Note 17 The Kinetic Theory of Gases Sections Covered in the Text: Chapter 18 In Note 15 we reviewed macroscopic properties of matter, in particular, temperature and pressure. Here we see how the temperature and pressure of a gas arises from the average movement of its constituent atoms and molecules. This amounts to making a connection between the microscopic and macroscopic worlds. The theory that achieves this is a marvel of classical physics; it is called the kinetic theory of gases. But we shall see that the kinetic theory, based as it is on classical Newtonian physics, is limited in what it can describe. We shall see the need of a new, modern, quantum, physics. In addition, in this note we shall encounter an explanation for the fact that heat always flows spon- taneously from a hot object to a cold object and never in the reverse. This fact is encapsulated in the second law of thermodynamics. Figure 17-1. The distribution of molecular speeds in a vol- A Microscopic View of a Gas ume of nitrogen gas at a temperature of 20 ˚C as deter- The history of studies into the nature of gases is very mined by experiment. long. We know from experiments too numerous to go into here (and the theory we shall present in this note) that an ordinary volume of gas is comprised of Kinetic Theory of Ideal Gases billions of molecules. At normal temperatures the molecules move in all directions at high speeds. We Assumptions infer that the motion of any one molecule, if it could We saw in Note 15 that bulk properties of an ideal gas be followed, would be quite random. At any instant are described exactly by the equation of state: the molecules do not all move at the same speed, but with a range or distribution of speeds, some faster pV = Nk T . …[17-1] some slower. The speed distribution of a typical gas B (N ) at 20 ˚C has been studied in detail; it is drawn in 2 To simplify our task of describing a gas theoretically Figure 17-1. we shall make these additional assumptions about it: The figure indicates that although we cannot state what speed a particular molecule moves at (because, € 1 It consists of many identical monatomic molecules in fact, we cannot follow its movements), we can state of mass m. what is its most likely or probable speed. You can see 2 Its molecules move randomly within the gas and from the figure that 20% of the molecules have a collide elastically with the walls of the container. speed in the range 500 – 600 m.s–1. Some, of the order of 0.5%, are moving very slowly, in the range 0 – 100 With these assumptions we begin with the macro- m.s–1. Others, of the order of 0.5%, are moving very scopic property of pressure. quickly, in the range 1100-1200 m.s–1. We shall see in this note that we can take the most probable speed as the average speed, to a good approximation. 1 1 Figure 17-1 bears a startling resemblance to the distribution of term test marks from a large class of PHYA10H3S students ! 17-1 Note 17 Kinetic Theory of Pressure = FavgΔtcoll = 2mvx Consider a cubic container of side L and volume V of a monatomic gas. Suppose that at an instant of clock using eq[17-2]. Thus time a molecule is moving with instantaneous speed v in the positive x direction toward the right wall x € 2mvx (Figure 17-2). The molecule collides with the wall and Favg = . …[17-3] rebounds. Δtcoll If we multiply F avg by the total number of collisions Ncoll that occur during the time interval ∆tcoll we get the net force€ exerted on the wall by all the molecules: Ncoll Fnet = Ncoll Favg = 2mvx . …[17-4] Δtcoll At the moment we do not have an expression for Ncoll. To find it we assume temporarily that all molecules € have the same x-component of velocity vx. These Figure 17-2. A molecule of gas collides with the right wall of molecules all travel a distance ∆x = vx∆tcoll along the x- the container and rebounds. axis during the elapsed time ∆tcoll. They are all con- tained in the shaded region of the container redrawn in Figure 17-3. In the collision the x-component of the molecule’s velocity is changed from +vx to –vx. This means that the molecule experiences an impulse of 2 (Jx )molecule = Δp = m(−vx ) − mvx = −2mvx . According to Newton’s third law the wall