f5ch20 - 558 CHAPTER 20 THE KINETIC THEORY OF GASES CHAPTER...

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Unformatted text preview: 558 CHAPTER 20 THE KINETIC THEORY OF GASES CHAPTER 20 Answer to Checkpoint Questions 1. 2. 3. 4. 5. all but c (a) all tie; (b) 3, 2, 1 gas A 5 (greatest change in T ), then tie of 1, 2, 3, and 4 1, 2, 3 (Q3 = 0, Q2 goes into work W2 , but Q1 goes into greater work W1 and increases gas temperature Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. increased but less than doubled the cooler room a, c, b tie of a and c, then b, then d 1180 J 1{4 d, tie of a and b, then c 20 J constant-volume process (a) 3; (b) 1; (c) 4; (d) 2; (e) yes (a) same; (b) increases; (c) decreases; (d) increases (a) 1, 2, 3, 4; (b) 1, 2, 3 4J a, 4; b, 8; c,1; d, 5; e, 3; f, 7; g, 2; h, 6 CHAPTER 20 THE KINETIC THEORY OF GASES 559 15. (a) 1, polyatomic; 2, diatomic; 3, monatomic; (b) more Solutions to Exercises & Problems (a) The number of moles is 1E m 2 5g n = M = 197 :g/mol = 0:0127 mol : Au (b) The number of atoms is m N = nNA = M NA = (0:0127 mol)(6:02 1023 = mol) = 7:65 1021 : 2E Each atom has a mass of m = M=NA , where M is the molar mass and NA is the Avogadro constant. The molar mass of arsenic is 74:9 g/mol or 74:9 10 3 kg/mol. 7:50 1024 arsenic atoms have a total mass of (7:50 1024 )(74:9 10 3 kg/mol)=(6:02 1023 mol 1 ) = 0:933 kg. 18 g of water has 6:021023 water molecules. So 1 gram of water has N = (1=18)(6:021023 ) water molecules. If these molecules were distributed uniformly over the surface of the Earth then the molecular number density would be 3P N = 6:02 1023 =18 = 6560= cm2 ; D = 4R2 4(6:4 108 cm)2 e i.e., there would be 6560 water molecules per cm2 . Let the volume of water in quesiton be v. The number of molecules it contains is then vNA =M , where is the density of water and M is the molar mass of water. If we denote the total volume of the water in all the ocaens as V , then 4P V = vNA : v M 560 CHAPTER 20 THE KINETIC THEORY OF GASES Solve for v: 6 2 3 m)(1:8 10 2 VM v = N ' (0:75)(4)(6::4 10 3m) (53 100 1023 /mol) kg/mol) (1 0 10 kg/m )(6: A 6 m3 = (2 cm)2 : ' 8 10 s s This is approximately the volume of water contained in a tablespoon. Since the molar mass of the ink is M = 18 g/mol, the number of molecules in m = 1 g = 10 6 g of ink is mNA = (10 6 g)(6:02 1023 mol 1 ) ' 3 1016 : N= M 18 g/mol The number of creatures in our galaxy, with the assumption made in the problem, is about N 0 = (5 109 )(1011 ) = 5 1020 : So the statement was wrong by a factor of about 20,000. (a) Use pV = nRT to solve for the volume V : V = nRT = (1:00 mol)(8:31 J/mol K)(273 K) = 2:24 10 3 m3 ; p 1:01 105 Pa which is 22:4 L. (b) The Loschmidt number is 23 :00 3 nL = (1(22:cm )(6:02 310 3/mol) = 2:69 1019 : 4 L/mol)(10 cm =L) 6E 5P (a) Solve pV = nRT for n: n = pV=RT = (100 Pa)(1:0 10 6 m3 )=(8:31 J/molK)(220 K) = 5:47 10 8 mol. (b) The number of molecules N is the product of the number of moles n and the number of molecules in a mole NA (the Avogadro constant). Thus N = nNA = (5:47 10 6 mol)(6:02 1023 mol 1 ) = 3:29 1016 molecules. Use pV = nRT . The number of molecules per cubic centimeter is 13 6 m3 )(6: 23 N = pV NA = (1:01 10 Pa)(1:00 10K)(293 K)02 10 /mol) = 25 : RT (8:31 J/mol 8E 7E CHAPTER 20 THE KINETIC THEORY OF GASES 561 Let T1 = 10:0 C = 283 K, T2 = 30:0 C = 303 K, p1 = 100 kPa; p2 = 300 kPa; and V1 = 2:50 m3 . (a) Use p1 V1 =T1 = nR to solve for n: 3 Pa)(2:50 3 n = p1 V1 = (100 10 :31 J/mol m ) = 106 mol : T1 R (283 K)(8 K) 9E (b) For the same amount of ideal gas, when the pressure changes to p2 , its temperature changes to T2 . So p2 V2 = nRT2 , which we solve for V2 : T V2 = nRT2 = T2 p2 1 303 K = 283 K 100 103 Pa (2:50 m3 ) = 0:892 m3 : 300 103 Pa p1 V p2 1 (a) Solve pV = nRT for n. First convert the temperature to the Kelvin scale: T = 40:0+273:15 = 313:15 K. Also convert the volume to m3 : 1000 cm3 = 100010 6 m3 . Then n = pV=RT = (1:01 105 Pa)(1000 10 6 m3 )=(8:31 J/molK)(313:15 K) = 3:88 10 2 mol. (b) Solve pV = nRT for T : T = pV=nR = (1:06 105 Pa)(1500 10 6 m3 )=(3:88 10 2 mol)(8:31 J/mol K) = 493 K = 220 C. Let V1 = 1000 in.3 ; V2 = 1020 in.3 ; p1 = 24:0 lb/in.2 ; T1 = 0:00 C = 273 K; and T2 = 27:0 C = 300 K. Since the number of moles of the gas remains the same, from pV = nRT we have n = p1 V1 =RT1 = p2 V2 =RT2 . Thus 11E 10E T p2 = T2 1 V1 p = 300:15 K V2 1 273:15 K 1000 in.3 (24:0 lb/in.2 ) = 27:0 lb/in.2 : 1020 in.3 The work done is 12E W= Z vf = (1:01 105 Pa)(22:4 L 16:8 L)(10 3 m3 =L) = 653 J : vi p dV = p(Vi Vf ) 562 CHAPTER 20 THE KINETIC THEORY OF GASES (a) The number of moles of the air molercules is n = pV=RT = N=NA . So 3 1 pV 1023 105 N = RT NA = (1:01 (8:31Pa)(1 m )(6:02 15 K)mol ) J/mol K)(293: 25 ; = 2:5 10 13P where T = 20 C = 293 K and P = 1 atm = 1:01 105 Pa were substituted. (b) Let Mair be the molar mass of the air. Then Mair = 3 MN2 + 1 MO2 . So the mass of 4 4 the air is pV 3 M + 1 M m = Mair n = RT 4 N2 4 O2 (1:01 105 Pa)(1 m3 ) 3 (14 g/mol) + 1 (32 g/mol) = (8:31 J/mol K)(293:15 K) 4 4 = 1:2 kg : 14P Since the pressure is constant the work is given by W = p(V2 V1 ). The initial volume is V1 = (AT1 BT12 )=p, where T1 is the initial temperature. The nal volume is V2 = (AT2 BT22 )=p. Thus W = A(T2 T1 ) B (T22 T12 ). 