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chap10 solutions - CHAPTEB TEN SOLUTIONS CHAPTER TEN...

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Unformatted text preview: CHAPTEB TEN SOLUTIONS CHAPTER TEN SOLUTIONS Chapter Ten Readings Berry, R.S., "When the Melting and Freezing Points are not the Same," Scientific American, August 1990, p.68. Greenslade, T.B., "The Maximum Density of Water," The Physics Teacher, November 1985, p. 474. ' Hall, M.B., "Robert Boyle," Scientific American,lAugust 1967, p. 84. Jones, R, "Fahrenheit and Celsius, A History," Physics Today, 18, 1980, p. 594. Otani, R. and Siegel, P., "Determining Absolute Zero in the Kitchen Sink”, The Physics Teacher, 29, 316, 1991. Romer, R.H., "Temperature Scales: Celsius, Fahrenheit, Kelvin, Reaumur, and Romer," The Physics Teacher, October 1982, p. 450. Ruthern, R., "Illuminating Zero", Scientific American, v. 268 (Apr. '93) p32-33. Wilson, R.E., "Standards of Temperature," Physics Today, January 1953, p. 10. 710.1 (a) T}: =1.8TC + 32 = 1.8(-273.15) + 32 = 459.67" (b) Tc =3 (T1: - 32) = 398.6 - 32) = 37“c, (c) r): =1.8TC + 32 :- 1.8(Tjg — 273.15) + 32 3: ~279.7° F. 10.2 22% (T1: - 32) = 3(136 - 32) = s7.8°c, and Tc=g (T1: - 32) = 31-127 -.32) = 288.3“ C. 10.3 (a) To convert from Celsius to Fahrenheit, we use 1"): =2 TC + 32 =2 {-252.87 °C) + 32 = ~423°P and to convert to Kelvin, we use TK = Tc + 273 =- -253°C + 273 -—- 20 K. (b) We usean approach here that is identical to that used in (a). T5 = 68‘} and Tg = 293 K 10.4 Let us use Tc=g (TE - 32) with T]: = 40°C. Tc=g(-40 - 32) = -40"C. 10.5 Since we have a linear graph, we know that the pressure is related to the temperature as P = A + 131“, where A and B are COnstants. To findA and B, we use the given data: 0.900 atm = A + (-80.0“C)B ‘ (1} 1.635 atm = A.+ (18.0“C)B a (2) Solving (1) and (2) simultaneously, we find: A: 1.272 atm, and B=4.652x10‘3 atm/“C. 141 CHAPTER TEN SOLUTIONS Therefore, P = 1.272 atm + (4.652 x10‘3 atm/°C)T. (a) At absolute zero: P e o = 1.272 atm + (4.652 x 10~3 arm/“(3) T, which gives 1": -273.5°C. (b) At the freezing point of water: P = 1.272 atm+ 0 = 1.272 atm, and at the boiling point: P = 1.272 atm + (4.652 X10'3 atm/°C)(100°C) = 1.737 atm. 10.6 When volume is constant, we have a linear relation between pres§ure and temperature; That is, at temperature T, the pressure is given by, P = A + BY: where A and B are constants. To find A and B, we use the given data: 0.700 atm = A + (100°C)B (1) 0.512 atm=A+0 (2) Thus, from (2): A : 0.512 atm. “ Then from (1 ): B = 1.88 x 10-3 atm/“C. Therefore, P = 0.512 atm + (1.88 x 10-3 aim/“OT. (a) When P 2 0.0400 atm, we have 0.04 atm = 0.512 atm + (1.88 1110‘3 atm/“QT, yielding T: 251°C. (b) When T: 450°C. P = 0.512 atm + (1.88 x10'3 atm/°C)(450°C_) : 1.358 atm. 10.? Apply 1}: =3 Tc + 32 to two different temperatures, which we will call 1 and 2. We have: 1121:; Tc1+32, (1) and #2:; Tt2+32. {2) Subtract (1) from (2) to obtain: TF2 - 'li21 = % (Tcz - Tel) 02', AT]: = g ATc. 10.8 Use L =L0(1+ «(T— Ya): I._20= 10+ aLo(—20.0“C) — L012), and L35=LO+ aLO(35.0°C) — LOT}; AL = L35 - L20: aL0(55"C), or AL =(11x10‘6 /°C)(518 n1)(55°C) = 0.313 m = 31 cm. 10.9 AI. = 610M: (1.42 x 10-5 °c-1)(2.168 cm)(85°C} = 2.62 x10‘3 m- Thus, L : £0 + AL = 2.168 cm + .00262 cm= 2.171 Cm. 10.10 (a) The change in length is AL = aLoAT= (19 x10*6 "c-1)(1.3000 m)(-20.0"C) = 4.94 x10‘4 m Thus, the final length of the pendulum is: Lf= 1.2995 m (b) From the expression for the period, T: hf; , We see that as the length decreases the period decreases. Thus, the clock runs fast. 10.11 We shall choose the radius as our linear dimension. L = + (I( T‘ 2.21 cm = 2.20 cmll + (130x 10‘6 C HAT], giving AT= 35°C. Therefore, T= 55”C. 10.24 10.26 P 10.27 10.28 10.29 CHAPTER TEN SOLUTIONS 10’11 Pa ———— =4.1.x10’21 01. 