9B-A
Assignment Physics #8
Cole
[Q18.5] Keeping the liquid under a higher pressure raises the boiling temperature. You don't want the cooling liquid in the car to turn to a gas. This would interrupt the flow of coolant. If, however, the pressure becomes to great it could damage the radiator, hoses or crack the engine block so a pressure release valve must be used. [Q18.14] The statement is incorrect. When two gases reach equilibrium, their temperatures are the same which means the average kinetic energy of the molecules of the two gases must be the same. Because one of the gases will have molecules with greater mass, their molecular speeds will be different. [18.14] a) Put all the things that are constant on one side of the ideal gas law and all the things that PV P V V PT change on the other. R are constant so: 1 1 = 2 2 Solving for the volume ratio: 2 = 1 2 T1 T2 V1 P2 T1 V (3.5 atm)(273 + 23 K) = 3.74 where P1 = P2 + gh = 3.5 atm 2 = V1 (1.0 atm)(273 + 4 K) b) No, the diver should not hold his breath because the expansion of the air in the divers lungs would cause the diver to explode. [18.32] a)
100
s = si pi = 10
i =10
11 12 24 15 19 10 12 20 17 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 150 150 150 150 150 150 150 150 150 150
s = 54.6 b)
100
s 2 = si 2 pi =
i =10
11 12 24 15 19 10 12 20 17 10 + 20 2 + 30 2 + 40 2 + 50 2 + 60 2 + 70 2 + 80 2 + 90 2 + 100 2 150 150 150 150 150 150 150 150 150 150 2 2 s = 3730 srms = s = 61.1 Note that the rms value is higher than the mean because the larger values squared (like 100) carry more weight in the rms calculation. 10 2 [18.41] a) Three rotational + three translational degrees of freedom = 6 degrees of freedom, so 6 1 mole CV = R = 3R, but mcV = CV cV = Cv = 3(8.3 J/mole K ) =1380 J/kgK 2 m 0.018 kg/mole b) Our value is below the 2000 J/kgK which means some other degrees of freedom are contributing somewhat. There is a contribution of 1380/6 = 231 per degree of freedom. To make up the difference we need slightly less than 3 additional degrees of freedom. However, there are two degrees of freedom for each vibrational mode, and water has many modes (4 modes 8 vibrational degrees of freedom - see additional problem 2 for CO2). Obviously, we are not exciting the vibrational modes -- 8 degrees of freedom is too much. At this temperature and pressure some "hot" molecules on the Boltzmann tail will have the vibrational modes kick in whereas the bulk will not be hot enough to excite the first excited state for the vibrational modes, so we see an increase in the heat capacity, but it is much smaller than it would be if all the vibrational modes were excited. [18.43] a) At room temperature most diatomic gases do not have the vibrational states excited, so Assigment #8, 11/22/05 Page 1 of 4 UC Davis
they have 3 translational + 2 rotational degrees of freedom = 5 CV = Q = CVT = (2.5 mol)(20.79 J/molK)(30 K) = 1.56 kJ 3 b) Now only 3 degrees of freedom contribute: Q = RT = 936 J. 2 [18.48] m = a) vmp = M 44 g/mol = = 7.3 10-26 kg N A 6.02 10 23 molecules/mol 2(1.38 10 -23 J/K )(300 K ) 7.3 10 kg
-26
5 R = 20.79 J/moleK 2
2 kT = m
= 3.4 102 m/s
8(1.38 10 -23 J/K )(300 K ) 8kT = = 3.8 102 m/s b) vav = m 7.3 10 -26 kg 3(1.38 10 -23 J/K )(300 K ) 3kT = = 4.1 102 m/s c) vrms = -26 m 7.3 10 kg Note that the most probable occurs at the peak in the M-B distribution, but because of the Boltzmann tail, both the average and the rms speeds are larger. [18.59] Use the ideal gas law and put all the constants on one side PV PV PV = R so that 2 2 = 1 1 T T2 T1 To use the ideal gas law we must use absolute pressure and temperature. P1 = 1.7 atm + 1.02 atm = 2.72 atm, T1 = 273 K + 5 K = 278 K, T2 = 273 K + 45 K = 318 K T V 318 0.0150 P2 = 2 1 P = 2.72 atm = 2.94 atm. P2gauge = 2.94 - 1.02 = 1.92 atm. (about 1 T1 V2 278 0.0159 28.2 lb/in2 -- not exactly going to blow the tire up. [18.62] Pressure is a mechanical thing so we need to sum forces. Draw a free body diagram for the mercury. Fy = PA - PA - Ayg = 0 where PV = PVo (ideal gas law). P(h - y) = Ph o o h Subbing into the force eqn: Po - Po = gy Solve for y: h-y P 1.01 10 5 Pa y = h - o = 0.9 m = 0.14 m g (13.6 103 kg/m3 )(9.8 m/s2 ) y h
PA
PA mg = Ayg
