In doing the numerical calculations for the exercises and problems for this chapter, the values of the ideal-gas constant have been used with the precision given on page 501 of the text, K.molatmL0.08206KmolJ3145.8 R Use of values of these constants with either greater or less precision may introduce differences in the third figures of some answers. 18.1: a) mol.3.56)molkg10400(kg)225.0( 3 tot   Mmn b) Of the many ways to find the pressure, Eq. (18.3) gives Pa.106.81atm67.2 L)(20.0 K)K)(291.15molatmL06mol)(0.0823.56( 6    V nRT p 18.2: a) The final temperature is four times the initial Kelvin temperature, or 4(314.15 K) –273.15= C983 to the nearest degree. kg.1024.5 )K15.314)(Katm/molL08206.0( )L60.2)(atm30.1)(kg/mol1000.4( b) 4 3 tot       RT MpV nMm 18.3: For constant temperature, Eq. (18.6) becomes atm.96.0)390.0110.0)(atm40.3()( 2112  VVpp 18.4: a) Decreasing the pressure by a factor of one-third decreases the Kelvin temperature by a factor of one-third, so the new Celsius temperatures is 1/3(293.15 K) – 273.15= C175 rounded to the nearest degree. b) The net effect of the two changes is to keep the pressure the same while decreasing the Kelvin temperature by a factor of one- third, resulting in a decrease in volume by a factor of one-third, to 1.00 L. 18.5: Assume a room size of 20 ft X 20 ft X 20 ft C.20ofre temperatuaAssume.m113ft4000 33 V molecules108.2 mol4685 )K293)(KJ/mol315.8( )m113)(Pa1001.1( so 27 35      A nNN RT pV nnRTpV b) 319 36 27 cmmolecules/105.2 cm10113 molecules108.2     V N 18.6: The temperature is K.15.295C0.22 T (a) The average molar mass of air is somol,kg108.28 3 M kg.1007.1 K)K)(295.15molatmL08206.0( mol)kg10L)(28.8atm)(0.90000.1( 3 3 tot       M RT pV nMm (b) For Helium mol,kg1000.4 3 M so kg.1049.1 K)K)(295.15molatmL08206.0( mol)kg10L)(4.00atm)(0.90000.1( 4 3 tot       M RT pV nMm 18.7: From Eq. (18.6), C.503K776 )cmPa)(4991001.1( )cmPa)(46.210(2.821 K)15.300( 35 36 11 22 12                      Vp Vp TT 18.8: a)        kg.373.0 K15.310KmolJ3145.8 m0750.0Pa10013.4molkg100.32 353 tot      RT MpV m b) Using the final pressure of Pa10813.2 5  and temperature of kg,0.275K,15.295   m so the mass lost is kg098.0 where extra figures were kept in the intermediate calculation of tot m . 18.9: From Eq. (18.6), Pa1036.3 m48.0 m750.0 K300.15 K430.15 Pa)1050.1( 5 3 3 5 2 1 1 2 12                                  V V T T pp . 18.10: a) mol.1078.5 K)(295.15K)molatmL(0.08206 L)10atm)(14000.1( 3 3     RT pV n b) kg.185mol)10mol)(5.78kg100.32( 33   18.11: L.0.159292.15)77.3L)(600.0()( 1212  TTVV 18.12: a) Pa.105.87gives(18.7)Eq. whilePa1028.7 66 VnRT b) The van der Waals equation, which accounts for the attraction between molecules, gives a pressure that is 20% lower. c) Pa,1028.7 5  2.1%. Pa,1013.7 5  d) As Vn decreases, the formulas and the numerical values are the same. 18.13: At constant temperature, atm.1.15.7)6.0atm)(0.1()( 2112  VVpp 18.14: a) .74.3))(50.3( 277K K296 1 2 2 1 1 2  T T p p V V b) Lungs cannot withstand such a volume change; breathing is a good idea. 18.15: a) C.70.3K343 K)molatmL0.08206mol)(0.11( L)atm)(3.10100( 2 2    nR Vp T b) This is a very small temperature increase and the thermal expansion of the tank may be neglected; in this case, neglecting the expansion means not including expansion in finding the highest safe temperature, and including the expansion would tend to relax safe standards. 