but of course there is an equal perpetual bombardment of atoms on the other side of the eardrum, so the net force on it is zero. If we were to take the air away from one side, or change the relative amounts of air on the two sides, the eardrum would then be pushed one way or the other, because the amount of bombardment on one side would be greater than on the other. We sometimes feel this uncomfortable effect when we go up too fast in an elevator or an airplane, especially if we also have a bad cold (when we have a cold, inflammation closes the tube which connects the air on the inside of the eardrum with the outside air 1

2 through the throat, so that the two pressures cannot readily equalize). (Taken from Feynman Lectures) Consider a container of dilute gas that consists of N independent molecules each of mass m. Pressure is due to molecular collisions with the walls of the container. We need to calculate the average rate of momentum transfer. Suppose we have a container in thermal equilibrium at temperature T with volume V. The molecules are moving in random directions and the average velocity of the molecules is zero (as many are moving in one direction as the other). v = 0 v x = v y = v z (3) While the average velocity is zero, the average speed is not zero. We will concentrate on v which is the root mean square (rms) speed. As a very simple example let us take helium gas in which all the molecules are single atoms, for which we may suppose that there is no internal motion in the atom. If we had a complex molecule, there might be some internal motion. For monoatomic gases, the kinetic energy is the total energy. The total energy (or internal energy) is denoted U. The total energy U is equal to the number of atoms times the average kinetic energy of each. The internal energy U consists mainly of the kinetic energies of the molecules so ( ) 1 U = N K = N m v (4) where K is the average kinetic energy per molecule and N is the total number of molecules. The average of a sum of terms is the sum of the averages of those terms so v = v x + v y + v z = v x + v y + v z (5) v x = v y = v z (6) So we can relate the average of the components of the velocity squared to the thermal energy of the gas: v = 3 v x (7)

3 vx = 1 3 v = U 3 mn (8) a. The Origin of Pressure Pressure arises from the multiple collisions the molecules of a gas have with the walls that contain the gas. First we compute the momentum transfer to a wall due to a single collision, and then find the number of molecules that strike the wall per unit time. A molecule colliding elastically with the right-hand wall of a box, only the x-component of velocity changes so the velocity before collision is: the velocity after collision is: The momentum change of the molecule P mol is v i = v x î + v y ĵ + v zˆk (9) v f = v x î + v y ĵ + v zˆk (10) P mol = m v f m v i = mv x î (11) and the momentum transfer to the wall is P = mv x î (1) The number of collisions with the wall that occur per unit time is calculated as follows. Consider the number of molecules that with an x-component velocity v x that strike an area A in a time interval dt is the number of molecules contained in an imaginary cylinder whose base is against the wall of area A and length v x dt. If the number of molecules per unit volume is N/V, the total number of molecules in the cylinder is (N/V (v x dt)a and the number of collisions in time dt will be half this number as 1/ of the molecules are moving to the right and 1/ to the left. N coll = 1 N V (v xdt)a (13) We can multiply this by the individual collision momentum transfer to find the momentum dp x transferred in time dt. 3

4 ( ) 1 N dp x = (mv x ) V (v xdt)a = mvx N Adt (14) V The force exerted on the area A is dp x dt = F x = mvx N V A (15) and the pressure on the wall p is the force per unit area p = F x A = N mv x (16) V Finally, recall that we have employed the average x-component of velocity squared for each molecule. We make this explicit by employing the notation vx and find p = m vx N ( ) U N V = m 3mN V = 3 U V (17) pv = 3 U (18) This is an important derivation, as we have used the microscopic properties of a gas to find a relation between macroscopic thermodynamic variables. For wider generality, let s write pv = U = (γ 1)U, where for a monoatomic 3 gas γ = 5/3. If we have an adiabatic compression. For an adiabatic compression, all the work done goes into changing the internal energy (du = pdα or if we multiply by mass P dv = du). Since U = P V/(γ 1): γ dv V du = P dv + V dp γ 1 (19) P dv P dv + V dp γ 1 (0) γp dv = V dp (1) + dp P = 0 () if we integrate the last equation with γ constant: 4