experiences the same magnitude of impulse but of opposite sign € (Jx )wall = −(Jx )molecule = +2mvx . …[17-2] From the definition of impulse, € Δtcoll Jx = ∫ Fx (t)dt , 0 where Fx(t) is the x-component of the force the mole- Figure 17-3. Half of the molecules contained in the shaded cule exerts on the wall during the collision and ∆tcoll is region will collide with the wall in the time interval ∆tcoll. the elapsed time for the collision. The details of the function€ Fx(t) are unknown to us. So to simplify matters (and to enable us to obtain a result at all), we On average half of the molecules are moving to the replace the actual unknown collision force with an left in the container, half to the right. The shaded average force Favg. Then region has volume A∆x where A is the surface area of the wall. Hence Δtcoll (Jx ) = Favgdt 1 N 1 N wall ∫ N = AΔx = Av Δt . …[17-5] 0 coll 2 V 2 V x coll From eq[17-5] the rate of collisions with the wall is 2 Recall that we defined impulse in Note 07. Impulse is the change in€ linear momentum. € 17-2 Note 17 N 1 N 2 1 2 rate of collisions = coll = Av . …[17-6] v = v . …[17-14] x ( x )avg rms Δtcoll 2 V 3 Substituting eq[17-6] into eq[17-4] the net force Using eq[17-14] in eq[17-8] the net force becomes becomes € € 1 N 2 Fnet = mvrmsA . 1 N N 2 3 V Fnet = (2mvx ) Avx = mvx A . …[17-7] 2 V V This means that the pressure on the wall is Recall that eq[17-7] is based on our earlier assumption that the molecules have the same speed. To make the F 1 N equation more accurate we now replace v 2 with its € net 2 x p = = mvrms. …[17-15] € average value and obtain: A 3 V N 2 Thus the theory predicts that the pressure p (a macro- F = m v A . …[17-8] net V ( x )avg scopic quantity) is proportional to the square of the average€ speed of the molecules (a microscopic quantity). Now let us turn to its counterpart Now, in general, a molecule can be moving in any expression for temperature. direction. Its speed is the magnitude of the velocity vector€ in 3D space: Kinetic Theory of Temperature 2 2 2 A molecule of mass m and velocity v has translational v = vx + vy + vz . …[17-9] kinetic energy 1 2 ε = mv . Squaring eq[17-9] and averaging we have 2 2 2 2 2 € v = v + v + v , Thus the average translational kinetic energy is ( )avg ( x y z )avg 2 2 2 € 1 2 1 2 = v + v + v . …[17-10] εavg = m v = mvrms , …[17-16] ( x )avg ( y )avg ( z )avg 2 ( )avg 2 € 2 Taking the square root of (v )avg we obtain what is using eq[17-11]. From eq[17-15] we can write called the root-mean-square speed vrms: € € 2 N 1 2 2 N 2 p = mvrms = εavg , …[17-17] vrms = v . …[17-11] 3 V 2 3 V ( ) avg Now the averages of each component squared are using eq[17-16]. We can rearrange eq[17-17] to get equal, i.e., 2 2 2 € 2 € v = v = v . …[17-12] pV = Nε . …[17-18] ( x )avg ( y )avg ( z )avg 3 avg 2 2 Hence v = v From eqs[17-1] and [17-18] we can write rms ( )avg € 2 2 2 2 € = v + v + v pV = NkBT = Nεavg , ( x )avg ( y )avg ( z )avg 3 € 2 = 3 v , …[17-13] from which it follows that ( x )avg € € 3 using eqs[17-10] and [17-12]. It follows therefore from εavg = kBT . …[17-19] eq[17-13] that 2 € 17-3 € Note 17 Turning eq[17-19] around we get an expression for Calculations of Thermal Energy and temperature: 2 Specific Heat T = εavg . …[17-20] 3kB Monatomic Ideal Gas (3 DFs) Recall that the thermal energy of a system is the sum Thus the temperature T (a macroscopic quantity) is of the kinetic and potential energies of its constituent proportional to εavg, the average translational kinetic particles: Eth = Kmicro + Umicro. In a monatomic gas such energy of€ the molecules (a microscopic quantity). as helium or neon Umicro = 0 and the kinetic energy is From eqs[17-16] and [17-19] we have entirely translational. This means that the thermal energy of a monatomic gas of N atoms is 1 2 3 εavg = mvrms = kBT , …[17-21] 2 2 E th = Kmicro = ε1 + ε2 + ...εN = Nεavg …[17-23] from which it follows that the rms speed of a molecule Substituting εavg from eq[17-21] we have is € 3kBT € 3 3 v = . …[17-22] E th = NkBT = nRT (monatomic) …[17-24] rms m 2 2 Let us consider a numerical example of molecular Thus the kinetic theory predicts that the thermal speeds predicted by the kinetic theory.