18.16: (a) The force of any side of the cube is ,)()( LnRTAVnRTpAF  since the ratio of area to volume is .K15.293C20.0For .1  TLVA N.103.66 m200.0 K)(293.15K)molJ(8.3145mol)3( 4    L nRT F b) For K,373.15C00.100 T .N1065.4 m200.0 )KK)(373.15molJ5mol)(8.3143( 4    L nRT F 18.17: Example 18.4 assumes a temperature of C0 at all altitudes and neglects the variation of g with elevation. With these approximations, RT/ 0 Mgy epp   We want y for and0.90so90.0 RT/ 0 Mgy epp   m850)90.0ln(  Mg RT y (We have used kg/mol108.28 3 M for air.) 18.18: From example 18.4, the pressure at elevation y above sea level is . RT/ 0 Mgy epp   The average molar mass of air is kg/mol,108.28 3 M so at an altitude of 100 m, ,01243.0 K)K)(273.15molJ3145.8( m)100)(sm80.9)(molkg108.28( 23 1      RT Mgy and the percent decrease in pressure is %.24.10124.011 01243.0 0   epp At an altitude of ,1243.0m,1000 2 RTMgy and the percent decrease in pressure is %.7.11117.01 1243.0   e These answers differ by a factor of ,44.91.24%11.7%  which is less than 10 because the variation of pressure with altitude is exponential rather than linear. 18.19: RTMyg epp   0 from Example 18.4. Eq. (18.5) says RT.Mρp )( Example 18.4 assumes a constant ρpT andso,K273 are directly proportional and we can write RTMgy eρρ   0 For 0 0124.0 0 988.0so,0124.0m,100 ρeρρ RT Mgy y   The density at sea level is 1.2% larger than the density at m.100 18.20: Repeating the calculation of Example 18.4 (and using the same numerical values for R and the temperature gives) Pa.1044.5)537.0( 4 atm  pp 18.21:      kg/mol)1028.8(56.5KK15.273KmolJ3145.8mkg364.0 33   MρRTp Pa.1028.2 4  18.22: mol.kg849molecule)kg10mol)(1.41molecules1002.6( 2123 A   mNM 18.23: Find the mass: kg0.1906mol)kg106mol)(63.5400.3( 3   nMm 335 33 cm4.21m1014.2 mkg108.9 kg1906.0     ρ m V 18.24: AA N RT pV nNN  mol)molecules10023.6( K)K)(300molJ3145.8( )m10Pa)(1.0010119.9( 23 369      molecules.1020.2 6  18.25: a) mol)molecules10(6.023 K)K)(7500molatmL08206.0( L molecules 1080 23 3 a          N RT V N V nRT p atm,102.8 17  about Pa.102.8 12  This is much lower, by a factor of a thousand, than the pressures considered in Exercise 18.24. b) Variations in pressure of this size are not likely to affect the motion of a starship. 18.26: Since this gas is at standard conditions, the volume will be 31633 m1023.2 )m104.22(   A N N V , and the length of a side of a cube of this volume is m.101.6)m1023.2( 6 3 1 316   18.27: mol,6.55 molg18.0 g1000  which is mol)molecules10mol)(6.0236.55( 23  = molecules.103.35 25  18.28: a) The volume per molecule is .m10091.4 Pa)10mol)(1.013molecules10(6.023 K)K)(300.15molJ3145.8( 326 523       ApA N RT nN pnRT N V If this volume were a cube of side L,   m,1045.3m10091.4 9 3 1 326  L which is (b) a bit more than ten times the size of a molecule. 18.29:      .m1000.9cm0.90g/cm00.1/g/mol0.18mol00.5) 3533   ρMnρmVa 18.28;ExcerciseSeeb)     3/1 23 35 3/1 A 3/1 molmolecules/10023.6mol00.5 m1000.9                            nN V N V m.1010.3 10  c) This is comparable to the size of a water molecule. 18.30: a) From Eq. (18.16), the average kinetic energy depends only on the temperature, not on the mass of individual molecules, so the average kinetic energy is the same for the molecules of each element. b) Equation (18.19) also shows that the rms speed is proportional to the inverse square root of the mass, and so and301.0 00.222 18.20 ,491.0 80.83 18.20 Nerms Rnrms Nerms Krrms  v v v v .614.0 00.222 80.83 Krrms Rnrms  v v 18.31: a) At the same temperature, the average speeds will be different for the different isotopes; a stream of such isotopes would tend to separate into two groups. .004.1 b) 349.0 352.0  18.32: (Many calculators have statistics functions that are preprogrammed for such calculations as part of a statistics application. The results presented here were done on such a calculator.) a) With the multiplicity of each score denoted by .1.61 150 1 b)6.54 150 1 isaverage the, 2/1 2 1                      iiii xnxnn (Extra significant figures are warranted because the sums are known to higher precision.) 