Chapter 19 1 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 Therefore 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 2 a Equation 19 3 yields n Msam M 2 5 197 0 0127 mol b The number of atoms is found from Eq 19 2 N nNA 0 0127 6 02 1023 7 64 1021 3 THINK We treat the oxygen gas in this problem as ideal and apply the ideal gas law EXPRESS In solving the ideal gas law equation pV nRT for n we first convert the temperature to the Kelvin scale and the volume to SI units ANALYZE a The number of moles of oxygen present is b Similarly the ideal gas law pV nRT leads to We note that the final temperature may be expressed in degrees Celsius as 220 C LEARN The final temperature can also be calculated by noting that or 881 40 0273 15 K313 15 KiT 3331000cm10miV 53321 0110Pa1 00010m3 8810mol 8 31J molK313 15KiipVnRT 53321 0610Pa1 50010m493K 3 8810mol8 31J molKffpVTnR ffiiifpVpVTT 53531 0610Pa1500cm 313 15K 493K1 0110Pa1000cmfffiiipVTTpV 882 CHAPTER 19 4 a With T 283 K we obtain b We can use the answer to part a with the new values of pressure and temperature and solve the ideal gas law for the new volume or we could set up the gas law in ratio form as where ni nf and thus cancels out which yields a final volume of 5 With V 1 0 10 6 m3 p 1 01 10 13 Pa and T 293 K the ideal gas law gives Consequently Eq 19 2 yields N nNA 25 molecules We can express this as a ratio with V now written as 1 cm3 N V 25 molecules cm3 6 The initial and final temperatures are and respectively Using the ideal gas law with we find the final pressure to be 7 a Equation 19 45 which gives 0 implies Q W Then Eq 19 14 with T 273 30 0 K leads to gives Q 3 14 103 J or Q 3 14 103 J b That negative sign in the result of part a implies the transfer of heat is from the gas 8 a We solve the ideal gas law pV nRT for n 3310010Pa2 50m106mol 8 31J molK283KpVnRTfffiiipVTpVT 33100kPa303K2 50m0 892 m300kPa283KfififiTpVVpT 1363231 0110 Pa1 010 m4 110mole 8 31 J molK293 KpVnRT 5 00C278 KiT 75 0C348 KfT ifVV 348K1 00 atm1 25 atm278KfffffiiiiipVTTpppVTT 638100Pa1 010m5 4710mol 8 31J molK220KpVnRT b Using Eq 19 2 the number of molecules N is 9 Since standard air pressure is 101 kPa then the initial absolute pressure of the air is pi 266 kPa Setting up the gas law in ratio form where ni nf and thus cancels out we have which yields Expressed as a gauge pressure we subtract 101 kPa and obtain 186 kPa 10 The pressure p1 due to the first gas is p1 n1RT V and the pressure p2 due to the second gas is p2 n2RT V So the total pressure on the container wall is 883 The fraction of P due to the second gas is then 11 THINK The process consists of two steps isothermal expansion followed by isobaric constant pressure compression The total work done by the air is the sum of the works done for the two steps EXPRESS Suppose the gas expands from volume Vi to volume Vf during the isothermal portion of the process The work it does is where the ideal gas law pV nRT was used to replace p with nRT V Now Vi nRT pi and Vf nRT pf so Also replace nRT with piVi to obtain 623116A5 4710mol6 0210mol3 2910molecules NnN fffiiipVTpVT 23231 6410m300K266kPa287 kPa1 6710m273KfififiTVppVT 121212 nRTnRTRTpppnnVVV 2221212 0 50 2 20 5pnRTVnpnnRTVnn 1ln ffiiVVfVViVdVWpdVnRTnRTVV fiifVVpp 1ln iiifpWpVp 884 CHAPTER 19 During the constant pressure portion of the process the work done by the gas is The gas starts in a state with pressure pf so this is the pressure throughout this portion of the process We also note that the volume decreases from Vf to Vi Now Vf piVi pf so ANALYZE For the first portion since the initial gauge pressure is 1 03 105 Pa pi 1 03 105 Pa 1 013 105 Pa 2 04 105 Pa The final pressure is atmospheric pressure pf 1 013 105 Pa Thus Similarly for the second portion we have The total work done by the gas over the entire process is LEARN The work done by the gas is positive when it expands and negative when it contracts 12 a At the surface the air volume is b The temperature and pressure of the air inside the submarine at the surface are we On the other hand at depth and have and Therefore using the ideal gas law the air volume at this depth would be 2 fifWpVV 2iififiifpVWpVppVp 5534152 0410Pa2 0410Pa0 14mln2 0010J 1 01310PaW 55342 1 01310Pa2 0410Pa 0 14m 1 4410J fiiWppV 443122 0010J 1 4410J 5 6010J WWW 2331 1 00 m 4 00 m 12 57 m12 6 mVAh 120C293 KT 101 00 atmpp 80 m h 230C243 KT 322051 00 atm1 00 atm 1024kg m 9 80m s 80 0 m 1 0110 Pa1 00 atm7 95 atm8 95 atm ppgh pVNkT c The decrease in volume is this volume corresponds to is Using Eq 19 5 the amount of air 885 Thus in order for the submarine to maintain the original air volume in the chamber of air must be released 13 a At point a we know enough information to compute n b We can use the answer to part a with the new values of pressure and volume and solve the ideal gas law for the new temperature or we could set up the gas law in terms of ratios note na nb and cancels out which yields an absolute temperature at b of Tb 1 8 103 K c As in the previous part we choose to approach this using the gas law in ratio form which yields an absolute temperature at c of Tc 6 0 102 K d The net energy added to the gas as heat is equal to the net work that is done as it progresses through the cycle represented as a right triangle in the pV diagram shown in triangle with Fig 19 20 This where we choose the plus sign because the volume change at to area is related inside turn that in the largest pressure is an increase Thus 331111221222211 00 atm243K12 57 m1 16 m8 95 atm293KpVTpTVVpVTpT 31211 44 mVVV 533 8 95 atm 1 0110Pa atm11 44m5 1010mol8 31J molK243KpVnRT 35 1010mol 32500Pa1 0m1 5mol 8 31J molK200KpVnRT 337 5kPa3 0m200K2 5kPa1 0mbbbbaaapVTTpVT 332 5kPa3 0m200K2 5kPa1 0mccccaaapVTTpVT 12area base height 886 CHAPTER 19 14 Since the pressure is constant …
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