a 2 [18.66] Look a van der Waals equation: P + 2 (V - b) = RT . Molecular volume should V have an effect when b 10% of V/ . b is the molecular volume of a mole, so take a molecule to RT and be a box of side = 2 10-10m b = 3 NA. Estimate V/ using the ideal gas law V/ P RT use 10 V/ b 10 l3 N A P RT (8.3J/mol K )(300 K ) = 5.2 107 Pa = 520 atm. P = 3 3 -10 23 10 l N A 10(2 10 m ) (6.02 10 / mol) You really have to squeeze the gas -- almost turn it to a liquid before the hard core becomes a major player. Note the hard core is critical to have a liquid phase. Assigment #8, 11/22/05 Page 2 of 4 UC Davis
[18.69] a), b) (See figure.) The solid curve is U(r ), in units of U, and with x = r/R. The dashed curve is F(r ) in units of U/R. Note that r1 < r2 . R = 2 0 , or r1 = R0 / 21 / 6. r1 13 7 Uo Ro U Ro = 0 = 12 From 13-26: F = - - r Ro r r r1 -1 / 6 r2 = R = 2 . F = 0 is the equilibrium position. r2 d) U(r2) = U(R) = -U. At r = , U = 0 W = U = 0- (-U) = U. [18.73] a) To escape the planet's gravitational pull requires that when you are at r = , you have GMm . no energy left over E = 0. Starting at the planet's surface Kescp = 1 mv 2 and U = - 2 Rp GMm GMm GMm =0 K= At the planet's surface mg = Conserving energy: K - Rp Rp Rp 2 Kescp = mgRp b) Three degrees of freedom contribute to the translation KE of the diatomic 3 kT = mgRp nitrogen molecule so 2 -3 2 6 2 mgRp 2 (28 10 kg/mol)(9.8 m/s )(6.38 10 m ) = 14 104 K. T= = 23 -23 3 k 3 (6.02 10 molecules/mol)(1.38 10 J/K ) 2.02 For hydrogen T = T = 1.01 104 K. The upper atmosphere of the earth has a 28.0 N temperature of 1000 K. The escape temperature of hydrogen is only ten times this temperature. Because of the Boltzmann tail on the distribution of molecular speeds, we would expect that over millions of years, the hydrogen gas would escape from the earth, but the nitrogen requires an escape temperature 100 times the temperature of the outer atmosphere and would be far less likely to evaporate from the earth. c) For nitrogen, T = 6.36 103 K and for hydrogen, T = 459 K. d) One the side facing the sun, the moon is about 400 K so hydrogen would quickly evaporate. The escape temp for nitrogen is only a factor of ten higher than this and would over millions of years, also evaporate away leaving the moon with no atmosphere. [18.83] a) We interpret probability as the frequency of occurance. For a discrete system, pi = ni N R When U = 0, 0 r1
12 6
where ni is the number of particles in the ith state, so ni = Npi . For the continuous system, the probability of finding the particle with a speedbetween v and v + dv is f(v) dv. Hence, the number of particles in this range is dni = N f(v) dv. Summing N = N
3/ 2 mv
2
v + v
v
f (v)dv.
mp 2 kT 4 m v 2 e 2 kT = 4 m 2 kT e -1 = b) vmp = f (vmp ) = 4 . mp 2kT 2kT m e vmp m For oxygen gas at 300 K, vmp = 3.95 102 m/s, and N = N f(v)v = 0.0421N. c) Increasing v by a factor of 7 changes f by a factor of 72e-48 , and N f(v)v = (2.94 10-21)N. d) Multiplying the temperature by a factor of 2 increases the most probable speed by a factor of Assigment #8, 11/22/05 Page 3 of 4 UC Davis
3/ 2
2 , and the answers are decreased by ; 0.0297 and 2.08 10-21. e) When the temperature is one-half what it was in parts (b) and (c ), the fractions increase by 2 to 0.0595 and 4.15 10-21. f) At lower temperatures, the distribution is more sharply peaked about the maximum (the most probable speed), as is shown in Fig. (16-17). Additional Problems: [1] If we write the average energy for the bead, we find E = < 1 mvx2 > + < 1 x2 > where the 2 2 bead could have kinetic energy along the wire, and rotational energy for spinning about an axis aligned with the wire. The bead must have a finite radius to fit around the wire so it will have a non-zero moment of inertia. The wire keeps the bead from spinning about any other axis. 2 degrees of freedom -- 1 translational and 1 rotational. There is 1 R contribution to the molar heat 2 capacity at constant volume for each degree of freedom Cv = 2( 1 R) = R. 2 [2] The two extremes of the oscillations are shown. Remember, since we count the translational degrees of freedom separately, the center of mass must be at rest for all modes. a)
Same type of oscillation as (1) but in the horizontal plane. (1) CM (2) CM
(3)
(4)
(b) There are 4 modes of vibration which give 8 vibrational degrees of freedom. Remember there are a kinetic and potential degree of freedom for each vibrational mode. In addition, there are 2 rotational degrees of freedom, and 3 translational degrees of freedom, and if the vibrational degrees of freedom are excited so are these others; thus, Cv = (13/2)R, and Cp = (15/2)R so that = Cp/Cv = 15/13.
Assigment #8, 11/22/05
Page 4 of 4
UC Davis