Chapter 21 Solutions

Transcription

1 Chapter 1 Solutions *1.1 One mole of helium contains Avogadro's number of molecules and has a mass of 4.00 g. Let us call m the mass of one atom, and we have N A m 4.00 g/mol 4.00 g/mol or m molecules/mol g/molecule m kg *1. We first find the pressure exerted by the gas on the wall of the container. P NkT V 3N A k B T V 3RT V 3(8.315 N m/mol K)(93 K) m Pa Thus, the force on one of the walls of the cubical container is F PA ( Pa)( m ) N 1.3 F Nm v t 500( kg) [8.00 sin 45.0 ( 8.00 sin 45.0 )]m/s 30.0 s N P F A 1.57 N/m 1.57 Pa 1.4 Consider the x axis to be perpendicular to the plane of the window. Then, the average force exerted on the window by the hail stones is v F Nm t Nm [v xf v xi ] t Nm [v sin θ ( v sin θ)] t Nm v sin θ t Thus, the pressure on the window pane is P F A Nm v sin θ At *1.5 F ( )( kg 300 m/s) 1.00 s 14.0 N and P F A 14.0 N m 17.6 kpa

2 Chapter 1 Solutions 1.6 Use Equation 1., P N mv 3V, so that K av mv 3PV N where N nn A N A K av 3PV (N A ) 3(8.00 atm)( Pa/atm)( m 3 ) ( mol)( molecules/mol) K av J/molecule 1.7 P N 3 V (KE ) Equation 1. N 3 PV (KE ) 3 ( )( ) ( ) molecules n N molecules N A mol molecules/mol Goal Solution G: The balloon has a volume of 4.00 L and a diameter of about 0 cm, which seems like a reasonable size for a typical helium balloon. The pressure of the balloon is only slightly more than 1 atm, and if the temperature is anywhere close to room temperature, we can use the estimate of L/mol for an ideal gas at STP conditions. If this is valid, the balloon should contain about 0. moles of helium. O: The average kinetic energy can be used to find the temperature of the gas, which can be used with PVnRT to find the number of moles. A: The gas temperature must be that implied by 1 mv 3 k BT for a monatomic gas like He. 1 T 3 mv k J a J/K 17.4 K Now PV nrt gives n PV RT ( N/m )( m 3 ) (8.315 J/mol K)(17.4 K) n 3.3 mol L: This result is more than ten times the number of moles we predicted, primarily because the temperature of the helium is much colder than room temperature! In fact, T is only slightly above the temperature at which the helium would liquify (4. K at 1 atm). We should hope this balloon is not being held by a child; not only would the balloon sink in the air, it is cold enough to cause frostbite!

3 Chapter 1 Solutions v 3k B T m v O v He M He M O m/s v O 477 m/s (a) PV Nk B T N PV ( Pa) π(0.150 m)3 k B T 3 ( J/K)(93 K) 3 atoms K 3 k BT 3 ( )(93) J J 1 3 mv k BT v rms 3k B T m 1.35 km/s 1.10 (a) PV nrt Nmv 3 K Nmv E trans E trans 3PV 3 ( )( ).8 kj mv 3k BT 3RT 3 (8.315)(300) N A ( ) J 1.11 (a) K 3 k BT 3 ( J/K)(43 K) J K 1 mv rms J so v rms For helium, J m (1) m 4.00 g/mol molecules/mol g/molecule m kg/molecule

4 4 Chapter 1 Solutions Similarly for argon, m 39.9 g/mol molecules/mol g/molecule m kg/molecule Substituting in (1) above, we find for helium, v rms 1.6 km/s ; and for argon, v rms 514 m/s *1.1 (a) 1 Pa (1 Pa) 1 N/m 1 Pa 1 J 1 N m 1 J m 3 For a monatomic ideal gas, E int 3 nrt For any ideal gas, the energy of molecular translation is the same, E trans 3 nrt 3 PV Thus, the energy per volume is E trans V 3 P 1.13 E int 3 nrt E int 3 nr T 3 (3.00 mol)(8.315 J/mol K)(.00 K) 75.0 J 1.14 The piston moves to keep pressure constant. Since V nrt P, then V nr T P for a constant pressure process. Q Q nc P T n(c V + R) T so T n(c V + R) Q n(5r/ + R) Q 7nR and V nr Q P 7nR Q 7P QV 7 nrt V ( J)(5.00 L) 7 (1.00 mol)(8.315 J/mol K)(300 K).5 L Thus, V f V i + V 5.00 L +.5 L 7.5 L