15P Suppose the gas expands from volume V1 to volume V2 during the isothermal portion of the process. The work it does is Z dV = nRT ln V2 ; p dV = nRT W= V1 V1 V1 V where pV = nRT was used to replace p with nRT=V . Now V1 = nRT=p1 and V2 = nRT=p2 , so V2 =V1 = p2 =p1 . Also replace nRT with p1 V1 to obtain p W = p1 V1 ln p1 : 2 Since the initial gauge pressure is 1:03 105 Pa, p1 = 1:03 105 Pa + 1:01 105 Pa = 2:04 105 Pa. The nal pressure is atmospheric pressure: p2 = 1:01 105 Pa. Thus 5 Pa)(0:14 m3 ) ln 2:04 Pa = 2:00 104 J : W = (2:04 10 1:01 Pa During the constant pressure portion of the process the work done by the gas is W = p2 (V1 V2 ). Notice that the gas starts in a state with pressure p2 , so this is the pressure V2 Z V2 CHAPTER 20 THE KINETIC THEORY OF GASES 563 W = p2 V1 p1 V1 = (p2 p1 )V1 p2 5 Pa 2:04 105 Pa)(0:14 m3 ) = 1:44 104 J : = (1:01 10 The total work done by the gas over the entire process is W = 2:00 104 J 1:44 104 J = 5:6 103 J. The diagrams are shown below. p cosntant pressure V constant volume isothermal isothermal constant volume V T cosntant pressure 16P throughout this portion of the process. Also note that the volume decreases from V2 to V1 . Now V2 = p1 V1 =p2 , so p cosntant pressure isothermal constant volume T (d) You can use pV = nRT to analyze each case. Note that n is proportional to the mass M of the ideal gas. For example, the slope of the constant-volume line in the p-T diagram is p=T = nR=V / mass of the ideal gas. The pressure p1 due to the rst gas is p1 = n1 RT=V , and the pressure p2 due to the second gas is p2 = n2 RT=V . So the total pressure on the container wall is 17P RT RT p = p1 + p2 = n1V + n2V = (n1 + n2 ) RT : V 564 CHAPTER 20 THE KINETIC THEORY OF GASES The fraction of P due to the second gas is then p2 = 0 n2 RT=V 1 = n n2 n = 2 +:5 :5 = 5 : p (n1 + n2 )(RT=V ) 1 + 2 0 18P (a) Use pV = nRT to solve for n: 3 3) (2 5 n = p1 V1 = (8::31 10 Pa)(1:0 mK) RT1 J/mol K)(200 = 1:5 mol : p (kN/m2) 5.0 b (b) TB = pB VB nR (7:5 103 Pa)(3:0 m3 ) = (1:5 mol)(8:31 J/mol K) = 1:8 103 K : (c) 2.0 a c V (m 3) 2.0 4.0 pC VC = (2:5 103 Pa)(3:0 m3 ) = 6:0 102 K : TC = nR (1:5 mol)(8:31 J/mol K) (d) Use Q = E int + W . For the cyclic process which starts at a and ends at a we have T = 0, so E int = 0. Thus 1 Q = W = 2 (Vc Va )(pb pc ) 1 = 2 (3 m3 1 m3 )(7:5 103 Pa 2:5 103 Pa) = 5:0 103 J : (a) The work done by the gas in the process is W = pV = (25 N/m3 )(1:8 m3 3:0 m3 ) = 30 J: Thus the change in internal energy of the gas is E int = Q W = 75 J ( 30 J) = 45 J : (b) The nal temperature is 3 Tf = pf Vf Ti = (1:8 m:0)(300 K) = 180 K : pi Vi 3 m3 19P CHAPTER 20 THE KINETIC THEORY OF GASES 565 Denote the lowerand higher elevations with subscripts 1 and 2, respectively. Then The gas volume V2 at the higher elevation is then p1 T2 V = (76 cmHg)(273 48) K (2:2 m3 ) = 3:4 m3 : V2 = p T 1 (38 cmHg)(273 + 20) K 2 1 Assume that the pressure of the air in the bubble is essentially the same as the pressure in the surrounding water. If d is the depth of the lake and is the density of water, then the pressure at the bottom of the lake is p1 = p0 + gd, where p0 is atmospheric pressure. Since p1 V1 = nRT1 , the number of moles of gas in the bubble is n = p1 V1 =RT1 = (p0 + gd)V1 =RT1 , where V1 is the volume of the bubble at the bottom of the lake and T1 is the temperature there. At the surface of the lake the pressure is p0 and the volume of the bubble is V2 = nRT2 =p0 . Substitute for n to obtain T2 p0 + gd V V = 2 3 2 5 998 103 293 K = 277 K 1:013 10 Pa + (0:1:013 10kg/m )(9:8 m/s )(40 m) (20 cm3 ) 5 Pa = 100 cm3 : 21P 20P p1 V1 = p2 V2 : T1 T2 T1 p0 1 Consider the open end of the pipe. The balance of the pressures inside and outside the pipe requires that p + w (L=2)g = p0 + w hg; where p0 is the atmospheric pressure, and p is the pressure of the air inside the pipe, which satis es p(L=2) = p0 L, or p = 2p0 . Solve for h: p 1 01 105 Pa 0 h = p g 0 + L = (1:00 10:3 kg/m3 )(9:80 m/s2 ) + 25:2 m = 22:8 m : 2 w Let the mass of the balloon be mb and that of the lift be ml . According to Archimedes's Principle the buoyant force is F = air gVb . The equilibrium of forces for the balloon requires ml g = F (mb + mair )g. Solve for the weight of the air in the balloon: mair g = air gVb ml g mb g = (7:56 10 2 lb/ft3 )(77; 000 ft3 ) 600 lb 500 lb = 4:7 103 lb = 2:1 104 N : 23P 22P 566 CHAPTER 20 THE KINETIC THEORY OF GASES So mair = 2:1 104 