558 CHAPTER 0 THE KINETIC THEORY OF GASES CHAPTER 0 Answer to Checkoint Questions. all but c. (a) all tie; (b) 3,, 3. gas A 4. 5 (greatest change in T ), then tie of,, 3, and 4 5.,, 3 (Q 3 0, Q goes into work W, but Q goes into greater work W and increases gas temerature Answer to Questions. increased but less than doubled. the cooler room 3. a, c, b 4. tie of a and c, then b, then d 5. 80 J 6. {4 7. d, tie of a and b, then c 8. 0 J 9. constant-volume rocess 0. (a) 3; (b) ; (c) 4; (d) ; (e) yes. (a) same; (b) increases; (c) decreases; (d) increases. (a),, 3, 4; (b),, 3 3. 4 J 4. a, 4; b, 8; c,; d, 5; e, 3; f, 7; g, ; h, 6
CHAPTER 0 THE KINETIC THEORY OF GASES 559 5. (a), olyatomic;, diatomic; 3, monatomic; (b) more Solutions to Exercises & Problems E (a) The number of moles is (b) The number of atoms is n m M Au :5 g 97 g/mol 0:07 mol : N nn A m M N A (0:07 mol)(6:0 0 3 mol) 7:65 0 : E Each atom has a mass of m MN A, where M is the molar mass and N A is the Avogadro constant. The molar mass of arsenic is 74:9 g/mol or 74:90 3 kg/mol. 7:500 4 arsenic atoms have a total mass of (7:500 4 )(74:90 3 kg/mol)(6:00 3 mol ) 0:933 kg. 3P 8 g of water has 6:00 3 water molecules. So gram of water has N (8)(6:00 3 ) water molecules. If these molecules were distributed uniformly over the surface of the Earth then the molecular number density would be D N 4R e 6:0 03 8 4(6:4 0 8 cm) 6560 cm ; i.e., there would be 6560 water molecules er cm. 4P Let the volume of water in quesiton be v. The number of molecules it contains is then vn A 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 V v vn A M :
560 CHAPTER 0 THE KINETIC THEORY OF GASES Solve for v: v s s V M (0:75)(4)(6:4 0 ' 6 m) (5 0 3 m)(:8 0 kg/mol) N A (:0 0 3 kg/m 3 )(6:0 0 3 /mol) ' 8 0 6 m 3 ( cm) : This is aroximately the volume of water contained in a tablesoon. 5P Since the molar mass of the ink is M 8 g/mol, the number of molecules in m g 0 6 g of ink is N mn A M (0 6 g)(6:0 0 3 mol ) 8 g/mol ' 3 0 6 : The number of creatures in our galaxy, with the assumtion made in the roblem, is about N 0 (5 0 9 )(0 ) 5 0 0 : So the statement was wrong by a factor of about 0,000. 6E (a) Use V nrt to solve for the volume V : V nrt which is :4 L. (b) The Loschmidt number is (:00 mol)(8:3 J/molK)(73 K) :0 0 5 Pa :4 0 3 m 3 ; n L (:00 cm3 )(6:0 0 3 /mol) (:4 L/mol)(0 3 cm 3 L) :69 0 9 : 7E (a) Solve V nrt for n: n VRT (00 Pa)(:00 6 m 3 )(8:3 J/molK)(0 K) 5:47 0 8 mol. (b) The number of molecules N is the roduct of the number of moles n and the number of molecules in a mole N A (the Avogadro constant). Thus N nn A (5:47 0 6 mol)(6:0 0 3 mol ) 3:9 0 6 molecules. 8E Use V nrt. The number of molecules er cubic centimeter is N V N A RT (:0 0 3 Pa)(:00 0 6 m 3 )(6:0 0 3 /mol) (8:3 J/molK)(93 K) 5 :
CHAPTER 0 THE KINETIC THEORY OF GASES 56 9E Let T 0:0 C 83 K, T 30:0 C 303 K, 00 kpa; 300 kpa; and V :50 m 3. (a) Use V T nr to solve for n: n V T R (00 03 Pa)(:50 m 3 ) (83 K)(8:3 J/mol K) 06 mol : (b) For the same amount of ideal gas, when the ressure changes to, its temerature changes to T. So V nrt, which we solve for V : V nrt 303 K 83 K T V 00 03 Pa 300 0 3 Pa T (:50 m 3 ) 0:89 m 3 : 0E (a) Solve V nrt for n. First convert the temerature to the Kelvin scale: T 40:0+73:5 33:5 K. Also convert the volume to m 3 : 000 cm 3 0000 6 m 3. Then n VRT (:00 5 Pa)(0000 6 m 3 )(8:3 J/molK)(33:5 K) 3:880 mol. (b) Solve V nrt for T : T VnR (:06 0 5 Pa)(500 0 6 m 3 )(3:88 0 mol)(8:3 J/molK) 493 K 0 C. E Let V 000 in. 3 ; V 00 in. 3 ; 4:0 lb/in. ; T 0:00 C 73 K; and T 7:0 C 300 K. Since the number of moles of the gas remains the same, from V nrt we have n V RT V RT. Thus 000 in. 3 T T V V 300:5 K 73:5 K (4:0 lb/in. ) 7:0 lb/in. : 00 in. 3 E The work done is Z vf W dv (V i V f ) v i (:0 0 5 Pa)(:4 L 6:8 L)(0 3 m 3 L) 653 J :
