= r. / cisely It was not isothermal, nor exactly adia- ! If / l/l /! i i \ i LjSj?

Transcription

1 376 Lea & Burke Physcs; The Nature of Thngs J»(Pa) At constant V A \ LjSj?! '/! If / l/l /!! j j J [ ^ 1 I X j j> 1 ' : / J! 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 = 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% ) a) P = atm b) W() 1660 J; W(z) J c) T (1) = 200K; T (2) = 256 K _d) Q (1) = J; Q (2) = J n.9.47tank Volume s V = m 3 Pressure P 5.00 atm Temperature T = 293 K Number of moles from PV = MPT pv \r RT (5.00 atm x x 1Q 5 ^) ( m 3 ] (8.314 J/mol-KX293K) = 3.33 moles of Argon Perform sothermal expanson => PM = Q.Q160 m = 5.00 atmx atm = m 3 Note, the temperature s stll 293 K. The fnal volume s Vo = Vt T 1 V f = 0.027m 3 pressure drops:, atm m 3 = 293 K x - x 3.00 atm = 97.7 K Work was done between the frst two ponts at constant T. None s done at constant volume 210K 0.6 m K X 0.5 m 3 = atnr\ 0.05 W,3 = I P(V) dv = r Jv, V

2 asreaured. From equaton 19.18: 5Jfc * = 2m T 2 Cp J/kg K Thus the atomc mass s x 10~ 26 / (1.66 x 10" 27 ) = 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 *) < 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 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 Snce the process s adabadc, the heat transfer s zero. Then from the frst law 1 mol of monatomc deal gas, W _ ( J/mol - K) (100 ^r^ x 10 3 J The mnus sgn means that work s done on the gas 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

3 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 kg.9.52 P = 1.02 x 10 5 Pa; Q = 150 J mol neon Ne, 2.00 mol H. The heat ca pacty of the neon s: CWe = 2^"^ = I (3.00 mol) ( J/mol-K) = J/K He; same Constant volume process, deal gas monatomc The heat capacty of the hydrogen s: C H = (2.00 mol)(8.3145j/mol-k) = 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) = moles The mass depends on the (unknown) molecular weght of the gas GMM = 40, whch corresponds to Argon Q = AI/ = 14.9J m = 1 g AT = 473 K 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 J/K 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 V z = 0.91 L = Adabatc: Q = 3.0 x 10 4 J = 1.03 atm

4 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 cm T-Q C 15.4 cm cm ~ 100 C - 0 C Then: The lab temperature s 22 C See solutons manual. **~^X )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 = L/mol. The volume per mole s: V m = j±- = L/mol 2.00 mo a\(v m -b) (_ 6.71 L 2 -atm/mol 2 \ 3.25 L/mol L/mol -( 7 - ( (3.25 L/mol) 2 )' J/mol - K = (7.00 atm atm), = ( L. atm. K/J) L 13 * ^ 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 x 1Q 5. ^ = XR ~ (2.00 mol) ( J/mol K) 10 3 L 1 atm whch s 20 K lower 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

5 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 _ X 10 6 _ Qwater ~ X 10 5 = The rato ncreases as the steam temperature ncreases or the skn temperature ncreases The latent heat released s: Q = Thus the energy released per square klometer s = (10 4 kj/m 2 ) f ^ V = kj/km 2 = 10 TJ/km 2 v ' 1km The energy requred s: Q = ML s = (0.30 kg) (399 kj/kg) = kj whch rounds to 120 kj. The tme requred, assumng no losses, s: 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 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) = m Thus the spacng between the rals should be at least 0.9 cm 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.