N=9:80 m/s2 = 2:1 103 kg: Now for the air inside the balloon pin Vb = nRTin = (mair =Mair )RT in , where Mair is the molar mass of the air and pin = pout = 1 atm. So the temperature inside the balloon is T in = pin Vb Mair Rmair (1:01 105 Pa)(77; 000 ft3 )(2:832 10 2 m3 =ft3 )(0:028 kg/mol) = (8:31 J/mol K)(2142:8 kg) C = 198 F : = 365 K = 92 (a) Use pV = nRT . The volume of the tank is 31 J/mol K)(273 C + 77 C) V = nRT = (300 g=17 g/mol)(8::35 106 Pa p 1 2 m3 = 38 L : = 3:8 10 (b) The number of moles of the remaining gas is 5 p0 V (8 7 10 2 m3 ) n0 = RT 0 = (8::31 10 Pa)(3:8 C + 22 C) = 13:5 mol : J/mol K)(273 24P The mass of the gas that leaked out is then m = 300 g (13:5 mol)(17 g/mol) = 71 g: When the valve is closed the number of moles of the gas in container A is nA = pA VA =RTA and that in container B is nB = 4pB VA =RTB : The total number of moles in both containers is then VA pA + 4pB = const. n=n +n = A B 25P TB After the valve is opened the pressure in container A is p0A = Rn0A TA =VA and that in container B is p0B = Rn0B TB =4VA . Equate p0A and p0B to obtain Rn0A TA =VA = Rn0B TB =4VA , or n0B = (4TA =TB )n0A . Thus p p T n = n0A + n0B = n0A 1 + 4T A = nA + nB = VA TA + 4T B : R A B B Solve the above eauation for n0A : R TA n0A = V (pA =TA 4+ 4pB =TB ) : R (1 + T =T ) A B CHAPTER 20 THE KINETIC THEORY OF GASES 567 Substitute this expression for n0A into p0 VA = n0A RTA to obtain the nal pressure p0 : 0 RT B A p0 = nAV A = pA1+ 44pT T=T=TB ; + A B A which gives p0 = 2:0 105 Pa. The rms speed of helium atoms at 1000 K is 26E Vrms = r 3RT = mN A s 3(8:31 J/mol K)(1000 K) = 2:50 103 m/s : 4:00 10 3 kg/mol 27E According to the kinetic theory the rms speed is RT vrms = 3M ; where T is the temperature and M is the molar mass. According to Table 20-1 the molar mass of molecular hydrogen is 2:02 g/mol = 2:02 10 3 kg/mol, so J/mol vrms = 3(8:3102 10K)(2:7 K) = 180 m/s : 3 kg 2: s r The rms speed of argon atoms at 313 K is 28E vrms = r 3RT = mN A s 3(8:31 J/mol K)(313 K) = 442 m/s : 39:9 10 3 kg/mol Use vrms = (3kT=me )1=2 , where me is the mass of the electron and k is the Boltzmann constant. So for electrons inside the Sun 3(1:38 10 23 J/K)(2:00 106 K) 1=2 = 9:53 106 m/s : vrms = 9:11 10 31 kg 29E 568 CHAPTER 20 THE KINETIC THEORY OF GASES (a) 30E vrms = r (b) Since vrms / T , the temperature at which vrms will be half that value is T1 = (1=2)2 T = 293 K=4 = 73:3 K = 200 C, and the temperature at which vrms will be twice that value is T2 = 22 T = 4(293 K) = 1:17 103 K = 899 C. Use vrms = 3RT=M , where M is the molar mass of the ideal gas. So 3RTH2 1=2 = 3RTHe 1=2 ; M M 31E p 3RT = mN A s 3(8:31 J/mol K)(273 C + 20:0 C) = 511 m/s : 28:0 10 3 kg/mol p H2 He which yields M 4 00 g/mol THe = MHe TH2 = 2::00 g/mol (20:0 + 273) K = 586 K = 313 C : H2 Use pV = nRT = (m=M )RT , where m is the mass of the gas and M is its molar mass. The density of the gas is then = m=V = pM=RT: (a) Now 32P vrms = = r s 3RT = M r 3p 3(1:00 10 2 atm)(1:01 105 Pa/atm) = 494 m/s : 1:24 10 2 kg/m3 (b) The molar mass of the ideal gas is 3 :31 J/mol K)(273 10 2 M = RT = (1:24: 10 kg/m )(801 105 Pa/atm) K) = 0:028 g/mol : 2 atm)(1: p (1 0 The gas is therefore nitrogen (N2 ). 33P On re ection only the normal component of the momentum changes, so for one molecule the change in momentum is 2mv cos , where m is the mass of the molecule, v is its speed, and is the angle between its velocity and the normal to the wall. If N molecules collide CHAPTER 20 THE KINETIC THEORY OF GASES 569 with the wall the change in their total momentum is P = 2Nmv cos and if the total time taken for the collisions is t then the average rate of change of the total momentum is P=t = 2(N=t)mv cos . This the average force exerted by the N molecules on the wall and the pressure is the force per unit area: 2 N p = A t mv cos 2 = 2:0 10 4 m2 (1023 s 1 )(3:3 10 27 kg)(1:0 103 m/s) cos 55 = 1:9 103 Pa : Notice that the value given for the mass was converted to kg and the value given for the area was converted to m2 . 2 (a) The average translational energy is given by K = 3 kT , where k is the Boltzmann 23 J=K ) and T is the temperature on the Kelvin scale. Thus K = constant (1:38 10 3 (1:38 10 23 J/K)(1600 K) = 3:31 10 20 J. Round to 3:3 10 20 J. 2 (b) Since 1 eV = 1:60 10 19 J, K = (3:31 10 20 J)=(1:60 10 19 J/eV) = 0:21 eV. 2 The translational kinetic energy of a particle is K = 3 kT and one electron-volt is equal to 19 J. 1:6 10 (a) At 0 C (273:15 K), 3 3 K = 2 kT = 2 (1:38 10 23 J/K)(273:15 K) 21 J = 5:65 10 21 J = 1:5:65 10 10 J/eV = 0:0353 eV ; 6 19 and at T = 100 C (373 K), 3 3 K = 2 kT = 2 (1:38 10 23 J/K)(373:15 K) 21 J = 7:72 10 21 J = 1:7:72 10 10 J/eV = 0:0483 eV: 6 19 35E 34E 2 (b) For one mole of ideal gas at T = 273 K, K = 3 RT = 3 (8:31 J/K mol)(273:15 K) = 2 2 3:40 103 J. At T = 373 K, K = 3 RT = 3 (8:31 J/K