376 Lea & Burke Physcs; The Nature of Thngs 19.44 J»(Pa) At constant V A \ LjSj?! '/! If / l/l /!! j j J [ ^ 1 I X j j> 1 ' : / J! 60 100 T('K) 200 Constant V lnes are mapped onto straght lnes on the P-T dagram. b) now WAB = 5.1 xlff'j W BC = -2.5 xlo 4! W DB = 0 = -2.0xl0 4 J 7 = 0.60 m 3 T = 2.0 x 10 2 K P = 0.050 atm =.VJT X RT (1.013xlO f ' -1^x0.050 atm)(0.60 m 3 ) J8.31 J/mol KKZOxlO* K) c / csely It was not sothermal, nor exactly ada-, batc. (The product PV changes by 8% ) 19.46 a) P = 0.164 atm b) W() 1660 J; W(z) -1900 J c) T (1) = 200K; T (2) = 256 K _d) Q (1) = -4160 J; Q (2) = -2990 J n.9.47tank Volume s V = 0.0160 m 3 Pressure P 5.00 atm Temperature T = 293 K Number of moles from PV = MPT pv \r RT (5.00 atm x 1.013 x 1Q 5 ^) (0.0160 m 3 ] (8.314 J/mol-KX293K) = 3.33 moles of Argon Perform sothermal expanson => PM = Q.Q160 m = 5.00 atmx 3 3.00 atm = 0.027 m 3 Note, the temperature s stll 293 K. The fnal volume s Vo = Vt T 1 V f = 0.027m 3 pressure drops:,... 1-00 atm 0.027 m 3 = 293 K x - x 3.00 atm 0.027 = 97.7 K Work was done between the frst two ponts at constant T. None s done at constant volume 210K 0.6 m 3 200 K X 0.5 m 3 = 6.063 atnr\ 0.05 W,3 = I P(V) dv = r Jv, V
asreaured. From equaton 19.18: 5Jfc * = 2m T 2 Cp 2 515 J/kg K Thus the atomc mass s 6. 699 x 10~ 26 / (1.66 x 10" 27 ) = 40. 4. From the perodc table, we conclude that the gas s argon. (Calcum also has an atomc mass of 40, but t s not a gas under normal laboratory condtons.) rom the free body dagram for the pston, we have: pa=m 9 =*p=? = (5-0 kg) ;; ""; ; /!. 02 x o 5 p a A 4.8 x 10~ 4 ~ 2 The pressure of the helum s l.oxlo 5 Pa, or 1 arm. The pressure remans constant because the pston remans n equlbrum (t rses slowly") Thus the heat transfer s: Q = p 2m 2 The number of molecules can be found from the deal gas law: PV AT N = W Q = 3 W kat = l pv - " 5 ( L02 x 1Q5 LO x 10 *) < 19.54 The sudden depresson means that no heal transfer occurs and the process s adabatc. Thus PV = constant So: For ar, 7 = 7/5, so PV? = ^= ( " L) = a 6095 (L5 L) = - 91 L 19.56 For the whole cycle, ACT = 0 and so Q = W. The work done equals the area of the trangle: W = \ (2.0m 3 ) (0.30 am) L013 x 10 " ^ = 3.0 x 10 4 J latm Thus the heat transfer s also Q = 3.0 x 10 4 J. 19.58 Snce the process s adabadc, the heat transfer s zero. Then from the frst law 1 mol of monatomc deal gas, W _ (8.3145 J/mol - K) (100 ^r^ 1. 25 x 10 3 J The mnus sgn means that work s done on the gas. 19.60 Snce the gas expands adabatcally, Q Q and so A 7 = W. But for any deal gas, AJ7 = Mc^AT = Me, (T 2 - TI) (equaton 19.16), Thus
19. Temperature and Thermal Energy 377 W = (3.33 mol) (8.31 -^^} (293 K) h = 4250 J P (Pa) am =* Corresponds to Krypton k_ m = 149 J/kg-K 4- = 248J/kg-K 1.38 x 1Q- 23 J/K 83.7 x 1.67 x 10-27 kg.9.52 P = 1.02 x 10 5 Pa; Q = 150 J 19.53 3.00 mol neon Ne, 2.00 mol H. The heat ca pacty of the neon s: CWe = 2^"^ = I (3.00 mol) (8.3145 J/mol-K) = 37.42 J/K 19.48 He; same 7. 19.49 Constant volume process, deal gas monatomc The heat capacty of the hydrogen s: C H = (2.00 mol)(8.3145j/mol-k) = 24.94 J/K by 1st Law, f no work done then AC/ = MdyT = -AfRAT A 7 = 10.0 J, AT = 10.0 K 10.0 J 3(8.314J/mol-K)(10.0K) = 0.080 moles The mass depends on the (unknown) molecular weght of the gas. 19.50 GMM = 40, whch corresponds to Argon 19.51 Q = AI/ = 14.9J m = 1 g AT = 473 K - 373 K = 100 K 14.9 J (10-3 kg)(100 K) = 149 J/kg-K For monatomc gas, GMM 3 fc _3 R 2m~ 2 GMM = =83.7g - C.1 total heat capacty total mass 37.42 J/K+24.94 J/K 2.00(l.OxlO~ :1 kg)+3.00(20.0xlo~ : kg) = 1010 J/ kg -K For Cp, the becomes =*cp = x xl006 J/kg-K = 1680 J/kg-K Note 7 = 1.66; no change n mxture. 19.54 V z = 0.91 L = 0.980. Adabatc: 19.56 Q = 3.0 x 10 4 J = 1.03 atm