56 CHAPTER 0 THE KINETIC THEORY OF GASES 3P (a) The number of moles of the air molercules is n VRT NN A. So N V RT :5 0 5 ; N A (:0 05 Pa)( m 3 )(6:0 0 3 mol ) (8:3 J/mol K)(93:5 K) where T 0 C 93 K and P atm :0 0 5 Pa were substituted. (b) Let M air be the molar mass of the air. Then M air 3 4 M N + 4 M O. So the mass of the air is m M air n V 3 RT 4 M N + 4 M O (:0 05 Pa)( m 3 ) (8:3 J/mol K)(93:5 K) : kg : 3 4 (4 g/mol) + 4 (3 g/mol) 4P Since the ressure is constant the work is given by W (V V ). The initial volume is V (AT BT ), where T is the initial temerature. The nal volume is V (AT BT ). Thus W A(T T ) B(T T ). 5P Suose the gas exands from volume V to volume V during the isothermal ortion of the rocess. The work it does is Z V Z V dv W dv nrt V nrt ln V ; V V where V nrt was used to relace with nrtv. Now V nrt and V nrt, so V V. Also relace nrt with V to obtain W V ln : V Since the initial gauge ressure is :03 0 5 Pa, :03 0 5 Pa + :0 0 5 Pa :04 0 5 Pa. The nal ressure is atmosheric ressure: :0 0 5 Pa. Thus :04 Pa W (:04 0 5 Pa)(0:4 m 3 ) ln :00 0 4 J : :0 Pa During the constant ressure ortion of the rocess the work done by the gas is W (V V ). Notice that the gas starts in a state with ressure, so this is the ressure
CHAPTER 0 THE KINETIC THEORY OF GASES 563 throughout this ortion of the rocess. Also note that the volume decreases from V to V. Now V V, so W V V ( )V (:0 0 5 Pa :04 0 5 Pa)(0:4 m 3 ) :44 0 4 J : The total work done by the gas over the entire rocess is W :00 0 4 J :44 0 4 J 5:6 0 3 J. 6P The diagrams are shown below. cosntant ressure V constant volume isothermal cosntant ressure isothermal constant volume V T cosntant ressure isothermal constant volume T (d) You can use V nrt to analyze each case. Note that n is roortional to the mass M of the ideal gas. For examle, the sloe of the constant-volume line in the -T diagram is T nrv / mass of the ideal gas. 7P The ressure due to the rst gas is n RT V, and the ressure due to the second gas is n RT V. So the total ressure on the container wall is + n RT V + n RT V (n + n ) RT V :
564 CHAPTER 0 THE KINETIC THEORY OF GASES The fraction of P due to the second gas is then n RT V (n + n )(RT V ) n n + n 0:5 + 0:5 5 : 8P (a) Use V nrt to solve for n: n V RT (:5 03 Pa)(:0 m 3 ) (8:3 J/molK)(00 K) :5 mol : (kn/m ) 5.0 b (b) T B BV B nr (7:5 03 Pa)(3:0 m 3 ) (:5 mol)(8:3 J/molK).0 a.0 4.0 c V (m3) :8 0 3 K : (c) T C CV C nr (:5 03 Pa)(3:0 m 3 ) (:5 mol)(8:3 J/molK) 6:0 0 K : (d) Use Q E int + W. For the cyclic rocess which starts at a and ends at a we have T 0, so E int 0. Thus Q W (V c V a )( b c ) (3 m3 m 3 )(7:5 0 3 Pa :5 0 3 Pa) 5:0 0 3 J : 9P (a) The work done by the gas in the rocess is W V (5 N/m 3 )(:8 m 3 3:0 m 3 ) 30 J: Thus the change in internal energy of the gas is (b) The nal temerature is E int Q W 75 J ( 30 J) 45 J : T f f V f i V i T i (:8 m3 )(300 K) 3:0 m 3 80 K :
CHAPTER 0 THE KINETIC THEORY OF GASES 565 0P Denote the lowerand higher elevations with subscrits and, resectively. Then V T V T : The gas volume V at the higher elevation is then (76 cmhg)(73 T V V T 48) K (38 cmhg)(73 + 0) K (: m 3 ) 3:4 m 3 : P Assume that the ressure of the air in the bubble is essentially the same as the ressure in the surrounding water. If d is the deth of the lake and is the density of water, then the ressure at the bottom of the lake is 0 + gd, where 0 is atmosheric ressure. Since V nrt, the number of moles of gas in the bubble is n V RT ( 0 + gd)v RT, where V is the volume of the bubble at the bottom of the lake and T is the temerature there. At the surface of the lake the ressure is 0 and the volume of the bubble is V nrt 0. Substitute for n to obtain V T 0 + gd V T 0 93 K :03 05 Pa + (0:998 0 3 kg/m 3 )(9:8 m/s )(40 m) (0 cm 3 ) 77 K :03 0 5 Pa 00 cm 3 : P Consider the oen end of the ie. The balance of the ressures inside and outside the ie requires that + w (L)g 0 + w hg; where 0 is the atmosheric ressure, and is the ressure of the air inside the ie, which satises (L) 0 L, or 0. Solve for h: h 0 w g + L :0 0 5 Pa (:00 0 3 kg/m 3 )(9:80 m/s ) + 5:0 m :8 m : 3P Let the mass of the balloon be m b and that of the lift be m l. According to Archimedes's Princile the buoyant force is F air gv b. The equilibrium of forces for the balloon requires m l g F (m b + m air )g. Solve for the weight of the air in the balloon: m air g air gv b m l g m b g (7:56 0 lb/ft 3 )(77; 000 ft 3 ) 600 lb 500 lb 4:7 0 3 lb : 0 4 N :