mol)(373:15 K) = 4:65 103 J. 2 Let the temperature be T , then K = 3kT=2 = 1:00 eV: Thus 19 J/eV) = 7:73 103 K : T = 2K = 2(1:00 eV)(1:610 10 J/K) 3k 3(1:38 23 36E 570 CHAPTER 20 THE KINETIC THEORY OF GASES The gravitational kinetic energy of one oxygen molecule is U = mgh, and the translational 2 kinetic energy of the same molecule is K = 3 kT . So 37E U = 3mgh = (0:032 kg=NA )gh = 2(0:032 kg)(9:80 m/s2 )(0:1 m) = 9:1 10 6 : 3kT=2 3(8:31 J/mol K)(273 K) K 3kT=2 Here the relation NA k = R was used. (a) See 32P. (b) pV = nRT = n(kNA )T = (nNA )kT = NkT: (a) Use = L=N , where L is the heat of vaporization and N is the number of molecules per gram. The molar mass of atomic hydrogen is 1 g/mol and the molar mass of atomic oxygen is 16 g/mol so the molar mass of H2 O is 1 + 1 + 16 = 18 g/mol. There are NA = 6:02 1023 molecules in a mole so the number of molecules in a gram of water is (6:02 1023 mol 1 )=(18 g/mol) = 3:341022 molecules/g. Thus = (539 cal/g)=(3:341022 g 1 ) = 1:61 10 20 cal. This is (1:61 10 20 cal)(4:186 J/cal) = 6:76 10 20 J. (b) The average translational kinetic energy is 3 K = 3 kT = 2 (1:38 10 23 J/K) [(32:0 + 273:13) K] = 6:32 10 21 J: 2 The ratio of K= = (6:76 10 20 J)=(6:32 10 21 J) = 10:7. They are not equivalent. Avogadro's law does not tell how the pressure, volume, and temperature are related, so you cannot use it, for example, to calculate the change in volume when the pressure increases at constant temperature. The ideal gas law, however, implies Avogadro's law. It yields N = nNA = (pV=RT )NA = pV=kT , where k = R=NA was used. If the two gases have the same volume, the same pressure, and the same temperature, then pV=kT is the same for them. This implies that N is also the same. The mean free path of a particle is given by = ( 2d2 n) 1 . Solve for d: 1 d= p 1 = p 2n 2(0:80 10 7 m)(2:7 1025 = m3 ) = 3:2 10 10 m = 0:32 nm : s s 41E 40P 39P 38P p CHAPTER 20 THE KINETIC THEORY OF GASES 571 42E (a) According to Eq. 20-22 the mean free path for molecules in a gas is given by ; = p 1 2d2 N=V where d is the diameter of a molecule and N is the number of molecules in volume V . Substitute d = 2:0 10 10 m and N=V = 1 106 molecules/m3 to obtain = p 2(2:0 1010 m)2 (1 106 m 3 ) 1 = 6 1012 m : (b) At this altitude most of the gas particles are in orbit around the Earth and do not suer randomizing collisions. The mean free path has little physical signi cance. 43E Substitute d = 1:0 10 2 m and N=V = 15=(1:0 10 3 m3 ) = 15 103 beans/m3 into = p 1 2d2 N=V to obtain 1 = 0:15 m : 2(1:0 10 2 m)2 (15 103 m 3 ) The conversion 1:00 L = 1:00 10 3 m3 was used. = p The average frequency is 44E 2 v f = = 2d vN : V p (a) Use pV = nRT = NkT , where p is the pressure, V is the volume, T is the temperature, n is the number of moles, and N is the number of molecules. The substitutions N = nNA and k = R=NA were made. Since 1 cm of mercury = 1333 Pa, the pressure is p = (10 7 )(1333) = 1:333 10 4 Pa. Thus 45P N= p = 1:333 10 4 Pa V kT (1:38 10 23 J/K)(295 K) = 3:27 1016 molecules/m3 = 3:27 1010 molecules/cm3 : 572 CHAPTER 20 THE KINETIC THEORY OF GASES (b) The molecular diameter is d = 2:00 10 10 m, so the mean free path is = p 1 =p 2d2 N=V 2(2:00 10 46P 1 10 m)2 (3:27 1016 m 3 ) = 172 m : Let the speed of sound be vs and its frequency be f . Then vs = = p 1 ; f 2d2 N=V which gives N = p2d2 v N = p2d2 v N pM f= s V s A s A RT 5 p = 2(3:0 10 10 m)2 (331 m/s)(6:02 1023 /mol) (1:01 10 Pa)(0:032 kg/mol) (8:31 J/mol K)(273 K) p 2d2 v = 3:7 109 Hz = 3:7 GHz : For an ideal gas pV = nRT . (a) When n = 1, V = Vm = RT=p, where Vm is the molar volume of the gas. So 47P Vm = RT = (8:31 J/molK)(273:15 K) = 22:5 L : p 1:01 105 Pa (b) Use vrms = 3RT=M: The ratio is given by p (c) Use He = ( 2d2 n ) 1 ; where n is the number of particles per unit volume given by n = N=V = NA n=V = NA p=RT = p=kT . So p vrms,He = MNe = 20 g = 2:25 : vrms,Ne MHe 4:0 g r r He = p = p kT 2 2d p 23 15 K) (1: = 1:41438 10 10 J/K)(273: 105 Pa) = 0:84 m : 2 (1:01 (1 10 m) 2d2 (p=kT ) 1 (d) Ne = He = 0:84 m: CHAPTER 20 THE KINETIC THEORY OF GASES 573 dAr = N2 = 27:5 10 6 cm = 1:7 : dN2 Ar 9:9 10 6 cm (b) Since / (N=V ) 1 / (p=T ) 1 ; 75 cmHg C; 75 cmHg) = 273 C 40 C (9:9 10 6 cm) = 7:9 10 6 cm : Ar ( 40 273 C + 20 C (a) Since / d 2 , 48P r r Ar (20 C; 15 cmHg) = 15 cmHg 1 (9:9 10 6 cm) = 5:0 10 5 cm ; This tells us that 1013 air molecules are needed to cover the period. (b) Assume that every second there are N air molecules which collide with the period. If each one of them bounces back elastically after the collision then the change in linear momentum per molecule per collision is 2mvx , where m is the molecular mass and vx is the component of the molecular velocity in the direction perpendicular to the surface of the paper containing the period. Thus the pressure exerted by the air molecules on the period is v p = 2mN