5 P V γ = C (3) Under adiabatic conditions, where the temperature rises as we compress because no heat is being lost, the pressure times the volume to the γ power is a constant for monoatomic gases! This is the same result as the Poisson s equations, but we arrived through a different pathway! b. The Meaning of Temperature Let us imagine a problem in which we have a box of gas with two different kinds of molecules in it, having masses m1 and m, velocities v1 and v, and so forth. If all of the No. molecules are standing still, that condition is not going to last, because they get kicked by the No. 1 molecules and so pick up speed. If they are all going much faster than the No. 1 molecules, then maybe that will not last eitherthey will pass the energy back to the No. 1 molecules. It can be shown that the average of the kinetic energy of one is the same as the average of the kinetic energy of the other, when they are both in the same gas in the same box in equilibrium. That means that the heavy ones will move slower than the light ones; this is easily shown by experimentation with atoms of different masses in an air trough. Now we would like to go one step further, and say that if we have two different gases separated in a box, they will also have equal average kinetic energy when they have finally come to equilibrium, even though they are not in the same box. The fact that the energy is the sum of the kinetic energy of all the molecules U = 1 M, v, and the fact that energy is related to pressure U = (3/)pV provides a link to temperature. U = 3 nrt = 3 NkT (4) to give the microscopic interpretation of temperature: kt = U 3 N = K (5) 3 The temperature of an ideal gas is a measure of the average kinetic energy of the constituents. Because the number of molecules canceled, T is independent of the amount of gas. 5

6 note: absolute zero in the Kelvin scale is the point at which the pressure drops to zero. In the microscopic view of an ideal gas, the temperature is zero when the average kinetic energy of the ideal gas is zero. Pressure vanished because the molecules no longer mover around and bounce against the walls. c. Specific Heat of the Perfect Gas In the examples above we were dealing with a monoatomic gas. For this type of gas, the total energy is only due to translation (Translatory Kinetic Energy) so < E >=< K >=< 3 kt >. However, when dealing with more complex molecules, there are other types of energy associated with these gases.the energy of the molecules of a gas can be divided into: 1. Translatory kinetic energy. Rotational kinetic energy 3. Energy of vibration of atoms relative to center of mass of whole molecule 4. Mutual potential energy (not relevant for ideal gasses) i. Equipartition The total average energy per molecule depends on how many independent motions a molecule can have. Point mass Energy has three terms propotional to v x,v y and v z Diatomic With rotational inertia about axes x and y we have two new terms in the energy (making it five). Vibration Brings two more terms related to the relative speed of the atomic constituents and separation (making it seven). Every term in the energy expression that is quadrati in an independent dynamical variable designates a degree of freedom. The contribution of each degree of freedom to the average energy of a molecule is kt/. This is the equipartition theorem. E = s kt (6) and E can contain contributions associated with translation, rotation and vibration. 6

7 Translatory Energy U = U t and in specific quantities u = u t = 3 kt = 3 RT. This is exactly the same as the example we had done above. u = 3 RT (7) c v = 3 R (8) c p = R + c v = 5 R (9) γ = c p /c v = 5/3 (30) Helium, Neon Argon, Krypton, Xenon. Rotational Energy For two atoms rigidly united we have two more degrees of freedom u = u t + u r = 5 kt = 5 RT u = 5 RT (31) c v = 5 R (3) c p = R + c v = 7 R (33) γ = c p /c v = 7/5 (34) H, HCl, N, CO, O, NO For three atoms rigidly united we have three more degrees of freedom u = u t + u r = 3kT = 3RT u = 3RT (35) c v = 3R (36) c p = R + c v = 4R (37) γ = c p /c v = 4/3 (38) However, classical theory breaks down for this. Vibration For the temperatures of our atmosphere, we will not deal with this. 7