566 CHAPTER 0 THE KINETIC THEORY OF GASES So m air : 0 4 N9:80 m/s : 0 3 kg: Now for the air inside the balloon in V b nrt in (m air M air )RT in, where M air is the molar mass of the air and in out atm. So the temerature inside the balloon is T in inv b M air Rm air (:0 05 Pa)(77; 000 ft 3 )(:83 0 m 3 ft 3 )(0:08 kg/mol) (8:3 J/mol K)(4:8 kg) 365 K 9 C 98 F : 4P (a) Use V nrt. The volume of the tank is V nrt (300 g7 g/mol)(8:3 J/molK)(73 C + 77 C) :35 0 6 Pa 3:8 0 m 3 38 L : (b) The number of moles of the remaining gas is n 0 0 V RT 0 (8:7 05 Pa)(3:8 0 m 3 ) (8:3 J/molK)(73 C + C) 3:5 mol : The mass of the gas that leaked out is then m 300 g (3:5 mol)(7 g/mol) 7 g: 5P When the valve is closed the number of moles of the gas in container A is n A A V A RT A and that in container B is n B 4 B V A RT B : The total number of moles in both containers is then n n A + n B V A A + 4 B const. R T A T B After the valve is oened the ressure in container A is 0 A Rn0 A T AV A and that in container B is 0 B Rn0 B T B4V A. Equate 0 A and 0 B to obtain Rn0 A T AV A Rn 0 B T B4V A, or n 0 B (4T AT B )n 0 A. Thus n n 0 A + n 0 B n 0 A Solve the above eauation for n 0 A : + 4T A T B n 0 A V R n A + n B V A A + 4 B : R T A T B ( A T A + 4 B T B ) ( + 4T A T B ) :
CHAPTER 0 THE KINETIC THEORY OF GASES 567 Substitute this exression for n 0 A into 0 V A n 0 A RT A to obtain the nal ressure 0 : 0 n0 A RT A V A A + 4 B T A T B + 4T A T B ; which gives 0 :0 0 5 Pa. 6E The rms seed of helium atoms at 000 K is V rms r 3RT mn A s 3(8:3 J/molK)(000 K) 4:00 0 3 kg/mol :50 0 3 m/s : 7E According to the kinetic theory the rms seed is r 3RT v rms M ; where T is the temerature and M is the molar mass. According to Table 0- the molar mass of molecular hydrogen is :0 g/mol :0 0 3 kg/mol, so s 3(8:3 J/molK)(:7 K) v rms 80 m/s : :0 0 3 kg 8E The rms seed of argon atoms at 33 K is v rms r 3RT mn A s 3(8:3 J/molK)(33 K) 39:9 0 3 kg/mol 44 m/s : 9E Use v rms (3kT m e ), where m e is the mass of the electron and k is the Boltzmann constant. So for electrons inside the Sun 3(:38 0 3 J/K)(:00 0 6 K) v rms 9:53 0 6 m/s : 9: 0 3 kg
568 CHAPTER 0 THE KINETIC THEORY OF GASES 30E (a) v rms r s 3RT 3(8:3 J/molK)(73 C + 0:0 C) mn A 8:0 0 3 kg/mol 5 m/s : (b) Since v rms / T, the temerature at which v rms will be half that value is T () T 93 K4 73:3 K 00 C, and the temerature at which v rms will be twice that value is T T 4(93 K) :7 0 3 K 899 C. 3E Use v rms 3RT M, where M is the molar mass of the ideal gas. So 3RTH 3RTHe ; M H M He which yields T He MHe M H T H 4:00 g/mol (0:0 + 73) K 586 K 33 C : :00 g/mol 3P Use V nrt (mm)rt, where m is the mass of the gas and M is its molar mass. The density of the gas is then mv MRT : (a) Now r r 3RT 3 v rms M s 3(:00 0 atm)(:0 0 5 Pa/atm) 494 m/s : :4 0 kg/m 3 (b) The molar mass of the ideal gas is M RT (:4 0 kg/m 3 )(8:3 J/molK)(73 K) (:0 0 atm)(:0 0 5 Pa/atm) 0:08 g/mol : The gas is therefore nitrogen (N ). 33P On reection only the normal comonent of the momentum changes, so for one molecule the change in momentum is mv cos, where m is the mass of the molecule, v is its seed, and is the angle between its velocity and the normal to the wall. If N molecules collide
CHAPTER 0 THE KINETIC THEORY OF GASES 569 with the wall the change in their total momentum is P Nmv cos and if the total time taken for the collisions is t then the average rate of change of the total momentum is Pt (Nt)mv cos. This the average force exerted by the N molecules on the wall and the ressure is the force er unit area: N mv cos A t :0 0 4 m (03 s )(3:3 0 7 kg)(:0 0 3 m/s) cos 55 :9 0 3 Pa : Notice that the value given for the mass was converted to kg and the value given for the area was converted to m. 34E (a) The average translational energy is given by K 3 kt, where k is the Boltzmann constant (:38 0 3 JK) and T is the temerature on the Kelvin scale. Thus K 3 (:38 0 3 J/K)(600 K) 3:3 0 0 J. Round to 3:3 0 0 J. (b) Since ev :60 0 9 J, K (3:3 0 0 J)(:60 0 9 J/eV) 0: ev. 35E The translational kinetic energy of a article is K 3 kt and one electron-volt is equal to :6 0 9 J. (a) At 0 C (73:5 K), and at T 00 C (373 K), K 3 kt 3 (:38 0 3 J/K)(73:5 K) 5:65 0 J 5:65 0 J :6 0 9 J/eV K 3 kt 3 (:38 0 3 J/K)(373:5 K) 7:7 0 J 7:7 0 J :6 0 9 J/eV 0:0353 ev ; 0:0483 ev: (b) For one mole of ideal gas at T 73 K, K 3 RT 3 (8:3 J/K mol)(73:5 K) 3:40 0 3 J. At T 373 K, K 3 RT 3(8:3 J/K mol)(373:5 K) 4:65 03 J. 36E Let the temerature be T, then K 3kT :00 ev: Thus T K 3k (:00 ev)(:6 0 9 J/eV) 3(:38 0 3 J/K) 7:73 0 3 K :
