General Physics(ppt) Ch 19

General Physics(ppt) Ch 19 - Chapter Chapter 19...

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Unformatted text preview: Chapter Chapter 19 Thermodynamics II kinetic theory of gases Thermodynamics II Ideal Ideal gas Kinetic Kinetic theory calculations of pressure and temperature Internal Internal energy and equipartition of energy Specific Specific heat of ideal gas The The kinetic theory of gases Treating Treating thermodynamics from a microscopic viewpoint. Clarify Clarify the microscopic basis of pressure and temperature temperature Macroscopic Macroscopic properties can be found by averaging over individual particles Mole and Avogadro’s number One One mole = 6.02 × 1023 elements = the number of atoms in a 12 g sample carbonsample of carbon-12 Avogadro’s Avogadro’s number NA = 6.02 × 1023 mol-1 10 A. A. Avogadro: suggested that all gases occupying the same volume under the same temperature and pressure with the same number of atoms or molecules. What What is ideal gas? All All gases approaches the ideal state at low density No No interaction between molecules Experiments Experiments showed that 1 mol different gases gases (hydrogen, oxygen, methane…) exhibit nearly the same pressure under identical volume and temperature Ideal gases pV = nRT p: absolute pressure V: volume n: number of moles of gas Macroscopic R: gas constant = 8.31 J/mol.K Microscopic Boltzmann Boltzmann constant k = R/NA = 1.38 × 10-23 J/K R/N pV = NkT , N: number of molecules T: temperature in Kelvins Please refer to Feynman lecture on Physics Ch 39 for derivation Work Work done at constant temp Isothermal Isothermal processes Vf nRT dV Vi V Vf 1 dV = nRT ln V Vi Vf W = ∫ pdV = ∫ Vi Vf = nRT ∫ Vi Which are isothermal processes? Work done at constant volume or constant pressure Constant Constant pressure i ↔ a Vf W = ∫ pdV = p (V f − Vi ) = p∆V Vi Constant Constant volume a ↔ f W =0 Pressure Pressure and molecular speeds What What is the connection between pressure and the speeds of molecules inside? Ignore Ignore molecule interaction and consider only elastic collisions with the wall Momentum delivered to the wall by the molecule during a collision ∆p x = 2mv x The average rate of momentum delivery 2 ∆p x 2mv x mv x = = ∆t 2 L / vx L Force on the wall Pressure and molecular speeds 2 2 2 pressure on Fx mv x1 / L + mv x 2 / L + ... + mvxN / L p= 2 = 2 the wall L L ( m2 2 2 = 3 v x1 + v x 2 + ... + v xN L ) () 2 = N vx avg 2 = nN Av x ,rms 2 = nN Avrms / 3 2 2 nmN Avrms nMvrms ⇒ p= = 3L3 3V Root-mean-square Link between microscopic and macroscopic properties vrms 3RT = M Kinetic Kinetic energy Average Average translational kinetic energy ( Kavg = 1 mv2 2 ) () = 1 m v2 2 avg 2 = 1 mvrms avg 2 = 3RT 3RT = M 2NA Kavg = 3 kT 2 Kavg = 1 m 2 3RT M At At a given temperature T, all ideal gas molecules— molecules—no matter what their mass— mass— have the same average translational kinetic energy. Measuring Measuring T = measuring Kavg MaxwellMaxwell-Boltzmann distribution What What is the speed distribution of gas molecules? Probability Probability function: 3/ 2 3/ 2 M 2 −Mv / 2RT m 2 −mv2 / 2kT P(v) = 4π = 4π ve ve 2πRT 2πkT 2 ∫ ∞ 0 P(v)dv = 1 Derivation MaxwellDerivation of Maxwell-Boltzmann distribution Boltzmann distribution P Normalization P For ideal gas in 1D P P In 3D space To find out the number (probability) of particle at v, we have to integrate over a shell dv = dvxdvydvz 3/ 2 m 2 −mv2 / 2kT P(v) = 4π ve 2πkT P Molecular speeds Average Average speed ∞ 8RT πM vavg = ∫ vP(v)dv = 0 RMS RMS speed vrms = ∞ ∫ 0 Most Most probable speed v2 P(v)dv = 3RT M dP(v) 2RT = 0; vP = dv M What is vavg, vrms, and vP, for oxygen and hydrogen gases? vrms 483 m/s 1932 m/s vavg 445 m/s 1780 m/s vP 395 m/s 1580 m/s The The internal energy Kavg = 3 kT 2 ⇒ Eint = NKavg = 3 NkT = 3 nRT 2 2 The The internal energy of an ideal gas is a function function of the gas temperature only. only Molar Molar specific heat At At constant volume Q = nCV ∆T ∆Eint = Q 3 2 nR∆T = nCV ∆T ⇒ CV = 3 R 2 At At constant pressure Q = nC p ∆T ∆Eint = Q − W W = p∆V = nR∆T ∆Eint = nCV ∆T ⇒ C p = CV + R Quick Quick quiz Rank Rank ∆Eint Equipartition of energy Each Each degree of freedom has associated with an energy of 1 kT per molecule (or RT RT 2 kT per mole). As Table 19-2 shows, the prediction that Cv=R agrees As 19with experiment for monatomic gases but fails for diatomic and polyatomic gases. Let us try to explain the gases. discrepancy ( 差異 ) by considering the possibility that molecules with more than one atom can store internal energy in forms other than translational kinetic energy. Fig. 19-11 shows common models of He (monatomic), Fig. 19oxygen (diatomic), and methane ( 甲 烷 , polyatomic). polyatomic). From such models, we would assume that all 3 types of molecules can have translational motions (x,y,z(x,y,zdirections) and rotational motions. motions. We would assume that the diatomic and polyatomic We molecules can have oscillatory motions (振動), with the atoms oscillating slightly toward and away from one another, as if attached to opposite ends of a spring. spring. To account of the various ways in which energy can be To stored in a gas, Maxwell introduced the theorem of the equipartition of energy (能量均分原理): Every kind of molecule has a certain number f of degrees of freedom ( 自由度 ), 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 kT per molecule (or RT per mole). Let us apply the theorem to the translational and Let Fig. 19rotational motions of the molecules in Fig. 19-11. For the translational motion, the molecules will have velocity components along x, y, and z axes. Thus, gas axes. molecules of all types have 3 degrees of translational freedom and, on average, an associated energy of kT per molecule. molecule. For the rotational motion, imagine the origin of our xyz For Fig. coordinate system at the center of each molecule (Fig. 19-11). In a gas, each molecule should be able to rotate 19with an angular velocity component along each of the 3 axes, so each gas should have 3 degrees of rotational freedom and, on average, an additional energy of kT per molecule. However, experiment shows this is true molecule. only for the polyatomic molecules. According According to quantum theory, the physics dealing with the allowed motions and energies of molecules and atoms, a monatomic gas molecule does not rotate and so has no rotational energy. energy. A diatomic molecule can rotate only about axes perpendicular to the line connecting the atoms (Fig. 19Fig. 1911b) and not about that line itself. Therefore, a diatomic 11b itself. molecule can have only 2 degrees of rotational freedom and a rotational energy of only 2(kT) per molecule. molecule. We replace Eq. 19-38 (Eint=nRT) with Eint=(f)nRT, We Eq. 19where f is the number of degrees of freedom listed in important Table 19-3. 19- Doing Doing so leads to the prediction which agrees with Eq. 19-43 for monatomic gases (f =3). As Table 19-2 shows, this prediction also agrees with experiment for diatomic gases (f=5), but it is too low for polyatomic gases (f=6 for molecules comparable to CH4). f=3 (三個移動 自由度) f=5 (三個移動、 三個移動、 二個轉動自由度) f=6 (三個移動、 三個移動、 三個轉動自由度) We can improve the agreement of kinetic theory with We experiment by including the oscillations of the atoms in a gas of diatomic or polyatomic molecules. molecules. For example, the two atoms in the O2 molecule of Fig. For Fig. 19-11b can oscillate toward and away from each other, with the interconnecting bond acting like a spring. spring. However, However, experiment shows that such oscillations occur only at relatively high temperatures of the gas — the motion is “turned on” only when the gas molecules have relatively large energies. Rotational motion is also energies. subject to such “turning on,” but at a lower temperature. temperature. Fig. 19-12 CV/R v.s. temperature for (diatomic) H2 gas. Because rotational and oscillatory motions begin at certain energies, only translation is possible at very low temperatures. As the temperature increases, rotational motion can begin. At still higher temperatures, oscillatory motion can begin. Figure 19Figure 19-12: Below about 80 K, we find that CV/R=1.5. /R=1 This result implies that only the three translational degrees of freedom of hydrogen are involved in the specific heat. heat. As the temperature increases, the value of CV/R As gradually increases to 2.5, implying that two additional degrees of freedom have become involved. Quantum involved. theory shows that these two degrees of freedom are associated with the rotational motion of the hydrogen molecules and that this motion requires a certain minimum amount of energy. At very low temperatures energy. (below 80 K), the molecules do not have enough energy to rotate. rotate. As the temperature increases from 80 K, first a few As molecules and then more and more obtain enough energy to rotate, and CV/R increases, until all of them are rotating and CV/R=2.5. /R=2 Similarly, Similarly, quantum theory shows that oscillatory motion of the molecules requires a certain (higher) minimum amount of energy. This minimum amount is not met energy. until the molecules reach a temperature of about 1000 K, Fig. 19as shown in Fig. 19-12. As the temperature increases beyond 1000 K, more As molecules have enough energy to oscillate and CV/R increases, until all of them are oscillating and CV/R=3.5. In Fig. 19-12, the plotted curve stops at 3200 K because In Fig. 19-12, there the atoms of a hydrogen molecule oscillate so much that they overwhelm (戰勝) their bond, and the molecule then dissociates ( 分 解 ) into two separate atoms. atoms. Gas Gas processes 定壓 等溫 絕熱 定容 Homeworks Ans: Ans: Ans: Ans: Ans: Ans: Ans: Extra ...
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This note was uploaded on 12/27/2011 for the course LSCI 103 taught by Professor K.y.liao during the Fall '11 term at National Cheng Kung University.

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