18.33: We known that . that and BABA TTVV  ;/a) VnRTp  we don’t know n for each box, so either pressure could be higher. A A A N RT pVN NRT N N pV where,sob)           is Avogadro’s number. We don’t know how the pressures compare, so either N could be larger.   .c) RTMmpV  We don’t know the mass of the gas in each box, so they could contain the same gas or different gases.   kTvm 2 3 av 2 2 1 d)  T A > T B and the average kinetic energy per molecule depends only on T, so the statement must be true. mkTv rms 3e)  We don’t know anything about the masses of the atoms of the gas in each box, so either set of molecules could have a larger . rms v 18.34: Box A has higher pressure than B. This could be due to higher temperature and/or higher particle density in A. Since we know nothing more about these gases, none of the choices is necessarily true, although each of them could be true. 18.35: K10300kg;10348.3) 627 nP   Tmmma %64.0s;m109.13 rms 6 rms  cvmkTv K103.7s,m100.3For 3 b) 107 mms 2 rms   Tv kmvT 18.36: From ,nRTpV  the temperature increases by a factor of 4 if the pressure and volume are each doubled. Then the rms speed MRTv 3 rms  increases by a factor of ,24  so the final rms speed is s.m500s)m250(2  18.37: J.106.21K)K)(300J10381.1)(23(a) 2123 2 3  kT .m1034.2 mol)molecules10023.6(mol)kg10(32.0 J)1021.6(22 b) 225 233 21 ave s m K       s,m1084.4 mol)kg10(32.0 K)K)(300molJ3145.8(33 sc) 2 3      M RT v which is of course the square root of the result of part (b). s)m1084.4( mol)molecules10(6.023 mol)kg100.32( ssd) 2 23 3               v N M mv A mkg1057.2 23   This may also be obtained from mol)molecules10(6.023 mol)kg10J)(32.01021.6(2 2 23 321 ave     mK e) The average force is the change in momentum of the atom, divided by the time between collisions. The magnitude of the momentum change is twice the result of part (d) (assuming an elastic collision), and the time between collisions is twice the length of a side of the cube, divided by the speed. Numerically, N.1024.1 m)(0.100 J)1021.6(2s2s s2 s2 19 212 ave      L K L mv vL mv F Pa.1024.1 f) 172 aveave   LFp     molecules.1015.8Pa1024.1Pa10013.1 g) 21175 ave   PP AA h) N RT pV NnN        . 22 1045.2 molmolecules 23 10023.6 K300K/molatmL08206.0 L00.1atm00.1                   i) The result of part (g) was obtained by assuming that all of the molecules move in the same direction, and that there was a force on only two of the sides of the cube. 18.38: This is the same calculation done in Example 16-9, but with .m106.1givingatm,1050.3 513   p 18.39: The rms speeds will be the same if the Kelvin temperature is proportional to the molecular mass;   )02.20.28(K15.293)( 2222 HNHN  MMTT C.1079.3K1006.4 33  18.40:   .sm1044.6 kg1000.3 )K300)(KJ10381.1(33 a) 3 16 23        m kT b) If the particle is in thermal equilibrium with its surroundings, its motion will depend only on the surrounding temperature, not the mass of the individual particles. 18.41: a) The six degress of freedom would mean a heat capacity at constant volume of       KJ/kg1039.1 K.J/mol9.2436 3 molkg100.18 KmolJ3145.83 3 2 1 3     M R RR , b) vibrations do contribute to the heat capacity. 18.42: a)        kg/mol018.0CJ/kg833so,massmolar  CC v CmolJ0.15        CJ/kg2060C,60at CJ/kg5.29kg/mol018.0CJ/kg1640C,180at   C.0.5at CJ/mol1.37kg/mol018.0  b) Vibrational degrees of freedom become more important. c) OHbecause3exceeds 2 RC V also has rotational degrees of freedom. 18.43:     kJ.56.1K0.30KmolJ79.20mol50.2(18.26),Eq. Usinga) Q b) From Eq. (18.25), 5 3 of the result of part (a), 936 J. [...]...  a) From Eq 18. 19, v s  3 (8.3145 J mol  K ) (300.15 K) (28.0  103 kg mol)  517 m s b) v s 3  299 m s 18. 72: b) a) 3kT 3(1.38  1023 J K ) (5800 K)   1.20  104 m s  27 (1.67  10 kg) m 2GM  R 2(6.673  10 11 N  m 2 kg 2 ) (1.99  1030 kg)  6 .18  105 m s 8 (6.96  10 m) c) The escape speed is about 50 times the rms speed, and any of Fig 18. 20 , Eq 18. 32 or Table 18. 