tx ; A 2 2 where t = 1 s and vx (vx ) 1 = vrms = 3. Also m = M=NA , where M is the average molar mass of the air molecules. Solve for N : (a) Using a ruler, we nd the diameter of the period D to be roughly 0:5 mm. So its area is A = D2 =4 2 10 7 m2 . Meanwhile, we estimate the diameter d of an air molecule to be roughly 2 10 10 m. So the area an air molecule covers is a = d2 =4 3 10 20 m2 . Thus A 2 10 7 1013 : a 3 10 20 49P p p N = 3pANA t = pANA t 2Mvrms 2 MRT 5 7 2 23 p = (1:01 10 Pa)(2 10 m )(6:02 10 /mol)(1 s) 7 1020 : 2 (0:028 kg/mol)(8:31 J/mol K)(300 K) 50P p (a) The average speed is v = Nv ; P 574 CHAPTER 20 THE KINETIC THEORY OF GASES where the sum is over the speeds of the particles and N is the number of particles. Thus 0 v = (2:0 + 3:0 + 4:0 + 5:0 + 6:0 + 7:10+ 8:0 + 9:0 + 10:0 + 11:0) km/s = 6:5 km/s : (b) The rms speed is given by vrms = Now X rP v2 : N i v2 = h (2:0)2 + (3:0)2 + (4:0)2 + (5:0)2 + (6:0)2 = 505 km2 /s2 ; so + (7:0)2 + (8:0)2 + (9:0)2 + (10:0)2 + (11:0)2 km2 /s2 s vrms = 51E 505 km2 /s2 = 7:1 km/s : 10 ni vi = 2(1:0) + 4(2:0) + 6(3:0) + 8(4:0) + 2(5:0) = 3:2 cm/s : 2+4+6+8+2 ni pP P (b) From vrms = ni vi 2 = ni we get v= P (a) P 2 2+ : 2 2 2 vrms = 2(1:0) + 4(2:0) + 46(360) 8+ 8(4:0) + 2(5:0) = 3:4 cm/s : 2 + + +2 r (c) There are eight particles at v = 4:0 cm/s, more than the number of particles at any other single speed. So 4:0 cm/s is the most probable speed. (a) 52E N 1 X v = 1 [4(200 m/s) + 2(500 m/s) + 4(600 m/s)] = 420 m/s ; v=N i 10 vrms = v u u t r N 1 X v2 = 1 [4(200 m/s)2 + 2(500 m/s)2 + 4(600 m/s)2 ] = 458 m/s : i N 10 i=1 i=1 Indeed we see that vrms > v. CHAPTER 20 THE KINETIC THEORY OF GASES 575 (b) You may check the validity of the inequality vrms v for any distribution you can think of. For example, imagine that the ten perticles are divided into two groups of ve particles each, with the rst group of particles moving at speed v1 and the second group at v2 . Obviously now v = (v1 + v2 )=2 and 2 2 vrms = v1 + v2 v1 + v2 = v : 2 2 r (c) If and only if all the individual molecular speeds are the same does vrms equal v. (a) v < vrms < vP . (b) For Maxwell distribution vP < v < vrms . Use vp = 2RT2 =M and vrms = 3RT1 =M: The ratio is 54E 53E p p T2 = 3 vp T1 2 vrms 2 3 = 2: (a) The rms speed of molecules in a gas is given by vrms = 3RT=M , where T is the temperature and M is the molar mass of the molecules. The speed required for escape p from the Earth's gravitational pull is v = 2gre , where g is the acceleration due to gravity at the Earth's surface and re (= 6:37 106 m) is the radius of the Earth. This expression can be derived easily. Take the zero of gravitational potential energy to be at in nity. Then energy of a particle with speed v and mass m at the Earth's surface is E = mgre + 1 mv2 . 2 If the particle is just able to travel far away its kinetic energy must tend toward 0 as its p distance from the Earth tends to 1. This means E = 0 and v = 2gre . p p Equate the expressions for the speeds to obtain 3RT=M = 2gre . The solution for T is T = 2gre M=3R. The molar mass of hydrogen is 2:02 10 3 kg/mol, so for that gas 2(9:8 m/s2 )(6:37 106 m)(2:02 10 3 kg/mol) = 1:0 104 K : T= 3 (8:31 J/mol K) The molar mass of oxygen is 32:0 10 3 kg/mol, so for that gas 2(9:8 m/s2 )(6:37 106 m)(32:0 10 3 kg/mol) = 1:6 105 K : T= 3 (8:31 J/mol K) 55P p 576 CHAPTER 20 THE KINETIC THEORY OF GASES (b) Now T = 2gm rm M=3R, where rm (= 1:74 106 m) is the radius of the moon and gm (= 0:16g) is the acceleration due to gravity at the moon's surface. For hydrogen 2(0:16)(9:8 m/s2 )(1:74 106 m)(2:02 10 3 kg/mol) = 4:4 102 K : T= 3(8:31 J/mol K) For oxygen 2 3 106 m)(32 T = 2(0:16)(9:8 m/s )(1:7431 J/mol K) :0 10 kg/mol) = 7:0 103 K : 3(8: (c) The temperature high in the Earth's atmosphere is great enough for a signi cant number of hydrogen atoms in the tail of the Maxwellian distribution to escape. As a result the atmosphere is depleted of hydrogen. Very few oxygen atoms, on the other hand, escape. 56P (a) The root-mean-square speed is given by vrms = 3RT=M . The molar mass of hydrogen is 2:02 10 3 kg/mol, so K)(4000 K) vrms = 3(8:31 J/mol 3 kg/mol) = 7:0 103 m/s : (2:02 10 s p (b) When the surfaces of the spheres that represent an H2 molecule and an A atom are touching the distance between their centers is the sum of their radii: d = r1 + r2 = 0:5 10 8 cm + 1:5 10 8 cm = 2:0 10 8 cm. (c) The argon atoms are essentially at rest so in time t the hydrogen atom collides with all the argon atoms in a cylinder of radius d and length vt, where v is its speed. That is, the number of collisions is d2 vtN=V , where N=V is the concentration of argon atoms. molr of collisions per unit time is d2 vN=V = (2:0 10 10 m)2 (7:0 103 m/s)(4:0 1025 m 3 ) = 3:5 1010 collisions/s. The formulas for the average speed and vrms are v = 8kT=m and vrms = 3kT=m: In this case r r 8kT = 2v 3kT v2 = m rms1 = 2 m ; 2 1 57E p p so m1 =m2 = 3=2 = 4:7: 58P (a) From the normalization condition Z 1 0 P (v) dv = Z v0 0 3 cv2 dv = 1 cv0 = N 3 CHAPTER 20 THE KINETIC THEORY OF GASES 577 3 we obtain c = 3N=v0 : (b) The average speed is 4 1 Z 1 P (v)v dv = 1 Z v0 cv3 dv = cv0 = 3 v : v=N N 0 4N 4 0 0 (c) The rms speed is s vrms = N 1 Z 1 s 0 P (v)v2 dv = N 1 Z v0 0 cv4 dv = r 5 cv0 = 3 ' 0:775v : 0 5N 5 r (a) The integral of the distribution function is the number of particles: P (v) dv = N . The area of the triangular portion is half the product of the base and altitude, or 1 av0 . 