5 Chapter 1 Solutions Use C P and C V from Table 1.. (a) Q nc P T (1.00 mol)(8.8 J/mol K)(40 300) K 3.46 kj E int nc V T (1.00 mol)(0.4 J/mol K)(10 K).45 kj W Q E int 3.46 kj.45 kj 1.01 kj 1.16 n 1.00 mol, T i 300 K Since V constant, W 0 (a) E int Q W 09 J 0 09 J E int nc V T n 3 R T so T ( E int) 3nR (09 J) 3(1.00 mol)(8.315 J/mol K) 16.8 K T T i + T 300 K K 317 K 1.17 (a) Consider heating it at constant pressure. Oxygen and nitrogen are diatomic, so C P 7R/ Q nc P T 7 nr T 7 PV T T Q 7 ( N/m )(100 m 3 ) 300 K (1.00 K) 118 kj U g mgy m U g gy J (9.80 m/s ).00 m kg *1.18 (a) C V 5 R 5 J mol mol K kg 719 J kg K kj kg K m nm M PV RT m kg ( Pa)(0.350 m 3 ) kg mol (8.315 J/mol K)(300 K)

6 6 Chapter 1 Solutions We consider a constant volume process where no work is done. Q mc v T (0.811 kg) kj (700 K 300 K) 33 kj kg K (d) We now consider a constant pressure process where the internal energy of the gas is increased and work is done. Q mc P T m(c V + R) T m(7r/) T m(7c V /5) T Q (0.811 kg) 7 5 kj (400 K) 37 kj kg K *1.19 Consider 800 cm 3 of (flavored) water at 90.0 C mixing with 00 cm 3 of diatomic ideal gas at 0.0 C: Q cold Q hot or m air C P,air (T f T i,air ) m w c w ( T) w ( T) w m airc P,air (T f T i,air ) m w c w [ρv] air C P,air (90.0 C 0.0 C) (ρ w V w )c w where we have anticipated that the final temperature of the mixture will be close to 90.0 C. C P,air 7 R 7 J mol mol C 8.9 g 1.01 J/g C ( T) w [( g/cm 3 )(00 cm 3 )](1.01 J/g C )(70.0 C ) [(1.00 g/cm 3 )(800 cm 3 )](4.186 J/g C ) or ( T) w C The change of temperature for the water is between 10 3 C and 10 C 1.0 Q (nc P T) isobaric + (nc V T) isovolmetric In the isobaric process, V doubles so T must double, to T i. In the isovolumetric process, P triples so T changes from T i to 6T i. Q n 7 R (T i T i ) + n 5 R (6T i T i ) Q 13.5nRT i 13.5PV

7 Chapter 1 Solutions In the isovolumetric process A B, W 0 and Q nc V T 500 J (500 J) 500 J n(3r/)(t B T A ) or T B T A + 3nR (500 J) T B 300 K + 3(1.00 mol)(8.315 J/mol K) 340 K In the isobaric process B C, Thus, Q nc P T 5nR (T C T B ) 500 J (500 J) (a) T C T B 5nR 1000 J 340 K 5(1.00 mol)(8.315 J/mol K) 316 K The work done by the gas during the isobaric process is W BC P B V nr(t C T B ) (1.00 mol)(8.315 J/mol K)(316 K 340 J) or W BC 00 J The total work done on the gas is then W on gas W by gas (W AB + W BC ) (0 00 J) or W on gas +00 J 1. (a) The heat required to produce a temperature change is Q n 1 C 1 T + n C T The number of molecules is N 1 + N, so the number of "moles of the mixture" is n 1 + n and Q (n 1 + n )C T, so C n 1C 1 + n C n 1 + n m Q n i C i T i 1 m i 1 n i C T m C i 1 m n i n i C i / i 1