2 will indicate... 1 mω2 A2 (sin 2 ωt ) ave 2 and mω2  k shows that K ace  U ave 18. 77: a) In the same manner that Eq (18. 27) was obtained, the heat capacity of the two-dimensional solid would be 2R = 16.6 J mol  K b) The heat capcity would behave qualitatively like those in Fig (18. 18), and heat capacity would decrease with decreasing temperature 18. 78: a) The two degrees of freedom associated with the rotation... maximum (the most probable speed), as is shown in Fig (18. 20) 18. 84: a) (0.60)(2.34  10 3 Pa)  1.40  10 3 Pa b) m  MpV (18. 0  10 3 kg mol)(1.40  103 Pa)(1.00 m 3 )   10 g RT (8.3145 J mol  K)(293.15 K) 18. 85: The partial pressure of water in the room is the vapor pressure at which condensation occurs The relative humidity is 1 81  42.6% 4 25 18. 86: a) The partial pressure is (0.35)(3.78  10... 2πkT  32 1 2 mv , 2 32 2ε  ε kT 8π  m   ε kT   e  εe m  2πkT  m 18. 47: Express Eq (18. 33) as f  Aε e  ε kT , with A a constant Then, df ε  ε kT  ε    ε kT   Ae  ε kT  e   Ae 1  kT  de kT     Thus, f will be a maximum when the term in square brackets is zero, or ε  1 2 mv  kT , 2 which is Eq (18. 34) 18. 48: Note that k R NA R   m M NA M a) 2(8.3145 J mol  K)(300 K) (44.0... such atmosphere 18. 75: a) From Eq (18. 19), m 3kT 3(1.381  1023 J K)(300 K)   1.24  1014 kg (0.001 m s) 2 v s2 b) mN A M  (1.24  10 14 kg)(6.023  10 23 molecules mol) (18. 0  10 3 kg mol)  4.16  1011 molecules 1 3  3V   3m ρ  c) D  2r  2   2   4π   4π  1 3 1  3(1.24  1014 kg)  6  2  4π (920 kg m3 )   2.95  10 m,    which is too small to see 3 18. 76: From x ... temperature, so 18. 58: The density ρ' of the hot air must be ρ  ρ   ρ ρ m    T 1  T  T   ρ ρ  (m V ) ρV    1   (290 kg)  (288.15 K)1   (1.23 kg m 3 )(500 m 3 )     1  545 K, which is 272C 18. 59: V T   (0.0150 m 3 )( 318. 15 K )  p 2  p1  1 2   (2.72 atm) V T   (0.0159 m 3 )(278.15 K )   2.94 atm,   2 1   so the gauge pressure is 1.92 atm 18. 60: (Neglect... integral (given in Problem18.81) has been used b) f (v)dv is the probability that a particle has speed between v and v  dv; the probability that the particle has some speed is unity, so the sum (integral) of f (v) dv must be 1 18. 81: With n  2 and a  m/2kT , the integral is  m  4π    2πkT  32   3 3kT π  3  2 (m 2kT ) 2  (m 2kT )  m ,    which is Eq (18. 16)   18. 82: 0  m  f (v)dv... factor of 0.733 , to 13 breaths min If the breathing rate is not increased, one would experience “shortness of breath.” 18. 65: 3N  3nN A  3(m M ) N A  3 (6.023  1023 molecules mol)(50 kg) (18. 0  103 kg mol)  5.0  10 27 atoms 18. 66: The volume of gas per molecule (see Problem 18. 28) is RT N Ap , and the volume of a 4 molecule is about V0  π (2.0  1010 m)3  3.4  10 29 m3 Denoting the ratio... From Table (18. 2), the speed is (1.60)v s, and so v s2  v2 3kT 3RT   m M (1.60) 2 (see Exercise 18. 48), and so the temperature is Mv 2 (28.0  103 kg mol) T  v 2  (4.385  10 4 K  s 2 m 2 )v 2 2 2 3(1.60) R 3(1.60) (8.3145 J mol  K) a) (4.385  104 K  s 2 m 2 )(1500 m s) 2  987 K b) (4.385  104 K  s 2 m 2 )(1000 m s) 2  438 K c) (4.385  104 K  s 2 m 2 )(500 m s) 2  110 K 18. 46: Making... to water to steam 18. 51: The temperature of 0.00 C is just below the triple point of water, and so there will be no liquid Solid ice and water vapor at 0.00C will be in equilibrium 18. 52: The atmospheric pressure is below the triple point pressure of water, and there can be no liquid water on Mars The same holds true for CO 2 V  βVo T  (3.6  10 5 C) (11 L)(21C)  0.0083 L 18. 53: a) V   . altitude is exponential rather than linear. 18. 19: RTMyg epp   0 from Example 18. 4. Eq. (18. 5) says RT.Mρp )( Example 18. 4 assumes a constant ρpT andso,K273. of freedom. 18. 43:     kJ.56.1K0.30KmolJ79.20mol50.2 (18. 26),Eq. Usinga) Q b) From Eq. (18. 25), 5 3 of the result of part (a), 936 J. 18. 44: a)