2 R The area of the rectangular portion is the product of the sides, or av0 . Thus P (v) dv = 1 av0 + av0 = 3 av0 and 3 av0 = N , so a = 2N=3v0 . 2 2 2 R (b) The number of particles with speeds between 1:5v0 and 2v0 is given by 12:v0 0 P (v) dv. 5v The integral is easy to evaluate since P (v) = a throughout the range of integration. Thus the number of particles is a(2:0v0 1:5v0 ) = 0:5av0 = N=3, where 2N=3v0 was substituted for a. (c) The average speed is given by 1 Z vP (v) dv : v=N For the triangular portion of the distribution P (v) = av=v0 and the contribution of this portion is 2 a Z v0 v2 dv = a v3 = av0 = 2 v ; 0 3N Nv 3Nv 9 0 0 0 0 59P R where 2N=3v0 was substituted for a. P (v) = a in the rectangular portion and the contribution of this portion is a Z 2v0 v dv = a (4v2 v2 ) = 3a v2 = v : N v0 2N 0 0 2N 0 0 Thus v = 2v0 =9 + v0 = 1:22v0 . (d) The mean-square speed is given by 2 vrms = N 1 Z v2 P (v) dv : 578 CHAPTER 20 THE KINETIC THEORY OF GASES The contribution of the triangular section is The contribution of the rectangular portion is Thus a Z v0 v3 dv = a v4 = 1 v2 : Nv0 0 4Nv0 0 6 0 a Z 2v0 v2 dv = a (8v3 v3 ) = 7a v3 = 14 v2 : N v0 3N 0 0 3N 0 9 0 2 2 vrms = 1 v0 + 14 v0 = 1:3v0 : 6 9 r (a) The internal energy is 3 3 E int = 2 NkT = 2 nRT = 3 (1:0 mol)(8:31 J/mol K)(273 K) = 3:4 103 J : 2 (b) For an ideal gas, as long as T is xed, so is E int , regardless the values of p and V . According to the rst law of thermodynamics E int = Q W . Since the process is isothermal E int = 0 (the internal energy of an ideal gas depends only on the temperature) and Q = W . The work done by the gas as its volume expands from Vi to Vf at temperature T is Z Vf Z Vf dV = nRT ln Vf : pdV = nRT W= V V For n = 1 mol, Q = W = RT ln(Vf =Vi ). 62E 61E 60E Vi Vi i The speci c heat is 3R= 3(8 31 J/mol 2 CV = mN 2 = (6:66 10 :27 kg)(6:02K)=1023 /mol) = 3:11 103 J/kg K : A (a) According to the rst law of thermodynamics Q = E int + W: When the pressure is a constant W = pV . So E int = Q pV = 20:9 J (1:0 105 Pa)(100 cm3 50 cm3 )(10 6 m3 =cm3 ) = 15:9 J : 63P CHAPTER 20 THE KINETIC THEORY OF GASES 579 (b) Q Q Cp = nT = n(pVQ ) = R V =nR p (8:31 J/mol K)(20:9 = (1:01 105 Pa)(50 10 J)m3 ) = 34:4 J/mol K : 6 (c) CV = Cp R = 34:4 J/mol K 8:31 J/mol K = 26:1 J: The straight line on the p-V diagram can be expressed as (p p1 )=(V V1 ) = p1 =V1 ; where p1 V1 = nRT1 . Solve for p: p = (p1 =V1 )V = (p2 =nRT1 )V: Thus 1 (a) 64P W= (b) Z 2V 1 V1 p dV = Z 2V1 V1 p2 V dV = p2 V 2 2V1 = 3 nRT : 1 1 nRT1 2nRT 1 V1 2 1 3 E int = 2 nRT2 (c) 3 nRT = 3 (2p )(2V ) 3 nRT 2 1 2 1 1 2 1 3 9 = (4 1) 2 nRT1 = 2 nRT1 : (d) The molar speci c heat would be 9 Q = W + E int = 3 nRT + 2 nRT1 = 6nRT1 : 2 Q 6 1 C = nT = T6RTT = (2p )(2V6RT1 ) T = 4T RT1T = 2R : 2 1 2 2 )=(nR 1 1 1 When the temperature changes by T the internal energy of the rst gas changes by n1 C1 T , the internal energy of the second gas changes by n2 C2 T , and the internal energy of the third gas changes by n3 C3 T . The change in the internal energy of the composite gas is E int = (n1 C1 + n2 C2 + n3 C3 ) T . This must be (n1 + n2 + n3 )C T , where C is the molar speci c heat of the composite. Thus 1 n2 C = n1 Cn ++ nC2 + n3 C3 : +n 1 2 3 65P 580 CHAPTER 20 THE KINETIC THEORY OF GASES (a) Use CV = 3R=(2mNA ) (see 62E). The mass of the argon atom follows as 3(8: m = 2C3R = 2(75 cal/kg C)(431 J/mol K)0 1023 /mol) :18 J/cal)(6: V NA 26 kg : = 6:6 10 (b) The molar mass of argon is 66P M = mNA = (6:6 10 26 kg)(6:0 1023 /mol) = 40 g/mol : For diatomic ideal gas CV = 5 R = 20:8 J. 2 (a) E int;ac = nCV (Tc Ta ) = nCV pc Vc pA VA R R = 8:3120:8 J K (2:0 103 Pa)(4:0 m3 ) (5:0 103 Pa)(2:0 m3 ) J/mol = 5:0 103 J : (b) The work done from a to c is 1 Wac = 2 (Vc Va )(pb + pc ) = 1 (4:0 m3 2:0 m3 )(5:0 103 Pa + 2:0 103 Pa) 2 = 7:0 103 J ; 67P and the heat added to the gas from a to c is Qac = E int;ac + Wac = 5:0 kJ + 7:0 kJ = 2:0 103 J: (c) Now Qabc = E int;abc + Wabc , where Wabc = pa (Vc Va ) = (5:0 102 Pa)(4:0 m3 20 m3 ) = 1:0 104 J and E int;abc = E int;ac = 5:0 103 J, since E is path-independent. So Qabc = 1:0 104 J 5:0 103 J = 5:0 103 J: CHAPTER 20 THE KINETIC THEORY OF GASES 581 As the volume doubles at constant pressure, so does the temperature in Kelvin. Thus