8 8 Chapter 1 Solutions 3RT 1.3 The rms speed of the gas molecules is v. Thus, to double the rms speed, the M temperature must increase by a factor of 4: T f 4T i. Since the pressure is proportional to volume in this process, P V P i constant or P P i V i V i V Then, PV nrt becomes P i V V i nrt or V V i P i nrt Therefore, V f V i T f T i 4 or V f V i. The work done by the gas is then W V fp P i dv V i V V iv 3 dv i V i P iv i The change in the internal energy of the gas is E int nc V T n 5 R (4T i T i ) 15 nrt i 15 P iv i and the energy transferred to the gas as heat is Q E int + W 15 P iv i + 3 P iv i 9P i V i *1.4 (a) P i V γ i P f V γ f so V f P i P f 1/γ / V i 0.0 T f P fv f P f V f T i P i V i P i (0.0)(0.118).35 V i Since the process is adiabatic, Q 0 Since γ 1.40 C P C V R + C V C V, C V 5 R, and T.35T i T i 1.35T i E int nc V T ( mol) 5 J mol K [1.35(300 K)] 135 J and W Q E int J 135 J

9 Chapter 1 Solutions 9 *1.5 (a) P i V γ i P f V γ f P f V i V f γ P i (5.00 atm) atm 30.0 T i P iv i nr (5.00)( Pa)( m 3 ) 366 K (.00 mol)(8.315 J/mol K) T f P fv f nr (1.39)( Pa)( m 3 ) 53 K (.00 mol)(8.315 J/mol K) The process is adiabatic: Q 0 γ 1.40 C P C V R + C V C V C V 5 R E int nc V T (.00 mol) 5 (8.315 J/mol K) (366 53) K 4.66 kj W Q E int kj 4.66 kj 1.6 V i π( m/) m m 3 The quantity of air we find from P i V i nrt i n P iv i ( Pa)( m 3 ) RT i (8.315 J/mol K)(300 K) n mol Adiabatic compression: P f kpa kpa kpa (a) P i V γ f P f V γ f V f V i (P i /P f ) 1/γ m 3 (101.3/901.3) 5/7 V f m 3 P f V f nrt f P i P f T f T P fv f i T P f P i V i i P i 1/γ Ti (P i /P f )(1/γ 1) T f 300 K(101.3/901.3) (5/7 1) 560 K

10 10 Chapter 1 Solutions The work put into the gas in compressing it is E int nc V T W ( mol) 5 (8.315 J/mol K)( ) K W 53.9 J Now imagine this energy being shared with the inner wall as the gas is held at constant volume. The pump wall has outer diameter 5.0 mm +.00 mm +.00 mm 9.0 mm, and volume [π( m) π( m) ] m m 3 and mass ρv ( kg/m 3 )( m 3 ) 53.3 g The overall warming process is described by 53.9 J nc V T + mc T 53.9 J ( mol) 5 (8.315 J/mol K)(T ff 300 K) + ( kg)(448 J/kg K)(T ff 300 K) 53.9 J (0.07 J/K J/K)(T ff 300 K) T ff 300 K.4 K 1.7 T f V i T i V f γ If T i 300 K, then T f 7 K Goal Solution G: The air should cool as it expands, so we should expect T f < 300 K. O: The air expands adiabatically, losing no heat but dropping in temperature as it does work on the air around it, so we assume that PV γ constant (where γ 1.40 for an ideal gas). A: Combine P 1 V γ 1 P V γ and P 1 nrt 1 V 1 with P nrt V to find T 1 V γ 1 1 T V γ 1 T T 1 V 1 γ K 1 V (1.40 1) 7 K L: The air does cool, but the rate is not linear with the change in volume (the temperature drops only 4% while the volume doubles)

11 Chapter 1 Solutions (a) The work done on the gas is P W ab V bp dv V a P b b Adiabatic, Q ab 0 For the isothermal process, W ab' nrt a V b' 1 V a V dv P b b Isothermal, T b T a nrt a ln V b' V a nrt a ln V a V b' P a V V b V b V a /10 Va Thus, W ab' (5.00 mol) J (93 K) ln (10.0) mol K 8.0 kj For the adiabatic process, we must first find the final temperature, T b. Since air consists primarily of diatomic molecules, we shall use γ air 1.40 and C V,air 5R/ 5(8.315)/ 0.8 J/mol K Then, from Equation 1.0, T b T a V a γ 1 (93 K)(10.0) K V b Thus, the work done on the gas during the adiabatic process is W ab (Q E int ) ab (0 nc V T) ab nc V (T b T a ) or W ab (5.00 mol)(0.8 J/mol K)(736 93) K 46.1 kj For the isothermal process, we have P b' V b' P a V a Thus, P b' P a V a' V b' (1.00 atm)(10.0) 10.0 atm For the adiabatic process, we have P b' V γ b P a V γ a Thus, P b P a V γ a (1.00 atm)(10.0) atm V b

