Chapter 18 - PHYS 132, WEEK 9 NOVEMBER 15-18 18.16. Solve:...

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Unformatted text preview: PHYS 132, WEEK 9 NOVEMBER 15-18 18.16. Solve: The average translational kinetic energy per molecule is e avg = 1 2 mv rms = 2 3 2 k B T fi vrms = 3k B T m Since we want the vrms for H2 and N2 to be equal, 3k B TH 2 mH2 = 3kB T mN2 fi TH 2 = mH2 mN2 TN 2 = 2u ^ (373 K ) = 26.6 K 28 u 18.20. Solve: (a) The average translational kinetic energy per molecule is e avg = 1 mv rms = 3 kB T 2 2 2 This means eavg doubles if the temperature T doubles. (b) The root-mean-square speed vrms increases by a factor of (c) The mean free path is 2 as the temperature doubles. l = 1 4 2 p ( N V )r 2 Because N/V and r do not depend on T, doubling temperature has no effect on l. 18.25. Solve: The volume of the air is V = 6.0 m 8.0 m 3.0 m = 144.0 m3, the pressure p = 1 atm = 1.013 105 Pa, and the temperature T = 20C = 293 K. The number of moles of the gas is n= pV RT = 5991 mols This means the number of molecules is N = nN A = (5991 mols ) 6.022 10 ( 23 mol -1 ) = 3.61 10 7 27 molecules. Since air is a diatomic gas, the room's thermal energy is E th = Ne avg = N ( 5 kB T ) = 3. 65 10 J 2 Assess: The room's thermal energy can also be obtained as follows: E th = nC V T = (5991 mols)(20.8 J/mol K)(293 K) = 3.65 107 J 18.29. Visualize: Refer to Figure 18.13. At low temperatures, CV = 3 R = 12 .5 J / mol K. At room temperature and modestly hot 2 temperatures, CV = R = 20 .8 J / mol K. At very hot temperatures, CV = 7 R = 29 .1 J / mol K. 2 Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is 0.20 g = 0.1 mol 2 g / mol 5 2 The heat needed to change the temperature of the gas from 50 K to 100 K at constant volume is Q = DE th = nC V DT = (0.1 mol)(12.5 J/mol K)(100 K 50 K) = 62.4 J (b) To raise the temperature from 250 K to 300 K, Q = DEth = (0.1 mol)(20.8 J/mol K)(300 K 250 K) = 104 J (c) To raise the temperature from 550 K to 600 K, Q = 104 J. (d) To raise the temperature from 2250 K to 2300 K, Q = DEth = nCVDT = (0.10 mol)(29.1 J/mol K)(50 K) = 145 J. 18.31. Solve: (a) The thermal energy of a monatomic gas is E th = fi TA = TB = 3 2 Nk B T = 3 2 nRT fi T = 2 E th 1 3 n R 1 2 ^ 5000 J ^ = 201 K 3 2.0 mol (8.31 J / mol K ) 1 2 ^ 8000 J ^ = 214 K 3 3.0 mol ( 8.31 J / mol K ) Thus, gas B has the higher initial temperature. (b) The equilibrium condition is (e A )avg = (e B ) avg = (e tot ) avg . This means E Af nA fi E Af = nA nA + nB E Bf = E tot = nB nA + nB = E Bf nB = E tot nA + nB ^ (5000 J + 8000 J ) = 5200 J 2 mols + 3 mols E tot = 3 mols ^ (13,000 J ) = 7800 J 5 mols 2 mols 18.40. Visualize: Please refer to Figure P18.40 in the textbook. Solve: (a) The most probable speed is 4 m/s. (b) The average speed is 2 2 m / s + 4 4 m / s + 3 6 m / s+ 1 8 m / s v avg = = 4.6 m / s 2 + 4 + 3 +1 (c) The root-mean-square speed is v rms = 2 (2 m / s ) + 4 (4 m / s ) + 3 (6 m / s ) + 1 (8 m / s ) 2 + 4 + 3 +1 2 2 2 2 = 4.94 m / s 18.41. Solve: (a) The cylinder volume is V = r2L = 1.571 103 m3. Thus the number density is N V = 2.0 10 22 1. 