T = (2 1)(273 K) = 273 K: The heat required is then 5 R + R T Q = nCp T = n(CV + R)T = n 2 7 = (1 mol) 2 (8:31 J/mol K)(273 K) = 8 103 J : (a) The molar mass of O2 is 32 g so n = 12 g=32 g = 0:375 mol. (b) For diatomic ideal gas without oscillation Cp = CV + R = 5 R + R = 7 R. Thus 2 2 Q = nCp T = 7 (0:375 mole)(8:31 J/mol K)(100 K) = 1090 J : 2 (c) Use E int = nCV T = 5 nRT: The ratio is 2 E int = 5nRT=2 = 5 = 0:714 : Q 7nRT=2 7 (a) Since the process is at constant pressure the heat added to the gas is given by Q = nCp T , where n is the number of moles in the gas, Cp is the molar speci c heat at constant pressure, and T is the increase in temperature. For a diatomic ideal gas Cp = 7 R. Thus 2 Q = 7 nR T = 7 (4:00 mol)(8:31 J/mol K)(60:0 K) = 6:98 103 J. 2 2 (b) The change in the internal energy is given by E int = nCV T . For a diatomic ideal gas CV = 5 R, so E int = 5 nR T = 5 (4:00 mol)(8:31 J/mol K)(60:0 K) = 4:99 103 J. 2 2 2 (c) According to the rst law of thermodynamics E int = Q W , so W = Q E int = 6:98 103 J 4:99 103 J = 1:99 103 J. (d) The change in the total translational kinetic energy is K = 3 nR T = 3 (4:00 mol)(8:31 J/mol K)(60:0 K) = 2:99 103 J : 2 2 71E 70P 69E 68E (a) Let pi , Vi , and Ti represent the pressure, volume, and temperature of the initial state of the gas. Let pf , Vf , and Tf represent the pressure, volume, and temperature of the nal state. Since the process is adiabatic pi Vi = pf Vf , so pf = (Vi =Vf ) pi = [(4:3 L)=(0:76 L)]1:4 (1:2 atm) = 14 atm. Round to 14 atm. Notice that since Vi and Vf have the same units, their units cancel and pf has the same units as pi . 582 CHAPTER 20 THE KINETIC THEORY OF GASES (b) The gas obeys the ideal gas law pV = nRT , so pi Vi =pf Vf = Ti =Tf and 6 atm)(0 76 L) Tf = pf Vf T1 = (13::2 atm)(4::3 L) (310 K) = 620 K : pi Vi (1 Note that the units of pi Vi and pf Vf cancel since they are the same. (a) For the adiabatic process from state 1 to state 2 p1 V1 = p2 V2 , so V1 = (1:0 atm) V1 1:3 = 2:5 atm : p2 = p1 V V1 =2 2 72E Now use the ideal gas equation pV = nRT to eliminate p in the equation for adiabatic processes above to get the temperature of the gas after the expansion: V1 1 = (273 K) V1 1:3 1 = 340 K : T2 = T1 V V1 =2 2 (b) At constant pressure V2 =T2 = V3 =T3 . So the nal volume is T 273 K 1 V3 = T3 V2 = 340 K V2 = 2 (0:8)V1 = 0:40 L : 2 73E Use the rst law of thermodynamics: E int = Q W . The change in internal energy is E int = nCV (T2 T1 ), where CV is the molar heat capacity for a constant volume process. Since the process is reversible Q = 0. Thus W = E int = nCV (T1 T2 ). For adiabatic processes pV = const. = C , and for any ideal gas pV = nRT . So the value of the constant is 74E C = pV = p = (1:0 105 Pa)1 [2(8:31 J/mol K)(300 K)] = (1:0 105 Pa) 0:4 [2(8:31 J/mol K)(300 K)]1:4 = 1500 N m2:2 : Here = Cp =CV = 1:4 for diatomic ideal gases without oscillation was substituted. nRT p CHAPTER 20 THE KINETIC THEORY OF GASES 583 (a) For adiabatic prosseses pV = const. = C . Thus 75E d C dp B = V dV = V dV V = CV = p : (b) The speed of sound is B = r p = (nRT=V ) = RT : vs = nM=V M s s r The speed of sound is vs = p=. So 2 2 3 3 3 3 3 = vs = (331 m/s) (1:29 10 1:0g/cm5)(10 kg/m )(1 cm =1 g) = 1:40 : p 10 Pa 76E p Use vs = RT=M to calculate the ratio: 77E p v1 = RT1 =M1 = M2 : v2 RT2 =M2 M1 s r (a) The speed of sound in an ideal gas is v = p= which, when combined with the ideal p gas law equation pV = nRT = (m=M )RT , can be re-written as v = RT=M: Thus 78P p = Mv : RT Since the nodes of the standing wave are 6:77 cm apart the wavelength of the sound wave is = 2(6:77 cm) = 13:5 cm: So the speed of sound can be obtained from v = f = (0:135 m)(1400 Hz) = 189 m/s: Thus 2 :127 C = C p = (0(8:31 kg/mol)(189 m/s) = 1:4 : J/mol K)(400 K) V 2 (b) Since = 1:4 for iodine, it is a diatomic gas. 584 CHAPTER 20 THE KINETIC THEORY OF GASES Use vs = RT=M and vrms = 3RT=M . The ratio is 79E p p vs = RT=M = = Cp = CV + R = 5:0R + R = 0:63 : vrms 3RT=M 3 3CV 3CV 3(5:0R) s r r r s In the free expansion from state 0 to state 1 we have Q = W = 0, so E int = 0, which means that the temperature of the ideal gas has to remain unchanged. Thus the nal pressure is V 1 0 p1 = pVV0 = p30V 0 = 3 p0 : (b) For the adiabatic process from state 1 to 2 we have p1 V1 = p2 V2 , i.e., 1 p (3V ) = (3:00) 1 p V ; 3 0 0 3 0 0 which gives = 4=3. The gas is therefore polyatomic. (c) From T = pV=nR we get K2 = T2 = p2 = (3:00) 1 = 1:44 : 3 K1 T1 p1 1 0 80P (a) Use pi Vi = pf Vf to compute : log(pi =pf ) = log(1:0 atm=1:0 105 atm) = 5 : = log(V =V ) log(1:0 103 L=1:0 106 L) 3 f i Therefore the gas is monatomic. (b) The nal temperature is 81P pf Vf = pf Vf = (1:0 105 atm)(1:0 103 L)(273 K) = 2:7 104 K : Tf = nR p V =T (1:0 atm)(1:0 106 L) i i i (c) The number of moles of gas present is 6 L) (1 0 n = pi Vi = (8::31atm)(1:0 10 K) = 4:5 