12 1 Chapter 1 Solutions 1.9 (a) See the diagram at the right. P P B V γ B P C V γ C 3P i B Adiabatic 3P i V γ i P i V γ C V C 3 1/γ V i 3 5/7 V i.19v i V C.19(4.00 L) 8.79 L (d) P B V B nrt B 3P i V i 3nRT i T B 3T i 3(300 K) 900 K After one whole cycle, T A T i 300 K 3P i A C V (L) V i 4L V C (e) In AB, Q AB nc V T n 5 R (3T i T i ) (5.00)nRT i Q BC 0 as this process is abiabatic P C V C nrt C P i (.19V i ).19nRT i so T C.19T i Q CA nc P T n 7 R (T i.19t i ) 4.17nRT i For the whole cycle, Q ABCA Q AB + Q BC + Q CA ( )nRT i 0.830nRT i ( E int ) ABCA 0 Q ABCA W ABCA W BACA Q ABCA 0.830nRT i 0.830P i V i W ABCA 0.830( Pa)( m 3 ) 336 J

13 Chapter 1 Solutions (a) See the diagram at the right. P P B V γ B P C V γ C 3P i B Adiabatic 3P i V γ i P i V γ C V C 3 1/γ V i 3 5/7 V i.19v i P B V B nrt B 3P i V i 3nRT i (d) T B 3T i After one whole cycle, T A T i P i A V i V C C V (L) (e) In AB, Q AB nc V T n 5 R (3T i T i ) (5.00)nRT i Q BC 0 as this process is abiabatic P C V C nrt C P i (.19V i ).19nRT i so T C.19T i Q CA nc P T n 7 R (T i.19t i ) 4.17nRT i For the whole cycle, Q ABCA Q AB + Q BC + Q CA ( )nRT i 0.830nRT i ( E int ) ABCA 0 Q ABCA W ABCA W ABCA Q ABCA 0.830nRT i 0.830P i V i

14 14 Chapter 1 Solutions 1.31 We suppose the air plus burnt gasoline behaves likes a diatomic ideal gas. We find its final absolute pressure: 1.0 atm(50.0 cm 3 ) 7/5 P f (400 cm 3 ) 7/ cm cm 3 P f 1.0 atm (1/8) 7/ atm Now Q 0, and W E int nc V (T f T i ) W n 5 RT f + 5 nrt i 5 ( P fv f + P i V i ) Before 5 [ (1.14 atm)(400 cm3 ) + (1.0 atm)(50.0 cm 3 )] N/m 1 atm 10 6 m 3 cm 3 W 150 J The time for this stroke is 1 1 min 60 s min s After So W t 150 J kw s 1.3 (1) E int Nf k BT () C V 1 de int n dt f nrt 1 fr (3) C P C V + R 1 (f + ) R (4) γ C P (f + ) C V f 1.33 (a) C ' V 5 nr 9.95 cal/k C' P 7 nr 13.9 cal/k C ' V 7 nr 13.9 cal/k C' P 9 nr 17.9 cal/k

15 Chapter 1 Solutions A more massive diatomic or polyatomic molecule will generally have a lower frequency of vibration. At room temperature, vibration has more chance of being excited than in a less massive molecule. Absorbing energy into vibration shows up in higher specific heats Rotational Kinetic Energy 1 Iω I mr, m kg, r m Cl I kg m ω s 1 K rot 1 Iω J Cl 1.36 The ratio of the number at higher energy to the number at lower energy is e E/k BT where E is the energy difference. Here, and at 0 C, E (10. ev)( J/1 ev) J k B T ( J/K)(73 K) J Since this is much less than the excitation energy, nearly all the atoms will be in the ground state and the number excited is ( )exp( J/ J) ( )e 433 This number is much less than one, so almost all of the time no atom is excited. At C, k B T ( J/K)1073 K J The number excited is ( )exp( J/ J) ( )e Call n 00 the sea-level number density of oxygen molecules, n N0 the sea-level number of nitrogen per volume, and n 0 and n N their respective densities at y 10.0 km. Then, n 0 n 00 exp ( m 0 gy/k B T) n N n NO exp ( m N gy/k B T) and n 0 n N n 00 n N0 exp ( m 0 gy/k B T + m N gy/k B T)

