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APgaslawsKMTinfo - 4 z a H" ﬂaw wpuwm.“ Description...

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Unformatted text preview: 4 z a H-" ﬂaw wpuwm.“ Description of an ideal gas Figure 21 —2 A cubical box of side 1, containing an ideal gas. A molecule is shown moving toward A1. 21 -3 Ideal Gas—A Microscopic Description From the microscopic point of view we deﬁne an ideal gas by making the following assumptions; it will then be our task to apply the laws of classical mechanics sta- tistically to the gas atoms and to show that our microscopic deﬁnition is consistent with the macroscopic deﬁnition of the preceding section: 1. A gas consists of particles, called molecules. Depending on the gas, each molecule consists of one atom or a group of atoms. If the gas is an element or a compound and is in a stable state, we consider all its molecules to be identical. 2. The. molecules are in random motion and obey Newton’s laws of motion. The molecules move in all directions and with various speeds. In computing the properties of the motion, we assume that Newtonian mechanics works at the microscopic level. As for all our assumptions, this one will stand or fall depending on whether or not the experimental facts it predicts are correct. 3. The total number of molecules is large. The direction and speed of motion of any one molecule may change abruptly upon collision with the wall or another molecule. Any particular molecule will follow a zigzag path because of these colli- sions. However, because there are so many molecules we assume that the re- sulting large number of collisions maintains the overall distribution of molecular velocities and the randomness of the motion. The volume of the molecules is a negligibly small fraction of the volume 0c— cup‘ied by the gas. Even though there are many molecules, they are extremely small. We know that the volume occupied by a gas can be changed through a large range of values with little difﬁculty, and that when a gas condenses the volume oc- cupied by the liquid may be thousands of times smaller than that of the gas. Hence, our assumption is plausible. 5. No appreciable forces act on the molecules except during a collision. To the extent that this is true a molecule moves with uniform velocity between collisions. Because we have assumed the molecules to be so small, the average distance between molecules is large compared to the size of a molecule. Hence, we assume that the range of molecular forces is comparable to the molecular size. 6. Collisions are elastic and are of negligible duration. Collisions between mol- ecules and with the walls of the container conserve momentum and (we assume) kinetic energy. Because the collision time is negligible compared to the time spent by a molecule between collisions, the kinetic energy that is converted to potential energy during the collision is available again as kinetic energy after such a brief time that we can ignore this exchange entirely. 21 —4 Kinetic Calculation of Pressure Let us now calculate the pressure of an ideal gas from kinetic theory. To simplify matters, we consider a gas in a cubical vessel whose walls are perfectly elastic. Let each edge be of length 1. Call the faces normal to the x-axis (Fig. 21—2) A1 and A2, each of area [2. Consider a molecule that has a velocity v. We can resolve v into components or, by, and v, in the directions of the edges. Since each wall is perfectly elastic, this particle will rebound, after collision with A1, with its x— i E g Pressure and molecular speed component of velocity exactly reversed. There will be no effect on v” or 122 , so that: the change Ap in the particle’s momentum will be " AP = p: - pa = (-mvx) — (map) = —2mv,, normal to A1. Hence, the momentum imparted to A1 will be 2mvx, since the total momentum is conserved. Suppose that this particle reaches A2 without striking any other particle on the way. The time required to cross the cube will be 1/ 12,. At A2 it will again have its x-component of velocity reversed and will return to Al'. Assuming no collisions in between, the round trip will take a time 21/ 1),. Hence, the number of collisions per unit time this particle makes with A1 is um /21, so that the rate at which it transfers momentum to Al is ' 2 Pi = ﬂ 2va 21 1 To obtain the total force on A1 , that is, the rate at which momentum is imparted to A1 by all the gas molecules, we must sum up mvxz/l for all the particles. Then, to ﬁnd the pressure, we divide this force by the area of A1, namely, 12. .If m is the mass of each molecule, we have , m _. P=7§(Ux12+9122 + ' ’ '), where v,r1 is the x-component of the velocity of particle 1, 12,2 is that of particle 2, and so on. If N is the total number of particles in the container, we can write this as \ But mN is simply the total mass of the gas and mN / V is the mass per unit volume. that is, the density p. The quantity (1);“2 + @122 + - - -)/Ij__is the average value of v,2 for all the particles in the container. Let us call this vac”. Then p = Putz- For any particle 122 = v32, + 12,,2 + 0,}. Because we have many particles and because they are moving entirely at random, the average values of 01,301,”, and 1);2 are equal and the value of each is exactly one-third the average value of 122. There is no pref_erence among the molecules for motion along any one of the three axes. Hence, of = 5?, so that p = b??? = ipﬁ. (21-3) Although we derived this result by neglecting collisions between particles, the result is true even when we consider collisions. Because of the exchange of ve- locities in an elastic collision between identical particles, there will always be some one molecule that will collide with A2 with momentum mvx corresponding to the one that left A1 with this same momentum. Also, the time spent during colli- sions is negligible compared to the time spent between collisions. Hence, our neglect of collisions is merely a convenient device for calculation. Likewise, we could have chosen a container of any shape —the cube merely simpliﬁes the cal- culation. Although we have calculated the pressure exerted only on the side A1 , if we neglect the weight of the gas, it follows from Pascal’s law that the pressure is the same on all sides and everywhere in the interior. The square root of 725 is called the root-mean-square speed of the molecules Speed of molecules and is a kind of average molecular speed (see Section 21—9). Using Eq. 21—3, we can calculate this root—mean-square speed from measured values of the pressure and density of the gas. Thus, um, = ﬁ = ‘1—353. 