104 mol : RTi J/mol K)(273 CHAPTER 20 THE KINETIC THEORY OF GASES 585 (d) The total translational energy per mole before the compression is 3 3 Ki = 2 RTi = 2 (8:31 J/mol K)(273 K) = 3:4 103 J : After the compression, 3 3 Kf = 2 RTf = 2 (8:31 J/mol K)(2:7 104 K) = 3:4 105 J : 2 (e) Since vrms / T , we have 2 vrms;i Ti K = T = 2:7273104 K = 0:01 : 2 vrms;f f (a) For the isothermal process the nal temperature of the gas is Tf = Ti = 300 K: The nal pressure is i pf = pVVi = (32 atm)(1:0 L) = 8:0 atm ; 4:0 L and the work done is 82P f W = nRTi ln Vf = pi Vi ln Vf Vi Vi 5 Pa/atm)(1:0 10 3 m3 ) ln 4:0 L = 4:4 103 J : = (32 atm)(1:01 10 1:0 L (b) For the adiabatic process pi Vi = pf Vf . Thus Vi = (32 atm) 1:0 L 5=3 = 3:2 atm ; pf = pf V 4:0 L f atm)(4:0 L)(300 f Tf = pfpVV Ti = (3:2 (32 atm)(1:0 L) K) = 120 K ; i i and 3 3 W = Q E int = E int = 2 nRT = 2 (pf Vf pi Vi ) = 3 [(3:2 atm)(4:0 L) (32 atm)(1:0 L)](1:01 105 Pa/atm)(10 3 m3 =L) 2 = 2:9 103 J : 586 CHAPTER 20 THE KINETIC THEORY OF GASES (c) Now = 1:4 so Vi = (32 atm) 1:0 L 1:4 = 4:6 atm ; pf = pi V 4:0 L f atm)(4:0 L)(300 f Tf = pfpVV Ti = (4:6 (32 atm)(1:0 L) K) = 170 K ; i i and W = Q E int = 5 nRT = 5 (pf Vf pi Vi ) 2 2 = 5 [(4:6 atm)(4:0 L) (32 atm)(1:0 L)](1:01 105 Pa/atm)(10 3 m3 =L) 2 = 3:4 103 J : Label the various states of the ideal gas as follows: it starts expanding adiabatically from state 1 until it reaches state 2, with V2 = 4 m3 ; then continues onto state 3 isothermally, with V3 = 10 m3 ; and eventually getting compressed adiabatically to reach state 4, the nal state. For the adiabatic process 1 ! 2 p1 V1 = p2 V2 , for the isothermal process 2 ! 3 p2 V2 = p3 V3 , and nally for the adiabatic process 3 ! 4 p3 V3 = p4 V4 . These equations yield V3 = p V2 V3 = p V1 V2 V3 : p4 = p3 V 2 V 1 V V V V Substitute this expression for p4 into the equation p1 V1 = p4 V4 (since T1 = T4 ) to obtain V1 V3 = V2 V4 . Now solve for V4 : 3 3 1 V4 = VVV3 = (2 m 4)(10 m ) = 5 m3 : m3 2 4 3 4 2 3 4 83P 900 cal, and the change in total translational kinetic energy is K = 3 nRT = 3 (3:0 mol)(8:31 J/mol K)(50 K) = 1:87 103 J = 450 cal : 2 2 For constant pressure we have E int = 900 cal; K = 450 cal, For the constant volume process W = 0. The change in internal energy is E int = nCV T = (3:0 mol)(6:00 cal/molK)(50 K) = 900 cal; the heat added is Q = W + E int = 84P Q = Cp T = n(CV + R)T = (3:0 mol)(6:00 cal/mol K + 8:31 J/mol K)(50 K) = 1:2 103 cal ; CHAPTER 20 THE KINETIC THEORY OF GASES 587 and W = Q E int = 1:2 103 cal 900 cal = 300 cal: For adiabatic process we have Q = 0, E int = 900 cal; K = 450 cal, and W = Q E int = 900 cal (where the minus sign indicate that the gas does negative work on the environment). In the following CV (= 3 R) is the molar speci c heat at constant volume, Cp (= 5 R) is 2 2 the molar speci c heat at constant pressure, T is the temperature increase, and n is the number of moles. (a) The process 1 ! 2 takes place at constant volume. The heat added is 3 Q = nCV T = 2 nR T = 3 (1:00 mol)(8:31 J/mol K)(600 K 300 K) = 3:74 103 J : 2 85P The change in the internal energy is E int = nCV T = 3:74 103 J. The work done by the gas is W = Q E int = 0. The process 2 ! 3 is adiabatic. The heat added is 0. The change in the internal energy is E int = nCV T = 3 nR T 2 3 (1:00 mol)(8:31 J/mol K)(455 K 600 K) = 1:81 103 J : =2 The work done by the gas is W = Q T = +1:81 103 J. The process 3 ! 1 takes place at constant pressure. The heat added is Q = nCp T = 5 nR T 2 The change in the internal energy is = 5 (1:00 mol)(8:31 J/mol K)(300 K 455 K) = 3:22 103 J : 2 E int = nCV T = 3 nR T 2 3 (1:00 mol)(8:31 J/mol K)(300 K 455 K) = 1:93 103 J : =2 The work done by the gas is W = Q E int = 3:22 103 J+1:93 103 J = 1:29 103 J. For the entire process the heat added is Q = 3:74 103 J + 0 3:22 103 J = 520 J, the change in the internal energy is E int = 3:74 103 J 1:81 103 J 1:93 103 J = 0, and the work done by the gas is W = 0 + 1:81 103 J 1:93 103 J = 520 J. 588 CHAPTER 20 THE KINETIC THEORY OF GASES (b) First nd the initial volume. Since p1 V1 = nRT1 , J/mol K)(300 K) = 2:46 10 2 m3 : V1 = nRT1 = (1:00 mol)(8:31 105 Pa) p (1:013 1 Since 1 ! 2 is a constant volume process V2 = V1 = 2:46 10 2 m3 . The pressure for state 2 is p2 = nRT2 = (1:00 mol)(8:31 J/molmK)(600 K) = 2:02 105 Pa : 2 3 V 2:46 10 2 This is equivalent to 2:00 atm. Since 3 ! 1 is a constant pressure process, the pressure for state 3 is the same as the pressure for state 1: p3 = p1 = 1:013 105 Pa (1:00 atm). The volume for state 3 is J/mol K)(455 K) = 3:73 10 2 m3 : V3 = nRT3 = (1:00 mol)(8:31 105 Pa p 1:013 3 86P (b) work done by environment: 7:72 104 J; heat absorbed: 5:47 104 J; (c) 5:17 J/mol K; (d) work done by environment: 4:32 104 J; heat absorbed: 8:87 104 J; molar speci c heat: 8:38 J/mol K ...
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