16 16 Chapter 1 Solutions So (n 0 /n N ) (n 00 /n N0 ) exp [ (m 0 m N )gy/k B T] exp ( ) u ( kg/u)(9.80 m/s )10 4 m ( J/K)300 K The ratio of oxygen to nitrogen molecules decreases to 85.5% of its sea-level value (a) V rms,35 V rms,37 3RT/M g/mol 1/ RT/M g/mol The lighter atom, 35 Cl, moves faster (a) v av n iv i N 1 15 [1() + (3) + 3(5) + 4(7) + 3(9) + (1)] 6.80 m/s (v ) av n iv i N 54.9 m /s so v rms (v ) av m/s v mp 7.00 m/s 1.40 Following Equation 1.9, v mp k B T m ( J/K)(4.0 K) m/s kg 1.41 Use Equation 1.6. Take dn v dv 0: 4πN m 3/ exp mv πk B T k B T v mv3 k B T 0 and solve for v mp to get Equation 1.9. Reject the solutions v 0 and v. Retain only mv k B T (a) From v rms 3k B T m, we find the temperature as T ( kg)( m/s) 3( J/mol K) K T ( kg)( m/s) 3( J/mol K) K

17 Chapter 1 Solutions At 0 C, 1 mv rms0 3 k BT 0 At the higher temperature, 1 m(v rms0) 3 k BT T 4T 0 4(73 K) 109 K 819 C 1.44 Visualize the molecules in liquid water at 0 C jostling about randomly. One happens to get kinetic energy corresponding to 430 J/g, and happens to be at the surface and headed upward. Then this molecule can break out of the liquid. ( a ) 430 J/g 430 J 18.0 g g 1 mol 1 mol molecules J/molecule J 1 mv v ( J) 18.0 u( km/s kg/1 u) If these were typical molecules in an ideal gas instead of exceptional molecules in liquid water, 1 mv 3 k BT T J J/K 3510 K These molecules got to be fast-moving in collisions that made other molecules slowmoving; the average molecular energy is unaffected (a) PV N RT and N PVN A N A RT so that N ( )(133)(1.00)( ) (8.315)(300) molecules 1 l n V πd 1/ V Nπd 1/ 1.00 m 3 ( molecules)π( m) () 1/ l 778 km f v l s 1

18 18 Chapter 1 Solutions Goal Solution G: Since high vacuum means low pressure as a result of a low molecular density, we should expect a relatively low number of molecules, a long free path, and a low collision frequency compared with the values found in Example 1.7 for normal air. Since the ultrahigh vacuum is 13 orders of magnitude lower than atmospheric pressure, we might expect corresponding values of N ~ 10 1 molecules/m 3, l ~ 10 6 m, and f ~ /s. O: The equation of state for an ideal gas can be used with the given information to find the number of molecules in a specific volume. The mean free path can be found directly from equation 1.30, and this result can be used with the average speed to find the collision frequency. A: (a) PV N RT and N PVN A N A RT so that N ( torr)(133 Pa/torr)( molecules/mol) (8.315 J/mol K)(300 K) molecules l 1 πd n v V Nπd 1.00 m 3 ( molecules)π( m) l m 778 km f v l 500 m/s m s 1 L: The pressure and the calculated results differ from the results in Example 1.7 by about 13 orders of magnitude as we expected. This ultrahigh vacuum provides conditions that are extremely different from normal atmosphere, and these conditions provide a clean environment for a variety of experiments and manufacturing processes that would otherwise be impossible The average molecular speed is v 8k B T/πm 8k B N A T/πN A m v v 8RT/πM 8(8.315 J/mol K)3.00 K/π( kg/mol) v 178 m/s

19 Chapter 1 Solutions 19 (a) The mean free path is l 1 πd n V 1 π( m) 1/m 3 l m The mean free time is l/v m/178 m/s s yr Now n V is 10 6 times larger, to make l smaller by 10 6 times: l m Thus, l/v s yr 1.47 From Equation 1.30, l 1 πd n V For an ideal gas, n V N V P k B T Therefore, l k B T πd P, as required l [ πd n V ] 1 n V P k B T d m n V 5 ( )(93) /m 3 l m, or about 193 molecular diameters 1.49 Using P n V k B T, Equation 1.30 becomes l k B T π Pd (1) (a) l ( J/K)(93 K) π( Pa)( m) m Equation (1) shows that P 1 l 1 P l. Taking P 1 l 1 from (a) and with l 1.00 m, we find P (1.00 atm)( m) 1.00 m atm