570 CHAPTER 0 THE KINETIC THEORY OF GASES 37E The gravitational kinetic energy of one oxygen molecule is U mgh, and the translational kinetic energy of the same molecule is K 3 kt. So U K 3mgh 3kT (0:03 kgn A)gh 3kT (0:03 kg)(9:80 m/s )(0: m) 3(8:3 J/mol K)(73 K) 9: 0 6 : Here the relation N A k R was used. 38P (a) See 3P. (b) V nrt n(kn A )T (nn A )kt NkT: 39P (a) Use LN, where L is the heat of vaorization and N is the number of molecules er gram. The molar mass of atomic hydrogen is g/mol and the molar mass of atomic oxygen is 6 g/mol so the molar mass of H O is + + 6 8 g/mol. There are N A 6:0 0 3 molecules in a mole so the number of molecules in a gram of water is (6:0 0 3 mol )(8 g/mol) 3:340 molecules/g. Thus (539 cal/g)(3:340 g ) :6 0 0 cal. This is (:6 0 0 cal)(4:86 J/cal) 6:76 0 0 J. (b) The average translational kinetic energy is K 3 kt 3 (:38 0 3 J/K) [(3:0 + 73:3) K] 6:3 0 J: The ratio of K (6:76 0 0 J)(6:3 0 J) 0:7. 40P They are not equivalent. Avogadro's law does not tell how the ressure, volume, and temerature are related, so you cannot use it, for examle, to calculate the change in volume when the ressure increases at constant temerature. The ideal gas law, however, imlies Avogadro's law. It yields N nn A (VRT )N A VkT, where k RN A was used. If the two gases have the same volume, the same ressure, and the same temerature, then VkT is the same for them. This imlies that N is also the same. 4E The mean free ath of a article is given by ( d n). Solve for d: s s d n 3: 0 0 m 0:3 nm : (0:80 0 7 m)(:7 0 5 m 3 )
CHAPTER 0 THE KINETIC THEORY OF GASES 57 4E (a) According to Eq. 0- the mean free ath for molecules in a gas is given by d NV ; where d is the diameter of a molecule and N is the number of molecules in volume V. Substitute d :0 0 0 m and NV 0 6 molecules/m 3 to obtain (:0 0 0 m) ( 0 6 m 3 ) 6 0 m : (b) At this altitude most of the gas articles are in orbit around the Earth and do not suer randomizing collisions. The mean free ath has little hysical signicance. 43E Substitute d :0 0 m and NV 5(:0 0 3 m 3 ) 5 0 3 beans/m 3 into d NV to obtain (:0 0 m) (5 0 3 m 3 ) 0:5 m : The conversion :00 L :00 0 3 m 3 was used. 44E The average frequency is f v d vn V : 45P (a) Use V nrt NkT, where is the ressure, V is the volume, T is the temerature, n is the number of moles, and N is the number of molecules. The substitutions N nn A and k RN A were made. Since cm of mercury 333 Pa, the ressure is (0 7 )(333) :333 0 4 Pa. Thus N V kt :333 0 4 Pa (:38 0 3 J/K)(95 K) 3:7 0 6 molecules/m 3 3:7 0 0 molecules/cm 3 :
57 CHAPTER 0 THE KINETIC THEORY OF GASES (b) The molecular diameter is d :00 0 0 m, so the mean free ath is d NV (:00 0 0 m) (3:7 0 6 m 3 ) 7 m : 46P Let the seed of sound be v s and its frequency be f. Then which gives f d v s N V v s f d NV ; d v s N A M d v s N A RT (3:0 0 0 m) (33 m/s)(6:0 0 3 /mol) (:0 05 Pa)(0:03 kg/mol) (8:3 J/molK)(73 K) 3:7 0 9 Hz 3:7 GHz : 47P For an ideal gas V nrt. (a) When n, V V m RT, where V m is the molar volume of the gas. So V m RT (8:3 J/mol K)(73:5 K) :0 0 5 Pa :5 L : (b) Use v rms 3RT M: The ratio is given by r r v rms,he MNe 0 g v rms,ne M He 4:0 g :5 : (c) Use He ( d n ) ; where n is the number of articles er unit volume given by n NV N A nv N A RT kt. So He (d) Ne He 0:84 m: d (kt ) kt d (:38 0 3 J/K)(73:5 K) :44( 0 0 m) (:0 0 5 Pa) 0:84 m :
