Chapter 12 The Kinetic Theory of GasesClassical thermodynamics has nothing to say about atoms or molecules. Its laws are concerned only with such macroscopic variables as pressure, volume, and temperature. However, we know that gas is made up of atoms or molecules (groups of atoms bound together). The pressure exerted by a gas must surely be related to steady drumbeat of its molecules on the walls of its container. The ability of a gas to take on the volume of its container must surely be due to the freedom of motion of its molecules. And the temperature and internal energy of a gas must surely be related to the kinetic energy of these molecules. Perhaps we can learn something about gases by approaching the subject from this direction. We call this molecular approach the kinetic theory of gases.12.1 Ideal Gases1、Our goal in this chapter is to explain the macroscopic properties of a gas, such as its pressure and its temperature, in terms of the behavior of the molecules that make it up.2、The experiments show that, at low enough densities, all real gases tend to obey the relation )pV , innRTgas(lawidealwhich p is the absolute pressure, A N N n /= is the number ofmoles of gas present, andR , the gas constant , has the same value for all gases, namely, K mol J R ⋅=/31.8. The temperature T must be expressed in kelvins. Above equation is called the ideal gas law . Provided the gas density is reasonably low, it holds for any type of gas, or a mixture of different types, with n being the total number of moles present.3、 Work done by an ideal gas at constant temperature :(1). Suppose that a sample of n moles of an ideal gas, confined to a piston-cylinder arrangement, is allowed to expand from an initial volume i V to a final volume f V .(2). Suppose further that the temperature T of the gas is held constant throughout the process. Such a process is called an isothermal expansion (and the reverse is called an isothermal compression ).4、Let us calculate the work done by an ideal gas during an isothermal expansion, we have⎰⎰==f i f i V V V V dV V nRT pdV W i fV V nRT ln =. Recall that the symbol ln specifies a naturallogarithm, that is, a logarithm to base e .12.2 Pressure, Temperature, RMS Speed, and Translational Kinetic Energy1、 Let n moles of an ideal gasbeing confined in a cubical box ofvolume V , as in the figure.2、 A typical gas molecule, of mass m and velocity v, is about to collide with the shaded wall. We assume that any collision of a molecule with a wall is elastic , so the change in the particle’s momentum is along the x axis and its magnitude is x x x x mv mv mv p 2)()(-=--=∆. Hence the momentum x p ∆ delivered to the wall by the molecule during the collision is x mv 2+.3、 The average rate at which momentum is delivered to the shaded wall by this single molecule is Lmv v L mv t p x x x x 2/22==∆∆. 4、 F rom Newton’s second law, the rate at which momentum is delivered to the wall is the force acting on that wall. To find the total force, we must add up the contributions of all molecules that strike the wall, allowing for the possibility that they all have different speeds. Dividing the total force x F by the area of the wall then gives the pressurep on that wall. Thus))((///2222132222212xN x x xN x x x v v v Lm L L mv L mv L mv L F p +++=+++== , where N is the number of molecules in the box.5、 Since A nN N = in which A N is Avogadro’s number ,above equation can be re-expressed as223x x A v V nM v L nmN p ==, where A mN M = is the molar mass of the gas , and V is thevolume of the box.6、 For any molecules, 2222z y x v v v v ++=. Because there aremany molecules and because they are all moving in random direction, the average values of the squares of their velocity components are equal, so that3/22v v x =. Thus V v nM p 32=. 7、 The square root of 2v is a kind of average speed, called the root-mean-square speed of the molecules and symbolized by rms v .8、 Using the relationnRT pV = for an ideal gas, it leads to M RT nM pV v v rms 332===.See the RMS speeds ofsome molecules in Table.9、 The averagetranslational kinetic energy of a single molecule of an ideal gas iskT T N R M RT m mv mv K A rms 23)(23)3)(21(212122=====, where the constantK J N R k A /1038.1/23-⨯== is the Boltzmann constant . So we come to the conclusion that at a given temperature T , all ideal gas molecules, no matter what their mass, have the same average translational kinetic energy . When we measure the temperature of a gas, we are also measuring the average translational kinetic energy of its molecules.12.3 Mean Free Path1、 Table gives us the RMS speed of some molecules, A question often arises: If molecules move so fast, why does it take as long as a minute or so before you can smell perfume if someone open a bottle across a room? This is because although the molecules move very fast between collisions, a given molecule will wander only very slowly away from its release point.2、 The right figure shows the path of atypical molecule as it moves through the gas,changing both speed and direction abruptlyas it collide elastically with other molecules.Between collisions, our typical moleculemoves in a straight line at constant speed.Although the figure shows all the other molecules as stationary,they too are moving in much the same way.3、 One useful parameter to describe this random motion is the mean free path λ. As its name implies, λ is the average distance traversed by a molecule between collisions.4、 The expression for the mean free path can be deduced from following steps:(1). First we assume that our molecule is traveling with a constant speed u and that all other molecules are at rest. We assume further that the molecules are spheres of diameter d. A collision will then take place if the centers of the molecules come within a distance d of each other. A help way to look at this situation is to consider our single molecule to have radius of d and all the other molecules to be points.