20 0 Chapter 1 Solutions For l m, we have P 3 (1.00 atm)( m) m 30 atm *1.50 (a) n PV RT ( Pa)(4.0 m 3.00 m.50 m) (8.315 J/mol K)(93 K) N nn A ( mol)( molecules/mol) mol N molecules (d) m nm ( mol)(0.089 kg/mol) 37.9 kg 1 m 0v 3 k BT 3 ( J/K)(93 K) J/molecule For one molecule, m 0 M kg/mol N A molecules/mol kg/molecule v rms ( J/molecule) kg/molecule 503 m/s (e) and (f) E int nc V T n 5 R T 5 PV E int 5 ( Pa)(31.5 m 3 ) 7.98 MJ *1.51 (a) P f 100 kpa T f 400 K V f nrt f (.00 mol)(8.315 J/mol K)(400 K) P f Pa m L E int 3.50nR T 3.50(.00 mol)(8.315 J/mol K)(100 K) 5.8 kj W P V nr T (.00 mol)(8.315 J/mol K)(100 K) 1.66 kj Q E int + W 5.8 kj kj 7.48 kj

21 Chapter 1 Solutions 1 T f 400 K V f V i nrt i (.00 mol)(8.315 J/mol K)(300 K) P i Pa m L P f P i T f T i (100 kpa) 400 K 300 K 133 kpa W P dv 0 since V constant E int 5.8 kj as in (a) Q E int + W 5.8 kj kj T f 300 K P f 10 kpa V f V i P i P f (49.9 L) 100 kpa 10 kpa 41.6 L E int 3.50nR T 0 since T constant W P dv nrt i V f dv V i V nrt i ln V f V i nrt i ln P i P f W (.00 mol)(8.315 J/mol K)(300 K) ln 100 kpa 10 kpa 910 J Q E int + W J 910 J (d) P f 10 kpa γ C P C V + R C V CV 3.50R + R 3.50R R P f V γ f P i V γ i so V f V i P i P f 1/γ (49.9 L) 100 kpa 7/ L 10 kpa T f T i P f V f P i V i (300 K) 10 kpa 100 kpa 43.3 L 49.9 L 31 K E int 3.50nR T 3.50(.00 mol)(8.315 J/mol K)(1.4 K) 7 J Q 0 (abiabatic process) W Q E int 0 7 J 7 J

22 Chapter 1 Solutions 1.5 (a) The average speed v av is just the weighted average of all the speeds. v av [(v) + 3(v) + 5(3v) + 4(4v) + 3(5v) + (6v) + 1(7v)] ( ) 3.65v First find the average of the square of the speeds, v av [(v) + 3(v) + 5(3v) + 4(4v) + 3(5v) + (6v) + 1(7v) ] v The root-mean square speed is then v rms v av 3.99v The most probable speed is the one that most of the particles have; i.e., five particles have speed 3.00v (d) PV 1 3 Nmv av Therefore, P 0 3 [m(15.95)v ] V 106 mv V (e) The average kinetic energy for each particle is K 1 mv av 1 m(15.95v ) 7.98mv 1.53 (a) PV γ k. So, W f P dv k f dv i i V γ P iv i P f V f γ 1 de int dq dw and dq 0 for an adiabatic process. Therefore, W E int 3 nr T nc V(T i T f ) To show consistency between these equations, consider that γ C P /C V and C P C V R. Therefore, 1/(γ 1) C V /R. Using this, the result found in part (a) becomes W (P i V i P f V f ) C V R Also, for an ideal gas PV R nt so that W nc V(T i T f )

23 Chapter 1 Solutions (a) Maxwell s speed distribution function is N v 4πN m 3/ v πk B T e mv /k B T With N , m M 0.03 kg N A kg T 500 K, and k B J/molecule K; this becomes N v ( )v e ( )v The following is a plot of this function for the range 0 v 1400 m/s. N v (s/m) v mp vavvrms The most probable speed occurs where N v is a maximum. v (m/s) From the graph, v mp 510 m/s v av 8k B T πm 8( )(500) π( m/s ) (d) 3k B T Also, v rms m 3( )(500) m/s 600 N v dv 300 The fraction of particles in the range 300 v 600 m/s is where N 10 N 4 and the integral of N v is read from the graph as the area under the curve. This is approximately 4400 and the fraction is 0.44 or 44%.