20.10 The change n length of the mercury column s proportonal to the change n temperature. L - LQ bolng ~~ LQ _ r - TO Tbolng ~ TQ Snce we are concerned wth temperature changes, the zero pont of the scale s rrelevant and we may use Cecus temperatures. So: 4.3 cm - 1.2 cm T-Q C 15.4 cm - 1.2 cm ~ 100 C - 0 C Then: The lab temperature s 22 C. 20.12 See solutons manual. **~^X 20.1 4)Vfe use the van der ty&als equaton of state (equaton 20.2): For sulfur doxde, Table 20.1 gves a = 6.71 L 2 -atm/mol 2 and b = 0.056 L/mol. The volume per mole s: V m = j±- = 3. 25 L/mol 2.00 mo a\(v m -b) (_ 6.71 L 2 -atm/mol 2 \ 3.25 L/mol -0.056 L/mol -( 7 - ( (3.25 L/mol) 2 )' 8.3145 J/mol - K = (7.00 atm 4-0. 6353 atm), = (2.. 9331 L. atm. K/J) L 13 * ^ 8.3145 J/mol K ' 10 3 L 1 atm =r 297 K\ For an4deal gas we would have: and so PV = T = = (7.00 atm) (6.50 L) 1 m 3 1.013 x 1Q 5. ^ = XR ~ (2.00 mol) (8.3145 J/mol K) 10 3 L 1 atm whch s 20 K lower. 20.16 In the van der \Vfcas model, the pressure v m -b v% would become nfnte once the molecules touch, that s, when V m b. Thus the volume per molecule s b/n* and we may estmate the dameter usng a sphercal model: *J~± 6 " N A
Because of the latent heat term, the steam carres more energy to the skn, and thus causes more severe burns. For example, f T^ = 70 F= f (70-32) C= 21 C, T 5team = 110 C, and m&aao = m^a, Qaeam _ 2. 62 X 10 6 _ Qwater ~ 3. 31 X 10 5 = The rato ncreases as the steam temperature ncreases or the skn temperature ncreases. 20.22 The latent heat released s: Q = Thus the energy released per square klometer s = (10 4 kj/m 2 ) f ^ V = 10 10 kj/km 2 = 10 TJ/km 2 v ' 1km 20.24 The energy requred s: Q = ML s = (0.30 kg) (399 kj/kg) = 119. 7 kj whch rounds to 120 kj. The tme requred, assumng no losses, s: 20.26 Every atom n the glass separates from every other at approxmately the same rate as the temperature s ncreased. Thus each atom on the surface of the bubble moves farther from ts neghbor, and the surface area ncreases. Thus the dameter of the bubble ncreases. 20.28. See solutons manual. 20JO/The spacng must be suffcent to allow for the expanson of the rals. The expecwfemperature change s 130 F -40 F = 90 F = 50 K. Each ral expands by: AX = alat = (11 x 10~ 6 /K) (16 m) (50 K) = 0.0088 m Thus the spacng between the rals should be at least 0.9 cm. 20.32 The steel expands when heated, so that the dameter of the hole s greater than the dameter of the copper rod. As the steel cools back to ts orgnal temperature, the dameter shrnks agan, so that t s slghtly less than the dameter of the rod. The rod s slghtly compressed and the assembly s frmly held together. As the temperature s decreased, both steel and copper shrnk. But a for copper (Table 20.3, 16.6 x 10~ 6 /K) s greater than that for steel (11 x 10~ 6 /K), so the copper shrnks more than the steel. Once the rod's dameter s less than the steel's dameter, the sleeve separates from the rod. For each metal.