CHAPTER 0 THE KINETIC THEORY OF GASES 573 48P (a) Since / d, r r d Ar N 7:5 0 6 cm d N Ar 9:9 0 6 cm :7 : (b) Since / (NV ) / (T ) ; Ar (0 5 cmhg C; 5 cmhg) (9:9 0 6 cm) 5:0 0 5 cm ; 75 cmhg Ar ( 40 73 C 40 C C; 75 cmhg) 73 C + 0 (9:9 0 6 cm) 7:9 0 6 cm : C 49P (a) Using a ruler, we nd the diameter of the eriod D to be roughly 0:5 mm. So its area is A D 4 0 7 m. Meanwhile, we estimate the diameter d of an air molecule to be roughly 0 0 m. So the area an air molecule covers is a d 4 3 0 0 m. Thus A a 0 7 3 0 0 03 : This tells us that 0 3 air molecules are needed to cover the eriod. (b) Assume that every second there are N air molecules which collide with the eriod. If each one of them bounces back elastically after the collision then the change in linear momentum er molecule er collision is mv x, where m is the molecular mass and v x is the comonent of the molecular velocity in the direction erendicular to the surface of the aer containing the eriod. Thus the ressure exerted by the air molecules on the eriod is mn v x At where t s and v x (vx) v rms 3. Also m MN A, where M is the average molar mass of the air molecules. Solve for N: ; N 3ANA t Mv rms AN At MRT (:0 05 Pa)( 0 7 m )(6:0 0 3 /mol)( s) (0:08 kg/mol)(8:3 J/molK)(300 K) 7 0 0 : 50P (a) The average seed is v P v N ;
574 CHAPTER 0 THE KINETIC THEORY OF GASES where the sum is over the seeds of the articles and N is the number of articles. Thus v (:0 + 3:0 + 4:0 + 5:0 + 6:0 + 7:0 + 8:0 + 9:0 + 0:0 + :0) km/s 0 6:5 km/s : (b) The rms seed is given by Now v rms rp v N : X h v (:0) + (3:0) + (4:0) + (5:0) + (6:0) i + (7:0) + (8:0) + (9:0) + (0:0) + (:0) km /s 505 km /s ; so v rms s 505 km /s 0 7: km/s : 5E (a) P ni v i v P ni (b) From v rms P ni v i P n i we get v rms (:0) + 4(:0) + 6(3:0) + 8(4:0) + (5:0) + 4 + 6 + 8 + r (:0) + 4(:0) + 6(3:0) + 8(4:0) + (5:0) + 4 + 6 + 8 + 3: cm/s : 3:4 cm/s : (c) There are eight articles at v 4:0 cm/s, more than the number of articles at any other single seed. So 4:0 cm/s is the most robable seed. 5E (a) v N v rms vu u t N NX i v i [4(00 m/s) + (500 m/s) + 4(600 m/s)] 40 m/s ; 0 NX i Indeed we see that v rms > v. v i r 0 [4(00 m/s) + (500 m/s) + 4(600 m/s) ] 458 m/s :
CHAPTER 0 THE KINETIC THEORY OF GASES 575 (b) You may check the validity of the inequality v rms v for any distribution you can think of. For examle, imagine that the ten erticles are divided into two grous of ve articles each, with the rst grou of articles moving at seed v and the second grou at v. Obviously now v (v + v ) and v rms r v + v v + v v : (c) If and only if all the individual molecular seeds are the same does v rms equal v. 53E (a) v < v rms < v P. (b) For Maxwell distribution v P < v < v rms. 54E Use v RT M and v rms 3RT M: The ratio is T T 3 v v rms 3 : 55P (a) The rms seed of molecules in a gas is given by v rms 3RTM, where T is the temerature and M is the molar mass of the molecules. The seed required for escae from the Earth's gravitational ull is v gr e, where g is the acceleration due to gravity at the Earth's surface and r e ( 6:37 0 6 m) is the radius of the Earth. This exression can be derived easily. Take the zero of gravitational otential energy to be at innity. Then energy of a article with seed v and mass m at the Earth's surface is E mgr e + mv. If the article is just able to travel far away its kinetic energy must tend toward 0 as its distance from the Earth tends to. This means E 0 and v gr e. Equate the exressions for the seeds to obtain 3RTM gr e. The solution for T is T gr e M3R. The molar mass of hydrogen is :0 0 3 kg/mol, so for that gas T (9:8 m/s )(6:37 0 6 m)(:0 0 3 kg/mol) 3 (8:3 J/mol K) :0 0 4 K : The molar mass of oxygen is 3:0 0 3 kg/mol, so for that gas T (9:8 m/s )(6:37 0 6 m)(3:0 0 3 kg/mol) 3 (8:3 J/molK) :6 0 5 K :
576 CHAPTER 0 THE KINETIC THEORY OF GASES (b) Now T g m r m M3R, where r m ( :74 0 6 m) is the radius of the moon and g m ( 0:6g) is the acceleration due to gravity at the moon's surface. For hydrogen T (0:6)(9:8 m/s )(:74 0 6 m)(:0 0 3 kg/mol) 3(8:3 J/molK) 4:4 0 K : For oxygen T (0:6)(9:8 m/s )(:74 0 6 m)(3:0 0 3 kg/mol) 3(8:3 J/molK) 7:0 0 3 K : (c) The temerature high in the Earth's atmoshere is great enough for a signicant number of hydrogen atoms in the tail of the Maxwellian distribution to escae. As a result the atmoshere is deleted of hydrogen. Very few oxygen atoms, on the other hand, escae. 