24 4 Chapter 1 Solutions 1.55 The pressure of the gas in the lungs of the diver will be the same as the absolute pressure of the water at this depth of 50.0 meters. This is: P P 0 + ρgh 1.00 atm + ( kg/m 3 )(9.80 m/s )(50.0 m) or P 1.00 atm + ( Pa) 1.00 atm Pa 5.84 atm If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere or less (or approximately one-sixth of the total pressure), oxygen molecules should make up only about one-sixth of the total number of molecules. This will be true if 1.00 mole of oxygen is used for every 5.00 moles of helium. The ratio by weight is therefore, (5.00 mol He)g (1.00 mol O )g (0.0 g)g (3.0 g)g 0.65 *1.56 n m M 1.0 kg kg/mol 41.5 mol (a) V i nrt i (41.5 mol)(8.315 J/mol K)(98 K) P i m Pa 3 P f P i V f P f P i so V f V i (0.514 m V i 3 ) m 3 00 T f P fv f nr ( Pa)(.06 m 3 ) (41.5 mol)(8.315 J/mol K) K (d) W V fp dv C V i Vi V fv 1/ dv P i V 1/ i V 3/ 3 V f P i 3 V i V 1/ i (V 3/ f V 3/ i ) W Pa m [(.06 m3 ) 3/ (0.514 m) 3/ ] J (e) E int nc V T (41.5 mol) 5 (8.315 J/mol K) ( ) K E int J Q E int + W J J J.8 MJ

25 Chapter 1 Solutions (a) Since pressure increases as volume decreases (and vice versa), dv dp < 0 and 1 dv V dp > 0 For an ideal gas, V nrt P and κ 1 1 d nrt V dp P If the compression is isothermal, T is constant and κ 1 nrt V 1 P 1 P For an adiabatic compression, PV γ C (where C is a constant) and κ 1 d C V dp P 1/γ (d) κ 1 1 P atm 1 (.00 atm) 1 V 1 γ C 1/γ P (1/γ)+1 P 1/γ γp 1/γ+1 1 γp γ C P C V and for a monatomic ideal gas, γ 5/3, so that κ 1 γp atm 1 (5/3)(.00 atm) *1.58 (a) The speed of sound is v B ρ where B VdP dv. According to Problem 57, in an adiabatic process, this is B 1 κ γp. Also, ρ m s V nm V (nrt)m V(RT) sound in the ideal gas is v PM RT where m s is the sample mass. Then, the speed of B ρ γp RT PM γrt M v 1.40(8.315 J/mol K)(93 K) kg/mol 344 m/s This nearly agrees with the 343 m/s listed in Table 17.1.

26 6 Chapter 1 Solutions We use k B R N A and M mn A : v γrt M γk B N A T mn A γk B T m The most probable molecular speed is k B T m, the average speed is 8k B T 3k B T πm, and the rms speed is m. All are somewhat larger than the speed of sound N v (v) 4πN m 3/ v exp( mv πk B T /k B T) Note that v mp (k B T/m) 1/ Thus, N v (v) 4πN m 3/ v πk B T e ( v /v mp ) and N v (v) N v (v mp ) v v mp e (1 v/v mp ) For v v mp /50, N v(v) N v (v mp ) 1 e [1 (1/50)] The other values are computed similarly, with the following results: v v mp N v (v) Nv(v mp ) 1/ / / To find the last value, note: (50) e e log500 e (ln10)( 499/ln10) 10 log /ln10 10 log /ln

27 Chapter 1 Solutions 7 *1.60 The ball loses energy 1 mv i 1 mv f 1 (0.14 kg) [(47.) (4.5) ]m /s 9.9 J The air volume is V π( m) (19.4 m) m 3, and its quantity is n PV RT ( Pa)( m 3 ) (8.315 J/mol K)(93 K) The air absorbs energy according to Q nc P T, so 3.47 mol T Q 9.9 J 0.96 C nc P (3.47 mol)(7/)(8.315 J/mol K) 1.61 (a) The effect of high angular speed is like the effect of a very high gravitational field on an atmosphere. The result is: The larger-mass molecules settle to the outside while the region at smaller r has a higher concentration of low-mass molecules. Consider a single kind of molecules, all of mass m. To supply the centripetal force on the molecules between r and r + dr, the pressure must increase outward according to F r ma r. Thus, PA (P + dp)a (nma dr)(rω ) where n is the number of molecules per unit volume and A is the area of any cylindrical surface. This reduces to dp nmω rdr. But also P nk B T, so dp k B T dn. Therefore, the equation becomes dn n mω k B T r dr giving n dn n 0 n mω k B T r r dr 0 or ln(n) n n 0 mω k B T r r 0 ln n n 0 mω k B T r and solving for n: n n 0 e mr ω /k B T