571 10 -3 m 3 = 1.273 10 25 m -3 (b) The mass of an argon atom is m = 40 u = 40(1.661 10-27 kg) = 6.64 10-26 kg fi vrms = 3k B T m 2 = 3 1.38 10 -23 J / K (323 K ) 6.64 10 -26 kg 2 ( ) = 449 m / s (c) vrms is the square root of the average of v2. That is, v rms = v ( ) avg = vx ( ) 2 avg + vy 2 ( ) 2 avg + vz 2 ( ) 2 avg An atom is equally likely to move in the x, y, or z direction, so on average (v x ) avg = v y v rms = 3 v x 2 ( ) 3 avg = vz ( ) 2 avg . Hence, ( ) 2 avg fi (v x )rms = (v ) 2 x avg = v rms = 259 m / s (d) When we considered all the atoms to have the same velocity, we found the collision rate to be 1 ( N V ) Av x (see Equation 18.10). Because 2 the atoms move with different speeds, we need to replace vx with (vx)rms. The end of the cylinder has area A = r2 = 7.85 103 m2. Therefore, the number of collisions per second is 1 2 ( N / V ) A(v x ) rms = 1 (1.273 10 25 m -3 )(7.85 10 -3 m 2 )(259 m / s) = 1.296 1025 s-1 2 1 3 (e) From kinetic theory, the pressure is N^ 2 N^ 2 p=1 mv = 1 mv rms = 3 3 V ( )avg V (1.273 10 25 m -3 )(6.64 10 -26 kg (449 m / s) = 56, 800 Pa ) 2 (f) From the ideal gas law, the pressure is V 1.571 10 m 3 The very slight difference with part (e) is due to rounding errors. -3 p= Nk B T = (2.0 10 )(1.38 10 22 -23 J / K (323 K ) ) = 56, 700 Pa Assess: 18.48. Solve: As the volume V1 of a gas increases to V2 = 2V1 at a constant pressure p1 = p2, the temperature of the gas changes from T1 to T2 as follows: p 2 V2 T2 = p 1 V1 T1 fi T2 = T1 V2 p 2 V1 p 1 = 2T1 Since the process occurs at constant pressure the heat transferred is Q = nC P DT = nC P (T2 - T1 ) = nC P (2T1 - T1 ) = nC P T1 For a monatomic gas, C P = CV + R = 3 2 R+R= R+R= 5 2 R R For the diatomic gas, C P = CV + R = 5 2 7 2 Thus Qdiatomic Q monatomic = n ( 7 R)T1 2 n ( 5 R)T1 2 = 1.40 18.53. Solve: (a) The rms speed is v rms = 3k B T m fi vrms hydrogen v rms oxygen = 32 u 2u =4 (b) The average translational energy is e = 3 kB T . Thus 2 e avg hydrogen e avg oxygen = Thydrogen Toxygen 5 2 =1 (c) The thermal energy is E th = fi E th hydrogen E th oxygen = n hydrogen n oxygen = nRT m hydrogen 32.0 g / mol m oxygen 2.0 g / mol = 16 18.60. Solve: (a) The escape speed is the speed with which a mass m can leave the earth's surface and escape to infinity (rf = ) with no left over speed (vf = 0). The conservation of energy equation Kf + Uf = Ki + Ui is 0+0= 1/2 1 2 mv esc - 2 GM e m Re fi vesc = 2GM e Re The rms speed of a gas molecule is vrms = (3kBT/m) . Equating vesc and vrms, and squaring both sides, the temperature at which the rms speed equals the escape speed is 3k B T m = 2GM e Re 2GM e ^ fi T = m ~ 3k B Re For a nitrogen molecule, with m = 28 u, the temperature is T = (28 1.661 10 -27 2( 6.67 10 -11 N m 2 / kg 2 ) (5. 98 10 24 kg) ^ kg) ~ = 141,000 K 3(1.38 10 -23 J / K) (6.37 10 6 m) (b) For a hydrogen molecule, with m = 2 u, the temperature is less by a factor of 14, or T = 10,100 K. (c) The average translational kinetic energy of a molecule is e avg = 3 kB T = 6.1 1021 J at a typical atmosphere temperature of 20C. The 2 kinetic energy needed to escape is Kesc = 1 mv esc . For nitrogen molecules, Kesc = 2.9 1018 J. Thus eavg/Kesc = 0.002 = 0.2%. Earth will retain 2 nitrogen in its atmosphere because the molecules are moving too slowly to escape. But for hydrogen molecules, with Kesc = 2.1 1019 J, the 2 ratio is eavg/Kesc = 0.03 = 3%. Thus a large enough fraction of hydrogen molecules are moving at escape speed, or faster, to allow hydrogen to leak out of the atmosphere into space. Consequently, earth's atmosphere does not contain hydrogen. ...
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This note was uploaded on 09/27/2008 for the course PHYS 131-133 taught by Professor All during the Spring '08 term at Cal Poly.

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