56P (a) The root-mean-square seed is given by v rms 3RTM. The molar mass of hydrogen is :0 0 3 kg/mol, so v rms s 3(8:3 J/molK)(4000 K) (:0 0 3 kg/mol) 7:0 0 3 m/s : (b) When the surfaces of the sheres that reresent an H molecule and an A atom are touching the distance between their centers is the sum of their radii: d r + r 0:5 0 8 cm + :5 0 8 cm :0 0 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 seed. That is, the number of collisions is d vtnv, where NV is the concentration of argon atoms. molr of collisions er unit time is d vnv (:0 0 0 m) (7:0 0 3 m/s)(4:0 0 5 m 3 ) 3:5 0 0 collisions/s. 57E The formulas for the average seed and v rms are v 8kT m and v rms 3kT m: In this case so m m 3 4:7: v r 8kT m v rms r 3kT ; m 58P (a) From the normalization condition Z 0 P (v) dv Z v0 0 cv dv 3 cv3 0 N
CHAPTER 0 THE KINETIC THEORY OF GASES 577 we obtain c 3Nv 3 0 : (b) The average seed is v N Z 0 P (v)v dv N Z v0 0 cv 3 dv cv4 0 4N 3 4 v 0 : (c) The rms seed is v rms s N Z P (v)v dv 0 s N Z v0 0 cv 4 dv r cv 5 0 5N r 3 5 ' 0:775v 0 : 59P R (a) The integral of the distribution function is the number of articles: P (v) dv N. The area of the triangular ortion is half the roduct of the base and altitude, R or av 0. The area of the rectangular ortion is the roduct of the sides, or av 0. Thus P (v) dv av 0 + av 0 3 av 0 and 3 av 0 N, so a N3v 0. R v (b) The number of articles with seeds between :5v 0 and v 0 is given by 0 :5v 0 P (v) dv. The integral is easy to evaluate since P (v) a throughout the range of integration. Thus the number of articles is a(:0v 0 :5v 0 ) 0:5av 0 N3, where N3v 0 was substituted for a. (c) The average seed is given by v N Z vp (v) dv : For the triangular ortion of the distribution P (v) avv 0 and the contribution of this ortion is Z a v0 v dv a v 3 0 av 0 Nv 0 3Nv 0 3N 9 v 0 ; 0 where N3v 0 was substituted for a. P (v) a in the rectangular ortion and the contribution of this ortion is a N Z v0 v 0 v dv a N (4v 0 v 0) 3a N v 0 v 0 : Thus v v 0 9 + v 0 :v 0. (d) The mean-square seed is given by v rms N Z v P (v) dv :
578 CHAPTER 0 THE KINETIC THEORY OF GASES The contribution of the triangular section is a Nv 0 Z v0 0 v 3 dv The contribution of the rectangular ortion is Thus a N Z v0 v 0 v dv a 3N (8v3 0 v rms a 4Nv 0 v 4 0 6 v 0 : v 3 0) 7a 3N v3 0 4 9 v 0 : r 6 v 0 + 4 9 v 0 :3v 0 : 60E (a) The internal energy is E int 3 NkT 3 nrt 3 (:0 mol)(8:3 J/molK)(73 K) 3:4 03 J : (b) For an ideal gas, as long as T is xed, so is E int, regardless the values of and V. 6E According to the rst law of thermodynamics E int Q W. Since the rocess is isothermal E int 0 (the internal energy of an ideal gas deends only on the temerature) and Q W. The work done by the gas as its volume exands from V i to V f at temerature T is W Z Vf V i dv nrt For n mol, Q W RT ln(v f V i ). 6E The secic heat is C V 3R mn A Z Vf V i dv V nrt ln Vf V i : 3(8:3 J/molK) (6:66 0 7 kg)(6:0 0 3 /mol) 3: 03 J/kgK : 63P (a) According to the rst law of thermodynamics Q E int + W: When the ressure is a constant W V. So E int Q V 0:9 J (:0 0 5 Pa)(00 cm 3 50 cm 3 )(0 6 m 3 cm 3 ) 5:9 J :
CHAPTER 0 THE KINETIC THEORY OF GASES 579 (b) C Q nt Q n(v nr) R Q V (8:3 J/molK)(0:9 J) (:0 0 5 Pa)(50 0 6 m 3 ) (c) C V C R 34:4 J/mol K 8:3 J/molK 6: J: 34:4 J/molK : 64P The straight line on the -V diagram can be exressed as ( )(V V ) V ; where V nrt. Solve for : ( V )V ( nrt )V: Thus (a) (b) W Z V V dv Z V V nrt V dv nrt V V V 3 nrt : E int 3 nrt 3 nrt 3 ( )(V ) (4 ) 3 nrt 9 nrt : 3 nrt (c) (d) The molar secic heat would be C Q W + E int 3 nrt + 9 nrt 6nRT : Q nt 6RT T T 6RT ( )(V )(nr) T 6RT 4T T R : 65P When the temerature changes by T the internal energy of the rst gas changes by n C T, the internal energy of the second gas changes by n C T, and the internal energy of the third gas changes by n 3 C 3 T. The change in the internal energy of the comosite gas is E int (n C + n C + n 3 C 3 ) T. This must be (n + n + n 3 )C T, where C is the molar secic heat of the comosite. Thus C n C + n C + n 3 C 3 n + n + n 3 :