28 8 Chapter 1 Solutions 1.6 First find v av as v av 1 N 0 v N v dv. Let a m k B T. Then, v av [4Nπ 1/ a 3/ ] N v 4 e av dv [4a 3/ π 1/ ] 0 3 8a π a 3k BT m The root-mean square speed is then v rms v av 3k BT m To find the average speed, we have v av 1 N vnv dv 0 (4Na 3/ π 1/ ) N v 3 e av dv 4a3/ π 1/ 0 a 8k B T π m P(atm) 3.00 B *1.63 (a) n PV RT ( Pa)( m 3 ) (8.315 J/mol K)(300 K).00 n 0.03 mol T B T P B A P A (300 K) K 1.00 A C T C T B 900 K V(L) V C V A T C T A (5.00 L) L E int,a 3 nrt A 3 (0.03 mol)(8.315 J/mol K)(300 K) 760 J E int,c E int,b 3 nrt B 3 (0.03 mol)(8.315 J/mol K)(900 K).8 kj (d) P(atm) V(L) T(K) E int (kj) A B C (e) For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For BC, keep the sample in the oven while gradually letting the gas expand to lift a

29 Chapter 1 Solutions 9 load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston.

30 30 Chapter 1 Solutions (f) For AB: W 0, E int E int,b E int,a ( ) kj 1.5 kj Q E int + W 1.5 kj For BC: E int 0, W nrt B ln(v C /V B ) W (0.03 mol)(8.315 J/mol K)(900 K) ln(3.00) 1.67 kj Q E int + W 1.67 kj For CA: E int E int,a E int,c ( ) kj 1.5 kj W P V nr T W (0.03 mol)(8.315 J/mol K)( 600 K) 1.01 kj Q E int + W 1.5 kj 1.01 kj.53 kj (g) We add the amounts of energy for each process to find them for the whole cycle. Q ABCA +1.5 kj kj.53 kj kj W ABCA kj 1.01 kj kj ( E int ) ABCA +1.5 kj kj 0

31 Chapter 1 Solutions With number-per-volume n 0 e ( mgy/kbt), the number of molecules above unit ground area is n(y)dy, and the number below altitude h is h n(y)dy. So, 0 0 h n(y)dy n h 0 e ( mgy/k B T) dy 0 o (a) f n(y)dy n 0 e ( mgy/k B T) dy 0 0 k B T/mg h e ( mgy/k B T) ( mg dy/k B T) 0 k B T/mg e ( mgy/k B T) ( mg dy/k B T) 0 e( mgy/k BT) h 0 e ( mgy/k BT) 0 e( mgh/k BT) e ( mgh/k BT) 1 1 e( mgh'/k BT) e ( mgh'/k BT) 1 or e (+mgh'/k BT) mgh'/k B T ln so h' k BT ln mg ( J/K)(70 K)(ln ) (8.9 u)( kg/u)(9.80 m/s ) 5.47 km *1.65 (a) ( g) 1.00 mol 18.0 g molecules 1.00 mol molecules After one day, 10 1 of the original molecules would remain. After two days, the fraction would be 10, and so on. After 6 days, only 3 of the original molecules would likely remain, and after 7 days, likely none. The soup is this fraction of the hydrosphere: 10.0 kg kg Therefore, today s soup likely contains this fraction of the original molecules. The number of original molecules likely in the pot again today is: 10.0 kg kg ( molecules) molecules 1.66 (a) For escape, 1 mv GmM R this can also be written as: 1 mv GmM R. Since the free-fall acceleration at the surface is g GM R, mgr For O, the mass of one molecule is kg/mol m molecules/mol kg/molecule

32 3 Chapter 1 Solutions Then, if mgr 10(3k B T/), the temperature is T mgr ( kg)(9.80 m/s )( m) 15k B 15( J/mol K) 4 K 1.67 (a) For sodium atoms (with a molar mass M 3.0 g/mol) 1 mv 3 k BT 1 M N A v 3 k BT (a) v rms 3RT M 3(8.315 J/mol K)( K) m/s kg d t m 0 ms v rms m/s

33 Chapter 1 Solutions 33