1.013 x 105 Pa m nr 2 {4.16 x 10'5 mol) P V P- V- We may use of the ideal gas equation expressed as = t . P1 3 Vi 3 Substitute Pf = g and. If}; Ti to obtain, = i . —TF —T_—, which becomes 1 (0.300 x 105 Pa)§0.700 1:13) _(0,200 x 105 Pa)(1.50 m3) '1; ' 300 K - From which, Ir : 560 K = 287 “C. P _noRTo ErianiVo 0_ V0 Po!floTon' 1 Pr 1 Ti 1 338 (l ( But V0 = Vf and Hf =5 no. Therefore, 2%) = 0.587. n fRTf Vr r= and, ,so 110:? E =2 Pf = 5.87 atm, since P0 = 10.0 atm. Applying the ideal gas law to both the initial and final states of the oxygen in the tank gives: PiV1=niRTi, {1) and Pfo=nfRTf (2) Dividing (2) by (1}, remembering that Vf= Vi and Tf = Ti , gives: P Hf: n; (Fir) = (0.40 monig = 0.25 mi. Thus, the quantity of oxygen withdravvn must be 0.4 mol - 0.25 mol = 0.15 mol. Since, for oxygen (02}, 1 mol = 32 g, the mass of oxygen withdrawn is found to be: m = (0.15 mi) fig: = 4.8 g Let us use V= gun? as the volume of the balloon, and the ideal gas law PrVr wPW: [ . ‘3_300 K .030 atm Tf ‘ 1; Wm, ‘1 ‘200 K 1 am This yields r1 = 7.1 m. in the form: (20 m)3. Vi =(1rr2)hi =::(1.50 m)2(4.00 m) = 28.3 1113 (original volume of gas inside) At 220 m down: P: Pm + pgh = 1.013 x105 Pa + (1025 kg/m3)(9.80 m/sz)(220 m) , or P = 2.311 x106 Pa. Pi Tr (1.013 x 105 Pa) (278) v =v- #— =2 . 3 = . 3 f ‘PfTi 83m (2.311 11:106 Pa) (293) 116m Also, Vf : (area)h‘=:n{1.50 m120'. Which gives: 11' = 0.164 m = 16.4 cm : height of the remaining air space Thus, the water has risen a distance of h — h' =4.000 III-0.164 m= 3.84 m inside the hull. 145 10.30 10.31 10.32 10.33 10.34 10.35 10.36 CHAPTER TEN SOLUTIONS We first find the pressure of the air in the bubble when at a depth of 100 m by use of P= Pam, + pgh: P a 1.013 x 105 Pa + (1000 kg/m3)(9.80 W32)(10() m) = 1.08 x 106 Pa. BE 5% Tr ’ T1 becomes: Pfo 2P; V1, or, (1.013 x105 PaH/f = (1.08 x106 Pa)(1.50 cm3), and Vf = 16.0 m3. We now use: . However, at constant temperature, this '3 PV= nRT=.(:—;)RT where m is the mass of the gas present and M is the molecular weight of this type 835' Thus, p=%:%¥ ‘ . _ __ . . , gt:_fi_ _ IL When pressure is_constant, this gives. pi — 71f — 01' Pf -Pi ( 1})- 273 K) = 0.131 kg/m3 We have pf 3p.- = 0.179 kg/m3(373 K We first find the pressure exerted by the gas on the wall of the container. . NkT 3NakT 3RT 3&31 N m/mol K)_(_293 Q .5 I P= V = v n v = 8‘00 x103 m3 =9.13x10~ Pa. Thus, the force on one of the walls of the cubical container is rep/1 = (9.13 x 105 P1111400 1(10‘2 m2) = 3.65 x 104 N. The average kinetic energy = ngregardless of the gas. (RE) = §1<r=gtt3s 1110-23 1/101300 10 = 6.21 x 10-21 J. Thus, 2 (a) The total random kinetic energy in one mole is the average kinetic energy of one atom times the number of atoms. Avogadro's number. K15: NA :3- kT= (6.02 x1023 atoms); {1.38 x10‘23 J/KMBOO K} : 3740]. (b) One mol of hydrogen (H2) has a mass of 2.00 x10‘3 kg, so K15: 3140 1%:200 1110‘3 kg) v2, giving v = 1930 m/s. One mole of helium has Avogadro's number of molecules and contains a mass of 4.00 g. Let us call 111 the mass of one atom, and we have 4.00 g/mol Nm=4.00 /1ntl,o = r a g ) r 6.02 x 1023 molecules/moi m = 6.64 x 10‘24 g/molecule = 6.64 x 10’27 kg/moiecule. Helium gas has one atom per molecule so this is also the , giving mass per atom. From the ideal gas equation, P: nRT _13.00 moll§8.31 ‘L/mol 'K1L300 £1 =3. 15?. V 2241110721113 34" 0 a I46 ...
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This homework help was uploaded on 02/05/2008 for the course PHYS 111 taught by Professor Shapiro during the Fall '01 term at Wesleyan.

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chap10 solutions - CHAPTEB TEN SOLUTIONS CHAPTER TEN...

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