580 CHAPTER 0 THE KINETIC THEORY OF GASES 66P (a) Use C V 3R(mN A ) (see 6E). The mass of the argon atom follows as m 3R C V N A 6:6 0 6 kg : (b) The molar mass of argon is 3(8:3 J/molK) (75 cal/kg C)(4:8 J/cal)(6:0 0 3 /mol) M mn A (6:6 0 6 kg)(6:0 0 3 /mol) 40 g/mol : 67P For diatomic ideal gas C V 5 R 0:8 J. (a) c V c A V A E int;ac nc V (T c T a ) nc V R R 0:8 J (:0 0 3 Pa)(4:0 m 3 ) 8:3 J/molK 5:0 0 3 J : (5:0 0 3 Pa)(:0 m 3 ) (b) The work done from a to c is W ac (V c V a )( b + c ) (4:0 m3 :0 m 3 )(5:0 0 3 Pa + :0 0 3 Pa) 7:0 0 3 J ; and the heat added to the gas from a to c is Q ac E int;ac + W ac 5:0 kj + 7:0 kj :0 0 3 J: (c) Now Q abc E int;abc + W abc, where W abc a (V c V a ) (5:0 0 Pa)(4:0 m 3 0 m 3 ) :0 0 4 J and E int;abc E int;ac 5:0 0 3 J, since E is ath-indeendent. So Q abc :0 0 4 J 5:0 0 3 J 5:0 0 3 J:
CHAPTER 0 THE KINETIC THEORY OF GASES 58 68E As the volume doubles at constant ressure, so does the temerature in Kelvin. T ( )(73 K) 73 K: The heat required is then 5 Q nc T n(c V + R)T n R + R T 7 ( mol) (8:3 J/molK)(73 K) 8 0 3 J : Thus 69E (a) The molar mass of O is 3 g so n g3 g 0:375 mol. (b) For diatomic ideal gas without oscillation C C V + R 5 R + R 7 R. Thus Q nc T 7 (0:375 mole)(8:3 J/molK)(00 K) 090 J : (c) Use E int nc V T 5 nrt: The ratio is E int Q 5nRT 7nRT 5 7 0:74 : 70P (a) Since the rocess is at constant ressure the heat added to the gas is given by Q nc T, where n is the number of moles in the gas, C is the molar secic heat at constant ressure, and T is the increase in temerature. For a diatomic ideal gas C 7 R. Thus Q 7 nr T 7(4:00 mol)(8:3 J/molK)(60:0 K) 6:98 03 J. (b) The change in the internal energy is given by E int nc V T. For a diatomic ideal gas C V 5 R, so E int 5 nr T 5(4:00 mol)(8:3 J/molK)(60:0 K) 4:99 03 J. (c) According to the rst law of thermodynamics E int Q W, so W Q E int 6:98 0 3 J 4:99 0 3 J :99 0 3 J. (d) The change in the total translational kinetic energy is K 3 nr T 3 (4:00 mol)(8:3 J/molK)(60:0 K) :99 03 J : 7E (a) Let i, V i, and T i reresent the ressure, volume, and temerature of the initial state of the gas. Let f, V f, and T f reresent the ressure, volume, and temerature of the nal state. Since the rocess is adiabatic i V i f V f, so f (V i V f ) i [(4:3 L)(0:76 L)] :4 (: atm) 4 atm. Round to 4 atm. Notice that since V i and V f have the same units, their units cancel and f has the same units as i.
58 CHAPTER 0 THE KINETIC THEORY OF GASES (b) The gas obeys the ideal gas law V nrt, so i V i f V f T i T f and f V f T f T i V i (3:6 atm)(0:76 L) (30 K) 60 K : (: atm)(4:3 L) Note that the units of i V i and f V f cancel since they are the same. 7E (a) For the adiabatic rocess from state to state V V, so V V (:0 atm) V V :3 :5 atm : Now use the ideal gas equation V nrt to eliminate in the equation for adiabatic rocesses above to get the temerature of the gas after the exansion: :3 340 K : T T V V (73 K) V V (b) At constant ressure V T V 3 T 3. So the nal volume is T3 73 K V 3 V V T 340 K (0:8)V 0:40 L : 73E Use the rst law of thermodynamics: E int Q W. The change in internal energy is E int nc V (T T ), where C V is the molar heat caacity for a constant volume rocess. Since the rocess is reversible Q 0. Thus W E int nc V (T T ). 74E For adiabatic rocesses V const. C, and for any ideal gas V nrt. So the value of the constant is nrt C V (:0 0 5 Pa) [(8:3 J/molK)(300 K)] (:0 0 5 Pa) 0:4 [(8:3 J/molK)(300 K)] :4 500 Nm : : Here C C V :4 for diatomic ideal gases without oscillation was substituted.
CHAPTER 0 THE KINETIC THEORY OF GASES 583 75E (a) For adiabatic rosseses V const. C. Thus B (b) The seed of sound is v s V d dv V d dv C V CV : s r s B (nrtv ) nmv r RT M : 76E The seed of sound is v s. So v s (33 m/s) (:9 0 3 g/cm 3 )(0 3 kg/m 3 )( cm 3 g) :0 0 5 Pa :40 : 77E Use v s RTM to calculate the ratio: v v s r RT M M : RT M M 78P (a) The seed of sound in an ideal gas is v which, when combined with the ideal gas law equation V nrt (mm)rt, can be re-written as v RTM: Thus Mv RT : Since the nodes of the standing wave are 6:77 cm aart the wavelength of the sound wave is (6:77 cm) 3:5 cm: So the seed of sound can be obtained from v f (0:35 m)(400 Hz) 89 m/s: Thus C C V (0:7 kg/mol)(89 m/s) (8:3 J/molK)(400 K) :4 : (b) Since :4 for iodine, it is a diatomic gas.