(2). As our single molecule zigzags through the gas, it sweeps out a short cylinder of cross-sectional area 2d π between successive collision. If we watch this molecule for a time interval t ∆, it moves a distance t u ∆. So the volume of the cylinder is t u d ∆2π. The number of collisions that occur is then equal to the number of (point) molecules that lie within this cylinder. It is t u d V N ∆2)/(π.(3). The mean free path is the length of the path divided by the numberof collisions)/()/(22V N u d v V N t u d t v collisions of numbers path of length ππλ=∆∆==.(4). The u in the denominator is the mean speed of our single molecule relative to the other molecules, which are moving. A detailed calculation, taking into account the actual speed distribution of the molecules, gives thatv u 2=. So )/(212V N d πλ=.5、The mean free path of air molecules at sea level is about m μ1.0. At an altitude of km 100, the mean free path is about 16cm. At 300 km, the mean free path is about 20 km.12.4 The Distribution of Molecular Speeds1、 In 1852, Scottish physics James Clerk Maxwell first solved the problem of finding the speed distribution of gas molecules. His result, known as Maxwell’s speed distribution law , isRT v M e v RT M v P 2/22/32)2(4)(-=ππ. Here v is the molecular speed, Tis the gas temperature, M is the molar mass of the gas, and R is the gas constant. The quantity )(v P is a probabilitydistribution function , defined as follow: The product dv v P )( isthe fraction of molecules whose speeds lie in the range v to dv v +.2、 As figure (a) shows, thisfraction is equal to the area of astrip whose height is)(v P and whose width is dv . The totalarea under the distribution curvecorresponds to the fraction ofthe molecules whose speed liebetween zero and infinity. Allmolecules fall into this category,so the value of this total area isunit.3、 There are two otherspeeds. The most probable speed P v is the speed at which )(v P is a maximum. The average speed v is a simple average of the molecular speeds. We will find thatM RT v π8= and M RT v P 2=，so we have relationsrms P v v v <<.12.5 Degree of Freedom and Molar Specific Heats1、 The right figure showskinetic theory models ofhelium (a monatomic gas),oxygen (diatomic), and methane (polyatomic). On the basis of their structure, it seems reasonable to assume that monatomic molecules, which are essentially point-like and have only a very small rotational inertia about any axis, can store energy only in their translational motion. Diatomic and polyatomic molecules, however, should be able to store substantial additional amount of energy by rotating or oscillating.2、To take these possibilities into account quantitatively, we use the theorem of the equipartition of energy, introduced by James Clerk Maxwell: Every kind of molecules has a certain number f of degree of freedom, which are independent ways in which the molecule can store energy. Each such degree of freedom has associated with it, on average, an energy of 2/kT per molecule (or 2/RT per mole).3、For translational motion, there are three degrees of freedom. For rotational motion, a monatomic molecule has no degree of freedom. A diatomic molecule has two rotational degrees of freedom. A molecule with more than two atoms has six degrees of freedom, three rotational and three translational.4、Internal energy is the energy associated with random motion of atoms and molecules. So a sample of n moles of a gas containsnN atoms. The internal energy of the sample is then AnRT f kT f nN E A 2)2)((int ==. Thus, the internal energy int E of anideal gas is a function of the gas temperature only; it does not depend on any other variable .5、 Molar specific heat at constant volume : (1) According to the definition of molar specific heat, we have T nC Q v ∆=. (2) Substituting it into the first law of thermodynamics, we find T nC T nC W Q E V v ∆=-∆=-=∆0int . (3) So we will have the relation R f T n RT f n T n E C v 2)2(int =∆∆=∆∆=. 6、 Molar specific heat at constant pressure : (1) the heat is T nC Q p ∆=. (2) According to the first law of thermodynamics, wehave T nC T nR T nC V p T nC W Q E V p p ∆=∆-∆=∆-∆=-=∆int .(3) So we have relationR C C V p +=.12.6 The Adiabatic Expansion of an Ideal Gas1、 Using the first law of thermodynamics, we have pdV Q dE -=int pdV -=. It means pdV dT nC V -=.2、 From the ideal gas law ()nRT pV = we have nRdT Vdp pdV =+ 3、 Combining these two equations, we have 0)(=+V dV C C p dp V p . Replacing the ratio of the molar specific heats with γ and integrating yieldt cons a pV tan =γ.11 4、 Using the ideal gas law, we also have t cons a TV tan 1=-γ, and t cons a T p tan 1=--γγ.5、 Free expansion: Since a free expansion of a gas is an adiabatic process that involves no work done on or by the gas, and no change in the internal energy of the gas. A free expansion is thus quite different from the type of adiabatic process described above, in which work is done and the internal energy changes. Since the int ernal energy isn’t change for free expansion, we must havef i RT f n RT f n 22=. It means f i T T =, andso f f i i V p V p =.。