8 3. The failure of classical physics (Taken from the Feynman Lectures) Experimentally, classical mechanics fails to explain the specific heats of more complex molecules... So, all in all, we might say that we have some difficulty. We might try some force law other than a spring, but it turns out that anything else will only make γ higher. If we include more forms of energy, γ approaches unity more closely, contradicting the facts. All the classical theoretical things that one can think of will only make it worse. The fact is that there are electrons in each atom, and we know from their spectra that there are internal motions; each of the electrons should have at least 1/kT of kinetic energy, and something for the potential energy, so when these are added in, γ gets still smaller. It is ridiculous. It is wrong. The first great paper on the dynamical theory of gases was by Maxwell in On the basis of ideas we have been discussing, he was able accurately to explain a great many known relations, such as Boyles law, the diffusion theory, the viscosity of gases, and things we shall talk about in the next chapter. He listed all these great successes in a final summary, and at the end he said, Finally, by establishing a necessary relation between the motions of translation and rotation (he is talking about the 1/kT theorem) of all particles not spherical, we proved that a system of such particles could not possibly satisfy the known relation between the two specific heats. He is referring to γ (which we shall see later is related to two ways of measuring specific heat), and he says we know we cannot get the right answer. Ten years later, in a lecture, he said, I have now put before you what I consider to be the greatest difficulty yet encountered by the molecular theory. These words represent the first discovery that the laws of classical physics were wrong. This was the first indication that there was something fundamentally impossible, because a rigorously proved theorem did not agree with experiment. About 1890, Jeans was to talk about this puzzle again. One often hears it said that physicists at the latter part of the nineteenth century thought they knew all the significant physical laws and that all they had to do was to calculate more decimal places. Someone may have said that once, and others copied it. But a thorough reading of the literature of the time shows they were all worrying about something. Jeans said about this puzzle that it is a very mysterious phenomenon, and it seems as though as the temperature falls, certain kinds of motions freeze out. If we could assume that the vibrational motion, say, did not exist at low temperature and did exist at high temperature, then we could imagine that a gas might exist at a temperature sufficiently low that vibrational motion does not occur, so 8

9 γ=1.40, or a higher temperature at which it begins to come in, so? falls. The same might be argued for the rotation. If we can eliminate the rotation, say it freezes out at sufficiently low temperature, then we can understand the fact that the γ of hydrogen approaches 1.66 as we go down in temperature. How can we understand such a phenomenon? Of course that these motions freeze out cannot be understood by classical mechanics. It was only understood when quantum mechanics was discovered. We will not cover quantum mechanics in this class. 4. Collisions and Transport Phenomena Molecules follow a tortuous path in their container, colliding with one another and the walls. At standard temperature and pressure an air molecule undergoes billions of collisions per second, but the average distance a molecule travels between collisions is a statistical quantity that can be calculated. By these collisions, molecules can carry thermal energy, odor, etc. The movement of molecules by random collisions is called diffusion. Molecules of diameter D collide when the path of the center of one molecule lies within an area πd presented by the second molecule. The area is the collision cross section σ. σ = πd. A molecule that moves with speed v and sweeps out an area σ sweeps out a volume V = σd = σvt in a time t. However, the molecule is not alone and encounters N collisions, but even if the path is bent, the volume V remains unchanged. There are on average a collision every τ = t/n seconds. τ is the mean collision time. Averaging over many molecules we replace v with v rms. τ = t N = t nv = t nσv rms t = 1 (39) nσv rms where n = N/V is the number density (not the number of moles). A more precise calculation takes into account the fact that the other molecules are also moving τ = 1 nσvrms (40) The average distance that a molecule travels before it is involved in a collision is the mean free path λ where 9

10 λ = τv rms = 1 nσ (41) λ is inversely proportional to both the density and the collision cross section. 4a. Random Walk and Diffusion Molecules move through a gas by diffusion. The average distance moved by a molecule is similar to the random walk problem (or drunkard s walk). In this problem, a drunkard starts at a lamppost and takes steps that are equal in length but random in direction. We can find the average displacement of the molecule after N steps as follows. Let the successive displacements be L 1, L... L n all of random directions but magnitude L. After N steps the displacement of the molecule is R N = L 1 + L L n. Squaring this quantity we have: R N = L 1 + L L N + L 1 L + L 1 L L N 1 L N (4) In the averaging process, all the dot products have an average value over time of zero, whereas the squared terms are equal to L. R N = NL (43) Connecting this with the properties of a gas. L = λ, after N steps the time is t = τn r = t τ λ (44) It is typical in random walk problems that the displacement squared is linear in time or that the displacement is proportional to the square root of the time. 10