584 CHAPTER 0 THE KINETIC THEORY OF GASES 79E Use v s RTM and v rms 3RTM. The ratio is v s v rms s r r r RTM 3RTM 3 C CV + R 3C V 3C V s 5:0R + R 3(5:0R) 0:63 : 80P In the free exansion from state 0 to state we have Q W 0, so E int 0, which means that the temerature of the ideal gas has to remain unchanged. Thus the nal ressure is 0V 0 V 0V 0 3V 0 3 0 : (b) For the adiabatic rocess from state to we have V V, i.e., 3 0(3V 0 ) (3:00) 3 0 V 0 ; which gives 43. The gas is therefore olyatomic. (c) From T VnR we get K K T T (3:00) 3 :44 : 8P (a) Use i V i f V f to comute : Therefore the gas is monatomic. (b) The nal temerature is log( i f ) log(v f V i ) log(:0 atm:0 05 atm) log(:0 0 3 L:0 0 6 L) 5 3 : T f f V f nr f V f i V i T i (:0 05 atm)(:0 0 3 L)(73 K) (:0 atm)(:0 0 6 L) (c) The number of moles of gas resent is n iv i RT i (:0 atm)(:0 06 L) (8:3 J/molK)(73 K) 4:5 04 mol : :7 0 4 K :
CHAPTER 0 THE KINETIC THEORY OF GASES 585 (d) The total translational energy er mole before the comression is After the comression, K i 3 RT i 3 (8:3 J/molK)(73 K) 3:4 03 J : K f 3 RT f 3 (8:3 J/molK)(:7 04 K) 3:4 0 5 J : (e) Since v rms / T, we have v rms;i v rms;f T i T f 73 K :7 0 4 K 0:0 : 8P (a) For the isothermal rocess the nal temerature of the gas is T f T i 300 K: The nal ressure is and the work done is f iv i V f W nrt i ln Vf V i i V i ln Vf V i (3 atm)(:0 L) 4:0 L 8:0 atm ; 4:0 L (3 atm)(:0 0 5 Pa/atm)(:0 0 3 m 3 ) ln :0 L 4:4 0 3 J : (b) For the adiabatic rocess i V i f V f. Thus f f Vi V f (3 atm) T f f V f T i i V i 53 :0 L 3: atm ; 4:0 L (3: atm)(4:0 L)(300 K) (3 atm)(:0 L) 0 K ; and W Q E int E int 3 nrt 3 ( f V f i V i ) 3 [(3: atm)(4:0 L) (3 atm)(:0 L)](:0 05 Pa/atm)(0 3 m 3 L) :9 0 3 J :
586 CHAPTER 0 THE KINETIC THEORY OF GASES (c) Now :4 so and f i Vi V f (3 atm) T f f V f T i i V i :4 :0 L 4:6 atm ; 4:0 L (4:6 atm)(4:0 L)(300 K) (3 atm)(:0 L) W Q E int 5 nrt 5 ( f V f i V i ) 70 K ; 5 [(4:6 atm)(4:0 L) (3 atm)(:0 L)](:0 05 Pa/atm)(0 3 m 3 L) 3:4 0 3 J : 83P Label the various states of the ideal gas as follows: it starts exanding adiabatically from state until it reaches state, with V 4 m 3 ; then continues onto state 3 isothermally, with V 3 0 m 3 ; and eventually getting comressed adiabatically to reach state 4, the nal state. For the adiabatic rocess! V V, for the isothermal rocess! 3 V 3 V 3, and nally for the adiabatic rocess 3! 4 3 V 3 4V 4. These equations yield 4 3 V3 V 4 V V 3 V3 V 4 V V V V 3 V3 V 4 : Substitute this exression for 4 into the equation V 4 V 4 (since T T 4 ) to obtain V V 3 V V 4. Now solve for V 4 : V 4 V V 3 V ( m3 )(0 m 3 ) 4 m 3 5 m 3 : 84P For the constant volume rocess W 0. The change in internal energy is E int nc V T (3:0 mol)(6:00 cal/molk)(50 K) 900 cal; the heat added is Q W + E int 900 cal, and the change in total translational kinetic energy is K 3 nrt 3 (3:0 mol)(8:3 J/molK)(50 K) :87 03 J 450 cal : For constant ressure we have E int 900 cal; K 450 cal, Q C T n(c V + R)T (3:0 mol)(6:00 cal/molk + 8:3 J/molK)(50 K) : 0 3 cal ;
CHAPTER 0 THE KINETIC THEORY OF GASES 587 and W Q E int : 0 3 cal 900 cal 300 cal: For adiabatic rocess 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). 85P In the following C V ( 3 R) is the molar secic heat at constant volume, C ( 5 R) is the molar secic heat at constant ressure, T is the temerature increase, and n is the number of moles. (a) The rocess! takes lace at constant volume. The heat added is Q nc V T 3 nr T 3 (:00 mol)(8:3 J/molK)(600 K 300 K) 3:74 03 J : The change in the internal energy is E int nc V T 3:74 0 3 J. The work done by the gas is W Q E int 0. The rocess! 3 is adiabatic. The heat added is 0. The change in the internal energy is E int nc V T 3 nr T 3 (:00 mol)(8:3 J/molK)(455 K 600 K) :8 03 J : The work done by the gas is W Q T +:8 0 3 J. The rocess 3! takes lace at constant ressure. The heat added is Q nc T 5 nr T 5 (:00 mol)(8:3 J/molK)(300 K 455 K) 3: 03 J : The change in the internal energy is E int nc V T 3 nr T 3 (:00 mol)(8:3 J/molK)(300 K 455 K) :93 03 J : The work done by the gas is W Q E int 3:0 3 J+:930 3 J :90 3 J. For the entire rocess the heat added is Q 3:74 0 3 J + 0 3: 0 3 J 50 J, the change in the internal energy is E int 3:74 0 3 J :8 0 3 J :93 0 3 J 0, and the work done by the gas is W 0 + :8 0 3 J :93 0 3 J 50 J.
588 CHAPTER 0 THE KINETIC THEORY OF GASES (b) First nd the initial volume. Since V nrt, V nrt (:00 mol)(8:3 J/molK)(300 K) (:03 0 5 Pa) :46 0 m 3 : Since! is a constant volume rocess V V :46 0 m 3. The ressure for state is nrt V This is equivalent to :00 atm. (:00 mol)(8:3 J/molK)(600 K) :46 0 m 3 :0 0 5 Pa : Since 3! is a constant ressure rocess, the ressure for state 3 is the same as the ressure for state : 3 :03 0 5 Pa (:00 atm). The volume for state 3 is V 3 nrt 3 3 (:00 mol)(8:3 J/molK)(455 K) :03 0 5 Pa 3:73 0 m 3 : 86P (b) work done by environment: 7:7 0 4 J; heat absorbed: 5:47 0 4 J; (c) 5:7 J/molK; (d) work done by environment: 4:3 0 4 J; heat absorbed: 8:87 0 4 J; molar secic heat: 8:38 J/molK