1 (21—40) In Eq. 21—3 we relate a macroscopic quantity (the pressure p) to an average value of a microscopic quantity (that is, to F or vmf). However, averages can be taken over short times or over long times, over small regions of space or large regions of space. The average computed in a small region for a short time might depend on the time or region chosen, so that the values obtained in this way may ﬂuctuate. This could happen in a gas of very low density, for example. We can ig- nore fluctUations, however, when the number of particles in the system is large enough. Example 3 Calculate the root-mean-square speed of hy- 3p 3 x 1.01 x 105 N /m2 drogen molecules at 000°C and 1.00 atm pressure, assuming hy- vrms = ‘p— = = 1340 m/S- ‘ drogen to be an ideal gas. Under these conditions hydrogen has a density p of 8.99 X 10‘2 kg/m". Then, since p = 1.00 atm = This is of the order of a mile per second, or 3600 mi/h. 1.01 x 105 N/m”, 1 \ \Table 21—1 gives the results of similar calculations for some gases at 0°C. These molecular speeds are roughly of the same order as the speed of sound at the same temperature. For example, in air at 0°C, um, = 485 m/s and the speed of sound is 331 m/s, and in hydrogen vrms = 1838 m/s and sound travels at 1286 m /s. These results are to be expected in terms of our model of a gas; see Problem 37. We must distinguish between the speeds of individual molecules, described by vrms , and the very much lower speed at which one gas diffuses into another. Thus, if we open a bottle of ammonia in one comer of a room, we smell ammonia in the opposite corner only after an easily measurable time lag. The diffusion speed is slow because large numbers of collisions with air molecules greatly -.reduce the tendency of ammonia molecules to spread themselves uniformly throughout the room. Table 21 —1 Some molecular speeds Translational Molecular kinetic energy weight," um.s (at 0°C), per mole (at 0°C), Gas g/mol m/s inmsz, J /mol H2 2.02 1838 3370 He 4.0 1311 3430 H20 18 615 3400 Ne 20.1 584 3420 N2 28 493 3390 CO 28 493 3390 Air 28.8 485 3280 02 32 461 3400 C02 44 393 3400 " The molecular weight and the mole are deﬁned on p. 375. We , will discuss the last column in this table in the next section. 21 —5 Kinetic Interpretation of Temperature If we multiply each side of Eq. 21—3 by the volume V, we obtain [7V = éhpvvrz'ms, ‘ where pV is simply the total mass of gas, p being the density. We can also write the. mass of gas as nM, in which n is the number of moles and M is the molecular weight. Making this substitution yields 1 Kinetic energy per mole Kinetic energy per molecule pV = énMvﬁms. The quantity. énMvEms is two-thirds the tolal kinetic energy of translation of the molecules, that is, iGnvams). We can ther\ write pV = glGnMvzz'ms) The equation of state of an ideal gas is pV = nRT. Combining these two expressions, we obtain worm. = m. ' (21—5) That is, the total translational kinetic energy per mole of the molecules of an ideal gas is proportional to the temperature. We may say that this result, Eq. 21—5, is ‘ necessary to ﬁt the kinetic theory to the equation of state of an ideal gas, or we may consider Eq. 21—5 as a deﬁnition of gas temperature on a kinetic theory or microscopic basis. In either case, we gain some insight into the meaning of tem- perature for gases. The temperature of a gas is related to the total translational kinetic energy measured with respect to the center of mass of the gas. The kinetic energy asso- ciated with the motion of the center of mass of the gas has no bearing on the gas temperature. The temperature of a gas in a container does not increase when we put the container on a moving train. Let us now divide each side of Eq. 21—5 by the Avogadro constant. NA, which (see p. 375, footnote) is the number of molecules per mole of a gas. Thus M/NA (=m) is the mass of a single molecule and we have 1.4L 2 -l _3 R 2(NA)vrms_2mv%ms_E(E)T- Now imvfms is the average translational kinetic energy per molecule. The ratio R/NA ——which we call k, the Boltzmann constant —plays the role of the gas con- stant per molecule. We have émvﬁms = ‘EkT (21-6) in which _ R _ ‘- 8.314 J/mol K _ _23 k — ————NA — ——-———-—-—-—-——-———6.022 x 1023 molecules/mo] — 1.381 x 10 J/molecule K. We shall return to the Boltzmann constant in. Section 21—9. In the last column of Table 21—1 we list calculated values of % vams. As Eq. 21—5 predicts, this quantity (the translational kinetic energy per mole) has (closely) the same value for all gases at the same temperatures, 0°C in this case. From Eq. 21—6 we conclude that at the same temperature T the ratio of the root-mean-square speeds of molecules of two different gases is equal to the square root of the inverse ratio of their masses. That is, from 2 _ _2_ mlvlrms _ l m2v§rms T ‘ 3k 2 ” 3k 2 we obtain 1 “m = ﬂ. (21—7) \ UZrms m1 \ We can apply Eq. 21—7 to the diffusion of two different gases in a‘container wi‘h porous walls placed in an evacuated space. The lighter gas, whose molecules mo e more rapidly on the average, will escape faster than the heavier one. The ratio of the numbers of molecules of the two gases that ﬁnd their way through the porous walls for a short time interval is equal to the square root of the inverse ratio of their masses, \/m2/m1. This diffusion process is one method of separating (readily ﬁssionable) 235U (0.7% abundance) from a normal sample of uranium con- taining mostly 238U (99.3% abundance). ...
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